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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

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Samacheer Kalvi 11th Chemistry Chapter 10 Chemical Bonding Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding Multiple Choice Questions

Question 1.
In which of the following compounds does the central atom obey the octet rule?
(a) XeF₄
(b) AICI₃
(c) SF₆
(d) SO₂
Answer:
(d) SCl₂
Solution:

Hence in (d) SCl₂ octet rule is followed

Question 2.
In the molecule O_A = C = O_B the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, +1
(b) +1, 0, – 1
(c) – 2, 0, +2
(d) 0, 0, 0
Answer:
(d) 0, 0, 0

Formal charge of
Formal charge of C = 4 – \(\left( 0+\frac { 8 }{ 2 } \right)\) = 4 – 4 = 0

Question 3.
Which of the following is electron deficient?
(a) PH₃
(b) (CH₃)₂
(c) BH₃
(d) NH₃
Answer:
(c) BH₃
Solution:
– electron rich,
CH₃ – CH₃ – Covalent neutral molecule,
BH₃ – electron deficient

Question 4.
Which of the following molecule contain no π bond?
(a) SO₂
(b) NO₂
(c) CO₂
(d) H₂O
Answer:
(d) H₂O
Solution:

Water (H₂O) contains only σ bonds and no π bonds.

Question 5.
The ratio of number of sigma (σ) and pi (π) bonds in 2- butynal is …………..
(a) 8/3
(b) 5/3
(c) 8/2
(d) 9/2
Answer:
(d) 9/2
Solution:

no. of σ bonds = 8 [4C – H; 3C – C; 1C— O]
no.of π bonds = 3 [2C – C; 1C – O]
∴ratio = \(\frac { 8 }{ 3 }\)

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?
(a) 120°, 80°
(b) 109°. 28
(c) 90°
(d) 89°, 117°
Answer:
(d) 89°, 117°

Solution:
Normal bond angle in regular trigonal bipyramidal are 90° and 120°. Due to l.p – b.p repulsion, bond angle is reduced to 89°, 117° option (d).

Question 7.
Assertion: Oxygen molecule is paramagnetic.
Reason: It has two unpaired electron in its bonding molecular orbital.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Correct statement: Oxygen molecule is paramagnetic
Correct Reason: It has two unpaired electrons in its antibonding molecular orbital.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when ……………….
(a) fully filled atomic orbitals overlap
(b) half filled atomic orbitals overlap
(c) non-bonding atomic orbitals overlap
(d) empty atomic orbitals overlap
Answer:
(b) half filled atomic orbitals overlap

Question 9.
In CIF₃, NF₃ and BF₃ molecules the chlorine, nitrogen and boron atoms are …………………….
(a) sp³ hybridised
(b) sp³, sp³ and sp² respectively
(c) sp² hybridised
(d) sp³d, sp³ and sp² hybridised respectively
Answer:
(d) sp³d, sp³ and sp² hybridised respectively
Solution:
CIF₃ – sp³d hybridisation
NF₃ – sp³ hybridisation
BF₃ – sp² hybridisation

Question 10.
When one s and three p orbitais hybridise,
(a) four equivalent orbitais at 90⁰ to each other will be formed
(b) four equivalent orbitais at 109⁰ 28’ to each other will be formed.
(c) four equivalent orbitals, that are lying the same plane will be formed
(d) none of these
Answer:
(b) four equivalent orbitals at 109° 28′ to each other will be formed.

Question 11.
Which of these represents the correct order of their increasing bond order?
(a) C₂ < C₂^2- < O₂
(b) C₂^2- < C₂⁺ < O² < O₂^2-
(c) O₂^2- < O₂⁺ < O₂ < C₂^2-
(d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Answer:
(d) O₂^2- < C₂⁺ < O₂ < C₂^2-
Solution:
bond order = (n_b – n_a)
bond order of O₂^2- = \(\frac { 1 }{ 2 }\) (8 – 6) = 1
bond order of C₂⁺ = \(\frac { 1 }{ 2 }\) (5 – 2) = 1.5
bond order of O₂ = \(\frac { 1 }{ 2 }\) (8 – 4) = 2
bond order of C₂^2- = \(\frac { 1 }{ 2 }\) (8 – 2) = 3

Question 12.
Hybridisation of central atom in PCl₅ involves the mixing of orbitais.
(a) s, p_x, p_y, d_x², d_x²-y²
(b) s, p_x, p_y, p_xy, d_x²-y²
(c) s, p_x, p_y, p_z, d_x²-y²
(d) p_x, p_y, p_xy, d_x²-y²
Answer:
(c) s, p_x, p_y, p_z, d_x²-y²
Solution:
PCl₅ – sp³d hybridisation s, p_x, p_y, p_z, d_x²-y²

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is ……………..
(a) H₂O₂ > O₃ > O₂
(b) O₂ > O₃ > H₂O
(c) O₂ > H₂O₂ > O₃
(d) O₃ > O₂ > H₂O₂
Answer:
(b) O₂ > O₃ > H₂O₂
Solution:
The bond order for O₂, O₃ and H₂O₂ decreases in the order 2 > 1.5 > 1

Question 14.
Which one of the following is diamagnetic?
(a) O₂
(b) O₂^2-
(c) O₂²⁺
(d) None of these
Answer:
(b) O₂^2-
Solution:
O₂^2- is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits π*_2py and π*_2pz

Question 15.
Bond order of a species is 2.5 and the number of electrons are in its bonding molecular orbital is found to be 8. The no. of electrons in its anti-bonding molecular orbital is ………………….
(a) three
(b) four
(c) zero
(d) cannot be calculated form the given information.
Answer:
(a) three
Solution:
Bond order = \(\frac { 1 }{ 2 }\) (n_b – n_a)
2.5 = \(\frac { 1 }{ 2 }\) (8 – n_a)
⇒ 5 = 8
⇒ n_a = 8 – 5 = 3

Question 16.
Shape and hybridisation of IF₅ are ………….
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Squai2e pyramidal, sp³d²
(d) Octahedral, sp³d²
Answer:
(c) Square pyramidal, sp³d²
Solution:
IF₅ – 5 bond pair + 1 lone pair
∴ hybridisation sp³d²

Question 17.
Pick out the incorrect statement from the following.
(a) sp³ hybrid orbitais are equivalent and are at an angle of 1 09°28’ with each other.
(b) dsp² hybrid orbitais are equivalent and bond angle between any two of them is 900.
(c) All five sp³d hybrid orbitais arc not equivalent. Out of these five sp³d hybrid orbitais, three are at an angle of 120°, remaining two are perpendicular to the plane containing the other three
(d) none of these
Answer:
(c) All five sp³d hybrid orbitals are not equivalent. Out of these five sp³d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three.

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are ………………..
(a) SeF₄, XeO₂F₂
(b) SF₄, XeF₂
(c) XeOF₄, TeF₄
(d) SeCI₄, XeF₄
Answer:
(a) SeF₄, XeO₂F₂
Solution:
SeF₄, XeO₂F₂ – sp³d hybridisation
T – shaped, one lone pair on central atom.

Question 19.
In which of the following molecules / ions BF₃, NO₂^–, H₂O the centrai atom is sp² hybridised?
(a) NO₂^– and H₂O
(b) NO₂^– and H₂O
(c) BF₃ and NO₂
(d) BF₃ and NH₂^–
Answer:
(c) BF₃ and NO₂
Solution:
H₂O – Central atom sp³ hybridised
NO₂^– – Central atom sp² hybridised
BF₃ – Central atom sp² hybridised
NH₂^–  – Central atom sp³ hybridised

Question 20.
Some of the following properties of two species, NO₃^– and H₃O⁺ are described below. Which one of them is correct?
(a) dissimilar in hybridisation for the central atom with different structure.
(b) isostnictural with same hybridisation for the Central atom.
(c) different hybridisation for the central atom with same structure
(d) none of these
Answer:
(a) dissimilar in hybridisation for the central atom with different structure.
Solution:
NO₃^– – sp² hybridisation, planar
H₃O⁺ – sp³ hybridisation, pyramidal

Question 21.
The types of hybridisation on the five carbon atom from right to left in the, 2,3 pentadiene.
(a) sp³, sp², sp, sp², sp³
(b) sp³, sp, sp, sp, sp³
(c) sp², sp, sp², sp , sp³
(d) sp³, sp³, sp², sp³, sp³
Answer:
(a) sp³, sp², sp, sp², sp³
Solution:

Question 22.
XeF₂ is isostructural with ………..
(a) SbCl₂
(b) BaCl₂
(c) TeF₂
(d) ICl₂^–
Answer:
(d) ICl₂^–
Solution:
XeF₂ is isostructural with ICI₂^–

Question 23.
The percentage of s-character of the hybrid orbitais in methane, ethane, ethene and ethyne are respectively ………………..
(a) 25, 25, 33.3, 50
(b) 50, 50, 33.3, 25
(c) 50, 25, 33.3, 50
(d) 50, 25, 25. 50
Answer:
(a) 25, 25, 33.3, 50
Solution:

Question 24.
Of the following molecules, which have shape similar to carbon dioxide?
(a) SnCI₂
(b) NO₂
(c) C₂H₂
(d) All of these
Answer:
(c) C₂H₂
Solution:
CO₂ – Linear
C₂H₂ – Linear

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order …………………
(a) l.p – l.p > b.p – b.p > l.p – b.p
(b) b.p – b.p > b.p – 1.p > l.p – b.p
(c) l.p – l.p > b.p – l.p > b.p – b.p
(d) b.p – b.p > l.p – l.p > b.p – l.p
Answer:
(c) l.p – l.p > b.p – l.p > b.p – b.p

Question 26.
Shape of CIF₃ is ……………………..
(a) Planar triangular
(b) Pyramidal
(c) ‘T’ Shaped
(d) none of these
Answer:
(c) ‘T’ Shaped
Solution:
dF₃ – sp³d hybridisation

Question 27.
Non- Zero dipole moment is shown by …………………
(a) CO₂
(b) p – dichlorobenzene
(c) carbon tetrachloride
(d) water
Answer:
(d) water
Solution:

Question 28.
Which of the following conditions is not correct for resonating structures?
(a) the contributing structure must have the same number of unpaired electrons.
(b) the contributing structures should have similar energies.
(c) the resonance hybrid should have higher energy than any of the contributing structure.
(d) none of these
Answer:
(c) the resonance hybrid should have higher energy than any of the contributing structure.
Solution:
Correct statement is – the resonance hybrid should have lower energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent and coordinate linkage is …………………
(a) NH₄Cl
(b) NH₃
(c) NaCl
(d) none of these
Answer:
(a) NH₄Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is ………….
(a) U
(b) 2U
(c) U/2
(d) 4U
Answer:
(d) 4U

Samacheer Kalvi 11th Chemistry Chemical Bonding Short Answer Questions.

Question 31.
Define the following

1. Bond order
2. Hybridisation
3. a- bond

Answer:
1. Bond order:
Bond order
The number of bonds formed between the two bonded atoms in a molecule is called bond order.

2. Hybridisation:
It is a process of mixing of atomic orbitais of the same atom with. comparable energy to form equal number of new equivalent orbitais with same energy. The resultant orbitais are called hybridised orbitais and they possess maximum symmetry and definite orientation in space so as to minimise the force of repulsion between their electrons.

3. σ – bond:
When two atomic orbitais overlap linearly along the axis, the resultant bond is called sigma (σ) bond.

Question 32.
What is a pi bond?
Answer:
Pi – bond:
When two atomic orbitals overlap sideways, the resultant covalent bond is called a pi (π) bond.

Question 33.
In CH₄, NH₃ and H₂O, the central atom undergoes sp³ hybridisation-yet their bond angles are different, why?
Answer:
1. In CH₄, NH₃ and H₂O the central atom undergoes sp³ hybridisation. But their bond angles are different due to the presence of lone pair of electrons.

2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs arc not same.

3. Bond pair-Bond pair < Bond pair – Lone pair < Lone pair – Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.

4. In case of CH₄, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle 109° 28’.

5. H₂O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of 104° 35’.

6. NH₃ has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 34.
Explain sp² hybridisation in BF₃
Answer:
1. sp² hybridisation in boron trifluoride – Boron atom – B. Electronic configuration [H₂]2s²2p².

2. In boron, the s orbital and two p orbitals in the valence shell hybridises to generate three equivalent sp² orbitais. These 3 orbitaIs lie in the same xy plane and the angle between any two orbitals is equal to 120°.

3. The 3 sp² hybridised orbitais of boron now overlap with the 2p_z orbitais of fluorine (3 atoms). This overlap takes place along the axis.

Question 35.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic.
Answer:

1. Electronic configuration of O atom is is 1s² 2s² 2P⁴
2. Electronic configuration of O, molecule is
   
3. Bond order =
4. Molecule has two unpaired electrons, hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:

1. Electronic configuration of C atom: 1s² 2s² 2p²
   Electronic configuration of O atom: 1s² 2s² 2p⁴
2. Electronic configuration of CO molecule is: σ1s² σ*1s² σ2s² σ*2² π2p_y² π2p_z² σ2p_x²
3. Bond order
4. Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
What do you understand by Linear combination of atomic orbitais in MO theory.
Answer:
Linear combination of atomic orbitais (LCAO):
1. The wave functions for the molecular orbitais can be obtained by solving Schrodinger wave equation for the molecule. Since solving Schrodinger wave equation is too complex, a most common method linear combination of atomic orbitais (LCAO) is used to obtain wave function for molecular orbitals.

2. Atomic orbitais are represented by wave functions ψ. Consider two atomic orbitals represented by the wave functions ψ_A and ψ_B with comparable energy that combines to form two molecular orbitals.

3. One is bonding molecular orbitai (ψ bonding) and the other is anti-bonding molecular orbital (ψ anti-bonding).

4. The wave function for molecular orbitais, ψ_A and ψ_B can be obtained by the LCAO as shown below:
ψ_bonding = ψ_A + ψ_B
ψ_anti-bonding = ψ_A – ψ_B

5. The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitais and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitais.

6. The formation of two molecular orbitals from two is orbitals is show below.

Constructive interaction:
The two is orbitals are in phase and have the same signs.

Destructive interaction:
The two is orbitals are out of phase and have opposite signs

Question 38.
Discuss the formation of N₂ molecule using MO Theory.
Answer:

1. Electronic configuration of N atom 1s² 2s² 2p³.
2. Electronic configuration of N, molecule is: σ1s² σ*1s² σ2s² σ*2² π2p_y² π2p_z² σ2p_x²
3. Bond order
4. Molecule has no unpaired electrons hence, it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: µ = q x 2d, where µ is the dipole moment, q is the charge, 2d is the distance between the two charges.

2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.

3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).

4. 1 Debye = 3.336 x 10^-30 Cm

Question 40.
Linear form of carbon dioxide molecule has two polar bonds. yet the molecule has zero dipole moment, why?
Answer:

1. The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
2. In CO₂, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
3. Hence the net dipole moment of the CO₂ is µ = µ₁ + µ₂ = µ₁ +( – µ₁) = 0
4. 
5. In this case

Question 41.
Draw the Lewis structures for the following species.

1. NO₃^–
2. SO₄^2-
3. HNO₃
4. O₃

Answer:
1. NO₃^–

2. SO₄^2-

3. HNO₃

4. O₃

Question 42.
Explain the bond formation in BeCl₂ and MgCl₂. BeCl₂ bond formation:
Answer:
1. Electronic confiuration of Be(Z = 4) is 1s² 2s² and electronic configuration of Cl (Z = 17) is 1s² 2s² 2p⁶ 3s² 3p⁵.

2. Beryllium has 2 electrons in its valence shell and chlorine atoms (2) have 7 electrons in their valence shell.

3. By losing two electrons, Beryllium attains the inert gas configuration of Helium and becomes a dipositive cation, Be²⁺ and each chlorine atom accepts one electron to become (Cl^–) uninegative anion and attains the stable electronic configuration of Argon.

4. Then Be²⁺ combine with 2Cl^– ions to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of 1 mole of BeCl₂, the amount of energy released is – 468 kJ/mol. This favours the formation of BeCl, and its stabilisation.

MgCI₂ bond formation:
1. Electronic configuration of Mg (z = 12) is 1s² 2s² 2p⁶ 3s².
Electronic configuration of Cl (z = 17) is 1s² 2p⁶ 3p⁶ 3p⁵

2. Magnesium has 2 electrons in its valence shell and chlorine has 7 electrons in its valence shell.

3. By losing two electrons, magnesium attains the inert gas configuration of Neon and becomes a dipositive cation (Mg²⁺) and two chlorine atoms accept these electrons to become two uninegative anions [2Cl^–] by attaining the stable inert gas configuration of Argon.

4. These ions, Mg²⁺ and 2Cl^– combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. The energy released during the formation of 1 mole of MgCl₂ is – 783 kJ/mole. This favours the formation of MgCI₂ and its stabilisation.

Question 43.
Which bond is stronger or π? Why?
Answer:
1. Sigma bonds (σ) are stronger than Pi bonds (π). Because, sigma bonds are formed from bonding orbitals directly between the nuclei of the bonding atoms, resulting in greater overlap and a strong sigma bond (axial overlapping).

2. π bonds results from overlap of atomic orbitals that are in contact through two areas of overlap (lateral overlapping). Pi bonds are more diffused bonds than sigma bonds.

Question 44.
Define bond energy.
Answer:
Bond energy:
Bond energy (or) Bond enthalpy is defined as the minimum amount of energy required to break one mole of a bond in molecules in their gaseous state. The unit of bond energy is kJ mol^-1

Question 45.
Hydrogen gas is diatomic whereas inert gases are monoatormic – explain on the basis of MO theory.
Answer:
1. Hydrogen gas is diatomic. According to MO theory. which is based on quantum mechanics H₂ molecule can be represented in terms of the following diagram called M.O. diagram.

H – H. i.e.. H₂ molecule has two atoms which are connected by 1 σ bond. So it is diatomic.

2. But in the case of inert gases. the valence shell is fully filled i.e.. an octet (8 electrons) (or) duplet (2 electrons) in case of Helium, due to which they are in monoatomic state and remain stable. So they do not combine with any atom (neither of same or of different elements). Due to this they do no exist in diatomic state and always exist in mono – atomic state.

Question 46.
What is Polar Covalent bond? Explain with example.
Answer:
1. If a covalent bound is formed between atoms having different electronegativities. the atom with higher electronegativity will have greater tendency to attract the shared pair of electrons towards itself than the other atom. As a result, the cloud of shared electron pair gets distorted and polar covalent bond is formed.

2. Example – HF – Hydrogen fluoride:
The electronegativities of hydrogen and fluorine on Pauling’s scale are 2.1 and 4 respectively. It means that fluorine attract the shared pair of

electrons approximately twice as much as hydrogen which leads to partial negative charge on fluorine atom and partial positive charge on hydrogen atom. Hence, the H – F bond is said to be a polar covalent bond.

Question 47.
Considering x-axis as molecular axis, which out of the following will form a sigma bond.

1. 1s and 2p_y
2. 2p_x and 2p_x
3. 2p_x and 2p_z
4. 1s and 2P_z

Answer:
Along X-axis as molecular axis, only 2p and 2p can form a sigma bond

Question 48
Explain resonance with reference to carbonate ion

1. For the above structure, we can draw two additional lewis structures by moving the lone pairs from the other two oxygen atoms O_B and O_C. and thus creating three similar structures in which the relative positive of the atoms are same.

2. They only differ in the position of bonding and lone pair of electrons. Such structures are called resonance structures and this phenomenon is called resonance.

3. It is evident from the experimental results that all carbon-oxygen bonds in carbonate ion are equivalent. The actual structure of the molecule is said to be a resonance hybrid, an average of these 3 resonance forms. The following structure gives a qualitative idea about the correct structure of CO₃^2- (carbonate) ion.

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in Ethylene, C₂H₄
1. Bonding in ethylene can he explained by hybridisation concept.

2. The valency of carbon is 4. The electronic configuration of carbon is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z⁰. One electron from 2s orbital is promoted to 2p_z. orbital in the excited state to satisfy the valency of carbon.

3. In ethylene both the carbon atoms undergo sp² hybridisation involving 2s, 2p_x and sp_y orbitals resulting in 3 equivalent sp² hybridised orbitals lying in the XY plane at an angle of 120⁰ to each other. The unhybridised 2p_z orbital lies perpendicular to the xy plane.

4. One of the sp² hybndised orbitals of each carbon atoms lying along the X – axis linearly overlaps with each other resulting in the formation of C – C sigma bond. The other two sp² hybridised orbitals of both carbon atom linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C – H sigma bonds on each carbon atom.

5. The unhybridised 2p_z orbital of both carbon atoms can overlap only sideways as they are not in the molecular axis. This lateral overlap results in the formation of a pi bond between the two carbon atoms.

Bonding in acetylene (C₂H₂):
1. The electronic configuration of valence shell of carbon atom in the ground state is [He] 2s² 2p_x¹ 2p_z⁰. One electron from 2s orbital is promoted to 2p_z orbital in the excited state to satisfy the valency of carbon.

2. In acetylene molecule, both the carbon atoms are in sp hybridised state. The 2s and 2p_x orbitals resulting in two equivalent sp hybridised orbitals are formed lying in a straight line along the X – axis. The unhybridised 2p_y, and 2p_z orbitais lie perpendicular to the X-axis.

3. One of the two sp hybridised orbitals of each carbon atom linearly overlaps with each other resulting in the formation of a C – C sigma bond. The other sp hybridised orbital of both carbon atoms linearly overlap with the two is orbitals of two hydrogen atoms leading to the formation of one C – H sigma bond on each carbon atom.

4. The unhybridised 2p_y and 2p_z orbitals of each carbon atom overlap sideways. This lateral overlap results in the formation of two pi bonds. (p_y – p_y) and (p_z – p_z) between the two carbon atoms.

Question 50.
What type of hybridisations are possible ¡n the following geometeries?

1. octahedral
2. tetrahedral
3. square planar.

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of F₇ and SF₆.
Answer:
VSEPR theory:
1. The shape of the molecules depend on the number of valence shell electron pair around the central atom.

2. There are two types of electron pairs namely, bond pairs and lone pairs.

3. Each pair of valence electrons around the central atom repel each other and hence they are located as far away as possible in three dimensional space to minimise the repulsion between them.

4. The repulsive interaction between the different types of electron pairs is in the following
order:

5. The lone pair of electrons are localised only on the central atom and interact with only one nucleus whereas the bond pairs are shared between two atoms and they interact with two nuclei. Because of this, the lone pairs occupy more space and have greater repulsive power than the bond pairs in a molecule.

IF₇:
It is an AB₇ type molecule. This molecule has 7 bond pair of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, IF₇ has pentagonal bipyramidal shape.

SF₆:
It is an AB₆ type molecule. This molecule has 6 bond pairs of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, SF₆ has octahedral shape.

