Text Book Back Questions and Answers - Chapter 10 Chemical Bonding 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Chemical Bonding
Textual Evaluation Solved
Question 1.
In which of the following compounds does the central atom obey the octet rule?
(a) $\mathrm{XeF}_4$
(b) $\mathrm{AICI}_3$
(c) $\mathrm{SF}_6$
(d) $\mathrm{SO}_2$
Answer:
(d) $\mathrm{SCl}_2$
Solution:
$\begin{array}{cc}\text { Compound } & N o . \text { of valance electrons on the central atom } \\ \mathrm{XeF}_2 & 10 \\ \mathrm{AlCl}_3 & 6 \\ \mathrm{SF}_6 & 12 \\ \mathrm{SCl}_2 & 8\end{array}$
Hence in (d) $\mathrm{SCl}_2$ octet rule is followed
Question 2.
In the molecule $\mathrm{O}_{\mathrm{A}}=\mathrm{C}=\mathrm{O}_{\mathrm{B}}$ the formal charge on $\mathrm{O}_{\mathrm{A}}, \mathrm{C}$ and $\mathrm{O}_{\mathrm{B}}$ 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
$
\begin{aligned}
& \qquad \mathrm{O}_{\mathrm{A}} / \mathrm{O}_{\mathrm{B}}=\mathrm{N}_{\mathrm{V}}-\left(\mathrm{N}_e+\frac{\mathrm{N}_b}{2}\right)=6-\left(4+\frac{4}{2}\right)=6-6=0 \\
& \text { Formal charge of } \\
& \text { Formal charge of } \mathrm{C}=4-\left(0+\frac{8}{2}\right)=4-4=0
\end{aligned}
$
Question 3.
Which of the following is electron deficient?
(a) $\mathrm{PH}_3$
(b) $\left(\mathrm{CH}_3\right)_2$
(c) $\mathrm{BH}_3$
(d) $\mathrm{NH}_3$
Answer:
(c) $\mathrm{BH}_3$
Solution:

$\ddot{\mathrm{NH}_3}, \ddot{\mathrm{P}} \mathrm{H}_3-$ electron rich,
$\mathrm{CH}_3-\mathrm{CH}_3$ - Covalent neutral molecule,
$\mathrm{BH}_3$ - electron deficient

Question 4.
Which of the following molecule contain no $\pi$ bond?
(a) $\mathrm{SO}_2$
(b) $\mathrm{NO}_2$
(c) $\mathrm{CO}_2$
(d) $\mathrm{H}_2 \mathrm{O}$
Answer:
(d) $\mathrm{H}_2 \mathrm{O}$
Solution:

Water $\left(\mathrm{H}_2 \mathrm{O}\right)$ contains only $\sigma$ bonds and no $\pi$ bonds.
Question 5.
The ratio of number of sigma $(\sigma)$ and pi $(\pi)$ bonds in 2- butynal is
(a) $8 / 3$
(b) $5 / 3$
(c) $8 / 2$
(d) $9 / 2$
Answer:
(d) $9 / 2$
Solution:

no. of $\sigma$ bonds $=8[4 \mathrm{C}-\mathrm{H} ; 3 \mathrm{C}-\mathrm{C} ; 1 \mathrm{C}-\mathrm{O}]$
no.of $\pi$ bonds $=3[2 \mathrm{C}-\mathrm{C} ; 1 \mathrm{C}-\mathrm{O}]$
$\therefore$ ratio $=\frac{8}{3}$
Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?
(a) $120^{\circ}, 80^{\circ}$
(b) $109^{\circ} .28$
(c) $90^{\circ}$
(d) $89^{\circ}, 117^{\circ}$
Answer:
(d) $89^{\circ}, 117^{\circ}$

Solution:
Normal bond angle in regular trigonal bipyramidal are $90^{\circ}$ and $120^{\circ}$. Due to 1.p - b.p repulsion, bond angle is reduced to $89^{\circ}, 117^{\circ}$ 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 $\mathrm{CIF}_3, \mathrm{NF}_3$ and $\mathrm{BF}_3$ molecules the chlorine, nitrogen and boron atoms are
(a) $\mathrm{sp}^3$ hybridised
(b) $\mathrm{sp}^3, \mathrm{sp}^3$ and $\mathrm{sp}^2$ respectively
(c) $\mathrm{sp}^2$ hybridised
(d) $\mathrm{sp}^3 \mathrm{~d}, \mathrm{sp}^3$ and $\mathrm{sp}^2$ hybridised respectively
Answer:
(d) $\mathrm{sp}^3 \mathrm{~d}, \mathrm{sp}^3$ and $\mathrm{sp}^2$ hybridised respectively
Solution:
$\mathrm{CIF}_3-\mathrm{sp}^3 \mathrm{~d}$ hybridisation

$\mathrm{NF}_3-\mathrm{sp}^3$ hybridisation
$\mathrm{BF}_3-\mathrm{sp}^2$ hybridisation
Question 10.
When one $s$ and three $p$ orbitais hybridise,
(a) four equivalent orbitais at $90^{\circ}$ to each other will be formed
(b) four equivalent orbitais at $109^{\circ} 28^{\prime}$ ' 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 $109^{\circ} 28^{\prime}$ to each other will be formed.
Question 11.
Which of these represents the correct order of their increasing bond order?
(a) $\mathrm{C}_2<\mathrm{C}_2{ }^{2-}<\mathrm{O}_2$
(b) $\mathrm{C}_2^{2-}<\mathrm{C}_2^{+}<\mathrm{O}^2<\mathrm{O}_2^{2-}$
(c) $\mathrm{O}_2^{2-}<\mathrm{O}_2^{+}<\mathrm{O}_2<\mathrm{C}_2^{2-}$
(d) $\mathrm{O}_2{ }^{2-}<\mathrm{C}_2{ }^{+}<\mathrm{O}_2<\mathrm{C}_2{ }^{2-}$
Answer:
(d) $\mathrm{O}_2^{2-}<\mathrm{C}_2^{+}<\mathrm{O}_2<\mathrm{C}_2^{2-}$
Solution:
bond order $=\left(\mathrm{n}_{\mathrm{b}}-\mathrm{n}_{\mathrm{a}}\right)$
bond order of $\mathrm{O}_2^{2-}=\frac{1}{2}(8-6)=1$
bond order of $\mathrm{C}_2^{+}=\frac{1}{2}(5-2)=1.5$
bond order of $\mathrm{O}_2=\frac{1}{2}(8-4)=2$
bond order of $\mathrm{C}_2{ }^{2-}=\frac{1}{2}(8-2)=3$

Question 12 .
Hybridisation of central atom in $\mathrm{PCl}_5$ involves the mixing of orbitais.
(a) $\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{d}_{\mathrm{x}}, \mathrm{d}_{\mathrm{x}}{ }^2-\mathrm{y}^2$
(b) $s, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{p}_{\mathrm{xy}}, \mathrm{d}_{\mathrm{x}}^2-\mathrm{y}^2$
(c) $\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{p}_{\mathrm{z}}, \mathrm{d}_{\mathrm{x}} 2 \mathrm{y}^2$
(d) $\mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{p}_{\mathrm{xy}}, \mathrm{d}_{\mathrm{x}}{ }^2-\mathrm{y}^2$
Answer:
(c) $\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{p}_{\mathrm{z}}, \mathrm{d}_{\mathrm{x}}{ }^2 \mathrm{y}^2$
Solution:
$\mathrm{PCl}_5-\mathrm{sp}^3 \mathrm{~d}$ hybridisation $\mathrm{s}, \mathrm{p}_{\mathrm{x}}, \mathrm{p}_{\mathrm{y}}, \mathrm{p}_{\mathrm{z}}, \mathrm{d}_{\mathrm{x}}{ }^2-\mathrm{y}^2$

Question 13.
The correct order of $\mathrm{O}-\mathrm{O}$ bond length in hydrogen peroxide, ozone and oxygen is
(a) $\mathrm{H}_2 \mathrm{O}_2>\mathrm{O}_3>\mathrm{O}_2$
(b) $\mathrm{O}_2>\mathrm{O}_3>\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{O}_2>\mathrm{H}_2 \mathrm{O}_2>\mathrm{O}_3$
(d) $\mathrm{O}_3>\mathrm{O}_2>\mathrm{H}_2 \mathrm{O}_2$
Answer:
(b) $\mathrm{O}_2>\mathrm{O}_3>\mathrm{H}_2 \mathrm{O}_2$