Question 52.
CO₂ and H₂O both are triatomic molecules but their dipole moment values are different. Why?
Answer:
1. Linear form of carbon dioxide has zero dipole moment. In CO₂ the dipole moment of two polar bonds are equal in magiitude but have opposite direction. Hence, the net dipole moment of the CO₂ molecule is
µ = µ₁ + µ₂
µ = µ₁ + (- µ₁) = 0

In this case

2. But in the case of water, net dipole moment is the vector sum µ₁ + µ₂ as follows:

Dipole moment in water is found to be 1.85 D.

3. CO₂ and H₂O both are triatomic molecules but their dipole moment values are zero and 1.85 D respectively.

Question 53.
Which one of the following has highest bond order? N₂, N₂⁺ or N₂^– ?
Answer:
N₂ (14 electrons)
Bond order = 3,
N₂⁺ (13 electrons)
Bond order = 2.5,
N₂^–(15 electrons)

So N₂ has the highest bond order.

Question 54.
Explain the covalent character in ionic bond.
Answer:
1. Ionic compounds like lithium chloride shows covalent character and it is soluble in organic solvents such as ethanol.

2. The partial covalent character in ionic compounds can be explained on the basis of a phenomenon called polarisation.

3. In an ionic compound, there is an electrostatic attractive force between the cation and anion. The positively charged cation attract the valence electrons of anion while repelling the nucleus.

This cause a distortion in the electron cloud of the anion and its electron density drills towards the cation, which results in some sharing of valence electrons between these ions. Thus, a partial covalent character is developed between them. This phenomenon is called polarisation.

4. Thus due to polarisation, ionic compounds shows covalent character.

Question 55.
Describe Fajan’s rule.
Answer:
1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the Cation greater will be the attraction on the electron cloud of the anion.

Similarly higher the magnitude of negative charge on anion, greater is its polansability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order – NaCI < MgCl₂ < AICI₃

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. e.g., LiCl is more covalent than NaCI.

4. Cation having ns²np⁶nd¹⁰ configuration shows greater polarising power than the cations with ns²np⁶ configuration. e.g., CuCI is more covalent than NaCl.

Samacheer Kalvi 11th Chemistry Chemical Bonding In Text Questions Evaluate yourself

Question 1.
Draw the lewis structures for

1. Nitrous acid (HNO₂)
2. Phosphoric acid
3. Sulphur troxide (SO₃)

Answer:
1. Nitrous acid (HNO₂)

2. Phosphoric acid

3. Sulphur troxide (SO₃)

Question 2.
Calculate the formal charge on each atom of carbonyl chloride (COCl₂)
Answer:
Formal charge 
Carbonyl chloride COCl₂

Formal charge on carbon atom = 4 – \(\left[ 0+\frac { 8 }{ 2 } \right]\) = 4 – 4 = 0
Formal charge on chlorine atom = 7 – \(\left[ 6+\frac { 2 }{ 2 } \right]\) = 7 – 7 = 0
Formal charge on oxygen atom = 6 – \(\left[ 4+\frac { 4 }{ 2 } \right]\) = 6 – 6 = 0

Question 3.
Explain the ionic bond formation in MgO and CaF₂
Magnesium oxide (MgO):
Answer:
Electronic configuration of Mg – 1s² 2s² 2p⁵ 3s²
Electronic configuration of O – 1s² 2s² 2p⁶ 3s⁶ 3p⁴.
1. Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.

2. By losing two electrons, Mg acquires the inert gas configuration of Neon and becomes a dipositive cation Mg²⁺
Mg → Mg²⁺ + 2e^–

3. Oxygen accepts the two electrons to become a dinegative oxide anion, O^2- thereby attaining the inert gas configuration of Neon.
O + 2e^– → O^2-

4. These two ions, Mg²⁺ and O^2- combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of magnesium oxide crystal 601.6 kJ mol^-1 energy is released . This favours the formation of magnesium oxide (MgO) and its stabilisation.

CaF₂, Calcium fluoride
1.  Calcium, Ca: [Ar] 4s², Fluorine F: [He] 2s² 2p⁵

2. Calcium has two electrons in its valence shell and fluorine has seven electrons in its valence shell.

3. By losing two electrons, calcium attains the inert gas configuration ofArgon and becomes a dipositive cation, Ca²⁺.

4. Two fluorine atoms, each one accepts one electron to become two unincgative fluoride ions (F^–) thereby attaining the stable configuration of Neon.

5. These three ions combine to form an ionic crystal in which they are held together by electrostatic attractive force.

6. During the formation of calcium fluoride crystal 1225.91 kJ mol^-1 of energy is released. This favours the formation of calcium fluoride, CaF₂ and its stabilisation.

Question 4.
Write the resonance structures for

1. Ozone molecule
2. N₂O

Answer:
1. Ozone molecule, O₃

2. Nitrous oxide, N₂O

Question 5.
Of the two molecules OCS and CS₂ which one has higher dipole moment value? Why?
Answer:
OCS and CS₂

The dipole moment µ_OCS = 0.7 149 ± 0.0003 Debye.
CS₂:
S = C = S
In CS₂, the bond dipoles of 2C = S have same values and bond dipoles cancel each other so dipole moment of CS₂ is zero. Among OCS and CS₂, OCS has a higher dipole moment because in OCS oxygen is more electronegative than sulphur and C = S and C = O bonds in OCS molecules do not cancel each other.

On the other hand CS₂, due to linear structure, the bond dipole of two C = S bonds cancel each other. On the other hand, CS₂ due to linear structure, the bond dipoles of two C = S bonds cancel each other and thc recultant dipo’c moment value is zero. So OCS has a higher dipole moment than CS₂.

Question 6.
Arrange the following in the decreasing order of Bond angle

1. CH₄, H₂O, NH₃
2. C₂H₂, BF₃, CCl₄

Answer:
1. CH₄, H₂O, NH₃:
NH₃ = 107°, Water= 104.5°, CH₄ = 109.5°
Decreasing order of bond angle: H₂O < NH₃ < CH₄

2. C₂H₂, BF₃, CCl₄:
C₂H₂ = 1800, BF₃ = 120°, CCl₄ = 109.5°
Decreasing order of bond angle: CCl₄ < BF₃ < C₂H₂

Question 7.
Bond angle in PH₄⁺ is higher than in PH₃. Why?
Answer:
Phosphorous in both PH₃ and PH₄⁺ is sp³ hybridised. Due to the absence of lone pair – bond pair repulsion and presence of four identical bond pair – bond pair interactions, PH₄⁺ assumes tetrahedral geometry with a bond angle of 109° 28’.

But PH₃ has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from 109° 28’ to 93.6°. As a result PH₃ is pyramidal.
PH₃ – Pyramidal with bond angle of 93.6°. PH₄⁺ Tetrahedral with bond angle of 109° 28’.

Question 8.
Explain the bond formation in SF₄ and CCl₄ using hybridisation concept.
Answer:
In SF₄, the central atom is sp³d hybridised.
1. The molecule SF³ will have a total 34 valence electron 6 form sulphur, 7 each from four fluorine atoms.

2. Sulphur atom will from 4 single bonds with fluorine atoms. These bonds account for the 8 electrons out of the 34 valence electrons. Each fluorine atom will have 3 lone pair of electrons in order to have a complete octet structure.

These lone pairs will use up 24 valence electrons. So the total used valence electrons, are 32. The remaining 2 electrons will be placed on the sulphur atom as a lone pair. Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair.

3. This is quite Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair. This is quite possible for sulphur because it has easy access to its 3d orbital which means that it can expand its octet and accommodate more than 8 electrons.

4. Sulphur forms 4 single bonds and has 1 lone pair which means that its steric number is equal to 5. In this case sulphur will use five hybrid orbitais, such as one 3s orbital three 3p orbitais and one 3d orbital. So the central atom is sp³d hybridised.

CCl₄:
1. It is not necessary to invoke hybridisation especially in CCl₄. It must be invoked for all tetrahedral bonds of carbon and other atoms.

2. The electronic configuration of an isolated carbon atom in its ground state is 1s² s² 2p².

3. CCl₄ is a tetrahedral molecule comprising of four single bonds known as a bonds between the carbon atom and the chlorine atoms. In this type of bonding, the 2s orbital and three 2p orbitals of carbon atoms are mixed to produce four identical orbitals, a process known as sp³ hybridisation.

Question 9.
The observed bond length of N₂⁺ is larger than N₂ while the bond length in NO⁺ is less than in NO. Why?
Answer:
(a)
(1) By molecular orbital theory, the bond order of both N₂⁺ is 2.5 whereas N₂ is 3.

2. N₂ has 5e^– in the antibonding molecular orbital whereas N₂⁺ has 4e^– in the antibonding molecular orbital. So N₂⁺ will make a stronger and shorter bond length.

3. More the bond order and bond strength, and lesser will be the bond length.

4. So we can easily conclude N₂ has more bond length than N₂
Bond order in
Bond order in
So, N₂ is more stable than N₂⁺. But bond length N₂⁺ is greater than N₂.

(b) NO⁺ & NO
Bond order of NO = 2.5
Bond order of NO⁺ = 3
Due to lesser bond order in NO, the bond length is greater than NO⁺ So, NO⁺ bond length is shorter than NO bond length.

Question 10.
Draw the MO diagram for acetylide ion C₂^2- and calculate its bond order.
Answer:
Acetylide ion, C₂^2- in acetylene

Electronic configuration of C² ion is:
σ1s² σ*1s² σ1s² π2p_x² π2p_y²

Bond order

Samacheer Kalvi 11th Chemistry Chemical Bonding Additional Questions Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding 1 Mark Questions and Answer:

I. Choose the correct answer.

Question 1.
Which is the correct Lewis structure of Helium?

Answer:

Solution:
Helium has only two electrons in its valence shell which is represented as a pair of dots (duplet).

Question 2.
Which one of the following form only covalent bonds?
(a) Alkali metals
(b) Metals
(c) Non metals
(d) Metalloids
Answer:
(c) Non metals

Question 3.
In which one of the following molecule triple bond is present?
(a) O₂
(b) H₂
(c) CO₂
(d) N₂
Answer:
(d) N₂
Solution:
N ≡ N

Question 4.
Which of the following is the lewis structure of water?

Answer:

Question 5.
Which one of the following element forms only one bond?
(a) Carbon
(b) Oxygen
(c) Fluorine
(d) Nitrogen
Answer:
(c) Fluorine

Question 6.
Which one is the preferred structure of CO₂?

Answer:

Question 7.
Which is the correct lewis structure of BF₃?

Answer:

Question 8.
Statement I: In sulphur hexafluoride, the central atom has more than eight valence electrons.
Statement II: The central atom can accommodate additional electron pairs by using outer vacant d orbitals.
(a) Statements I and II are correct and statement II is the correct explanation of statement I.
(b) Statements I and II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statements I and II are correct and statement II is the correct explanation of statement I.

Question 9.
Which one of the following molecule has complete octet?
(a) BF₃
(b) BeCl₂
(c) BCl₃
(d) CCI₄
Answer:
(d) CCI₄

Question 10.
Which one of the following does not have electrovalent bond?
(a) KCI
(b) NaI
(c) MgO
(d) CCI₄
Answer:
(d) CCI₄

Question 11.
Which one of the following has an ionic bond?
(a) CO₂
(b) CH₄
(c) CaF₂
(d) BeCI₂
Answer:
(c) CaF₂

Question 12.
During the formation of 1 mole of KCI crystal, the amount of energy released is ………….
(a) 418.81 kJ
(b) 348.56 kJ
(c) 718 kJ
(d) 70.25 kJ
Answer:
(c) 718 kJ

Question 13.
Which one of the following has coordinate covalent bond?
(a) CaF₂
(b) MgO
(c) [Fe(CN)₆]^4-
(d) KCI
Answer:
(c) [Fe(CN)₆]^4-

Question 14.
The distance between the nuclei of the two covalently bonded atoms is called …………….
(a) bond order
(b) bond length
(c) bond angle
(d) bond enthalpy
Answer:
(b) bond length

Question 15.
The length of a bond can be determined by ……………
(a) spectroscopic method
(b) x – ray diffraction method
(c) electron-diffraction method
(d) all the above
Answer:
(d) all the above

Question 16.
The value of carbon-carbon single bond length is ………..
(a) 1.43A
(b) 1.54Å
(c) 1.33A
(d) 1.20A
Answer:
(b) 1.54Å

Question 17.
The value of carbon – carbon double bond length is …………..
(a) 1.43Å
(b) 1.20Å
(c) 1.54A
(d) 1.33A
Answer:
(d) l.33A

Question 18.
The value of carbon – carbon triple bond length is ……………..
(a) 1.33A
(b) l.20Å
(C) 1.54A
(d) 1.43A
Answer:
(b) 1.20Å

Question 19.
Among the following which one has bond order as 3?
(a) N₂
(b) O₂
(c) HCHO
(d) CH₄
Answer:
(a) N₂

Question 20.
Which one of the flowing has bond order as 2?
(a) N₂
(b) C₂ – H₄
(c) CH₄
(d) HCN
Answer:
(b) C₂H₄

Question 21.
Identify the molecule with bond order 1.
(a) N₂
(b) O₂
(c) H₂
(d) C₂H₄
Answer:
(c) H₂

Question 22.
Which one of the following has zero dipole moment?
(a) HF
(b) H₂
(c) CO
(d) NO
Answer:
(b) H₂

Question 23.
Which one of the following is called polar molecule?
(a) H₃
(b) O₂
(c) F₂
(d) NO
Answer:
(d) NO

Question 24.
Statement I: CuCl is more covalent than NaCI.
Statement II: As compared to Na⁺. Cu⁺ is small and have 3s² 3p⁶ 3d¹⁰ configuration and show greater polarisation.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not the correct explanation of I
(c) Statement I & II are correct but II is wrong
(d) Statement I & II are wrong and II is the correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 25.
Which of the following has see saw shape?
(a) PCl₅
(b) IO₂F₂^–
(c) SOF₄
(d) ClO₃³
Answer:
(b) IO₂F₂^–

Question 26.
Which one of the following has trigonal bipyramidal shape?
(a) SF₆
(b) IF₄⁺
(c) AsF₅
(d) SF₄
Answer:
(c) AsF₅

Question 27.
Which one of the following does not have tetrahedral shape?
(a) NH₄⁺
(b) ClO₄^–
(c) HCHO
(d) CH₄
Answer:
(c) HCHO

Question 28.
Which one of the following has linear shape?
(a) O₃
(b) CO₃^2-
(c) NO_3–
(d) BCl₃
Answer:
(a) O₃

Question 29.
Which of the following has linear shape?
(a) PCI₅
(b) SnBr₂
(c) BeCl₂
(d) CCl₂F₂
Answer:
(c) BeCl₂

Question 30.
Which one of the following has tetrahedral shape?
(a) HCHO
(b) BeCl₂
(c) PbCl₂
(d) CF₂Cl₂
Answer:
(d) CF₂CI₂

Question 31.
Which one of the following pair has T – shapcd structure?
(a) BrF₃, CIF₃
(b) SF₄, IF₄⁺
(c) PCl₅, AsF₅
(d) NH₃, PF₃
Answer:
(a) BrF₃, CIF₃

Question 32.
Which one of the following has pentagonal bipyramidal shape?
(a) XeF₄
(b) XeOF₄
(c) IF₇
(d) IOF₅
Answer:
(c) IF₇

Question 33.
Which one of the following has linear shape?
(a) I₃^–
(b) ICI₄^–
(c) BrF₅
(d) IOF₅
Answer:
(a) I₃^–

Question 34.
Which one of the following is the correct increasing order of bond angle?
(a) H₂O < CH₄ < BF₃ < BeCI₂
(b) BeCI₂ < BF₃ < CH₄ < H₂O
(c) BF₃ < CH₄ < BeCI₂ < H₂O
(d) CH₄ < BeCI₂ < H₂O < BF₃
Answer:
(a) H₂O < CH₄ < BF₃ < BeCI₂

Question 35.
Which one of the following hybridisation takes place in the formation of BeCI₂?
(a) sp²
(b) sp
(c) sp³
(d) dsp²
Answer:
(b) sp

Question 36.
Which hybridisation is possible in BF₃?
(a) sp²
(b) sp
(c) sp³
(d) sp³d
Answer:
(a) sp²

Question 37.
Which one of the following has bond order as 2.5?
(a) O₂
(b) NO
(c) CO
(d) H₂
Answer:
(b) NO

Question 38.
Which one of the following is an electron deficient compound?
(a) Al₂Cl₆
(b) AlBr₃
(c) SF₆
(d) BF₃
Answer:
(d) BF₃

Question 39.
Apply the VSEPR model to XeF₄, which of the following molecular shape is consistent with the model?
(a) Square planar
(b) Tetrahedral
(c) Square pyramidal
(d) Octahedral
Answer:
(a) Square planar

Question 40.
On the basis of molecular orbital theory, select the most appropriate option.
(a) The bond order of O₂ is 2.5 and it is paramagnetic
(b) The bond order of O₂ is 1.5 and it is paramagnetic
(c) The bond order of O₂ is 2 and it is diamagnetic
(d) The bond order of O₂ is 2 and it is paramagnetic
Answer:
(d) The bond order of O₂ is 2 and it is paramagnetic

Question 41.
Which of the following molecule does not exist due to its zero bond order?
(a) H₂^–
(b) He₂⁺
(c) He₂
(d) H₂⁺
Answer:
(c) He₂

Question 42.
Which of the following molecules have bond order equal to 1?
(a) NO, HF, HCl, Li₂, CO
(b) H₂, Li₂, HF, Br₂, HCI
(c) Li₂, B₂, CO, NO, He₂⁺
(d) B₂, CO, He₂⁺, NO, HF
Answer:
(b) H₂, Li₂, HF, Br₂, HCI
Solution:
Bond order of He₂⁺ = 0.5
Bond order of NO = 2.5
Bond order of CO = 3

Question 43.
Arrange the following molecules in decreasing order of bond length.
(a) O₂ >O₂^– > O₂⁺ > O₂^2-
(b) O₂^2- > O₂^– > O₂ > O₂^–
(c) O₂^2- > O₂^– > O₂⁺ > O₂
(d) O₂⁺ > O₂⁺ > O₂^2- > O₂
Answer:
(b) O₂^2- > O₂^– > O₂ > O₂^–
Solution: Since the bond length is inversely proportional to the bond order so option ‘b’ is correct.

Question 44.
Among the following which shows the maximum covalent character?
(a) MgCI₂
(b) FeCI₂
(c) SnCI₂
(d) AICI₃
Answer:
(d) AICI₃

Question 45.
Which of the following has maximum number of lone pairs associated with Xe?
(a) XeF₂
(b) XeO₃
(c) XeF₄
(d) XeF₆
Answer:
(a) XeF₂

Question 46.
During the formation of a chemical bond …………………
(a) energy decreases
(b) energy increases
(c) energy remains zero
(d) energy remains constant
Answer:
(a) energy decreases

Question 47.
Using MO theory, predict which of the following species has the shortest bond length?
(a) O₂⁺
(b) O₂^–
(c) C₂^2-
(d) O₂²⁺
Answer:
(d) O₂²⁺

Question 48.
Identify the incorrect statement regarding the molecule XeO₄.
(a) XeO₄ molecule is tetrahedral
(b) XeO₄ molecule is square planar
(c) There are four pπ – dπ bonds
(d) There are four sp³ – p, s bonds
Answer:
(b) XeO₄ molecule is square planar

Question 49.
Which of the following contains maximum number of lone pairs on the central atom?
(a) ClO₃^–
(b) XeF₄
(c) SF₄
(d) I₃^–
Answer:
(d) I₃^–

Question 50.
Which one of the following is a correct set?
(a) H₂O, sp³, bent
(b) H₂O, sp², linear
(c) NH₄⁺, dsp², square planar
(d) CH₄⁺, dsp² tetrahedal
Answwer:
(a) H₂O, sp³, bent

II. Match the following

Question 1.

Answer:
(b) 4 3 1 2

Question 2.

Answer:
(a) 3 4 1 2

Question 3.

Answer:
(a) 3 4 2 1

Question 4.

Answer:
(b) 3 4 1 2

Question 5.


Answer:
(c) 2 3 4 1

Question 6.

Answer:
(b) 2 4 1 3

Question 7.

Answer:
(c) 4 3 1 2

Question 8.

Answer:
(d) 2 3 4 1

Question 9.

Answer:
(a) 2 4 1 3

Question 10.

Answer:
(a) 4 3 2 1

Question 11.

Answer:
(a) 4 1 2 3

III. Fill in the blanks.

Question 1.
The electrovalent bond is present in ………..
Answer:
NaCI
Solution:
Na⁺ cation and Cl^– anion are held together by electrostatic attractive forces and this is called electrovalent bond.

Question 2.
The structure (or) shape of water molecule is …………
Answer:
inverted ‘V’ shape
Solution:

Question 3.
The structure of CO₂ is ………..
Answer:
linear
Solution:

Question 4.
In the formation of a chemical bond between Na and C[, they attain the stable configuration of …………….
Answer:
Ne, Ar
Solution:
Na⁺: 1s² 2s² 2p⁶ = [Ne]
Cl^–: 1s² 2s² 3s² 3p⁶ = [Ar]

Question 5.
The mutual sharing of one or more pair of electrons between the two combining atoms results in the formation of …………..
Answer:
Covalent bond

Question 6.
Formal charge of an atom can be calculated by the formula ………….
Answer:

Question 7.
The formal charge on the carbon atom in the following structure
is …………………
Answer:
zero
Solution:
Formal charge on carbon atom

Question 8.
The formal charge on both oxygen atoms in the structure
is …………..
Answer:
0
Solution:
Formal charge on oxygen

Question 9.
The formal charge on singly bonded oxygen atom in the structure
is …………..
Answer:
-1
Solution:
Formal charge on singly bonded oxygen atom

Question 10.
The formal charge on the triply bonded oxygen atom in the structure
is …………
Answer:
+ 1
Solution:
Formal charge on triply bonded oxygen atom

Question 11.
The complete transfer of one or more valence electron from one atom to another leads to the formation of ………….
Answer:
Ionic bond

Question 12.
The shape of the molecule is determined approximately by ……………
Answer:
bond angle

Question 13.
The unit of bond enthalpy is …………..
Answer:
kJ mol^-1

Question 14.
The electronegativity of hydrogen and fluorine on Pauling’s scale are …………..
Answer:
2.1 and 4

Question 15.
The unit of dipole moment is ……………
Answer:
Coulomb^-1 m²

Question 16.
The dipole moment of CO₂ is ……………
Answer:
0
Solution:

Question 17.
The shape of sulphur hexafluoride is …………
Answer:
Octahedral

Question 18.
The type of hybridisation takes place in methane is ………………
Answer:
sp³

Question 19.
The type of hybridisation takes place in SF₆ is …………..
Answer:
sp³d²

Question 20.
The number of lone pair of electrons on C – atom present in CO₂ are …………
Answer:
4

Qustion 21.
In SF₆, the bond angle is …………….
Answer:
90⁰

Question 22.
The ions have noble gas electronic configuration was suggested by ……………….
Answer:
Kossel

Question 23.
Tetrachlorornethane is a molecule …………..
Answer:
non polar

Question 24.
In C₂H₄, type of bonds present are ……………
Answer:
Covalent bonds only

Question 25.
Molecule with bond of shape trigonal pyramid is …………..
Answer:
BF₃

Question 26.
When magnesium reacts with oxygen, nature of bond formed is …………….
Answer:
ionic

Question 27.
The number of lone pair of electrons in water molecule is …………..
Answer:
2

Question 28.
Double bonds as compared to single bonds are ……………….
Answer:
Shorter

Question 29.
Number of chlorine atoms which form equatorial bonds in PCI₅ molecule are/is ………..
Answer:
3

Question 30.
The hybridisation of B in BF₃ is ……………
Answer:
sp²

Question 31.
Bond order of O₂, F₃, N₂ respectively are ……………
Answer:
2, 1, 3

Question 32.
Hybridisation which takes place in acetylene is …………….
Answer:
sp

Question 33.
Bond order of O₂, F₂, N₂ respectively are ………….
Answer:
2, 1,3

Question 34.
Hybridisation which takes place in acetylene is …………..
Answer:
sp

Question 35.
The hybndisation of orbitais of N atom in NO₃^–, NO₃⁺ and NH₄⁺ are respectively ……………….
Answer:
sp², sp, sp³

Question 36.
Malleability and ductility of metals can be accounted due to the capacity of layers of …………………. to slide over one another.
Answer:
metal ions

Question 37.
For a stable molecule, the value of bond order must be ……………
Answer:
positive

Question 38.
In acetylene molecule between the carbon atoms there are ………….. σ and ……………. bonds.
Answer:
one, two

IV. Choose the odd one out.

Question 1.
(a) Hydrogen
(b) Chlorine
(c) Neon
(d) Argon
Answer:
(c) Neon. It is mono atomic whereas others are diatomic.