Solution:
The bond order for $\mathrm{O}_2, \mathrm{O}_3$ and $\mathrm{H}_2 \mathrm{O}_2$ decreases in the order $2>1.5>1$
Question 14.
Which one of the following is diamagnetic?
(a) $\mathrm{O}_2$
(b) $\mathrm{O}_2^{2-}$
(c) $\mathrm{O}_2^{2+}$
(d) None of these
Answer:
(b) $\mathrm{O}_2^{2-}$
Solution:
$\mathrm{O}_2{ }^{2-}$ is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits $\pi^*{ }_{2 \mathrm{py}}$ and $\pi^*{ }_{2 \mathrm{pz}}$
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 $=\frac{1}{2}\left(\mathrm{n}_{\mathrm{b}}-\mathrm{n}_{\mathrm{a}}\right)$
$2.5=\frac{1}{2}\left(8-\mathrm{n}_{\mathrm{a}}\right)$
$\Rightarrow 5=8$
$\Rightarrow \mathrm{n}_{\mathrm{a}}=8-5=3$
Question 16.
Shape and hybridisation of $\operatorname{IF}_5$ are
(a) Trigonal bipyramidal, $\mathrm{sp}^3 \mathrm{~d}^2$
(b) Trigonal bipyramidal, $\mathrm{sp}^3 \mathrm{~d}$
(c) Squai2e pyramidal, $\mathrm{sp}^3 \mathrm{~d}^2$
(d) Octahedral, $\mathrm{sp}^3 \mathrm{~d}^2$
Answer:
(c) Square pyramidal, $\mathrm{sp}^3 \mathrm{~d}^2$
Solution:
$\mathrm{IF}_5-5$ bond pair +1 lone pair
$\therefore$ hybridisation $\mathrm{sp}^3 \mathrm{~d}^2$

Question 17.
Pick out the incorrect statement from the following.
(a) $\mathrm{sp}^3$ hybrid orbitais are equivalent and are at an angle of $109^{\circ} 28^{\prime}$ with each other.
(b) $\mathrm{dsp}^2$ hybrid orbitais are equivalent and bond angle between any two of them is 900 .
(c) All five $\mathrm{sp}^3 \mathrm{~d}$ hybrid orbitais arc not equivalent. Out of these five $\mathrm{sp}^3 \mathrm{~d}$ hybrid orbitais, three are at an angle of $120^{\circ}$, remaining two are perpendicular to the plane containing the other three
(d) none of these
Answer:
(c) All five $\mathrm{sp}^3 \mathrm{~d}$ hybrid orbitals are not equivalent. Out of these five $\mathrm{sp}^3 \mathrm{~d}$ hybrid orbitals, three are at an angle of $120^{\circ}$ 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) $\mathrm{SeF}_4, \mathrm{XeO}_2 \mathrm{~F}_2$
(b) $\mathrm{SF}_4, \mathrm{XeF}_2$
(c) $\mathrm{XeOF}_4, \mathrm{TeF}_4$
(d) $\mathrm{SeCI}_4, \mathrm{XeF}_4$
Answer:
(a) $\mathrm{SeF}_4, \mathrm{XeO}_2 \mathrm{~F}_2$
Solution:
$\mathrm{SeF}_4, \mathrm{XeO}_2 \mathrm{~F}_2-\mathrm{sp}^3 \mathrm{~d}$ hybridisation
$\mathrm{T}$ - shaped, one lone pair on central atom.
Question 19.
In which of the following molecules / ions $\mathrm{BF}_3, \mathrm{NO}_2{ }^{-}, \mathrm{H}_2 \mathrm{O}$ the centrai atom is $\mathrm{sp}^2$ hybridised?
(a) $\mathrm{NO}_2{ }^{-}$and $\mathrm{H}_2 \mathrm{O}$
(b) $\mathrm{NO}_2{ }^{-}$and $\mathrm{H}_2 \mathrm{O}$
(c) $\mathrm{BF}_3$ and $\mathrm{NO}_2$
(d) $\mathrm{BF}_3$ and $\mathrm{NH}_2^{-}$
Answer:
(c) $\mathrm{BF}_3$ and $\mathrm{NO}_2$
Solution:
$\mathrm{H}_2 \mathrm{O}-$ Central atom $\mathrm{sp}^3$ hybridised