Question 2.
(a) NaCl
(b) CO₂
(c) LiF
(d) MgO
Answer:
(b) CO₂. It contains covalent bond whereas others have ionic bond.

Question 3.
(a) Methane
(b) Ceasium chloride
(c) Ammonia
(d) Boron trifluoride
Answer:
(b) Ceasium chloride. It is an ionic compound whereas others are covalent compounds.

Question 4.
(a) H₂
(b) O₂
(c) Cl₂
(d) F₂
Answer:
(b) O₂. It’s bond order is 2 whereas in others bond order is 1.

Question 5.
(a) BeCI₂
(b) CS₂
(c) BF₃
(d) HCN
Answer:
(c) BF₃. It is AB₃ type whereas others are AB₂ type.

Question 6.
(a) XeO₂F₂
(b) PCI₅
(c) AsF₅
(d) SOF₄
Answer:
(a) XeO₂F₂. It is AB₄L type whereas others are AB₅ type.

V. Choose the correct pair.

Question 1.
(a) NaCI – ionic compound
(b) NH₃ – coordinate compound
(c) BF₃ – ionic compound
(d) H₂ – ionic compound
Answer:
(a) NaCl – ionic compound

Question 2.
(a) O₂ – Bond order 3
(b) H₂ – Bond order 2
(c) N₂ – Bond order 3
(d) Cl₂ – Bond order 2
Answer:
(c) N₂ – Bond order 3
[N = N] Bond order is 3.

Question 3.
(a) CH₄ – ionic bond
(b) BF₃ – dative bond
(c) NH₃ – metallic bond
(d) CCI₄ – covalent bond
Answer:
(d) CCI₄ – covalent bond

Question 4.
(a) CH₄ – 107° 18’
(b) H₂O – 109°28’
(c) NH₃ – 104°35’
(d) BF₃ – 120°
Answer:
(d) BF₃ – 120°

Question 5.
(a) AB₃ – Linear
(b) AB₃ – V-shape(or)bent
(c) AB₄ – Trigonal planar
(d) AB₅ – T-shape
Answer:
(a) AB₃ – Linear

VI. Choose the incorrect pair.

Question 1.
(a) CS₂ – Linear
(b) BF₁ – Trigonal planar
(c) CH₄ – T-shape
(d)NH₃ – Pyramidal
Answer:
(c) CH₄ – T-shape

Question 2.
(a) AB₃ – Trigonal planar
(b) AB₃L₂ – T-shape
(c) AB₅ – Trigonal bipyramidal
(d) AB₃L – Bent
Answer:
(a) AB₃L : Bent.
Actually AB₃L is pyramidal.

Question 3.
(a) AB₇ – IF₇
(b) AB₄L₂ – ICI₄^–
(c) AB₆ – XeOF₄
(d) AB₅L – IF₅
Ans.
(c) AB₆ – XeOF₄
Actually XeOF₄ is AB₅L type.

Question 4.
(a) Fluorine – Bond order 1
(b) Oxygen – Bond order 2
(c) Nitrogen – Bond order 2
(d) Cyanide – Bond order 3
Answer:
(c) Nitrogen – Bond order 2.
Actually N = N bond order is 3.

Question 5.
(a) CH₄ – sp³
(b) PCI₅ – sp³d
(c) BeCl₂ – sp
(d) BF₃ – sp³d²
Answer:
(d) BF₃ – sp³d²
Actually BF⁴ is sp² hybridised.

VII. Assertion and Reason.

Question 1.
Assertion (A): Diatomic molecules such as H₂, O₂, F₂ are non-polar molecules.
Reason (R): H₂, O₂, F₂ have zero dipole moment.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A): HF, HCl, CO and No are polar molecules.
Reason (R): They have non zero dipole moments and so they are polar molecules.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 3.
Assertion (A): H₂, Li₂, C₂, N₂ are diamagnetic.
Reason (R): All have no unpaired electrons and so they are diamagnetic.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A): B₂, O₂, NO are paramagnetic in nature.
Reason (R): They have unpaired electrons and are paramagnetic.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 5.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Absence of bond gap is the main reason for high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).

Question 6.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Due to thermal excitation of many electrons from the valence band to the conductance band, metals have high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is wrong but (R) is correct.
(d) (A) is correct but (R) is wrong.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

VIII. Choose the correct statement.

Question 1.
(a) The metallic luster is due to reflection of light by the electron cloud.
(b) Metals have low inciting point and low boiling point.
(c) Metals have low thermal conductivity.
(d) Electrical conductivity of metals is low.
Answer:
(a) The metallic luster is due to reflection of light by the electron cloud.

Question 2.
(a) NO molecules is diamagnetic
(b) O₂ molecules is paramagnetic
(c) N₂ molecules is paramagnetic
(d) Li₂ molecules is paramagnetic
Answer:
(b) O₂ molecules is paramagnetic

Question 3.
(a) BeCl₂ undergoes sp³ hybridisation
(b) 8F₃ undergoes sp³d hybridisation
(c) CH₄ undergoes sp³d² hybridisation
(d) PCl₅ undergoes sp³d hybridisation
Answer:
(d) PCl₅ undergoes sp³d hybridisation

Samacheer Kalvi 11th Chemistry Chemical Bonding 2 Mark Questions and Answers

I. Write brief answer to the following questions.

Question 1.
What are chemical bonds?
Answer:
The interatomic attractive forces which holds the constituent atoms/ions together in a molecule are called chemical bonds.

Question 2.
State octet rule.
Answer:
The atoms transfer or share electrons so that all the atoms involved in chemical bonding obtain eight electrons in their outer shell (valence shell). It is called octet rule.

Question 3.
What is meant by covalent bond?
Answer:
The mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

Question 4.
Draw the lewis structure of

1. H₂O
2. SO₃.

Answer:

Question 5.
Calculate the formal charge on the carbon atom and oxygen atom in the structure

Answer:
Formal charge on carbon

Formal charge on oxygen

Question 6.
Calculate the formal charge on the carbon atom and oxygen atom in the structure

Answer:
Formal charge on carbon

Formal charge on singly bonded oxygen

Formal charge on triply bonded oxygen

Question 7.
Among and  which is a preferable structure for CO₂ molecule why?
Answer:
Structure I of
Formal charge on carbon

Formal charge on oxygen

Structure II of
Formal charge on carbon

Formal charge on singly bonded oxygen

Formal charge on triply bonded oxygen

A structure in which all formal charges are zero is preferred over the one with non – zero charges. In case of CO₂ structure , structure I is preferred over the structure II as it has zero formal charge for all the atoms.

Question 8.
Draw the lewis structures of a few molecules containing odd electrons.
Answer:
Few molecules have a central atom with an odd number of valence electrons. For example, in nitrogen dioxide and nitric oxide all the atoms does not have octet configuration.

Question 9.
Draw the lewis structure of PCl₅ and SF₆
Answer:

Question 10.
Define bond length.
Answer:
The distance between the nuclei of two covalently bonded atoms is called bond length. For e.g., in a covalent molecule A – B. the bond length is equal to the sum of the radii of bonded atoms. i.e., r_A + r_ß = bond length.

Question 11.
Prove that bond order is inversely proportional to bond length.
Answer:
1. Bond order ∝

2. An example for illustrating the above equation is Carbon – carbon single bond length (I .54Å) is longer than the carbon-carbon double bond length (1 .34Å) and the carbon- carbon triple bond length (1 .20Å).

Question 12.
Define Bond angle.
Answer:
Covalent bonds are directional in nature and are oriented in specific direction in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.

Question 13.
Define Resonance.
Answer:
The similar structures in which the relative position of the atoms are same but they differ in the position of bonding and lone pair of electrons are called resonance structures and this phenomenon is called resonance.

Question 14.
What are polar and non-polar molecules?
Answer:
1. Diatomic molecules such as H₂, O₂, F₂ have zero dipole moment and are called non polar molecules.

2. Molecules such as HF, HCl, CO, NO have non zero dipole moment values and are called polar molecules.

Question 15.
What is meant by polarisaion?
Answer:
The ability of a cation to polarise an anion is called its polarising ability and the tendency of the anion to get polarised is called its polarisability. This phenomenon is known as polarisation.

Question 16.
Among NaCI, MgCI₂ and AICI₃ which shows more covalent character? Why?
Answer:
Among, the ionic compounds NaCI, MgCl₂ and AICI₃ the charge of the cation increases in the order Na⁺ < Mg²⁺ < Al³⁺, thus the covalent character also follows the same order NaCl < MgCI₂ < AlCI₃. So AICI₃ shows more covalent character.

Question 17.
Lithium chloride is more covalent than sodium chloride. Justify this statement.
Answer:
1. The smaller cation and larger anion shows greater covalent character due to greater extent of polarisation.

2. The size of Li⁺ ion is smaller than Na⁺ ion and hence the polarising power of Li⁺ ion is more. So lithium chloride is more covalent than sodium chloride.

Question 18.
Lithium iodide is more covalent than Lithium chloride. Give reason.
Answer:
Lithium iodide is more covalent than Lithium chloride as the size of I^– ion is larger than Cl^– ion. Hence I^– ion will be more polarised than Cl^– ion by the cation Li⁺. So LiI is more covalent than LiCl.

Question 19.
Draw the structure of AB₄L and AB₃L₂ type of molecules with example.
Answer:
AB₄L:
Bond pairs = 4
Lone pair = 1
Shape = see saw
e.g., SF₄ Strucutre

AB₃L₂:
Bond pairs = 3
Lone pair = 2
Shape = T shaped
e.g., CIF₃ Strucutre

Question 20.
Draw the structure of AB₄L₂ and AB₇ type of molecules with example
Answer:
1. AB₄L₂:
Bond pairs = 4
Lone pairs = 2
Shape = Square planare.g., XeF₄

2. AB₇:
Bond pairs = 7
Lone pairs = Nil
Shape = pentagonal bipyramidal
e.g., IF₇

Question 21.
Explain the bond formation of hydrogen molecule.
Answer:
1. Electronic configuration of hydrogen atom is 1s¹.

2. During the formation of H₂ molecule, the 1 s orbitais of two hydrogen atoms containing one unpaired electron with opposite spin overlaps with each other along the internuclear axis.

This overlap is called s – s overlap. Such axial overlap results in the formation of a sigma (σ) covalent bond.

Question 22.
Explain the bond formation of fluorine molecule.
Answer:
1. Valence shell electronic confIguration of fluorine atom is 2s² 2p_x² 2p_y² 2p_z¹

2. When the half filled p_z orbitais of two fluorine atoms overlap along the z – axis, a σ covalent bond is formed between them.

Question 23.
How is HF molecule formed?
Answer:

1. Electronic configuration of hydrogen atom is 1s¹.
2. Valence shell electronic configuration of fluorine atom is 2s² 2p_x² 2p_y² 2p_z¹.
3. When half filled is orbital of hydrogen linearly overlaps with a halt filled 2p_z orbital of fluorine, a σ covalent bond is formed between hydrogen and fluorine.

Question 24.
What is meant by metallic bond?
Answer:
The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is generally known as the metallic bond.

Question 25.
Why metallic bonding is referred as electronic bonding?
Answer:
1. Metallic crystals are an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to the ionisation of the valence electrons of the atom of the metal.

2. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is also referred to as electronic bonding.

Question 26.
Metals have high density. Give reason.
Answer:
The electrostatic attraction between the metal ions and the free electrons yields a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

Question 27.
Metals are ductile in nature. why?
Answer:
In the close packed structure of metallic crystal. it contains many slip planes along which movement can occur during mechanical loading, so the metal acquires ductility.

Question 28.
Give reason behind the lustrous nature, high melting point and boiling point of metals?
Answer:
1. The metallic lustre is due to the reflection of light by the electron cloud.

2. As the metallic bond is strong enough. the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

Question 29.
Metals are very good electrical conductors. Why?
Answer:
1. The bonding in metal is explained by molecular orbital theory. As per this theory. the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each, and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

Question 30.
Metals have high thermal conductivity. Give reason.
Answer:
High thermal conductivity of metals is due to thermal excitation of many electrons from the valence band to the conduction band.

Question 31.
Except Cu, Ag and Au, most metals are black. Why?
Answer:
Most metals are black except copper, silver and gold. It is due to the absorption of light of all wavelengths. Absorption of light of all wavelengths is due to the absence of band gap in metals.

Question 32.
Write the favourable factors for the formation of ionic bond.
Answer:

1. Low ionisation enthalpy of metal atoms.
2. High electron gain enthalpy of non-metal atoms.
3. High lattice enthalpy of compound formed.

Question 33.
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:

Because of two lone pair of electrons on O – atom, repulsion on bond pairs is greater in H₂O in comparison to NH₃. Thus, the bond angle is less in H₂O molecules.

Question 34.
Write the significance/applications of dipole moment.
Answer:

1. In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those with zero dipole moments are non-polar in nature.
2. In the determination of shapes of molecules.
3. In calculating the percentage of ionic character.

Question 35.
Although both CO₂ and H₂O are triatomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
Answer:
In CO₂, there are two C = O bonds. Each C = O bond is a polar bond. The net dipole moment of CO₂ moleculc is zero. This is possible only if CO₂ is a linear molecule. (O = C = O). The bond dipoles of two C = O bonds cancels the dipole moment of each other.

Whereas, H₂O molecule has a net dipole moment (1.84 D). H₂O molecule has a bent structure because here the O – H bonds are oriented at an angle of 104.5° and do not cancel the bond dipole moments of each other.

Question 36.
What is the total number of sigma and pi bonds in the following molecules?

1. C₂H₂
2. C₂H₄

Answer:

1. H – C = C – H
Sigma bond = 3
π bond = 2

2.
Sigma bond = 5
π bond = 1

Question 37.
Use molecular orbital theory to explain why the Be₂ molecule does not exist
Answer:
E.C of Be = 1 s² 2s²
M.O.E.C of Be₂ = σ1s² σ2s² σ*2s²
Bond order = \(\frac { 1 }{ 2 }\) (4 – 4) = 0
Hence, Be₂

Question 38.
Compare the relative stabililty of the following species and indicate their magnetic properties O₂, O₂⁺, O₂^– (superoxide), O₂^2-(peroxide)
Answer:
O₂ – Bond order = 2 paramagnetic
O₂⁺ – Bond order = 2.5, paramagnetic
O₂^– – Bond order = 1.5, paramagnetic
O₂^2- – Bond order = 1, diagmagnetic
Order of relative stability is:

Question 39.
Account for the following:

1. Water is a liquid while H₂S is a gas
2. NH₃ has higher boiling point than PH₃.

Answer:

1. In case of water, hydrogen bonding causes association of the H₂O molecules. There is no such hydrogen bonding in H₂S, that is why it is a gas.
2. There is hydrogen bonding in NH₃ but not in PH₃.

Question 40.
Why B₂ is paramagnetic in nature while C₂ is not?
Answer:
The molecular orbital electronic configuration of both B₂ and C₂ are:
B₂:

C₂:

Since, B₂ has two unpaired electrons, therefore, B₂ is paramagnetic C₂ has no unpaired electron, therefore, C₂ is diamagnetic.

Samacheer Kalvi 11th Chemistry Chemical Bonding 3 Mark Questions and Answers

Question 1.
Draw the lewis structure of

1. Nitrogen
2. Carbon
3. Oxygen

Answer:

Question 2.
Draw the lewis structure of

1. Ammonia
2. Methane
3. Dinitrogen pentoxide.

Answer:
Lewis dot structures of Molecules

Question 3.
Calculate the bond enthalpy of OH bond in water.
Answer:
1. In the case of polyatomic molecules with two or more same bond types, the arithmetic mean of the bond energy value of the same type of bonds is considered as average bond enthalpy.

2. For e.g., in water, there are two OH bonds present and the energy needed to break them are not same.

3. H₂O(g) → H(g) + OH(g)
∆H₁ = 502 kJ mol^-1
OH(g) → H(g) + O(g)
∆H = 427 kJ mol
The average bond enthalpy of OH bond in water = \(\frac { 502 + 427 }{ 2 }\) = 464.5 kJ mol^-1

Question 4.
Explain how the ionic character in a covalent bond is related to electronegativity?
Answer:
1. The extent of ionic character in a covalent bond can be related to the electronegativity difference of the bonded atoms.

2. In a typical polar molecule A^δ- – B^δ+ the electronegativity difference (X_A – X_B) can be used to predict the percentage of the ionic character as follows

3. If the electronegativity difference X_A – X_B is equal to 1.7, then the bond A – B has 50% ionic character.

4. If it is greater than 1.7, then the bond X_A – X_B has more than 50% ionic character.

5. If it is greater than 1.7, then the bond A – B has more than 50% ionic character.

6. If it is lesser than 1.7, then the bond A – B has less than 50% ionic character.

Question 5.
CuCI is more covalent than NaCl. Give reason.
Answer:
1. Cations having ns²np⁶nd¹⁰ configuration show greater polansing power than the cations with ns² np⁶ configuration. Hence they show greater covalent character.

2. CuCI is more covalent than NaCI. As compared to Na⁺ (1.13Å), Cu⁺(0.6Å)is small and has 3s²3p⁶3d¹⁰ configuration.

3. Electronic configuration of Cu⁺: [Ar] 3d¹⁰
Electronic configuration of Na⁺: [He] 2s²2p⁶
So CuCI is more covalent than NaCI

Question 6.
Draw the structure of AB₂, AB₃, AB₃L type of molecules with example.
Answer:
1. AB₂
Number of bond pairs = 2
Shape = Linear
Example – BeCl₂

2. AB₃
Number of bond pairs = 3
Shape = Tirgonal planar
Example – BF₃

3. AB₂L
Number of bond pairs = 2
Number of lone pairs = 1
Shape = Bent (or) inverted V shape
Example – O₃

Question 7.
Give example and structure of

1. AB₃L
2. AB₅
3. AB₂L₂

type of molecules with example.
Answer:
1. AB₃L
Number of bond pairs = 3
Number of lone pairs = 1
Shape = Pyramidal
Example – NH₃

2. AB₅
Number of bond pairs = 5
Number of lone pairs = 0
Shape = Trigonal bipyramidal
Example = PCIC₅

3. AB₂L₂
Number of bond pairs = 2
Number of lone pairs = 2
Shape = Bent
Example – H₂O

Question 8.
Draw the shape of

1. XeF₂
2. IOF₅
3. XeOF₄

Answer:
1. XeF₂

2. IOF₅

3. XeOF₄

Question 9.
Explain the bonding in oxygen molecule.
Answer:
1. Valence shell electronic configuration of oxygen atom is
2s² 2p_x² 2p_y¹2p_z¹

2. When the half filled p_z orbitaIs of two oxygen atoms overlap along the z – axis a σ covalent bond is formed between them. Other two hail filled p_y orbitais of two oxygen atoms overlap laterally to form a π – cova1ent bond between the oxygen atoms.

3. Thus in oxygen molecule, two oxygen atoms are connected by two covalent bonds (double bond). The other two pair of electrons present in the 2s and 2p_x orbital do not involve in bonding and remains as lone pair on the respective oxygen atoms.

Question 10.
Explain about the molecular orbital diagram of hydrogen molecule.
Answer:

1. Electronic configuration of H atom 1s¹
2. Electronic configuration of H, molecule – σ1s¹
   Bond order
3. Molecule (H₂) has no unpaired electrons, hence it is diamagnetic.

Question 11.
Draw and explain the M.O. diagram of lithium molecule.
Answer:

1. Electronic configuration of Li atom – 1s¹
2. Electronic configuration of Li₂ molecule is ais2 σ*1s² σ*1s² σs²
3. Bondorder = N_b – N_b/2 = 4 – 2/2
4. Li₂ molecule has no unpaired electrons, hence it is diamagnetic.

Question 12.
Draw and explain the M.O. diagram of Boron molecule.

Answer:

1. Electronic configuration of B = 1s² 2s² 2p³
2. Electronic configuration of B, = σ1s² σ*1s²σ2s² σ*2s² π2p_x¹ π 2p_z¹
3. Bond order
4. B₂ molecule has two unpaired electrons hence it is paramagnetic.

Question 13.
Draw and explain the molecular orbital diagram of carbon molecule.
Answer:

1. Electronic configuration of C atom – 1s² 2s² 2p²
2. Electronic configuration of C₂ molecule is σ1s² σ*1s² σ*2s² σ*2s² π 2p_x² π 2p_y²
3. Bond order

Question 14.
Write Lewis dot symbols for atoms of the following elements: Mgq Naq B O, N, Br.
Answer:

Question 15.
write Lewis symbols for the following atoms and ions: S and S^2-; Al and Al³⁺; H and OH^–
Answer:

Question 16.
Draw the Lewis structures for the following molecules and ions H₂S, SiCl₄, BeF₂, CO₃^2-, HCOOH
Answer:

Question 17.
Define Octet rule. Write its significance and limitations.
Answer:
Octet rule:
Atoms of elements combine with each other in order to complete their respective octet so as to acquire the stable nearest noble gas configuration.

Significance:
It helps to explain why dilfferent atoms combine with each other to form ionic compounds or covalent compounds.

Limitations of Octet rule:
1. According to octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements.

However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: XeF₂, XeF₄, XeO₃ etc. Therefore, validity of the octet rule has been challenged.

2. This theory does not account for the shapes of molecules.

Question 18.
Write the resonance structure for SO₃, NO₂ and NO₃
Answer:

Question 19.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.
For example, in
there are only 4 bond pairs, but in  there are two bond pairs and two lone pairs.

Question 20.
Distinguish between a sigma bond and a pi bond
Answer:
Sigma (σ) Bond

1. σ – bond is formed by the axial overlap of the atomic orbitais.
2. The bond is quite strong.
3. Only one lobe ofthep-orbitals is involved in the overlap.
4. Electron cloud of the molecular orbital is symmetrkal around the internuclear axis.

Pi (π) Bond

1. π – bonnd is formed by the sidewise overlap of atomic orbitais.
2. It is comparatively a weaker bond.
3. Both lobes of the p-orbitais are involved in the overlap.
4. The electron cloud is not symmetrical.

Question 21.
Write the Important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
1. The combining atomic orbitals should have comparable energies. For example, is orbital of one atom can combine with 1s atomic orbital of another atom, 2s orbitai can combine with 2s orbital and so on.

2. The combining atomic orbitals must have proper orientations so that they are able to overlap to a considerable extent.

3. The extent of overlapping should be large.

Question 22.
What are Lewis structures? Write the Lewis structure of H₂, BeF₂ and H₂O.
Answer:
The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures. The Lewis structure of

Question 23.
What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory?
Answer:

1. The shape of a molecule depends upon the no. of electron pairs around the central atom.
2. There is a repulsive force between the electron pairs, which tend to repel one another.
3. The electron pairs in space tend to occupy such positions that they arc at maximum distance, so that the repulsive force will be minimum.
4. A multiple bond is treated as lilt is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.

Question 24.
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with four H atoms at the corners of the square and C atom at its centre. Explain why CH₄ is not square planar?
Answer:
Electronic configuration of carbon atom: C: σ1s²2s²2p².
in the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridisation in CH₄ molecule and takes tetrahedral shape.

For a square planar shape, the hybridisation of the central atom has to be dsp³. However, an atom of carbon does not have d – orbitals to undergo dsp³ hybridisation. Hence, the structure of CH₄ is tetrahedral.

Question 25.
Explain why BeH₂ molecule has a zero dipole moment although the Be – H bonds are polar.
Answer:
The Lewis structure for BeH₂ molecule is as follows:. There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH₂, is of the type AB₂. It has a linear structure.

Dipole moments of each Be – H bond are equal and opposite in direction. Therefore, they nullify each other. Hence, BeH₂ has a net zero dipole moment.

IV. Answer the following questions in detail:

Question 1.
Explain about Kossel-Lewis approach to chemical bonding.
Answer:
1. Kossel and Lewis approach to chemical bonding is based on the inertness of the noble gases which have little or no tendency to combine with other atoms.

2. They proposed that noble gases are stable due to their completely filled outer electronic configuration.

3. Elements other than noble gases try to attain the completely filled outer electronic configuration by losing, gaining or sharing one or more electrons from their outer shell.

4. For e.g., sodium loses one electron to form Na ion and chlorine accepts that electron to give chloride ion, Cl^–.These two ions are held together by electrostatic attractive forces, a bond known as an electrovalent bond.

5. In diatomic molecules such as nitrogen and oxygen, they achieve the stable noble gas electronic configuration by mutual sharing of electrons.

6. Lewis introduced a scheme to represent the chemical bond and the electrons present in the outer shell of the atom called Lewis dot structure.

7. For example, the electronic configuration of nitrogen is 1s²2s²2p³. It has 5 electrons in its outer shell. The lewis structure of nitrogen is

8. In N, molecule, equal sharing of 3 electrons from each nitrogen atom takes place as fol lows

Question 2.
What is meant by covalent bond?
Explain the covalent bonding in H₂, O₂, N₂.
Answer:
1. Mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

2. If two atoms share just one pair of electron, a single covalent bond is formed as in the çase of hydrogen molecule (H₂).

3. If two or three electron pairs are shared benveen the two combining atoms, then the covalent bond is called double bond and triple bond respectively, as in the case of O₂ and N₂ molecules respectively.

Question 3.
What is an ionic bond?
Explain about the formation of ionic bond with a suitable example.
Answer:
1. The complete transfer of electrons leads to the fomiation of a cation and an anion. Both these ions are held together by electrostatic attractive forces which is known as ionic bond.

2. KCl: Potassium chloride
Electronic configuration of K [Ar] 4s
Eleçtronic configuration of Cl = [Ne] 3s² 3p⁵

3. Potassium has 1 electron in its valence shell and chlorine has 7 electrons in its valence shell.

4. By losing one electron potassium attains the nearest inert gas configuration of Argon and becomes a unipositive cation (K) and chlorine accepts this electron to become uninegative chloride ion (CI) to attain the stable configuration of nearest noble gas, Argon.

5. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

6. During the formation of one mole of potassium chloride crystal from its constituent ions, 718 kJ of energy is released. This favours the formation of KCl and its stabilisation.

Question 4.
Define coordinate covalent bond. Illustrate the formation of coordinate covalent bond with a suitable example.
Answer:
1. In the bond formation, one of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation and these electrons are shared by both the combining atoms, and the bond formed is called coordinate covalent bond.

2. The combining atom which donates the pair of electron is called the donor atom and the other atom is called the acceptor atom. This bond is denoted by an arrow starting from the donor atom pointing towards the acceptor atom.

3. For example, in ferricyanide ion [Fe(CN)₆]^4- each cyanide ion (CN^–) donates a pair of electrons to form a coordinate bond with iron (Fe²⁺) and these electrons are shared by Fe²⁺ and CN^– ions.

4.

5. Ammonia having a lone pair of electrons donates its pair to an electron deficient molecule such as BF₃ to form a coordinate covalent bond

Question 5.
Explain about valence bound theory for the formation of H₂ molecule.
Answer:
1. Two hydrogen atoms H_a and H_b are separated by infinite distance. At this stage, there is no interaction between these two atoms and the potential energy of this system is arbitrarly taken as zero.

2. As these two atoms approach each other, in addition to electrostatic attractive forces between the nucleus and its own electrons, the following new forces begins to operate

3. The new attractive forces  arise between:

• nucleus of H_a and valence electron of H_b
• nucleus of H_b and the valence electron of H_a

4. The new repulsive forces  arise between:

• the nucleus Of H_a and H_b
• the valence electrons of H_a and H_b

5. The attractive forces tend to bring H_a and H_b together whereas the repulsive forces tends to push them apart.

6. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than repulsive forces and the potential energy decreases.

7. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.

8. At this stage, there is a maximum overlap between the atomic orbitals of H_a and H_b and atoms H_a and H_b are now said to be bonded together by a covalent bond.

Question 6.
What arc the salient features of Valence Bond (VB) theory?
Answer:
1. When half filled orbitals of two atoms overlap, a covalent bond will be formed between them.

2. The resultant overlapping orbitals are occupied by the two electrons with opposite spins. For example when H₂ is formed, the two is electron of two hydrogen atoms get paired up and occupy the overlapped orbitals.

3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.

4. Each atomic orbital has a specific direction (excepts-orbital which is spherical) and hence orbital overlap takes place in the direction that maximises overlap.

5. Depending upon the nature of overlap, the bonds are classified as σ covalent bond and π it covalent bond.

6. When two atomic orbitals overlap linearly along the axis, the resultant bend is called a sigma (σ) bond. This overlap is also called or “axial overlap”.

7. When two atomic orbitals overlap sideways the resultant covalent bond is called a pi (π) bond.

Question 7.
Explain about sp hybridisation with suitable example.
Answer:

1. bond rormation in Beryllium chloride takes place by sp hybridisation.
2. The valence shell of Beryllium has the electronic configuration as follows:

3. In BeCl₂, both the Be – Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explains this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital.

4. Now the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction.

5. Each of the sp hybridised orbitals linearly overlap with p orbital of the chlorine to form a covalent bond between Be and Cl atoms as follow:

Question 8.
Explain the formation of methane using VB theory?
Answer:
1. Methane is formed by sp³ hybridisation. In CH₄ molecule, the central carbon atom is bounded to four hydrogen atoms.

2. The ground state valence shell electronic configuration of carbon is [He] 2s²2p_x² 2p_y¹ 2p_x⁰

3. En order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in 2s orbital of carbon is promoted to its 2P_z orbital in the excited state.

4. The one 2s orbital and three 2p orbitals of carbon atom mixes to give four equivalent sp³ hybridised orbitals. The angle between any of the two sp³ hybridised orbitals is 109°•28’

5. The Is orbital of the four hydrogen atoms overlap linearly with the four sp³ hybridised orbitais of carbon to form four C – H σ bonds in the methane molecule as follows

Question 9.
Explain sp³d hybridisation with a suitable example.
Answer:
1. In the PCl₅ molecule, the central atom phosphorous is covalently bonded to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes sp³d² hybridisation which involves its one 3s orbital, three 3p orbitals and one vacant 3d orbital (d_z²)

2. The ground state electronic configuration of phosphorous is [Ne] 3s²3p_x²3p_y¹3p_z¹

3. One of the paired electrons in the 3s orbital of phosphorous ¡s promoted to one of its vacant 3d orbital (dz²) in the excited state.

4. The 3p_z orbitals of the five chlorine atoms linearly overlap along the axis with the five sp³d hybridised orbitals of phosphorous to form the five P – CI bonds as follows.

Question 10.
Explain about. sp³d² hybridisation with an example.
Answer:
1. In sulphur hexafluoride SF₆, the central atom sulphur extend its octet to undergo sp³d hybridisation to generate six sp³d² hybridised orbitals which accounts for six equivalent S – F bonds.

2. The ground state electronic configuration of sulphur is [Ne] 3s²3p_x¹3p_y¹3p_z¹

3. One electron each form 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals d_z² and d_x²-y² in the excited state.

4. A total of six valence orbitals from sulphur (one 3s orbital, three 3p orbitals and two 3d orbitals) (d_x² and d_x²-y²) which mixes to give six equivalent sp³d² hybridised orbitals. The orbital geometry is octahedral.

5. The six sp³d² hybridised orbitals of sulphur overlaps linearly with 2p_z orbital of six fluorine atoms to form the six S – F bonds in sulphur hexa fluoride.

Question 11.
Explain about the salient features of molecular orbital theory.
Answer:
1. When atoms combine to form molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.

2. The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.

3. The number of molecular orbitals formed is the same as the number of combining atomic orbitals. half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.

4. The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called σ*, π* and δ*.

5. The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Autbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.

6. Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using
the foLlowing equation:

N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.

7. A bond order of zero value indicates that the molecule does not exist.

Question 12.
Explain the MO diagram for NO molecule.

Answer:

1. Electronic configuration of N atom is 1s² 2s² 2p³
2. Electronic configuration of O atom is 1s² 2s² 2p⁴
3. Electronic configuration of NO molecule is
   
4. Bond order
5. NO molecule has one unpaired electron, hence it is paramagnetic.

Question 13.
Explain about metallic bonding.
Answer:
1. The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.

2. According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.

3. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.

4. The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

5. As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.

6. As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no and thermal conductivity. The metallic lustre is due to the reflection of light by the electron cloud.As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

7. High thermal conductivity of metals is due. to thermal excitation of many electrons from the valence bond to the conduction band.

Question 14.
Explain about the bonding in metals by molecular orbital theory.
Answer:
1. According to molecular orbital theory the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

4. High thermal conductivity is due to thermal excitation of many electrons from the valence band to the conduction band.

5. With an increase in temperature, the electrical conductivity decreases due to vigorous thermal motion of lattice ions that disrupts the uniform lattice structure. that is required for free motion of electrons within the crystal.

Common Errors

1. The number of bonds formed by elements may go wrong.
2. When writing Lewis structure, electrons may be written in an irregular way.
3. Coordinate covalent bond should not be written as a line

Rectifications

1. Always hydrogen and fluorine form 1 bond Oxygen 2 bonds, Nitrogen 3 bonds, Carbon 4 bonds
2. When writing lewis structure, each atom should be surrounded by eight electrons in such a way as 4 pairs of electrons.
3. Coordinate covalent bond should be written from donor atom to acceptor atom as an arrow mark Donor→ Acceptor

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Samacheer Kalvi 11th Chemistry Chapter 9 Solutions Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Solutions Multiple Choice Questions

Question 1.
The molality of a solution containing 1 .8g of glucose dissolved in 250g of water is …………
(a) 0.2 M
(b) 0.01 M
(c) 0.02 M
(d) 0.04 M
Answer:
(d) 0.04 M
Solution:

Question 2.
Which of the following concentration terms is/are independent of temperature?
(a) molality
(b) molarity
(c) mole fraction
(d) (a) and (c)
Answer:
(d) (a) and (c)
Solution:
Molality and mole fraction are independent of temperature.

Question 3.
Stomach acid, a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide
Al(OH)₃ + 3HCl(aq) → AlCl₃ + 3H₂O
How many millilitres of 0.1 M Al(OH)₃ solution are needed to neutralise 21 mL of 0.1 M HCl
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) none of these
Answer:
(b) 7 mL
Solution:
M₁ x V₁ = M₂ x V₂
∵ 0.1 M Al(OH)₃ gives 3 x 0.1 = 0.3 M OH^– ions .
0.3 x V₁ = 0.1 x 21
V₁= \(\frac { 0.1 x 21 }{ 0.3 }\) = 7ml

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 x 10⁴ atm at 300K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300K?
(a) 1 x 10^-4
(b) 1 x 10^-6
(c) 2 x 10^-5
(d) 1 x 10^-5
Answer:
(d) 1 x 10^-5
Solution:
P_N2 = 0.76atm
K_H = 7.6 x 10⁴
x = ?
P_N2 = K_H . x
0.76 = 7.6 x 10⁴x x
x = \(\frac { 0.76 }{ 7.6\times { 10 }^{ 4 } }\) = 1 x 10^-5

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350K is 8 x 10⁴ atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350K and 4 atm pressure is ………….
(a) 4 x 10^-4
(b) 4 x 10⁴
(c) 2 x 10^-2
(d) 2.5 x 10^-4
Answer:
(d) 2.5 x 10^-4
Solution:
K_H = 8 x 10⁴
(x_N2 )_{in air} = 0.5
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction Total pressure
= O.5 x 4 = 2

Question 6.
Which one of the following is incorrect for ideal solution?
(a) ∆H_mix = 0
(b) ∆U_mix = o
(c) ∆P = P_Observed – P_{Calculated by raoults law} = 0
(d) ∆G_mix = 0
Answer:
(d) ∆G_mix = 0
Solution:
For an ideal solution, ∆S_mix \(\neq\) 0; Hence ∆G_mix \(\neq\) 0
∴ Incorrect is ∆G_mix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Answer:
(c) CO₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.

Question 8.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………
(a) P₁ + x₁(P₂ – P₁)
(b) P₂ – x₁(P₂ + P₁)
(c) P₁ – x₂(P₁ – P₂)
(d) P₁ + x₂(P₁ – P₂)
Answer:
(c) P₁ – x₂(P₁ – P₂)
Solution:
P_total = P₁ + P₂
= P₁ x₁ + P₂x₂
= P₁(1 – x₂) + P₂x₂
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂(P₁ – P₂)
[∵x₁ + x₂ = 1
x₁ = 1 – x₂]

Question 9.
Osomotic pressure (π) of a solution is given by the relation ……………
(a) π = nRT
(b) πV = nRT
(c) πRT = n
(d) none of these
Answer:
(b) πV = nRT
Solution:
n = CRT
n = \(\frac { n }{ V }\)
π V = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
(a) Acetone + chloroform
(b) Water + nitric acid
(c) HCI + water
(d) ethanol + water
Answer:
(d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A dissolved in water will be …………
(a) \(\frac { 2x }{ y }\)
(b) \(\frac { y }{ 0.2x }\)
(c) \(\frac { 0.2x }{ y }\)
(d) \(\frac { 5x }{ y }\)
Answer:
(d) \(\frac { 5x }{ y }\)
Solution:
Given,

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If K_b = 0.52, the boiling point of this solution will be …………..
(a) 102°C
(b) 100°C
(c) 101°C
(d) 100.52°C
Answer:
(c) 101°C
Solution:
\(\frac { ΔP }{ P° }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } }\)
W₂ = 6.5g
W₁ = 100g
K_b = 0.52

Question 13.
According to Raoults law, the relative Lowering of vapour pressure for a solution is equal to….
(a) molefraction of solvent
(b) mole fraction of solute
(c) number of moles of solute
(d) number of moles of solvent
Answer:
(b) mole fraction of solute
Solution:
\(\frac { ∆P }{ P° }\) = x₂ (Mole fraction of the solute)

Question 14.
At same temperature. which pair of the following solutions are isotonic?
(a) 0.2 M BaCl₂ and 0.2M urea
(b) 0.1 M glucose and 0.2 M urea
(c) 0.1 MNaCl and 0.1 MK₂SO₄
(d) 0.1 MBa(NO₃)₂ and 0.1 MNa₂ SO₄
Answer:
(d) 0.1 M Ba (NO₃)₂ and 0.1 M Na₂ SO₄
Solution:
0.1 x 3 ion [Ba² + 2NO₃^–], 0.1 x 3 ion [2Na⁺, SO₄^–]

Question 15.
The empirical formula of a non-electrolyte(X) is CH₂O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
(a) C₂H₄O₂
(b) C₈H₁₆O₈
(c) C₄H₈O₄
(d) CH₂O
Answer:
(b) C₈H₁₆O₈
Solution:
(π₁)_{non electrolute} = (π₂)_glucose
C₁RT = C₂RT

\(\frac { 6 }{ n(30) }\) = 0.025
n = \(\frac { 6 }{ 0.025 x 30 }\) = 30
∴ Molecular formula C₈H₁₆O₈

Question 16.
The K_H for the solution of oxygen dissolved in water is 4 x 10⁴ atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is …………..
(a) 4.6 x 10³
(b) 1.6 x 10⁴
(c) 1 x 10^-5
(d) 1 x 10⁵
Answer:
(c) 1 x 10^-5
Solution:
K_H = 4 x 10⁴ atm,
(P_O2)_air = 0.4 atm,
(x_o2)_{in solution} = ?
air – in solution
(P_O2)_air = K_H(x_o2)_{in solution}
0.4 = 4 x 10⁴(x_o2)_{in solution}
(x_o2)_{in solution} = \(\frac { 0.4 }{ 4\times { 10 }^{ 4 } }\) = 1 x 10^-5

Question 17.
Normality of 1.25M sulphuric acid is …………
(a) 1.25 N
(b) 3.75 N
(c) 2.5 N
(d) 2.25 N
Answer:
(c) 2.5 N
Solution:
Normality of H₂SO₄ = (No. of replacable H⁺) x M = 2 x 1.25 = 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is …………..
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non – ideal and shows negative deviation from Raoults Law
Answer:
(d) non – ideal and shows negative deviation from Raoults Law
Solution:
∆H_mix is negative and show negative deviation from Raoults law.

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 3.5 x 10^-3. The mole fraction of water in that solution is …………
(a) 0.0035
(b) 0.35
(c) 0.0035/18
(d) 0.9965
Answer:
(d) 0.9965
Solution:

Question 20.
The mass of a non-volatile solute (molar mass 80 g mol^-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90% ………..
(a) 10g
(b) 20g
(c) 9.2 g
(d) 8.89g
Answer:
(d) 8.89g
Solution:

Question 21.
For a solution, the plot of osmotic pressure (π) verses the concentration (e in mol L^-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is ………..
(a) 310 x 0.082 K
(b) 3 10°C
(c) 37°C
(d) \(\frac { 310 }{ 20.082}\)
Answer:
(c) 37°C
Solution:
π = CRT
y = x(m)
m = RT
310 R = RT
T = 310 K
= 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1 .26g of protein. At 300K, the osmotic pressure of this solution is found to be 2.52 x 10^-3 bar. The molar mass of protein will be (R =0.083 Lhar mol^-1 K^-1) ……………
(a) 62.22 Kg mol^-1
(b) 12444 g mol^-1
(c) 300g mol^-1
(d) none of these
Answer:
(a) 62.22 Kg mol^-1
Solution:
π = CRT
M = \(\frac { WRT }{ π1 }\) = \(\frac { 1.26\times 0.083\times 300 }{ 2.52\times { 10 }^{ -3 }\times 0.2 }\) = 62.22Kg mol^-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………..
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Solution:
Ba(OH)₂ dissociates to form Ba²⁺ and 2OH^-1 ion
α = \(\frac { (i – 1) }{ (n – 1) }\)
i = α (n – 1) + 1
n = i = 3 ( for Ba (OH)₂, α = 1 )

Question 24.
What is the molality of a 10% w/w aqueous sodium hydroxide solution?
(a) 2.778
(b) 2.5
(c) 10
(a) 0.4
Answer:
(b) 2.5
Solution:
100% \(\frac { w }{ w }\) aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution.

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is ………
(a) α = \(\frac { n(i – 1) }{ n – 1 }\)
(b) α² = \(\frac { n(1 – i) }{ n – 1 }\)
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
(d) α = \(\frac { n(1 – i) }{ n(1 – i) }\)
Answer:
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
Solution:
α = \(\frac { (i – 1)n }{ (n – 1) }\) (or) \(\frac { n(i – 1) }{ (1 – n) }\)

Question 26.
Which of the following aqueous solutions has the highest boiling point?
(a) 0.1 M KNO₃
(b) 0.1 M Na₃PO₄
(c) 0.1 M BaCl₂
(d) 0.1 M K₂SO₄
Answer:
(a) 0.1 M KNO₃
Solution:
Elevation of boiling point is more in the case of Na₃PO₄(no. of ions 4; 3 Na⁺, PO₄^3-)

Question 27.
The freezing point depression constant for water is 1.86° k kg mo1^-1 . If 5g Na₂SO₄ is dissolved in 45g water, the depression in freezing point is 3.64°C. The van’t Hoff factor for Na₂SO₄ is ……..
(a) 2.50
(b) 2.63
(c) 3.64
(d) 5.50
Answer:
(a) 2.50
Solution:
K_f = 1.86
W₂ = 5g
∆T_f = 3.64
M₂ = 142
W₁ = 45g
ΔT_f = i x K_f

Question 28.
Equimolal aqueous solutions of NaCI and KCI are prepared. If the freezing point of NaCI is – 2°C, the freezing point of KCI solution is expected to be ………
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Answer:
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Solution:
Equimolal aqueous solution of KCI also shows 2° C depression in freezing point.