$\mathrm{NO}_2^{-}$- Central atom $\mathrm{sp}^2$ hybridised
$\mathrm{BF}_3-$ Central atom $\mathrm{sp}^2$ hybridised
$\mathrm{NH}_2^{-}-$Central atom $\mathrm{sp}^3$ hybridised
Question 20 .
Some of the following properties of two species, $\mathrm{NO}_3{ }^{-}$and $\mathrm{H}_3 \mathrm{O}^{+}$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:
$\mathrm{NO}_3{ }^{-}-\mathrm{sp}^2$ hybridisation, planar
$\mathrm{H}_3 \mathrm{O}^{+}-\mathrm{sp}^3$ hybridisation, pyramidal
Question 21.
The types of hybridisation on the five carbon atom from right to left in the, 2,3 pentadiene.
(a) $\mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$
(b) $\mathrm{sp}^3, \mathrm{sp}, \mathrm{sp}, \mathrm{sp}, \mathrm{sp}^3$
(c) $\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3$
(d) $\mathrm{sp}^3, \mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}^3, \mathrm{sp}^3$
Answer:
(a) $\mathrm{sp}^3, \mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^2, \mathrm{sp}^3$
Solution:

Question 22.
$\mathrm{XeF}_2$ is isostructural with
(a) $\mathrm{SbCl}_2$
(b) $\mathrm{BaCl}_2$
(c) $\mathrm{TeF}_2$
(d) $\mathrm{ICl}_2^{-}$
Answer:
(d) $\mathrm{ICl}_2^{-}$
Solution:
$\mathrm{XeF}_2$ is isostructural with $\mathrm{ICI}_2^{-}$
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) $\mathrm{SnCI}_2$
(b) $\mathrm{NO}_2$
(c) $\mathrm{C}_2 \mathrm{H}_2$
(d) All of these
Answer:
(c) $\mathrm{C}_2 \mathrm{H}_2$
Solution:

$
\begin{aligned}
& \mathrm{CO}_2 \text { - Linear } \\
& \mathrm{C}_2 \mathrm{H}_2 \text { - Linear }
\end{aligned}
$
Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order
(a) 1.p - 1.p > b.p - b.p $>1 . p-$ b.p
(b) b.p - b.p $>$ b.p - 1.p $>1 . p-$ b.p
(c) 1.p - 1.p $>$ b.p $-1 . p>$ b.p - b.p
(d) b.p - b.p $>1 . p-1 . p>$ b.p - 1.p
Answer:
(c) 1.p - 1.p > b.p $-1 . p>$ b.p - b.p
Question 26.
Shape of $\mathrm{CIF}_3$ is
(a) Planar triangular
(b) Pyramidal
(c) ' $\mathrm{T}$ ' Shaped
(d) none of these
Answer:
(c) ' $\mathrm{T}$ ' Shaped
Solution:
$\mathrm{dF}_3-\mathrm{sp}^3 \mathrm{~d}$ hybridisation
Question 27.
Non- Zero dipole moment is shown by
(a) $\mathrm{CO}_2$
(b) $\mathrm{p}$-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) $\mathrm{NH}_4 \mathrm{Cl}$
(b) $\mathrm{NH}_3$
(c) $\mathrm{NaCl}$
(d) none of these
Answer:
(a) $\mathrm{NH}_4 \mathrm{Cl}$
Question 30.
$\mathrm{CaO}$ and $\mathrm{NaCl}$ have the same crystal structure and approximately the same radii. If $\mathrm{U}$ is the lattice energy of $\mathrm{NaCl}$, the approximate lattice energy of $\mathrm{CaO}$ is
(a) $\mathrm{U}$
(b) $2 \mathrm{U}$
(c) $U / 2$
(d) $4 \mathrm{U}$
Answer:
(d) $4 \mathrm{U}$
Question 31.
Define the following
1. Bond order
2. Hybridisation
3. a- bond

Answer:
1. Bond order:
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}
$
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. $\sigma$ - 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 $(\pi)$ bond.
Question 33.
In $\mathrm{CH}_4, \mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$, the central atom undergoes $\mathrm{sp}^3$ hybridisation-yet their bond angles are different, why?
Answer:
1. In $\mathrm{CH}_4, \mathrm{NH}_3$ and $\mathrm{H}_2 \mathrm{O}$ the central atom undergoes $\mathrm{sp}^3$ 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 $\mathrm{CH}_4$, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle $109^{\circ} 28^{\prime}$.

5. $\mathrm{H}_2 \mathrm{O}$ has 2 bond pairs and 2 lone pairs. There is large repulsion between $\mathrm{lp}-\mathrm{lp}$. Again repulsion between $1 \mathrm{p}$ - $b p$ 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 $104^{\circ} 35^{\prime}$.
6. $\mathrm{NH}_3$ has 3 bond pairs and 1 lone pair. There is repulsion between $1 \mathrm{p}-\mathrm{bp}$. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to $107^{\circ} 18$.
Question 34.
Explain $\mathrm{sp}^2$ hybridisation in $\mathrm{BF}_3$
Answer:
1. $\mathrm{sp}^2$ hybridisation in boron trifluoride - Boron atom - B. Electronic configuration $\left[\mathrm{H}_2\right] 2 \mathrm{~s}^2 2 \mathrm{p}^2$.

2. In boron, the $\mathrm{s}$ orbital and two $\mathrm{p}$ orbitals in the valence shell hybridises to generate three equivalent $\mathrm{sp}^2$ orbitais. These 3 orbitaIs lie in the same xy plane and the angle between any two orbitals is equal to $120^{\circ}$.
3. The $3 \mathrm{sp}^2$ hybridised orbitais of boron now overlap with the $2 p_z$ 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 $\mathrm{O}_2$ is paramagnetic.
Answer:

1. Electronic configuration of $\mathrm{O}$ atom is is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{P}^4$
2. Electronic configuration of $\mathrm{O}$, molecule is
$
\begin{gathered}
\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_x^2 \pi 2 p_y^2 \pi 2 p_z^2 \pi^* 2 p_y^1 \pi * 2 p_z^1 \\
\quad=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{10-6}{2}=2
\end{gathered}
$
3. Bond order $=$
4. Molecule has two unpaired electrons, hence it is paramagnetic.
Question 36.
Draw MO diagram of $\mathrm{CO}$ and calculate its bond order.
Answer:

1. Electronic configuration of $\mathrm{C}$ atom: $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^2$
Electronic configuration of $\mathrm{O}$ atom: $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^4$
2. Electronic configuration of $\mathrm{CO}$ molecule is: $\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2^2 \pi 2 \mathrm{p}_{\mathrm{y}}^2 \pi 2 \mathrm{p}_{\mathrm{z}}^2 \sigma 2 \mathrm{p}_{\mathrm{x}}^2$
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2} \frac{10-4}{2}=3
$
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 $\psi$. Consider two atomic orbitals represented by the wave functions $\psi_A$ and $\psi_{\mathrm{B}}$ with comparable energy that combines to form two molecular orbitals.
3. One is bonding molecular orbitai ( $\psi$ bonding) and the other is anti-bonding molecular orbital ( $\psi$ antibonding).
4. The wave function for molecular orbitais, $\psi_A$ and $\psi_B$ can be obtained by the $L C A O$ as shown below:
$
\begin{aligned}
& \psi_{\text {bonding }}=\psi_{\mathrm{A}}+\psi_{\mathrm{B}} \\
& \psi_{\text {anti-bonding }}=\psi_{\mathrm{A}}-\psi_{\mathrm{B}}
\end{aligned}
$
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 $\mathrm{N}_2$ molecule using MO Theory.
Answer:

1. Electronic configuration of $\mathrm{N}$ atom $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^3$.
2. Electronic configuration of $N$, molecule is: $\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 2 \mathrm{~s}^2 \sigma^* 2^2 \pi 2 \mathrm{p}_{\mathrm{y}}^2 \pi 2 \mathrm{p}_{\mathrm{z}}^2 \sigma 2 \mathrm{p}_{\mathrm{x}}^2$
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2} \frac{10-4}{2}=3
$
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: $\mu=q \times 2 \mathrm{~d}$, where $\mu$ is the dipole moment, $\mathrm{q}$ is the charge, $2 \mathrm{~d}$ 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 ( $\mathrm{Cm}$ ). It is usually expressed in Debye unit (D).
4. 1 Debye $=3.336 \times 10^{-30} \mathrm{Cm}$
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 $\mathrm{CO}_2$, there are two polar bonds $[\mathrm{C}=\mathrm{O}]$, which have dipole moments that are equal in magnitude but have opposite direction.
3. Hence the net dipole moment of the $\mathrm{CO}_2$ is $\mu=\mu_1+\mu_2=\mu_1+\left(-\mu_1\right)=0$