Question 29.
Phenol dimerises in henzene having van’t Hoff factor 0.54. What is the degree of association?
(a) 0.46
(b) 92
(c) 46
(d) 0.92
Answer:
(d) 0.92
Solution:
α = \(\frac { (1-i)n }{ (n-1) }\) = \(\frac { (1 – 0.54)2 }{ (2 – 1) }\) = 0.46 x 2 = 0.92
Question 30.
Assertion : An ideal solution obeys Raoults Law
Reason : In an ideal solution, solvent-solvent as well as solute-solute interactions are similar to solute-solvent interactions.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Chemistry Solutions Short Answer Questions

Question 31.
Define

1. Molality
2. Normality

Answer:
1. Molality (m):
It is defined as the number of moles of the solute present in 1 kg of the solvent

2. Normality (N):
It is defined as the number of gram equivalents of solute in I litre of the solution.

Question 32.
What is a vapour pressure of liquid? What is relative lowering of vapour pressure?
Answer:
1. The pressure of the vapour in equilibrium with its liquid ¡s called vapour pressure of the liquid at the given temperature.

2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure

Question 33.
State and explain Henry’s law.
Answer:
Henry’s law:
This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.”
P_solute ∝ x_{solute in solution}
P_solute = K_H. x_{solute in solution}
x_solute = mole fraction of solute in the solution
K_H = empirical constant.
P_solute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of K_H depends on the nature of the gaseous solute and solvent.

Question 34.
State Raoult law and obtain expression for lowering of apour pressure when nonvolatile solute is dissolved in solvent.
Answer:
Raoult’s law:
This law states that “in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
P_A ∝ x _A
when x_A = 1,
then k = P°_A
(P°A = vapour pressure of pure component)
P_A = P°_A . xa
P_B = P°_B . xb
when a non volatile is dissolved in pure water, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.
P_solution ∝ x_A
P_solution = k . x_A
x_A = 1, k = P°_solvent
P°_solution = P°_solvent – P_solution
Lowering of vapour pressure = P°_solvent – P_solution
Relative lowering of vapour pressure = \(\frac { P° – P }{ P° }\) = x_B
where x_B = Mole fraction of solute.

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
K_f = molar freezing point depression constant or cryoscopic constant.
∆T_f = K_f . m,
where
∆T_f = depression in freezing point.
m = molality of the solution
K_f = cryoscopic constant
If m = I
∆T_f = K_f
i.e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. K_f is directly proportional to the molal concentration of the solute particles.

W_B = mass of the solute
W_A = mass of solvent
M_B = molecular mass of the solute.

Question 36.
What is osmosis?
Answer:
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Question 37.
Define the term bisotonic
Answer:
1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e.. the net solvent flow between two isotonic solutions is zero.

Samacheer Kalvi 11th Chemistry Solutions Long Answer Questions

Question 38.
You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
1. Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

2. Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

3. Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCI in 1 litre of watcr at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCI in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCI in 1 litre of water at 25°C.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
1. The change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However the solubility of gases generally increases with increase of pressure.

2. According to Le – chatlier’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent.

3. If pressure increases, solubility of gas also increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 gL^-1. Calculate the molality.
Answer:
Given:
Molarity = 12 M HCI
Density of the solution = 1.2 g L^-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.

Calculate mass of water (solvent)
Mass of 1 litre HCI solution = density x volume
= 1.2gmL^-1 x 1000 mL = 1200g
Mass of LICI = No. of moles of HCI x molar mass of HCI
= 12mol x 36.5 g mol^-1 = 438g
Mass of waler = mass of HCI solution – mass of HCI
Mass of waler = 1200 – 438 = 762 g
Molalily =\(\frac { 12 }{ 0.762 }\) = 15.75m

Question 41.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood. What is the osmotic pressure of blood?
Solution.
C = 0.25 M
T = 37O.28 K
(π)_gIucose = CRT
(π) = 0.25 mol L^-1 x 0.082 L atm K^-1 morl^-1 x 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5g of glycine (NH₂ – CH₂ – COOH) dissolved in 500g of water.
Solution:

Question 43.
Which solution has the lower freezing point? 10g of methanol (CH₃OH) in 100g of water (or) 20g of ethanol (C₂H₅HO) In 200g of water.
Solution:
∆T_f = K_f . m i.e. ∆T_f ∝ m

∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one litre of 10^-4 M potassium sulphate?
Solution:
In 10^-4M K₂SO₄ solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2 K⁺ and 1SO₄^2-)
1 mole of K₂SO₄ contains 3 x 6.023 x 10²³ ions
10 mole of K₂SO₄ contains 3 x 6.023 x 10² x 10^-4 ions = 18.069 x 10¹⁹

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 x 10^-5 mm Hg at a particular constant temperature. At this temperature, calculate the solubiiitv of methane at

1. 75O mm Hg
2. 84OnimHg

Solution:
(K_H)_Benzene = 4.2 x 10^-5 mm Hg. Solubility of methane = ? P = 750 mm Hg, p = 840 mm Hg
According to Henry’s Law,
P = K_H . x_solution
750 mm Hg = 4.2 x 10^-5 mm Hg . x_solution

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C. Calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K kg mol^-1.
Solution:
T₁= 0.093°C = 0.093K
m = ?
K_f = 1.86K kg mol^-1
∆T_f = k_f . m

Question 47.
The vapour pressure of pure henzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P⁰_C6H6 = 640 mm Hg
W₂ = 2.2 g (non volatile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ = ?

Samacheer Kalvi 11th Chemistry Solutions In Text Questions – Evaluate Yourself

Question 1.
If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.
Solution.
Mass of KOH = 5.6g
No. of moles = \(\frac { 5.6 }{ 5.6 }\) = 0.1 mol
1. Volume of the solution = 500 ml = 0.5 L

2. Volume of the solution = IL

3. Volume of the solution = IL
Molarity = \(\frac { 0.1 }{ 1 }\) M

Question 2.
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.
Solution:
Mass of glucose = 2.82 g
No. of moles of glucose = \(\frac { 2.82 }{ 180 }\) = 0.0 16
Mass of water = 30g = \(\frac { 30 }{ 18 }\) = 1.67
x_H2O = \(\frac { 1.67 }{ 1.67 + 0.016 }\) = \(\frac { 1.67 }{ 1.686 }\) = 0.99
x_H2O + x_glucose = 1
0.99 + x_glucose = 1
x_glucose = 1 – 0.99 = 0.01

Question 3.
The antiseptic solution of iodopovidone for the use of external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL.
Solution:
10% \(\frac { w }{ v }\) means that 10 g of solute in 100 ml solution
∴ Amount of iodopovidone in 1.5 ml = \(\frac { 10g }{ 100ml }\) x 1.5 ml = 0.15 g

Question 4.
A litre of sea water weighing about 1.05 kg contains 5 mg of dissohed oxygen (O₂). Express the concentration of dissolved oxygen in ppm.
Solution:

Question 5.
Describe how would you prepare the following solution from pure solute and solvent

1. 1 L of aqueous solution of 1.5 M COCI₂.
2. 500 mL of 6.0 % (v/v) aqueous methanol solution.

Solution:

1. mass of 1.5 moles of COCI₂ = 1.5 x 129.9 = 194.85g
2. 194.85g anhydrous cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.

Question 6.
How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250 M NaOH solution.
Solution:
6% \(\frac { v }{ v }\) aqueous solution contains 6g of methanol in 100 ml solution. To prepare 500 ml of 6% v/v solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

Question 7.
Calculate the proportion of O₂ and N₂ dissolved in water at 298 K. When air containing 20% O₂ and 80% N₂ by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O₂) = 4.6 x atm and K_H (N₂) 8.5 x 10⁴ atm.
Solution:
C₁V₁ = C₂V₂
6M (V₁) = 0.25M x 500 ml
V₁ = \(\frac { 0.25 x 500 }{ 6 }\)
V₁ = 20.3 mL

Question 8.
Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer.
Solution:
Total pressure = 1 atm
P_N2 = \((\frac { 80 }{ 100 })\) x Total pressure = \(\frac { 80 }{ 100 }\) x 1 atm = 0.8 atm
P_O2 = \((\frac { 20 }{ 100 })\) x 1 = 0.2 atm
According to Henry’s Law
P_solute = K_H x _{solute in solution}
P_N2 = (K_H)_Nitrogen x Mole fraction of Nitrogen in solution

Question 9.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Solution:
P⁰_{pure benzene} = 50.71 mm Hg
P⁰_nepthalene = 32.06 mm Hg
Number of moles of benzene = \(\frac { 39 }{ 78 }\) = 0.5 mol
Number of moles of naphthalcne = \(\frac { 128 }{ 128 }\) =1 mol
Mole fraction of benzene = \(\frac { 0.5 }{ 1.5 }\) = 0.33
Mole fraction of naphthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene =P⁰_benzene x Mole fraction of benzene
= 50.71 x 0.33 = 16.73 mm Hg
Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg
Mole fraction of benzene in vapour phase = \(\frac { 16.73 }{ 16.73 + 21.48 }\) = \(\frac { 16.73 }{ 38.21 }\) = 0.44
Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56

Question 10.
Vapour pressure of a pure liquid A is 10.0 torr at 27°C. The vapour pressure is lowered to 9.0 torr on dissolving one grani of B in 20g of A. If the molar mass of A is 200 g mol^-1 then calculate the molar mass of B.
Solution:
P⁰_A = 10 torr
P_solution = 9 torr
W_A = 20 g
W_B = 1 g
M_A = 200 g mol^-1
M_B = ?

Question 11.
2.56g of Sulphur is dissolved in 100g of carbon disuiphide. The solution boils at 319.692K. What is the molecular formula ofSulphur in solution? The boiling pointof CS₂ is 319. 450K. Given that Kb for CS₂ = 2.42 K kg mol^-1
Solution:
W₂ = 2.56g
W₁ = 100g
T = 319.692 K
K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K

M₂ = 256g mol^-1
Molecular mass of sulphur in solulion = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac { 256 }{ 32 }\) = 8
Hence, molecular tòrmula of sulphur is S₈.

Question 12.
2g of a non electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5.12 K Kg mol^-1. Find the molar mass of the solute.
Solution:
W₂ = 2g
W₁ = 75g
∆T_f = 0.2 K
k_f = 5.12 K kg mol^-1
M₂ = ?

Question 13.
What is the mass of glucose (C₆H₁₂O₆) in it one litre solution is isotonic with 6g L^-1 of urea (NH₂CONH₂)?
Solution:
Osmotic pressure of urea solution (π₁) = CRT
\(\frac { { W }_{ 2 } }{ { M }_{ 2 }V }\)RT = \(\frac { 6 }{ 60 x 1 }\) x RT
Osmotic pressure of glucose solution
(π₂) \(\frac { { W }_{ 2 } }{ 180\times 1 }\) x RT
For isotonic solution, π₁ = π₂
\(\frac { 6 }{ 60 }\) = \(\frac { { W }_{ 2 } }{ 180\times 1 }\) RT ⇒ W₂ = \(\frac { 6 }{ 60 }\) x 180
W₂ = 18 g

Question 14.
0.2m aqueous solution of KCI freezes at – 0.68°C calculate van’t Hoff factor. K_f for water is 1.86 K kg mol^-1.
Solution:

Given,
∆T_f = 0.680 K
m = 0.2 m,
∆T_f (observed) = 0.680K
∆T_f(Calculated) = k_f
m = 1.86 K kg mol^-1 x 0.2 mol kg^-1 = 0.372K

Samacheer Kalvi 11th Chemistry Solutions Example problems Solved

Question 1.
What volume of 4M HCI and 2M HCI should be mixed to get 500 mL of 2.SM HCI?
Solution:
Let the volume of 4M HCl required to prepare 500 mL of 2.5 M HCI = x mL
Therefore, the required volume of 2M HCI = (500 – x) mL
We know from the equation x = \(\frac { 250 }{ 2 }\) = 125 mL
Hence, volume of 4M HCI required = 125 mL
Volume of 2M HCl required = (500 – 125) mL = 375 mL

Question 2.
0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
Solution:
P_solute = K_H x_{solute in solution}
At pressure 1.5 atm, p₁ = K_H x₁ ………..(1)
At pressure 6.0 atm, p₂ = K_Hx₂ …………..(2)
Dividing equation (1) by (2)
Weget
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } }\) = \(\frac { { x }_{ 1 } }{ { x }_{ 2 } }\)
\(\frac { 1.5 }{ 6.0 }\) = \(\frac { { 0.24 } }{ { x }_{ 2 } }\)
Therefore
x₂ = \(\frac { 0.24×6.0 }{ 1.5 }\) = 0.96 g/L

Question 3.
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when P_A° is 1.013 bar?
Solution:

In a 2% solution weight of the solute is 2g and solvent is 98g
ΔP = P_A⁰ – P_solution = 1.013 – 1.004 bar = 0.009 bar

Question 4.
0.75 g of an unknown substance is dissolved in 200 g solvent. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5K kg more then, calculate the molar mass of unknown substance.
Solution:
∆T_b = K_b m = K_b x W₂ x 1000/M₂ x W₁
M₂ = K_b x W₂ x 1000/∆T_b x W₁
= 7.5 x 0.75 x 1000/0.15 x 200 = 187.5g mol^-1

Question 5.
Ethylene glycol (C₂H₆O₂) can be used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. K_f for water = 1.86 K kg mol^-1 and molar mass of ethylene glycol is 62g mol^-1.
Solution:
Weight of solute (W₂) = 20 mass percent of solution means 20g of ethylene glycol
Weight of solvent (water) W₁ = 100 – 20 = 80g

The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e. 7.5 K lower than the normal freezing point of water (273 – 7.5)K = 265.5K

Question 6.
At 400K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.
Solution:

Question 7.
The depression in freezing point is 0.24K obtained by dissolving 1g NaCI in 200g water. Calculate van’t – Hoff factor. The molal depression constant is 1.86 K kg mol^-1.
Solution:
Sol. Molar mass of solute

Samacheer Kalvi 11th Chemistry Solutions Additional Questions Solved

Samacheer Kalvi 11th Chemistry Solutions 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Among the following, which one is mostly present in sea water?
(a) NaCI
(b) Nal
(c) KCI
(d) MgBr₂
Answer:
(a) NaCI

Question 2.
Statement I: The most common property of sea water and air is homogeneity.
Statement II: The homogeneity implies uniform distribution of their constituents through the mixture.
(a) Statements I and II arc correct and II is the correct explanation of I.
(b) Statements I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) StatementI I and II are correct and II is the correct explanation I.

3. Which one of the following is a homogeneous mixture?
(a) Sea water
(b) Air
(c) Alloys
(d) All the above
Answer:
(d) All the above

Question 4.
Statement I: Salt solution is an aqueous solution.
Statement II: If water is used as the solvent, the resultant solution is called an aqueous solution.
(a) Statements I and II are correct but II is not the correct explanation of I.
(b) Statements I and II are correct and II is the correct explanation of I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statements I and II are correct and II is the correct explanation of I.

Question 5.
Statement I: The dissolution of ammonium nitrate increases steeply with increase in temperature.
Statement II: The dissolution process of ammonium nitrate is endothermic in nature.
(a) Statement I and II are correct and statement II is the correct explanation of statement I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement I.

Question 6.
In which of the following compound the solubility decreases with increase of temperature?
(a) sodium chloride
(b) ammonium nitrate
(c) cerie sulphate
(d) calcium chloride
Answer:
(c) ceric sulphate

Question 7.
Which of the following is not an ideal solution?
(a) Benzene & toluene
(b) n – Hexane & n – Heptane
(c) Ethyliodide & ethyl bromide
(d) Ethanol and water
Answer:
(d) Ethanol and water

Question 8.
Which one of the following shows positive deviation from Raoult’s law?
(a) Ethyliodide and Ethyl bromide
(b) Ethyl alcohol and cyclohexane
(c) Chioro benzene & bromo benzene
(d) Benzene & toluenc
Answer:
(b) Ethyl alcohol and cyclohexane

Question 9.
Which one of the following is not an non-ideal solution showing positive deviation?
(a) Benzene & acetone
(b) CCl₄ & CHCI₃
(c) Acetone & ethyl alcohol
(d) Benzene and toluene
Answer:
(d) Benzene and toluene

Question 10.
Which of the following shows negative deviation from Raoults law?
(a) Phenol and aniline
(b) Benzene and toluene
(c) Acetone and ethanol
(d) Bcnzene and acetone
Answer:
(a) Phenol and aniline

Question 11.
Which of the following is not an non-ideal solution showing negative deviation?
(a) Phenol and aniline
(b) Ethanol and water
(c) Acetone + Chlorotorm
(d) n – Heptane and n – Hexane
Answer:
(d) n – Heptanc and n – Hexane

Question 12.
Statement I: A solution of potassium chloride in water deviates from ideal behavior.
Statement II: The solute dissociates to give K and Cl ion which form strong ion dipole interaction with water molecules.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not correct explanation of I
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 13.
Statement I: Acetic acid deviates from ideal behaviour.
Statement II: Acetic acid exists as a dimer by forming inter molecular hence deviates from Raoults law.
(a) Statement I & II are correct and II is the correct explanation of I.
(b) Statement I & II are correct but II is not the correct explanation of I.
(c) Statement I is true but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I & II are correct but II is the correct explanation of I.

Question 14.
Which one of the following has found to have abnormal molar mass? hydrogen bonds and
(a) NaCl
(b) KCI
(c) Acetic acid
(d) all the above
Answer:
(d) All the above

Question 15.
What would be the value of van’t Hoff factor for a dilute solution of K₂SO₄ in water.
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(a) 3
Solution:
ions produced = n = 3
Since
K₂SO₄ → 2K⁺ + SO₄^2-
K₂SO₄ is completely dissociated so
∝ = \(\frac { i – 1 }{ n – 1 }\) = \(\frac { i – 1 }{ 3 – 1 }\) = 1
i – 1 = 1 x 2
i – 1 = 2
i = 2+1 = 3

Question 16.
In the determination of molar mass of AB using a colligative property, what may be the value of van’t Hoff factor if the solute is 50% dissociates?
(a) 0.5
(b) 1.5
(c) 2.5
(d) 1
Answer:
(b) 1.5
Solution:
∝ = \(\frac { i – 1 }{ n – 1 }\) = 0.5
\(\frac { i – 1 }{ 2 – 1 }\) = 0.5
i – 1 = 0.5
i = 0.5 + 1 = 1.5

Question 17.
Which of the following solution has the highest boiling point?
(a) 5.85% solution of NaCI
(b) 18.0% solution of glucose
(c) 6.0% solution of urea
(d) All have same boiling point
Answer:
(a) 5.85% solution of NaCl

Question 18.
Which one of the following pair is called an ideal solution?
(a) nicotine – water
(b) water – ether
(c) water – alcohol
(d) Chiorobenzene – bromobenzene
Answer:
(d) Chiorobenzene – bromobenzene

Question 19.
Which of the following is not a colligative property?
(a) optical activity
(b) osmotic pressure
(c) elevation boiling point
(d) depression in freezing point
Answer:
(a) optical activity

Question 20.
On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
(a) Sugar crystals in cold water
(b) Sugar crystals in hot water
(c) powdered sugar in cold water
(d) powdered sugar in hot water
Answer:
(d) powdered sugar in hot water

II. Match the following.
Question 1.

Answer:
(a) 3 4 1 2

Question 2.

Answer:
(d) 3 4 2 1

Question 3.

Answer:
(c) 2 4 1 3

Question 4.

Answer:
(a) 4 3 1 2

Question 5.

Answer:
(b) 2 4 1 3

III. Fill in the blanks.

Question 1.
……… covers more than 70% of the earth’s surface.
Answer:
Seawater

Question 2.
……… is an important naturally occurring solution.
Answer:
Air

Question 3.
An example of solid homogeneous mixture is ……….
Answer:
Brass

Question 4.
A mixture of N₂, O₂, CO₂ and other traces of gases is known as ………
Answer:
Air

Question 5
……… a non – aqueous solution.
Answer:
Br₂ in CCl₄

Question 6.
……… is an example for gaseous solution.
Answer:
Camphor in nitrogen gas

Question 7 .
……… is used for dental filling.
Answer:
Amalgam of potassium

Question 8.
Carbonated water is an example for ………
Answer:
Liquid solution

Question 9.
Humid oxygen is an example of ………
Answer:
Gaseous solution

Question 10.
The concentration of commercially available H₂O₂ is ………
Answer:
3%

Question 11.
The molality of the solution containing 45g of glucose dissolved in 2kg of water is ………
Answer:
0.125m

Question 12.
5.845 g of NaCl is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the solution in molarity is ………
Answer:
0.2 M
Solution:

Question 13.
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is ………
Answer:
0.5N
Solution:

Question 14.
5.85 g of NaCI is dissolved in water and the solution was made upto 500 ml using a standard flask. The strength of the solution in formality is ………
Answer:
0.2 F
Solution:

Question 15.
Neomycin, amino glycoside antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of oinment base. The mass percentage of neomycin is ………
Answer:
1%
Solution:
The mass percentage of neomycin

Question 16.
0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are ……….
Answer:
0.25, 0.75
Solution:

= \(\frac { 0.5 }{ 1.5 + 0.5 }\) = \(\frac { 0.5 }{ 2.0 }\) = 0.25
Mole fraction of water = \(\frac { 1.5 }{ 2.0 }\) = 0.75

Question 17.
50 mL of tincture of benzoin, an antiseptic solution contains 10 ml of benzoin. The volume percentage of benzoin is ……….
Answer:
20%
Solution:
Volume percentage of benzoin

= \(\frac { 10 }{ 50 }\) x 100 = 20%

Question 18.
A 60 ml of paracetamol pediatric oral suspension contains 3g of paracetamol. The mass percentage of paracciamol is …………
Answer:
5%
Solution:
Mass percentage of paracetamol =

= \(\frac { 3 }{ 60 }\) x 100 = 5%

Question 19.
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is ………..
Answer:
400 ppm
Solution:

Question 20.
The concentration term used in the neutralisation reactions is …………
Answer:
Normality

Question 21.
The concentration term is used in the calculation of vapour pressure of solution is …………..
Answer:
Mole fraction

Question 22.
The term used to express the active ingredients present in therapeutics is ………
Answer:
Percentage units

Question 23.
When maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called ………..
Answer:
Saturated solution

Question 24.
The solvent in which sodium chloride readily dissolves is …………
Answer:
Water

Question 25.
………… is used by deep-sea divers.
Answer:
Helium, nitrogen and oxygen

Question 26.
The mathematical expression of Raoult’s law is ………..
Answer:
P_A = P_A⁰ . X_A

Question 27.
……….. is an ideal solution?
Answer:
Chloro benzene & bromo benzene

Question 28.
………….. is important in some vital biological systems.
Answer:
osmotic pressure

Question 29.
………. is not a colligative property.
Answer:
vapour pressure

Question 30.
According to van’t Hoff equation. the value of osmotic pressure t is equal to …………
Answer:
π = CRT

Question 31.
The osmotic pressure of the blood cells is approximately equal to at 37°C.
Answer:
7 atm.