5. In this case $\mu=\vec{\mu}_1+\vec{\mu}_2=\vec{\mu}_1+\left(-\vec{\mu}_1\right)=0$
Question 41.
Draw the Lewis structures for the following species.
1. $\mathrm{NO}_3^{-}$
2. $\mathrm{SO}_4^{2-}$
3. $\mathrm{HNO}_3$
4. $\mathrm{O}_3$
Answer:
1. $\mathrm{NO}_3^{-}$

$\text { 2. } \mathrm{SO}_4^{2-}$

$\text { 3. } \mathrm{HNO}_3$

$\text { 4. } \mathrm{O}_3$

Question 42.
Explain the bond formation in $\mathrm{BeCl}_2$ and $\mathrm{MgCl}_2 \cdot \mathrm{BeCl}_2$ bond formation:
Answer:
1. Electronic confiuration of $\mathrm{Be}(\mathrm{Z}=4)$ is $1 \mathrm{~s}^2 2 \mathrm{~s}^2$ and electronic configuration of $\mathrm{Cl}(\mathrm{Z}=17)$ is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2$ $3 \mathrm{p}^5$
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, $\mathrm{Be}^{2+}$ and each chlorine atom accepts one electron to become $\left(\mathrm{Cl}^{-}\right)$uninegative anion and attains the stable electronic configuration of Argon.
4. Then $\mathrm{Be}^{2+}$ combine with $2 \mathrm{Cl}^{-}$ions to form an ionic crystal in which they are held together by electrostatic attractive forces.
5. During the formation of 1 mole of $\mathrm{BeCl}_2$, the amount of energy released is $-468 \mathrm{~kJ} / \mathrm{mol}$. This favours the formation of $\mathrm{BeCl}$, and its stabilisation.
$\mathrm{MgCI}_2$ bond formation:
1. Electronic configuration of $\mathrm{Mg}(\mathrm{z}=12)$ is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2$.
Electronic configuration of $\mathrm{Cl}(\mathrm{z}=17)$ is $1 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{p}^6 3 \mathrm{p}^5$
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 $\left(\mathrm{Mg}^{2+}\right)$ and two chlorine atoms accept these electrons to become two uninegative anions $\left[2 \mathrm{Cl}^{-}\right]$by attaining the stable inert gas configuration of Argon.
4. These ions, $\mathrm{Mg}^{2+}$ and $2 \mathrm{Cl}^{-}$combine to form an ionic crystal in which they are held together by electrostatic attractive forces.
5. The energy released during the formation of $1 \mathrm{~mole}$ of $\mathrm{MgCl}_2$ is $-783 \mathrm{~kJ} / \mathrm{mole}$. This favours the formation of $\mathrm{MgCI}_2$ and its stabilisation.

Question 43.
Which bond is stronger or $\pi$ ? Why?
Answer:
1. Sigma bonds $(\sigma)$ are stronger than Pi bonds $(\pi)$. 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. $\pi$ 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 $\mathrm{kJ} \mathrm{mol}^{-1}$
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 $\mathrm{MO}$ theory. which is based on quantum mechanics $\mathrm{H}_2$ molecule can be represented in terms of the following diagram called M.O. diagram.

$\mathrm{H}-\mathrm{H}$. i.e.. $\mathrm{H}_2$ molecule has two atoms which are connected by $1 \mathrm{\sigma}$ 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 $-\mathrm{HF}-$ 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 $\mathrm{H}-\mathrm{F}$ bond is said to be a polar covalent bond.
Question 47.
Considering $\mathrm{x}$-axis as molecular axis, which out of the following will form a sigma bond.
1. 1s and $2 \mathrm{p}_{\mathrm{y}}$
2. $2 \mathrm{p}_{\mathrm{x}}$ and $2 \mathrm{p}_{\mathrm{x}}$
3. $2 \mathrm{p}_{\mathrm{x}}$ and $2 \mathrm{p}_z$
4. 1 s and $2 \mathrm{P}_z$
Answer:
Along $\mathrm{X}$-axis as molecular axis, only $2 \mathrm{p}$ and $2 \mathrm{p}$ 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 $\mathrm{O}_{\mathrm{B}}$ and $\mathrm{O}_{\mathrm{C}}$. 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 $\mathrm{CO}_3{ }^{2-}$ (carbonate) ion.