Question 32.
Which one of the following is applied in water purification?
Answer:
reverse osmosis

Question 33.
In commercial reverse osmosis process, the semi permeable membrane used is ………..
Answer:
cellulose acetate

Question 34.
The degree of dissociation α is equal to ……….
Answer:
\(\frac { i – 1 }{ n – 1 }\)

Question 35.
The degree of association a is equal to ……….
Answer:
\(\frac { (i – 1)n }{ n – 1 }\)

Question 36.
The estimated vantt Hoff factor for acetic acid solution in benzene is ………..
Answer:
0.5

Question 37.
The estimated van’t Hoff factor for sodium chloride in water is ………..
Answer:
2

Question 38.
Number of moles of the solute dissolved per dm3 of solution is ……….
Answer:
molarity

Question 39.
Molarity of pure water is ………….
Answer:
55.55
Solution:

Question 40.
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………..
Answer:
0.1
Solution:
\(\frac { P° – P }{ P° }\) = x₂
x₂ = No. of moles of glucose
\(\frac { 18 }{ 180 }\) = 0.1
\(\frac { P° – P }{ P° }\) = 0.1

Question 41.
When NaCl is dissolved in water, boiling point ………..
Answer:
increases

Question 42.
Use of glycol as antifreezer in automobile is an important application of …………….
Answer:
Colligative property

Question 43.
Ethylene glycol is mixed with water and used as antifreezer in radiators because …………..
Answer:
it lowers the freezing point of water

Question 44.
Colligative properties of a solution depend on ………… present in it.
Answer:
Number of solute particles

Question 45.
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ………….
Answer:
low atmospheric pressure

IV. Choose the odd one out.

Question 1.
(a) Air
(b) Camphor in nitrogen gas
(c) Humid oxygen
(d) Salt water
Answer:
(d) Salt water.
a, b and e are gaseous solution whereas d is a liquid solution.

Question 2.
(a) CO₂ dissolve in water
(b) Salt water
(c) Solution of H₂ in palladium
(d) Ethanol dissolved in water
Answer:
(c) Solution of H₂ in palladium
a, b and d are liquid solutions whereas c is a solid solution.

Question 3.
(a) Amalgam of potassium
(b) Camphor in nitrogen gas
(c) Solution of H₂ in palladium
(d) Gold alloy
Answer:
(b) Camphor in nitrogen gas
a, b and d arc solid solutions whereas b is gaseous solution.

Question 4.
(a) Vapour pressure
(b) Lowering ofvapour pressure
(c) Osmotic pressure
(d) Elevation of boiling point
Answer:
(a) Vapour pressure
b, e and dare colligative properties whereas a is a physical property.

Question 5.
(a) Benzene and tolucne
(b) Chlorobenzene and Bromobenzene
(c) Benzene and acetone
(d) n – hexane and n – heptane
Answer:
(a) Benzene and acetone
a, b and dare ideal solutions whereas c is a non-ideal solution.

Question 6.
(a) Ethyl alcohol and cyclohexane
(b) Ethyl bromide and ethyl iodide
(c) Acetone and ethyl alcohol
(d) Benzene and acetone
Answer:
(a) Ethyl bromide and ethyl iodide
a, e and dare non-ideal solutions whereas b is an ideal solution.

V. Choose the correct pair.

Question 1.
(a) Humid oxygen – Liquid solution
(b) Gold alloy – Solid solution
(c) Salt water – Gaseous solution
(d) Solution of H₂ in palladium – Gaseous solution
Answer:
(b) Gold alloy – Solid solution

Question 2.
(a) Air – Gaseous solution
(b) Amalgam of potassium – Liquid solution
(c) Salt water – Solid solution
(d) Carbonated water – Solid solution
Answer:
(a) Air – Gaseous solution

Question 3.
(a) Benzene and toluene – Non-ideal solution
(b) Benzcnc and acetone – Non-ideal solution
(c) Chlorobenzene and bromo henzene – Non-ideal solution
(d) Carbon tetrachloride and Chloroform – ideal solution
Answer:
(b) Benzene and acetone – Non-ideal solution

Question 4.
(a) Benzene and toluene – Ideal solution
(b) n-hexane and n-heptane – Non-ideal solution
(c) Ethyl iodide and ethyl bromide – Non-ideal solution
(d) Chiorobenzene and bromo benzene – Non-ideal solution
Answer:
(a) Benzene and toluene – Ideal solution

VI. Choose the incorrect pair.
Question 1.

Answer:

Question 2.
(a) Benzene and acetone – Ideal solution
(b) Ethyl alcohol and cyclohexane – Non-ideal solution
(C) n-hexane and n-heptanc – Ideal solution
(d) Chioro benzene – Ideal solution
Answer:
(a) Benzene and acetone – Ideal solution

VII Assertion & Reason.

Question 1.
Assertion (A) : When NaCI is added to water, a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes the depression in freezing poi nl.
(a) Assertion and Reason are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Assertion and Reason are correct and R is the correct explanation of A.

Question 2.
Assertion (A): Ammonia reacts with water does not obey Henry’s law.
Reason (R): The gases reacting with the solvent does not obey Henry’s law.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct hut (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Acetic acid solution deviates from Raoult’s law.
Reason (R): Association of solute molecules exists as a dimer by forming intermolecular. hydrogen bonds and hence deviates from Raoult’s law.
(a) Both (A) and (R) arc wrong.
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

VIII. Choose the correct statement.

Question 1.
(a) Raoult’s law is applicable to volatile solid solute in liquid solvent
(b) Henry’s law is applicable to solution containing solid solute in liquid solvent
(c) For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.
(d) For saturated solution containing volatile solid solute in liquid solvent both laws are obeyed.
Answer:
(c) For very dilute solutions. the solvent obeys Raoult’s law and the solute obeys LIenrys law.

Samacheer Kalvi 11th Chemistry Solutions 2 Marks Questions and Answers

I. Write brief answer to the following questions:

Question 1.
What is the common property observed in naturally existing solution? Explain it.
Answer:

1. Sea water, air are the naturally existing homogeneous mixture. The common property observed in these is homogeneity.
2. The homogeneity implies uniform distribution of their constituents or components through out the mixture.

Question 2.
Define solution with an example.
Answer:
1. A solution is a homogeneous mixture of two or more substances consisting of atoms. ions or molecules.

2. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained. In this solution, Na⁺ and C^– ions are uniformly distributed in water. Here NaCI is the solute and water is the solvent.

Question 3.
What are aqueous and non aqueous solution? Give example.
Answer:

1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution. e.g., salt in water.
2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl₄ etc, the resultant solution is called a non aqueous solution. e.g., Br, in CCI₄.

Question 4.
Define molality.
Answer:
Molality is defined as the number of moles of solute present in 1 kg of the solvent.

Question 5.
Define molaritv.
Answer:
Molarity is defined as the number of moles of solute present in 1 litre of the solution.

Question 6.
Define normality.
Answer:
Normality is deflncd as the number of gram equivalents of solute present in 1 litre of the solution.

Question 7.
Define forniality.
Answer:
Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution.

Question 8.
Define mole fraction.
Answer:
Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

Question 9.
Show that the sum of mole fraction of a solution is equal to one.
Answer:
Consider a solution containing two components A and 13 whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A and n_B respectively.

Question 10.
Define mass percentage.
Answer:
Mass percentage is defined as the ratio of the mass of the solute in g to the mass of solution in g multiplied by 100.

Question 11.
Define volume percentage.
Answer:
Volume percentage is defined as the ratio of volume of solute in mL to the volume of solution in ml multiplied by 100.

Question 12.
Define mass by volume percentage.
Answer:
It is defined as the ratio of the mass of the solute in g to the volume of the solution in ml multiplied by 100.

Question 13.
What is meant by ppm? Where is it used?
Answer:
1. part per million =

2. ppm is used to express the quantity of solutes present in small amounts in solutions.

Question 14.
What is meant by stock solution (or) standard solution? What is meant by working standard?
Answer:
1. A standard solution or a stock solution is a solution whose concentration is accurately known.

2. At the time of experiment, the solution with required concentration is prepared by diluting the stock solution. This diluted solution is called working standard.

Question 15.
Define solubilitv.
Answer:
The solubility of a substance is defined as the amount of the solute that can be dissolved in loo g of the solvent at a given temperature to form a saturated solution.

Question 16.
Ammonia is more soluble than oxygen in water. Why?
Answer:
Ammonia forms hydrogen bonding with water molecules, this intermolecular bonds arc very strong and thus the ammonia is more soluble in water. Ammonia is strongly interact with water to form ammonium hydroxide. But oxygen is more electronegative it is not able to interact with water more. So NH₃ is more soluble than O₂ in water.

Question 17.
Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement.
Answer:
When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

Question 18.
Dissolution of ammonium nitrate increases with increase in temperature. Why?
Answer:
The dissolution process of ammonium nitrate is endothermic. So the solubility increases with increase in temperature.

Question 19.
What is the relationship between the solubility of eerie sulphate with temperature?
Answer:
The dissolution of eerie sulphate is exothermic and the solubility decreases with the increase in temperature.

Question 20.
Why in the dissolution of CaCl₂, the solubilit increases moderately with high temperature?
Answer:
Even though the dissolution of CaCI₂, is cxothcrmic, the soluhility increases moderately with increase in temperature. Here the entropy factor plays a significant role in deciding the position of equilibrium.

Question 21.
Why the carbonated drinks are stored in pressurized container?
Answer:
1. The carbonated beverages contain CO₂ dissolved in them. To dissolve the CO₂ in these drinks, CO₂ gas is bubbled through them under high pressure.

2. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO₂ drops to the atmospheric pressure level and hence bubbles of CO₂ rapidly escape from the solution and show effervescence.

Question 22.
Define

1. Evaporation
2. Condensation.

Answer:
1. Evaporation:
If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process in called evaporation.

2. Condensation:
The vapour molecules are in random motion during which they collide with each other and also with the walls of the container. As the collision is inelastic, they lose their energy and as a result the vapour returns back to liquid state. This process is called as condensation.

Question 23.
State Dalton’s law of partial pressure.
Answer:
According to Dalton’s law of partial pressure, the total pressure in a closed vessel will be equal to the sum of the partial pressure of the individual components.
P_total = P_A + P_B

Question 24.
Give the reason behind the lowering of vapour pressure in the dissolution of NaCl in water?
Answer:
NaCI is a non volatile solute. When a non volatile solute is dissolved in pure solvent, the vapour pressure of pure solvent will decrease. In such solution, vapour pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.

Question 25.
What are ideal solution? Give example.
Answer:
An ideal solution is a solution in which each component i.e., the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration.

Question 26.
What are non-ideal solution? Give example.
Answer:

1. The solutions which do not obey Raoult’s law over the entire range of concentration are called non-ideal solutions.
2. The deviation of the non-ideal solution from the Raoult’s law may be positive (or) negative.
3. Example, Ethyl alcohol and cyclohexane.

Question 27.
What are colligative properties? Give example.
Answer:
The properties which do not depend on the chemical nature of the solute but depends only on the number of solute particles present in the solution are called colligative properties. e.g.,

1. Relative lowering of vapour pressure – \(\frac { P° – P}{ P° }\)
2. Osmotic pressure – π
3. Elevation of boiling point – ∆T_b
4. Depression in freezing point – ∆T_f

Question 28.
What is meant by elevation of boiling point?
Answer:
1. The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure.

2. When a non-volatile solute is added to pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to 1 atm, the temperature of the solution has to be increased.

3. As a result, the solution boils at a higher temperature (T_b) then the boiling point of pure solvent (T_b°). This increase in the boiling point is known as elevation of boiling point.

Question 29.
Define ebullioscopic constant.
Answer:
Ebullioscopic constant k_b, is equal to the elevation in boiling point for 1 molal solution.

Question 30.
Define osmotic pressure.
Answer:
Osmotic pressure can be defined as the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semipermeable membrane.

Question 31.
Write the Van’t Hoff equation of osmotic pressure.
Answer:
Van’t Hoff equation states that for dilute solutions, the osmotic pressure is directly proportional to the molar concentration of the solute and the temperature of the solution.
π = CRT
where
π = Osmotic pressure
C = concentration
T = Temperature
R = gas constant

Question 32.
Define Van’t Hoff factor.
Answer:
van’t Hoff factor (I) is defined as the ratin of the actual molar mass to the abnormal molar mass of the solute.

Question 33.
How is degree of dissociation and degree of association are related with van’t Hoff factor?
Answer:
The degree of dissociation or association can be related to van’t Hoff factor
1. using the following relationship

• α_dissociation = \(\frac { i – 1 }{ n – 1 }\)
• α_association = \(\frac { (1 – i)n }{ n – 1 }\)

where n = number of solute particles

Question 34.
Give an example of a solid solution ¡n which the solute is a gas.
Answer:
Solution of hydrogen in palladium.

Question 35.
What role does the molecular interaction play in solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. The intermolecular forces both in alcohol and water are H-bonds. When alcohol and water are mixed,

they form solution because of formation of H-bonds between alcohol and H₂O molecules hut these interactions are weaker and less extensive than those in pure water. Hence, they show positive deviation from ideal behaviour.

Question 36.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. it is because the fact that this process involves decrease of entropy. Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent \(\rightleftharpoons\) Solution + Heat)

Question 37.
Why is the freezing point depression of 0.1 M NaCl solution nearly twice that of 0.1M glucose solution?
Answer:
NaCl is an electrolyte and it dissociates completely whereas glucose being a non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double for NaCI solution than that for glucose solution of same molarity.

Therefore depression in freezing point being a colligative property ¡s nearly twice for NaCl solution than that for glucose solution of same molarity.

Question 38.
Why a person suffering from high blood pressure is advised to take minimum quantity of common salt?
Answer:
Osmotic pressure is directly proportional to the concentration of solutes. Our body fluid contains a number of solutes. On taking large amount of salt, ions entering into the body fluid thereby raises the concentration of solutes. As a result, osmotic pressure increases which may rupture the blood cells.

Samacheer Kalvi 11th Chemistry Solutions 3 Marks Questions and Answers

Question 1.
What are gaseous solution ? Give its various types with example
Answer:

Question 2.
What are liquid solutions ? Explain with example
Answer:

Question 3.
What are solid solution? Give example.
Answer:

Question 4.
How will you prepare a standard solution?
Answer:

1. A standard solution or a stock solution is a solution whose concentration is accurately known.
2. A standard solution of required concentration can be prepared by dissolving a required amount of a solute in a suitable amount of solvent.
3. It is done by transforming a known amount of solute to a standard flask of definite volume. A small amount of water is added lo the flask and shaken well to dissolve the salt.
4. Then water is added to the flask to bring the solution level lo the mark indicated at the top end of the flask.
5. The flask is stoppered and shaken well to make concentration uniform.

Question 5.
What are the advantages of standard solution.
Answer:
1. The error due to weighing the solute can be minimised by using concentrated stock solution that requires large quantities of solute.

2. We can prepare working standards of different concentrations by diluting the stock solution which is more efficient since consistency is maintained.

3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 6.
Explain the solubilities of ammonium nitrate, calcium chloride, ceric sulphate and sodium chloride in water at different temperature with a graph.
Answer:
1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10% increase in solubility between 0°C to 100°C.

2. The dissolution process of ammonium nitrate is endothermic, the solubility increases with

3. In the case of eerie sulphate. the dissolution is exothermic and the solubility decreases with increase in temperature.

4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here the entropy factor also plays a significant role in deciding the position of equilibrium.

Question 7.
Explain the effect of temperature gaseous solute ¡n liquid solvent.
Answer:
1. In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature.

2. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak inter molecular forces when the temperature increases, the average. kinetic energy of the molecules present in the solution also increases.

3. The increase in kinetic energy breaks (he weak inter molecular forces between the gaseous solute and liquid solvent with results in the release of the dissolved gas molecules to gaseous state.

4. The dissolution of most of the gases in Liquid solvents is an endothermic process, the increase in temperature decreases the dissolution of gaseous molecules.

Question 8.
Give reason why aquatic species are less sustained in hot water?
Answer:
There will be decrease in solubility of gases in solution with increase in temperature. During summer, in hot water rivers, due to high temperature. the availability of dissolved oxygen decreases. So the aquatic species are less sustained in hot water.

Question 9.
Deep – sea divers use air diluted with helium gas in their tanks. Why? (or) Justify this statement.
Answer:
1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure.

2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank.

3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream.

These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life.

4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen.

Question 10.
What are the limitations of Henry’s law?
Answer:

1. Henry’s law is applicable at moderate temperature and pressure only.
2. Only the less soLuble gases obey Henry’s law.
3. The gases reacting with solvent do not obey Henry’s law.
4. The gases obeying Henrys law should not associated or dissociated while dissolving in the solvent.

Question 11.
Explain how benzene in toluene obeys Raoult’s law.

Answer:
The variation of vapour pressure of pure benzene and toluenc with its mole fraction is given in the graph.
1. The vapour pressure of pure toluene and pure benzene are 22.3 and 74.7 mm Hg respectively.

2. The graph shows the partial vapour pressure of pure components increases linearly with the increase of the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the straight line.

3. P_solution = P⁰_toluene + x_benzene (P⁰_benzene – P⁰_toluene)

Question 12.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute.
Answer:
P_solution ∝ x_A by Raoult’s law. where x_A is the mole fraction of the solvent.
P_solution = k . x_A
When
x_A = 1
k = P⁰_solvent
P⁰_solvent = partial pressure of pure solvent
P_solution = P⁰_solvent . x_A

where
x_B = mole fraction of the solute
x_A + x_B = 1
x_B = 1 – x_A

Question 13.
How would you compare Raoult’s law and Henry’s law.
Answer:
1. According to Raou It’s law, for a solution containing a non volatile solute.
P_solution = P⁰_solvent . x_solute

2. According to henry’s law, P_solution = K_H . x_solute in solution

3. The difference between the above two laws is the proportionality constant P° (Raoult’s law) and K_H (Heniys law).

4. henry’s law is applicable to solution containing gaseous solute in liquid solvent, while Raoult’s law is applicable to non volatile solid solute in the liquid solvent.

5. If the solute is non volatile then the Henry’s law constant will become equal to the vapour pressure of pure solvent Po. thus Raoult’s law becomes a special case of Henry’s law.

6. For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.

Question 14.
What are the necessary conditions for an ideal solution? Give two example. For an ideal solution
1. There is no change in volume on mixing two components (solute and solvent)
∆V_mixing = O

2. There is no exchange of heat when the solute is dissolved in solvent (∆H_mixing = 0)

3. Escaping tendency of the solute and the solvent present in it should be same as in pure liquids.

4. Examples – For ideal solution: Benzene and toluene, n-Hexane and n-Heptane, ethyl bromide and ethyl iodide, chlorobenzene and bromo benzene.

Question 15.
Explain how non-ideal solutions shows positive deviation from Raoult’s law.
Answer:

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water.
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interaction).
3. This results in the increased evaporation of both componeins from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoult’s law.
5. Here, the mixing OCCSS is endothermic i.e., (∆H_mixing > O) and there will be a slight increase in volume (∆V_mixing > O)

Question 16.
Explain with suitable example about negative deviation from law.
Answer:
1. Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves.

2. When mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are Stronger than the hydrogen bonds formed amongst themselves.

3. Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution.

4. Asa result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆V_mixing < 0) on mixing.

5. During this process evolution of heat takes place i.e., ∆V_mixing < 0 (exothermic).

6. Examples – Acetone + Chloroform, Chloroform + Diethyl ether

Question 17.
The vapour pressure of a solution containing a non volatile, non-electrolyte solute is always lower than that of pure solvent. Give reason.
Answer:
1. The vapour pressure of a solution (P) containing flOfl volatile solute is lower than that of pure solvent (P°).

2. Consider a closed system is which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibbs free energies of solvent in a liquid and gaseous phase are equal (∆G = O).

3. When a solute is added to this solvent the dissolution takes place and its free energy (G) decreases due to increase in entropy.

4. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease.

5. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure.

6. Thus the vapour pressure of the solution must decrease to maintain the equilibrium.

Question 18.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
According to Raoult’s law,
P_solution x_A, where x_A = mole fraction of solvent.
P_solution = k . x_A, where k = proportionality constant
For a pure solvent,
Vapour pressure = P°, x_A = 1
P°_solution = k x 1 = k
Substituting P°_solvent in Raoult’s law
P_solution = P°_solvent . x_A
Relative lowenng of vapour pressure

substituting P_solution as P°x_B in the above eaquation
Relative lowering of vapour pressure

x_A + x_B = I
where x_B = mole fraction of solute. It is clear that the relative lowering of vapour pressure depends only on the mole fraction ofthe solute (x_B) and is independent of its nature. Therefore relative lowering of vapour pressure is a colligative property.

Question 19.
Explain why boiling point of solution is greater than that of pure solvent?
Answer:
When a non volatile solute is added to a pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to I atm the temperature of the solution has to be increased.

As a result, the solution boils at a higher temperature (T_b) than the boiling point of the pure solvent (T°_b). This increase in the boiling point is known as elevation of boiling point ∆T_b = T_b – T°_b.

Question 20.
Graphically prove that Tb ¡s greater than T°_b.
Answer:

1. The vapour pressure of the solution increases with increase in temperature. The variation of vapour pressure with respect to temperature of pure water is given by the curve – A.

2. At 100°C, the vapour pressure of water is equal to I atm. Hence, the boiling point of water is 100°C (T°_b).

3. When a solute is added to water, the vapour pressure of the resultant solution is lowered. The variation of vapour pressure with respect to temperature for the solution is given by curve-B.

4. From the graph, it is evident that the vapour pressure of the solution is equal to 1 atm. pressure at the temperature T_b which is greater than T°_b. The difference between these two temperatures (T_b – T°_b) gives the elevation of boiling point.
∆T_b = T_b T°_b.

Question 21.
Derive the relationship between the elevation of boiling point and molar mass of non volatile solute.
Answer:
The elevation of boiling point ∆T_b = T_b T°_b.
∆T_b is directly proportional to the concentration of the solute particles.
∆T_b ∝ m, (m = molaLiiy)
∆T_b = k_b. m, where k_b = ebullioscopic constant

Question 22.
Define

1. freezing point
2. Depression in freezing point.

Explain with graph.
Answer:

1. Freezing point is defined as the temperature at which the solid and the liquid states of the substances have the same vapour pressure.

2. When a non volatile solute is added to water at its freezing point, the freezing point of water is lowered from 0°C. The lowering of freezing point of the solvent when a solute is added is called depression in freezing point AT1..

3. ∆T_f = T⁰_f – T_f

Question 23.
Define

1. cryoscopic constant
2. ebullioscopic constant

Answer:

1. ∆T_f = k_f. m, where k_f = cryoscopic constant. If m = 1. then AT_f = k_f
k_f is defined as depression in freezing point for 1 molal solution.

2. ∆T_f = k_f. m where k_f ebullioscopic constant. If m = 1 then AT_f = k_f
k_b is defined as elevation in boiling point for 1 molal solution.

Question 24.
What are the significances of osmotic pressure over other colligative properties ?
Answer:
1. Unlike elevation of boiling point and the depression in freezing point, the magnitude of osmotic pressure is large.

2. The osmotic pressure can be measured at room temperature enables to determine the molecular mass of biomolecules which are unstable at higher temperature.

3. Even for a very dilute solution, the osmotic pressure is large.

Question 25.
What is haemolysis ? intravenous fluid are isotonic to blood?
Answer:
1. The osmotic pressure of the blood cells is approximately equal to 7 atm at 37°C.

2. The intravenous injections should have saine osmotic pressure as that of the blood (isotonic vith blood).

3. If the intravenous solutions are too dilute that is hypotonie, the solvent from outside of the cells flow into the cell to normalise the osmotic pressure and this process is called haernolysis causes the cells to burst.

4. On the other hand, if the solution is too concentrated, that is hypertonic. the solvent molecules will flow out of the cells,which causes the cells to shrink and die.

5. For this reason, the intravenous fluids are prepared such that they are isotonic to blood (0.9% mass/volume sodium chloride solution).

Question 26.
Explain reverse osmosis.
Answer:
1. The pure water moves through the semipermeable membrane to the NaCl solution due to osmosis.

2. This process can be reversed by applying pressure greater than the osmotic pressure to the solution side. Now the pure water moves from the solution side to the solvent side and this process is called reverse osmosis.

3. Reverse osmosis can be defined as a process in which a solvent passes through a semipermeable membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater than the osmotic pressure.

Question 27.
Explain about the application of reverse osmosis in water purification.
Answer:
1. Reverse osmosis is used in the desalination of sea water and also in the purification of drinking water.

2. When a pressure higher than the osmotic pressure is applied on the solution side (sea water) the water molecules moves from solution side to the solvent side through semi permeable membrane (opposite to osmotic flow). The pure water can be collected.

3. Cellulose acetate (or) polyamide membranes are commonly used in commercial system.

Question 28.
Acetic acid is found to have molar mass as 120 g mol^-1. Prove it.
Answer:
1. In certain solvent, solute molecules associate to form a dimer. This reduces the total number of molecules formed in solution and as a result the calculated molar mass will be higher than the actual molar mass.

2. Acetic acid in benzene exist as a dimer

3. The molar mass of acetic acid calculate using colligative properties is found to be around 120 g mol^-1 is two times of the actual molar mass 60 g mol^-1.

Question 29.
Depression in freezing point of NaCI is twice that of in urea. Why?
Answer:
1. The electrolyte NaCI dissociates completely into its constituent ions in their aqueous solution. This causes an increase in the total number olparticles present in the solution.

2. When we dissolve 1 mole of NaCI in water it. dissociates and gives 1 mole of Na⁺ and 1 mole of Cl^–. Hence the solution will have 2 moles of particles.

But when we dissolve 1 mole of urea (non electrolyte) in water it appears as 1 mole only. So the colligative property value would be double in NaCl than in urea.

Question 30.
What is van’t Hoff factor? Calculate the van’t Hoff factor value for

1. acetic acid
2. NaCl

Answer:
1. van’t Hoff factor is defined as the ratio ofthe actual molar mass to the abnormal (calculated) molar mass of the solute.

2.

van’t Hoff factor (1) for acetic acid = \(\frac { 60 }{ 120 }\) = 0.5

3. van’t Hoff factor (2) for NaCl = \(\frac { 117 }{ 58.5 }\) = 2

Question 31.
Differentiate between ideal solution and non-ideal solution.
Answer:
Ideal solution
An ideal solution is a solution in which each component obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

• ∆H_mixing = 0
• ∆V_mixing = 0

Example: Benzenc and toluene n – Hexane and n – Heptane

Non-ideal solution
The solutions which do not obey Raoult’slaw over the entire range of concentrationsare called non-ideal solution.
For a non-ideal solution.

• ∆H_mixing \(\quad \neq\) 0
• ∆V_mixing \(\quad \neq\) 0

Example: Ethyl alcohol and Cyclo hexane, Benzene and acetone .

Question 32.
Explain the factors when i = 1, i < 1 and i >1 ?
Answer:
1. For a solute that does not dissociate or associate the vant’s hoff factor is equal to 1 (i = 1) and the molar mass will be close to the actual molar mass.

2. For that solute that associate to form higher oligomers in solution, the van’t Hoff factor will be less than 1 (i < 1) and the observed molar mass will be greater than the actual molar mass.

3. For solutes that dissociates into their constituent ions the van’t Hoff factor will be more than one (i > 1) and the observed molar mass will be less than the normal molar mass.

Question 33.
State Henry’s law and mention some of its important applications.
Answer:
Henry’s law: The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Application of Henry’s law:

1. In the production of carbonated beverages
   (as solubility of CO₂ increase at high pressure).
2. In the deep sea diving.
3. In the function of lungs.
4. For climbers or people living at high altitudes,

Question 34.
What type of non – idealities are exhibited by cyclohexane – ethanol and acetone – chloroform mixture? Give reason for your answer.
Answer:
Ideal solutions are those which obey Raoult’s law over extreme range of concentration. Ideal solutions have another important properties:

• ∆H_mix = 0
• ∆V_mix = 0

Here-forces of attraction between A – A. B – B and A – B are of the same order. Non ideal solutions do not obey Raoult’s law over the entire range of concentration.
∆H_mixing\(\quad \neq\) 0 and ∆V_mixing\(\quad \neq\) 0

Cyclohexane – ethanol mixture shows positive deviation from Raoult’s law because forces of attraction between cyclohexane and ethanol are less than in between pure cyclohexane as well as pure ethanol.

Acetone-Chloroform mixture shows negative deviation from Raoults law because forces of attraction between acetone and chloroform are higher than that in between pure acetone and pure chloroform molecules.

Question 35.
Given below is the sketch of a plant for carrying out a process.

1. Name the process occurring in the above plant.
2. To which container does the net flow of solvent take place?
3. Name one SPM which can he used in this plant.
4. Give one practical use of the plant.

Answer:

1. Reverse osmosis
2. In fresh water container from salt water container.
3. Cellulose acetate is semipermeable membrane (SPM)
4. Purification of water

Question 36.
Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?
Answer:
π = CRT
π = \(\frac { n }{ V }\)RT
πV = nRT

Osmotic pressure is inversely proportional to the molecular mass of the soLute.

Question 37.
1. Menthol is a crystalline substance with peppermint taste. A 6.2% solution of menthol in cyclohexane freezes at – 1.95°C.

Determine the formula mass of menthol. The freezing point and molal depression constant of cyclohexane are 6.5°C and 20.2 K m^{-1, respectively.}

2. State Henry’s Law and mention its two important applications.

3. Which of the following has higher boiling point and why? 0.1 M NaCl or 0.1 M Glucose
Answer:
1.

M_B = 158 g mol^-1

2. Henry’s Law:
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Applications:

• Solubility of CO₂ is increased at high pressure.
• Mixture of He and O₂ are used by deep sea divers because he is less soluble than nitrogen.

3. 0.1 M NaCI, because it dissociates in solution and furnishes greater number of particles per unit volume while glucose being a non-electrolyte does not dissociate.

Question 38.
Water is a universal solvent. But alcohol also dissolves most of the substances soluble in water and also many more. Boiling point of water is 100°C and that of alcohol is 80°C. The specific heat of water is much higher than the specific heat of alcohol.

1. List out three possible differences if instead of water as the liquid ¡n our body we had alcohol.
2. What value can you derive from this special property of water and its innumerable uses in sustaining life on earth?

Answer:
1.
(i) Even a small rise in temperature in the surroundings will raise the temperature of’ the body because the specific heat of alcoholis much less than the specific heat of water. So, in order to cool the body, more sweating will take place.

(ii) As there is less H bonding in alcohol, it will gel evaporated faster. The alcohol will be evaporated at such a faster rate that the liquid has to be ingested all the time.

(iii) Ice which floats on water helps aquatic life to exist even in winter as water insulates the heat from liquid below it to go back to the surroundings. Solid alcohol does not have such special properties.

2. Praise is to the almighty that has so thoughtfully given such special properties to water and made it a liquid that could sustain life on earth.

Question 39.
State Henry’s law for solubility of a gas in a liquid. Explain the significance of Henry’s law constant (K_H). At the same temperature, hydrogen is more soluble in water than helium. Which of theni will have a higher value of K_H and Why?
Answer:
Henry’s law states that the solubility of a gas in liquid at a given temperature is directly proportional to the partial pressure of the gas.
P = K_H x
where P is the pressure of the gas, x is the mole fraction of the gas in the solution and K_H is the Henry’s law constant. KH is a function of the nature oIgas.

Higher the value of K_H at a given temperature. lower is the solubility of the gas in the liquid. As helium is less soluble in water, so it has a higher value of K_H than hydrogen.

Henry’s Law:
As dissolution of agar in liquid is an exothermic process, therefore, the solubility should decrease with in increase in temperature.

Question 40.
What is meant by positive and negative deviations from Raoult’s law and how is the sign ∆H_mix of related to positive and negative deviations from Raoult’s law?
Answer:
Negative deviations:
In these type of deviations, the partial vapour pressure of each component A and B of solution is higher than the vapour pressure calculated from Raoult’s law. For example -Water and ethanol, chloroform and water.

Positive deviations:
In case of positive deviation A – B interactions are weaker than those between A – A or B – B. This means that in such solutions molecules or A (or B) will find it easier to positive deviation from Raoult’s law.

Samacheer Kalvi 11th Chemistry Solutions 5 Marks Questions and Answers

II. Answer the following questions in detail:

Question 1.

1. Define solution.
2. Explain the types of solutions with suitable example.

Answer:
1. A solution is a homogeneous mixture of two or more substances, consisting of atoms. ions or molecules.

2. Types and examples of solution.

Question 2.

1. Define Solubility
2. Explain about the factors that influences the solubility

Answer:

1. The solubility of a substance at a given temperature is defined as the amount of the solute that can be dissolved in 100 g of the solvent at a given temperature to form a saturated solution.

2. Factors influencing solubility
(a) Nature of solute and solvent: Sodium chloride, an ionic compound readily dissolves in polar solvent such as water but it does not dissolve in non polar solvent such as benzene. Most of the organic compounds dissolve in organic solvent and do not dissolve in water.

(b) Effect of temperature: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. The dissolution of NaCl does not vary as the maximum solubility is achieved at normal temperature.

The dissolution of ammonium nitrate is endothermic, the solubility increases with increase in temperature. The dissolution of eerie sulphate is exothermic and the solubility decreases with increase of temperature. In the case of gaseous solute in liquid solvent the soluhility decreases with increase in temperature.

Effect of pressure:
Generally the change in pressure does not have any significant effect in the solubility of solids and l?quids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Question 3.
Explain about the factors that are responsible for deviation from Raoult’s law.
Answer:
1. Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A – A), the solute molecules (B – B) and between the solvent and solute molecules (A – B) are expected to be similar. if these interactions are dissimilar, there will be a deviation from ideal behaviour.

2. Dissolution of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and causes deviation from Raoult’s law. e.g., KCI in water deviates from ideal behaviour due to dissociation as K⁺ and Cl^– ion which form strong ion-dipole interaction with water molecules.

3. Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example in solution acetic acid exists as a dimer by forming intermolecular hydrogen bonds and hence deviates from Raoult’s law.

4. Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which cause decrease in the attractive force between them. As result, the solution deviates from Raoult’s law.

5. Pressure:
At high pressure, the molecules tends to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus a solution deviates from Raoult’s law at high pressure.

6. Concentration:
When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from Raoult’s law.

Question 4.
How would you determine molar mass from relative lowering of vapour pressure.
Answer:
1. The measurement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.

2. A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.

3. According to Raoult’s law, the relative lowering of vapour pressure is

W_A = weight of solvent
W_B = weight of solute
M_A = Molar mass of solvent
M_B = molar mass of solute

where n_A = number of moles of solvent
n_B = number of moles of solute.
For dilute solution, n_A > > n_B, n_A + n_B = n_A
Then,
Number of moles of solvent and solute arc


From the above equation, molar mass of the solute M_B can be calculated using the known values of W_A, W_B, M_A and the measured relative lowering of vapour pressure.

Question 5.
How would you determine the molar mass from osmotic pressure.
Answer:
According to van’t Hoff equation
π = CRT
C = \(\frac { n }{ V }\)
Here n = number of moles of solute dissolved in ‘V’ litre of the solution.
π = \(\frac { n }{ V }\) . RT = πV = nRT
If the solution is prepared by dissolving W_B of the non-volatile solute in W_A g of solvent, then the number of moles of ‘n’ is

where
M_B = molar mass of the solute
Substituting n value, we get

From the above equation, we can calculate the molar mass of the solute.

Question 6.
What are ideal and non-ideal solutions? Explain with suitable diagram the behaviour of ideal solutions.
Answer:
Ideal solutions:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Ideal solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces.

Examples:

1. Benzene and toluene
2. n – Hexane and n-Heptane
3. Chiorobenzene and bromobenzene.

Characteristics:

1. They must obey Raoult’s law.
2. ∆H mixing should be zero.
3. ∆W mixing should be zero, i.e. volume change on mixing is zero.

Non – ideal solutions:
The solutions which do not obey Raoult’s law are called non-ideal solutions. In case of non – ideal solutions there is a change in volume and heat energy when the two components are mixed.

Characteristics:

1. They does not obey Raoult’s law.
2. ∆Vmix \(\quad \neq\) 0
3. ∆Hmix \(\quad \neq\) 0

Behaviour of Ideal Solutions:
A plot of P₁ or P₂ versus the mole fraction x₁ and x₂ for an ideal solution gives a linear plot. These Lines (I and II)pass through the points and respectively whenx1 and x₂ is equal to unity.

Similarly the plot (Line III) of P_total versus x₂ is also linear. The minimum value of is P₁° and the maximum value is P₂°, assuming that component 1 is less volatile than component 2, i.e. P₁° < P₂₀.

Question 7.
Explain with a suitable diagram and appropriate example, why some non-ideal solution shows positive deviation from Raou It’s law.
Answer:
Some non-ideal solutions show positive deviation from Raoult’s law. Consider a solution of two components A and B.  If A-B interactions in the solution are weaker than the A – A and B – B interactions in the two liquids forming the solution, then the escaping tendency of molecules A and B from the solution become more than in pure liquids.

The total vapour pressure will be greater than the corresponding vapour pressure as expected on the basis of Raoult’s law. This type of behaviour of solution is called positive deviation from Raoult’s law. The boiling point of such solutions are lowered. Mathematically,
P_A < P_A⁰ x_A
P_B < P_B⁰ x_B
The total vapour pressure is less than P_A + P_B
P < P_A + P_B
P < P°_A x x_A + P°_B x_B
Hence
P₁ = P_A
P₂ = P_B
Examples of solutions showing positive deviations

1. Ethyl alcohol and water
2. Benzene and acetone
3. Ethyl alcohol and cyclohcxanc
4. Carbon tetrachloride and chloroform.

Question 8.

1. What is Osmotic pressure and how is it related to the molecular mass of a non volatile substances?
2. What advantage the osmotic pressure method has over the elevation of boiling point method for determining the molecular mass?

Answer:
1. Osmotic pressure:
It is the pressure of the solution column that can prevent the entry of solvent molecules through a semi-permeable membrane, when the solution and the solvent are separated by the same. It is denoted by π. Its unit is mm 11g or atmosphere.
We know that, π = CRT
where π is the osmotic pressure and R is the gas constant
π = \(\frac { { n }_{ 2 } }{ V }\) RT
where V is volume of solution per litre containing n₂ moles of solute.

By the above relation molar mass of solute can be calculated.

2. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and molarity of the solution is used instead of molality.

The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperature and polymers have poor solubility.

III. Numerical Problems

Question 1.
Calculate the mole fraction of benzene ¡n solution containing 30% by mass in carbon tetrachioride.
Solution:
30% of benzene in carbon tetrachloride by mass means that Mass of benzene in the solution = 30g
Mass of solution = 100g
Mass of carbon tetrachloride = 100g – 30g = 70g
Molar mass of benzene (C₆H₆) = 78 g mol^-1
Molar mass of CCl₄ = 12 + (4 x 35.5) 154g mol^-1

Question 2.
Calculate (he molarity of each of the following solutions:
Solution:

1. 30 g of CO(NO₃)₂. 6H₂O = in 4.3 L of solution
2. 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

1. Molar mass of CO(NO₃)₂. 6H₂O = 58.7 + 2(14 + 48) + (6 x 8)g mol^-1
= 58.7 + 124 + 108g mol^-1 = 290.7 gmol^-1

Volume of solution = 4.3 L

2. 1000 mL of 0.5 M H₂SO₄ Contain H₂SO₄ = 0.5 moles
30 mL of 0.5 M H₂SO₄ contain H₂SO₄ = \(\frac { 0.5 }{ 1000 }\) x 30 mole = 0.015 mole
Volume of solution = 500 mL = 0.500 L

Question 3.
Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
0.25 molal aqueous solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 6Og mol^-1
0.25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the solution = 1000 + 15 g = 1015 g = 1.015 g
Thus, 1.015 kg of solution contain urea = 15 g
2.5 kg of solution will require urea = \(\frac { 15 }{ 1.015 }\) x 2.5 kg = 37g

Question 4.
H₂S, a toxic gas with rotten egg like smell is used for the qualitative analysis. 1f the solubility of H₂S in water at STP is 0.195 m. Calculate Henrvs law constant.
Solution:
Solubility of H₂S gas = 0.195 m
= 0.195 mole in 1 kg of the Solvent (water)
1 kg of the solvent (water) = 1000 g
Mole fraction of H₂S gas in the solution (x) = \(\frac { 0.195 }{ 0.195 + 55.55 }\) = 0.0035
Pressure at STP = 0.98 7 bar
Applying Henry’s law
P_H2S = K_H x x_H2S

Question 5.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Here
P°₁ = 23.8 mm
W₂ = 50g
M₂ (urea) = 60g mol^-1
W₁ = 850g
M₁(Water) = 18g mol^-1
Here we have to calculate P_s
Applying Raoult’s law,

Thus, relative lowering of vapour pressure = 0.017
Substituting P° = 23.8 mm Hg

We get,
23.8 – P_s = 0.017 P_s
P_s = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg

Question 6.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500g of water such that it boils at 100°C? .
Solution:
Elevation in boiling point required, ∆T_b = 100 – 99.63° = 0.37°
Mass of solvent (water) W₁ = 500g
Mass of solute, C₁₂H₂₂O₁₁ = 342 g mol^-1
Molar mass of solvent M₁ = 18 g mol^-1
Applying the formula,

Question 7.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL^-1?
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of HNO₃ = 63g mol^-1
68g HNO₃ = mole = 1.079 mole
Density of solution = 1.504 g mL^-1
Volume of solution = \(\frac { 100 }{ 1.504 }\) mL = 66.5 mL = 0.0665 L

Question 8.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate tile mass percentage of the resulting solution.
Solution:
300 g of 25% solution contains solute = 75g
400g of 40% solution contains solute = 160 g
Total mass of solute = 160 + 75 = 235 g
Total mass oÍ’ solution = 300 + 400 = 700 g
% of solute in the final solution = \(\frac { 235 }{ 700 }\) x 1oo = 33.5%
% of water in the finaI solution = 100 – 33.5 = 66.5%

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCI₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass)

1. express this in percentage by mass
2. determine the molality of chloroform in the water sample.

Solution:
1. 15 ppm means 15 parts in million (10₆) parts by mass in the solution
% of mass = \(\frac { 15 }{ { 10 }^{ 6 } }\) x 100 = 1.5 x 10^-4

2. Taking 15g chloroform in 106g of the solution
Mass of the solvent = 10⁶ g
Molar mass of CHCl₃ = 12 + 1 + (3 x 35.5) = 119.5 g mol^-1

Question 10.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Vapour pressure of pure water at the boiling point
(P°) = 1 atm 1.013 bar
Vapour pressure of solution P_s = 1.004 bar
M₁ = 18 g mol^-1
M₂ = ?
Mass of solute = W₂ = 2g
Mass of solution = 100g
Mass of solvent W₁ = 98 g
Applying Raoult’s law for dilute solution

Question 11.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.
Solution:
Molar mass of cane sugar
C₁₂H₂₂O₁₁ = 342 g mol^-1
Molality of sugar = \(\frac { 5 x 1000 }{ 342 x 100 }\) = 0.146
∆T₂ for sugar solution = 273.15 – 271 = 2.15°
∆T_f = K_f x m
K_f = \(\frac { 2.15 }{ 0.146 }\)
Molality of glucose solution = \(\frac { 5 }{ 180 }\) x \(\frac { 1000 }{ 100 }\) = 0.278 m
∆T_f (Glucose) = \(\frac { 2.15 }{ 0.146 }\) x 0.278 = 4.09°K
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K

Question 12.
Calculate the amount of benzoic acid (C₆HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
0.15 M solution means that 0.15 moles of C₆H₅COOH is present in IL
= 1000 mL of the solution
Molar mass of C₆H₅COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol^-1
Thus, 1000 mL of solution contains benzoic acid = 18.3g
250 mL of solution will contain benzoic acid
= \(\frac { 18.3 }{ 1000 }\) x 250 = 4.575 g

Question 13.
A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60 °C. Determine the molecular mass of the solute (For ether K_b 2.02 K kg mol^-1)
Solution:
We have, mass of solute, W₂ = 8 g
Mass of solvent, W₁ = 100 g
Elevation of boiling point
∆T_b = 36.86 – 35.60 = 1.26⁰C
K_b = 2.02
Molecular mass of the solute

= 128.25g mol^-1

Question 14.
Ethylene glycol (molar mass = 62 g mol^-1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 of this substance in 100 g of water. Would it be advisable to keep this substance ¡n the car radiator during summer? (K_f for water = 1.86 k kg/mol^-1) (K_b for water = 0.512 K kg/mol^-1)
Solution:

Since water boils at 100°C, so a solution containing ethylene glycol will boil at 101.024 °C, SO it is advisable to keep this substance in car radiator during summer.

Question 15.
15.0 g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material? K_f for water = 1.86 K kg mol^-1.
Solution:
W_(solute) = 1 5.0 g
W_(solvent) = 450 g
∆T_f = T°_f – T_f = 0 – ( – 0.34) = 0.34 °C
∆T_f = k_f m
0.34 = 1.86 x \(\frac { 15 }{ M }\) x \(\frac { 1000 }{ 450 }\)
M = \(\frac { 1.86 x 15 x 1000 }{ 0.34 x 450 }\) = 182.35 g mol^-1

Common Errors
Common Errors:

1. Students are writing solute, solvent, students get confused to write A (or) B, 1 (or)2.
2. Mole, mole fraction may be confused.
3. In writing osmosis definition Students get confused in mentioning concentration terms.
4. van’t Hoff factor I formula may be con fused.
5. Students may get contused rhen they write solute and solvent.
6. Mole and mole fraction may be confused by students.
7. Standard solutions must be known.
8. When students write Raoult’s law, they get confused with solute and solvent.
9. When they write the definition of osmosis, the conc. term may be little con fused.
10. van’t Hoff equation may be written wrongly.