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in Ethylene, $\mathrm{C}_2 \mathrm{H}_4$
1. Bonding in ethylene can he explained by hybridisation concept.
2. The valency of carbon is 4 . The electronic configuration of carbon is $1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}{ }^1 2 \mathrm{p}_{\mathrm{y}}{ }^1 2 \mathrm{p}_z{ }^0$. One electron from $2 \mathrm{~s}$ orbital is promoted to $2 \mathrm{p}_{\mathrm{z}}$. orbital in the excited state to satisfy the valency of carbon.

3. In ethylene both the carbon atoms undergo $\mathrm{sp}^2$ hybridisation involving $2 \mathrm{~s}, 2 \mathrm{p}_{\mathrm{x}}$ and $\mathrm{sp}_{\mathrm{y}}$ orbitals resulting in 3 equivalent $\mathrm{sp}^2$ hybridised orbitals lying in the $\mathrm{XY}$ plane at an angle of $120^{\circ}$ to each other. The unhybridised $2 \mathrm{p}_z$ orbital lies perpendicular to the xy plane.

4. One of the $\mathrm{sp}^2$ hybndised orbitals of each carbon atoms lying along the $\mathrm{X}$ - axis linearly overlaps with each other resulting in the formation of $\mathrm{C}-\mathrm{C}$ sigma bond. The other two $\mathrm{sp}^2$ hybridised orbitals of both carbon atom linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two $\mathrm{C}-\mathrm{H}$ sigma bonds on each carbon atom.
5. The unhybridised $2 \mathrm{p}_z$ 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 $\left(\mathrm{C}_2 \mathrm{H}_2\right)$ :
1. The electronic configuration of valence shell of carbon atom in the ground state is $[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}_{\mathrm{x}}{ }^1 2 \mathrm{p}_z{ }^0$. One electron from $2 s$ orbital is promoted to $2 \mathrm{p}_z$ orbital in the excited state to satisfy the valency of carbon.
2. In acetylene molecule, both the carbon atoms are in $s p$ hybridised state. The $2 \mathrm{~s}$ and $2 \mathrm{p}_{\mathrm{x}}$ orbitals resulting in two equivalent $s p$ hybridised orbitals are formed lying in a straight line along the $\mathrm{X}$ - axis. The unhybridised $2 \mathrm{p}_{\mathrm{y}}$, and $2 \mathrm{p}_z$ orbitais lie perpendicular to the $\mathrm{X}$-axis.

3. One of the two $s p$ 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 $\mathrm{C}-\mathrm{H}$ sigma bond on each carbon atom.
4. The unhybridised $2 \mathrm{p}_{\mathrm{y}}$ and $2 \mathrm{p}_z$ orbitals of each carbon atom overlap sideways. This lateral overlap results in the formation of two pi bonds. $\left(p_y-p_y\right)$ and $\left(p_z-p_z\right)$ between the two carbon atoms.
Question 50.
What type of hybridisations are possible in the following geometeries?
1. octahedral
2. tetrahedral
3. square planar.
Answer:
1. Octahedral geometry is possible by $\mathrm{sp}^3 \mathrm{~d}^2$ (or) $\mathrm{d}^2 \mathrm{sp}^3$ hybridisation.
2. Tetrahedral geometry is possible by $\mathrm{sp}^3$ hybridisation.
3. Square planar geometry is possible by $\mathrm{dsp}^2$ hybridisation.
Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of $F_7$ and $\mathrm{SF}_6$.
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.
$\mathrm{IF}_7$
It is an $\mathrm{AB}_7$ 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, $\operatorname{IF}_7$ has pentagonal bipyramidal shape.

$\mathrm{SF}_6$ :
It is an $\mathrm{AB}_6$ 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, $\mathrm{SF}_6$ has octahedral shape.