Rectifications:

1. Always solvent is first so it is denoted as A (or) 1 solute is second so, it is denoted as B (or)2.
2. Mole = n ; Mole fraction x
3. Osmosis-movenient of solvent from low concentration to high concentration through a semipermeable membrane.
4. 
5. For e.g.. solid in liquid means solid is the solute and liquid is the solvent.
6. 
7. 1 N = 1 normal solution, 0.1 N Decinormal solution, 0.01 N = Centinormal solution
8. In Raoult’s law, “A” is always solvent and “B” is always solute.
9. In osmosis, always solvent rnoes through semi permeable membrane from low concentration to high concentration.
10. van’t Hoff factory = i
    

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Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Multiple Choice Questions

Question 1.
If K_b and K_f for a reversible reactions are 0.8 x 10^-5 and 1.6 x 10^-4 respectively, the value of the equilibrium constant is
(a) 20
(b) 0.2 x 10^-1
(c) 0.05
(d) None of these
Answer:
(a) 20
Solution:

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria

The relation between K₁ and K₂ is

Answer:

Solution:

Question 3.
The equilibrium constant for a reaction at room temperature is K₁ and that at 700 K is K₂. If K₁ > K₂ then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The reaction does not attain equilibrium
(d) The reverse reaction is exothermic
Answer:
(a) The forward reaction is exothermic
Solution:
T₁ = 25 + 273 = 298 K, T₂ = 700 K

In this case, T₂ > T₁ and K₁ > K₂

ΔH° is – ve i.e., forward reaction is exothermic.

Question 4.
The formation of ammonia from N₂(g) and H₂(g) is a reversible reaction
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) + Heat
What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer:
(c) equilibrium is shifted to the left
Solution:
Increase in temperature, favours the endothermic reaction, given that formation of NH₃ is exothermic i.e., the reverse reaction is endothermic.
∴ Increase in temperature, shift the equilibrium to left

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by ………….
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure
Solution:

increase in pressure, favours the forward reaction.

Question 6.
Which one of the following is incorrect statement?
(a) for a system at equilibrium, Q is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer:
(a) for a system at equilibrium, Q is always less than the equilibrium constant
Solution:
Correct statement is, for a system at equilibrium, Q = K_eq

Question 7.
K₁ and K₂ are the equilibrium constants for the reactions respectively.

What is the equilibrium constant for the reaction NO₂(g) \(\rightleftharpoons\) \(\frac { 1 }{ 2 }\)N₂(g) + O₂(g)

Answer:

Solution:
Let equilibrium constant for the required reaction be K. Then,

Question 8.
In the equilibrium, 2A(g) \(\rightleftharpoons\) 2B(g) + C₂(g) the equilibrium concentrations of A, B and C, at 400 K are 1 x 10⁴ M, 2.0 x 10^-3 M, 1.5 x 10^-4 M respectively. The value of K_c. for the equilibrium at 400 K is
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 x 10^-2
Answer:
(a) 0.06
Solution:
[A] = 1 x 10^-4 M
[B] = 2 x 10^-3 M
[C] = 1.5 x 10^-4 M

Question 9.
An equilibrium constant of 3.2 x 10^-6 for a reaction means, the equilibrium is ………
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of these
Answer:
(b) largely towards reverse direction
Solution:

Question 10.
\(\frac { { K }_{ c } }{ { K }_{ p } }\) for the reaction, N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g) is ………
(a) \(\frac { 1 }{ RT }\)
(b) \(g\sqrt { RT } \)
(c) RT
(d) (RT)²
Answer:
(d) (RT)²
Solution:
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)
∆n_g = 2 – 4 = – 2
K_p = K_c (RT)^-2
\(\frac { { K }_{ c } }{ { K }_{ p } }\) = (RT)²

Question 11.
For the reaction, AB(g) \(\rightleftharpoons\) A(g) + B(g), at equilibrium, AB is 20% dissociated at a total pressure of R The equilibrium constant K is related to the total pressure by the expression ………..
(a) P = 24 K_p
(b) P = 8 K_p
(c) 24 p = K_p
(d) none of these
Answer:
(a) P =24 K_p
Solution:

Question 12.
In which of the following equilibrium, K and K are not equal?
(a) 2NO(g) \(\rightleftharpoons\) N₂(g) + O₂(g)
(b) SO₂(g) + NO₂(g) \(\rightleftharpoons\) SO₃(g) + NO(g)
(c) H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
(d) PCI₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
Answer:
(d) PCI₅(g) \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
Solution:
For reactions given in options (a), (b) and (c) ∆n_g = 0
For option (d) ∆n_g = 2 – 1 = 1
∴ K_p = K_c (RT)

Question 13.
If x is the fraction of PCI₅ dissociated at equilibrium in the reaction,
PCl₅ \(\rightleftharpoons\) PCl₃ + Cl₂
then starting with 0.5 moIe of PCI₅ the total number of moles of reactants and products at equilibrium is
(a) 0.5 – x
(b) x + 0.5
(e) 2x + 0.5
(d) x ± 1
Answer:
(b) x + 0.5
Solution:

Total no. of moles at equilibrium = 0.5 – x + x + x = 0.5 + x

Question 14.
The values of K_p1 and K_p2 for the reactions X \(\rightleftharpoons\) Y + Z and A \(\rightleftharpoons\) 2B are in the ratio 9 : 1 if degree of dissociation and initial concentration of X and A be equal then total pressure at equilibrium P₁, and P₂ are in the ratio
(a) 36 : 1
(b) 1: 1
(c) 3 : 1
(d) 1: 9
Answer:
(a) 36: 1
Solution:

Question 15.
In the reaction, Fe(OH)₃(s) \(\rightleftharpoons\) Fe³⁺(aq) + 3OH^–(aq), if the concentration of OH^– ions is decreased by ¼ times, then the equilibrium concentration of Fe³⁺ will …………
(a) not changed
(b) also decreased by ¼ times
(c) increase by 4 times
(d) increase by 64 times
Answer:
(d) increase by 64 times
Solution:

[∵ Concentration of solids is constant)
When concentration of OH^– ions decreased by \(\frac { 1 }{ 4 }\) times, then

To maintain K_C as constant, concentration of Fe³⁺ will increase by 64 times.

Question 16.
Consider the reaction where K_p = 0.5 at a particular temperature
PCI₅(g) \(\rightleftharpoons\) PCI₃(g) + Cl₂(g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true?
(a) more PCI₃ will be produced
(b) more Cl₂ will be produced
(c) more PCI₅ will be produced
(d) none of these
Answer:
(c) more PCI₅ will be produced .
Solution:
K_p = 0.5

Q = \(\frac { 1 x 1 }{ 1 }\)
Q > K_p
∴Reverse reaction is favoured; i.e., more PCI₅ will be produced.

Question 17.
Equimolar concentrations of H₂ and I₂ are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H₂ has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) 33%
(b) 66%
(c)(33)² 0/0
(d) 16.5%
Answer:
(a) 33%
Solution:

As degree of dissociation cannot be negative. Therefore, degree of dissociation
= \(\frac { a }{ 3 }\) x 100 = 33.33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 1 and the equilibrium constant is 50. The rate constant for the reverse reaction is ……….
(a) 11.5
(b) 5
(c) 2 x 10²
(d) 2 x 10³
Answer:
(b) 5
Solution:

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Answer:
(c) All the physical processes stop at equilibrium
Solution:
Correct statement – Physical processes occurs at the same rate at equilibrium.

Question 20.
For the fórmation of two moles of SO₃(g) from SO₂ and O₂, the equilibrium constant is K₁. The equilibrium constant for the dissociation of one mole of SO₃ into SO₂ and O₂ is …………

Answer:

Solution:
2SO₂ +O₂ \(\rightleftharpoons\) 2SO₃

Dissociation of 1 mole of 2SO₃

Question 21.
Match the equilibria with the corresponding conditions …………
(i) Liquid \(\rightleftharpoons\) Vapour – 1. Melting point
(ii) Solid \(\rightleftharpoons\) Liquid – 2. Saturated solution
(iii) Solid \(\rightleftharpoons\) Vapour – 3. Boiling point
(iv) Solute (s) \(\rightleftharpoons\) Solute (Solution) – 4. Sublimation point
5. Unsaturated solution

Answer:
(b) 3, 1, 4, 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B \(\rightleftharpoons\) C, if the concentration of the reactants A and B are doubled, then the equilibrium constant will ………
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer:
(d) remain the same
Solution:
A + B \(\rightleftharpoons\) C
K_C = \(\frac { [C] }{ [A] [B] }\)
If [A] and [B] are doubled, [C] increases 4 times to maintain K_C as constant.
∴ Equilibrium constant will remain the same.

Question 23.
[CO(H₂O)₆]²⁺ (aq) (pink) + 4C^– (aq) \(\rightleftharpoons\) [COCI₄]^2- (aq) (blue) + 6H₂O (1) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one
of the following is true?
(a) ∆H > 0 for the forward reaction
(b) ∆H = 0 for the reverse reaction
(c) ∆H < 0 for the forward reaction
(d) Sign of the ∆H cannot be predicted based on this information.
Answer: (a) ∆H > 0 for the forward reaction
Solution:
On cooling, reverse reaction predominates and the solution is pink in colour. Decrease in temperature, favours the reverse reaction i.e. reverse reaction is exothermic (∆H < 0)and for the forward reaction is endothermic (∆H > 0).

Question 24.
The equilibrium constants of the following reactions are:
N₂ + 3H₂ \(\rightleftharpoons\) 2NH₃ ; K₁
N₂ + O₂ \(\rightleftharpoons\) 2NO ; K₂
H₂+ \(\frac { 1 }{ 2 }\)O₂ \(\rightleftharpoons\) H₂O ; K₃
The equilibrium constant (K) for the reaction;  will be

Answer:

Solution:

Question 25.
A 20 litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value will be.
Given that: SrCO₃(s) \(\rightleftharpoons\) SrO(s) + CO₂(g)
(a) 2 litre
(b) 5 litre
(c) 10 Litre
(d) 4 litre
Answer:
(b) 5 litre
Solution:
Given that K_p = 1.6 atm
V₁ = 20 L
V₂ = ?
T₁ = 400 K
T₂ = 400 K
K_p = P_co2
P_co2 = 1.6 atm
P₁ = 0.4 atm
P₂ = 1.6 atm

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Short Answer Questions

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer:
In the chemical reaction, as the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. So even at particular temperature, Q is not constant. Even once the equilibrium is achieved then change in concentration of reactants or products, pressure, volume will change the value of Q.

Question 28.
What is the relation between K_p and K_C. Give one example for which K_p is equal to K_C.
Answer:
The relation between K_p and K_c is K_p = K_C (RT)^∆n_g
K_p = equilibrium constant is terms of partial pressure.
K_c = equilibrium constant is terms of concentration.
R = gas constant
T = Temperature.
∆n_g = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆n_g = 0
K_p = K_C(RT)⁰ = K_C i.e., K_p = K_C
Example: H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)
∆n_g = 2 – 2 = 0
∴ K_p = K_C for the synthesis of HI.

Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K_C is larger or smaller than K_p ?
Answer:
For a homogeneous reaction at equilibrium, number of moles of products (n_p) are greater than the number of moles of reactants (n_R) then ∆n_g = + ve
n_p > n_R
∆n_g = + ve
If ∆n_g is ve, K_p value is greater than K_C
K_p = K_C. (RT)^+ve
K_p > K_C
Example: PCl₅ \(\rightleftharpoons\) PCl₃(g) + Cl₂(g)
2 – 1 = 1
K_p = K_C (RT)¹
K_p > K_C

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?
Answer:
When Q > K_C the reaction will proceed in the reverse direction, i.e, formation of reactants.

Question 31.
For the reaction: A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g); ∆H is – ve.
Answer:
The following molecular scenes represent different reaction mixture
(A – dark grey, B – light grey)

1. Calculate the equilibrium constant K_p and K_C
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:
1.
At equilibrium

2.

3.  Since ∆n_g = 2 – 2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 32.
State Le – Chatelier principle.
Answer:
It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.

Question 33.
Consider the following reactions

1. H₂(g) + l₂(g) \(\rightleftharpoons\) 2HI(g)
2. CaCO₃(s) \(\rightleftharpoons\) CaO(s) + CO₂(g)
3. S(s) + 3F₂(g) \(\rightleftharpoons\) SF₆(g)

In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
1.  H₂(g) + l₂(g) \(\rightleftharpoons\) 2HI(g)
In the above equilibrium reaction, volume of gaseous molecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

2.
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

3.
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 34.
State law of mass action.
Answer:
The law states that “At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.”

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:

1. A large value of K_C indicates that the reaction reaches equilibrium with high product yield.
2. A low value of K_C indicate that the reaction reaches equilibrium with low product formed.
3. In general, if the K is greater than I the reaction proceeds nearly to completion. If is less than 10^-3, the reaction rarely proceeds.
4. If K < 10^-3, reverse reaction is favoured. If K_c > 10³, forward reaction is favoured.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Long Answer Questions

Question 36.
Derive a general expression for the equilibrium constant K_p and K_C for the reaction.
3H₂(g) + N₂(g) \(\rightleftharpoons\) 2NH₃(g)
Answer:
In the formation of ammonia, ‘a’ moles of Nitrogen and ‘b’ moles of hydrogen gas are allowed to react in a container of volume of ‘V’. Let ‘x’ moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
N₂(g) + 3H₂(g) \(\rightleftharpoons\) 2NH₃(g)

Applying law of mass action

Total number of moles at equilibrium n = a – x + b -3x + 2x = a + b – 2x

Total number of moles at equilibrium
n = a – x + b – 3x + 2x = a + b – 2x

Question 37.
Write a balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression?
Answer:

Question 38.
What is the effect of added inert gas on the reaction at equilibrium at constant volume?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.

Question 39.
Derive the relation between K_p and K_C.
Answer:
Consider a general reaction in which all reactants and products are ideal gases.
x A + y B \(\rightleftharpoons\)  lC+mD
The equilibrium constant K_C is

The ideal gas equation is
PV = nRT or P = \(\frac { n }{ V }\) RT
Since,
Active mass = molar concentration = \(\frac { n }{ V }\)
P = Active mass x RT
Based on the above expression, the partial pressure of the reacants and products can be expressed as

On substitution in equation (2),

By comparing equation (1) and (4), we get
K_p = K_C (RT) _∆ng
where ∆n_g is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If ∆n_g = 0, K_b = K_C(RT)⁰
K_p = K_C
Example: H₂(g) + I₂ \(\rightleftharpoons\) 2HI(g)

(ii) where
∆n_g = +Ve
K_p = K_C (RT)^+ve
K_p = K_C
Example: 2NH₃(g) N₂(g) + 3H₂(g)

(iii) When
∆n_g = -Ve
K_p = K_C (RT)^-ve
K_p < K_C
Example: 2SO₂(g) +O₂(g) \(\rightleftharpoons\) 2SO₃(g)

Question 40.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.
Answer:
Given that

Question 41.
For the reaction: SrCO₂(s) \(\rightleftharpoons\) SrO (s) + CO₂(g), the value of equilibrium constant K_p = 2.2 x 10^-4 at 1002K. Calculate K_C for the reaction.
Answer:
For the reaction, SrCO₂(s) \(\rightleftharpoons\) SrO (s) + CO₂(g)
∆n_g = 1 – 0 = 1
K_p = K_p (RT)
2.2 x 10^-4 = K_C (0.0821)(1002)

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate K_p and K_C for this reaction.
Answer:

Question 43.
Oxidation of nitrogen monoxide was studied at 200°C with initial pressures of 1 atm NO and 1 atm of O₂. At equilibrium partial pressure of oxygen is found to be 0.52 atm. Calculate K_p value.
Answer:
2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)

Question 44.
1 mol of CH₄, 1 mole of CS₂ and 2 mol of H₂S are 2 moI of H₂ are mixed in a 500 mL flask. The equilibrium constant for the reaction K_C = 4 x 10^-2 mol² lit^-2. In which direction will the reaction proceed to reach equilibrium?
Answer:

∴ The reaction will proceed in the reverse direction to reach the equilibrium.

Question 45.
At particular temperature K_C = 4 x 10^-2 for the reaction
H₂S(g) \(\rightleftharpoons\) 2H₂(g) + \(\frac { 1 }{ 2 }\) S₂(g). Calculate K_C for each of the following reaction
(i). H₂S(g) \(\rightleftharpoons\) 2H₂(g) + S₂(g)
(ii). 3H₂S(g) \(\rightleftharpoons\) 3H₂(g) + \(\frac { 3 }{ 2 }\) S₂(g)
Answer:
K_C = 4 x 10^-2 for the reaction

For the reaction 3H₂S(g) \(\rightleftharpoons\) 3H₂(g) + \(\frac { 3 }{ 2 }\) S₂(g)

Question 46.
28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17g NH₃ was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer:
Given

[NH₃] = \((\frac { 17 }{ 17 })\) = 1 mol = 2x
x = 0.5 mol
At equilibrium, [N₂] = 1 – x = 0.5 mol
[H₂] = 3 – 3x = 3 – 3 (0.5) = 1.5 mol
Weight of N₂ (no. of moles of N₂) x molar mass of N₂ = 0.5 x 28 = 14g
Weight of H₂ = (no. of moles of H₂) x molar mass of H₂= 1.5 x 2 = 3g

Question 47.
The equilibrium for the dissociation of XY₂ is given as,
2XY₂ (g) \(\rightleftharpoons\) 2XY (g) + Y₂(g).
If the degree of dissociation x is so small compared to one. Show that 2 K_p = Px³ where P is the total pressure and K is the dissociation equilibrium constant of XY₂.
Answer:
2XY₂ (g) \(\rightleftharpoons\) 2XY (g) + Y₂(g)


Question 48.
A sealed container was filled with I mol of A, (g), I mol B, (g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction:
A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g)
Answer:
A₂(g) + B₂(g) \(\rightleftharpoons\) 2AB(g)


[B₂]_eq = 1 – x = 1 – 0.33 = 0.667
[AB]₂ = 2x = 2 x 0.33 = 0.66
Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives the quantitative temperature dependance of equilibrium constant K. The relation between standard free energy change ∆G° and equilibrium constant is
∆G° = – RT ln K  …………(1)
We know that,
∆G° = ∆H° – T∆S°        ………….(2)
Substituting (1) in equation (2)
– RT lnK = ∆H° – T∆S°

Rearranging, ln K = \(\frac { – ∆H° }{ RT }\) + \(\frac { ∆S° }{ R }\)   ………….(4)
Differentiating equation (3) with respect to temperature
d(lnK) ∆H° \(\frac { d(lnK) }{ dT }\) = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)
Equation (4) is known as differential form of van’t Hoff equation. On integrating the equation 4, between T₁ and T₂ with their respective equilibrium constants and K₂

Equation is known as integrated form of van’t Hoff equation.

Question 50.
The equilibrium constant K for the reaction
N₂(g) + 3H₂(g) = 2NH₃(g) is 8.19 x 10²
at 298 K and 4.6 x 10^-1 at 498 K. Calculate ∆H° for the reaction.
Answer:

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO₃(s) \(\rightleftharpoons\) CaO (s) + CO₂(g) is 1.017 x 10^-3 atm at 500° C. Calculate K_p, at 600° C for the reaction. ∆H for the reaction is 181 kJ mol^-1 and does not change in the given range of temperature.
Answer:

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
Consider the following reaction
Fe³⁺(aq) + SCN^–(aq) [Fe(SCN)]²⁺(aq)
A solution is made with initial Fe³⁺, SCN^– concentration of 1 x 10^-3 M and 8 x 10^-4M respectively. At equilibrium [Fe(SCN)]²⁺ concentration is 2 x 10^-4M. Calculate the value of equilibrium constant.
Answer:

Question 2.
The atmospheric oxidation of NO
2NO(g) + O₂(g) \(\rightleftharpoons\) 2NO₂(g)
was studied with initial pressure of 1 atm of NO and I atm of O₂. At equilibrium, partial pressure of oxygen is 0.52 atm. Calculate K_p of the reaction.
Answer:

As,
1 – x = 0.52
x = 0. 48
= At equilibrium,
P_NO = 1 – 2x = 1 – 2(0.48) = 0.04
P_NO2 = 2x = 2(0.48) = 0.96

Question 3.
The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
CO(g) + H₂O(g) \(\rightleftharpoons\) CO₂(g) + H₂(g) At a given temperature K_p = 2.7.  If 0.13 moI of CO, 0.56 moI of water, 0.78 mol of CO₂ and 0.28 mol of H₂ are introduced into a 2L flask, find out in which direction must the reaction proceed to reach equilibrium.
Answer:
CO(g) + H₂O(g) \(\rightleftharpoons\) CO₂(g) + H₂(g)
Given
K_p = 2.7
[CO] = 0.13 mol, [H₂O] = 0.56 mol
[CO₂] = 0.78 mol; [H₂] = 0.28 mol
V = 2L
K_p = K_C (RT)^Δn_g
2.7 = K_C(RT)°
K_C = 2.7

Q = 3
Q > K_C Hence, the reaction proceed in the reverse direction.

Question 4.
1 moI of PCl₅, kept in a closed container of volume 1 dm³ and was allowed to attain equilibrium at 423 K. Calculate the equilibrium composition of reaction mixture. (The K_C value for PCl₅ dissociation at 423 K is 2)
Answer:
PCl₅ \(\rightleftharpoons\) PCL₃ + Cl₂
Given that [PCl₅]_initial = 1 mol
V = 1 dm³
K_C = 2

∴ Equilibrium concentration of
[PCl₅]_eq = \(\frac { 1-x }{ 1 }\) = 1 – 0.732 = 0.268 M
[PCl₃]_eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732
[Cl₂]_eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732

Question 5.
The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.
Answer:
N2O₄(g) \(\rightleftharpoons\) 2NO₂(g)
T₁ = 298 K
K_p1 = 0.15
T₂ = 100⁰C = 100 + 273 = 373 K
K_p2 = ?

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
One mole of H₂ and one mole of I₂ are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data: [H₂] = I mole, [I₂] = 1 mole
At equilibrium, [HI] = 0.4 mole, K_C = ?
H₂(g) + I₂(g) \(\rightleftharpoons\) 2HI(g)

Question 2.
The equilibrium concentrations of NH₃, N₂ and H₂ are 1.8 x 10² M, 1.2 x 10^-2 M and 3 x 10^-2 M respectively. Calculate the equilibrium constant for the formation of NH₃ from N₂ and H₂. [Hint: M = mol lit^-1]
Answer:
Given data:
[NH₃] 1.8 x 10^-2 M, [N₂] = 1.2 x 10^-2M, [H₂] 3 x 10^-2 M, K_C = ?