Question 52 .
$\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ both are triatomic molecules but their dipole moment values are different. Why?
Answer:
1. Linear form of carbon dioxide has zero dipole moment. In $\mathrm{CO}_2$ the dipole moment of two polar bonds are equal in magiitude but have opposite direction. Hence, the net dipole moment of the $\mathrm{CO}_2$ molecule is
$
\begin{aligned}
& \mu=\mu_1+\mu_2 \\
& \mu=\mu_1+\left(-\mu_1\right)=0
\end{aligned}
$

$
\mu=\vec{\mu}_1+\left(-\vec{\mu}_1\right)=0
$
In this case
2. But in the case of water, net dipole moment is the vector sum $\mu_1+\mu_2$ as follows:

Dipole moment in water is found to be $1.85 \mathrm{D}$.
3. $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ both are triatomic molecules but their dipole moment values are zero and $1.85 \mathrm{D}$ respectively.
Question 53.
Which one of the following has highest bond order? $\mathrm{N}_2, \mathrm{~N}_2{ }^{+}$or $\mathrm{N}_2^{-}$?
Answer:
$\mathrm{N}_2$ (14 electrons)
Bond order $=3$
$
\text { B.O. }=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{10-4}{2}=3
$
$\mathrm{N}_2^{+}$(13 electrons)
Bond order $=2.5$,
$
\text { B.O. }=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{9-4}{2}=2.5
$
$\mathrm{N}_2^{-}(15$ electrons $)$
$
\text { B.O. }=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{10-5}{2}=2.5
$
So $\mathrm{N}_2$ 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, $\mathrm{Na}^{+}<$ $\mathrm{Mg}^{2+}<\mathrm{Al}^{3+}$, the covalent character also follows the order $-\mathrm{NaCI}<\mathrm{MgCl}_2<\mathrm{AICI}_3$
3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. e.g., $\mathrm{LiCl}$ is more covalent than $\mathrm{NaCI}$.
4. Cation having $n s^2 n p^6 n d^{10}$ configuration shows greater polarising power than the cations with $n s^2 n p^6$ configuration. e.g., $\mathrm{CuCI}$ is more covalent than $\mathrm{NaCl}$.
Question 1.
Draw the lewis structures for
1. Nitrous acid $\left(\mathrm{HNO}_2\right)$
2. Phosphoric acid
3. Sulphur troxide $\left(\mathrm{SO}_3\right)$
Answer:
1. Nitrous acid $\left(\mathrm{HNO}_2\right)$
Lewis dot structure

2. Phosphoric acid 

$\text { 3. Sulphur troxide }\left(\mathrm{SO}_3\right)$

Question 2.
Calculate the formal charge on each atom of carbonyl chloride $\left(\mathrm{COCl}_2\right)$

Answer:
$
=\mathrm{N}_{\mathrm{V}}-\left[\mathrm{N}_l+\frac{\mathrm{N}_b}{2}\right]
$
Formal charge Carbonyl chloride $\mathrm{COCl}_2$

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 $\mathrm{MgO}$ and $\mathrm{CaF}_2$
Magnesium oxide ( $\mathrm{MgO})$ :
Answer:
Electronic configuration of $\mathrm{Mg}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5 3 \mathrm{~s}^2$
Electronic configuration of $O-1 s^2 2 s^2 2 p^6 3 s^6 3 p^4$.
1. Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.
2. By losing two electrons, $\mathrm{Mg}$ acquires the inert gas configuration of Neon and becomes a dipositive cation $\mathrm{Mg}^{2+}$
$\mathrm{Mg} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{e}^{-}$
3. Oxygen accepts the two electrons to become a dinegative oxide anion, $\mathrm{O}^{2-}$ thereby attaining the inert gas configuration of Neon.
$
\mathrm{O}+2 \mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}
$
4. These two ions, $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$ combine to form an ionic crystal in which they are held together by electrostatic attractive forces.
5. During the formation of magnesium oxide crystal $601.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ energy is released. This favours the formation of magnesium oxide $(\mathrm{MgO})$ and its stabilisation.
$\mathrm{CaF}_2$, Calcium fluoride
1. Calcium, $\mathrm{Ca}:[\mathrm{Ar}] 4 \mathrm{~s}^2$, Fluorine $\mathrm{F}:[\mathrm{He}] 2 \mathrm{~s}^2 2 \mathrm{p}^5$
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, $\mathrm{Ca}^{2+}$
4. Two fluorine atoms, each one accepts one electron to become two unincgative fluoride ions ( $\mathrm{F}^{-}$) 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 $1225.91 \mathrm{~kJ} \mathrm{~mol}^{-1}$ of energy is released. This favours the formation of calcium fluoride, $\mathrm{CaF}_2$ and its stabilisation.

Question 4.
Write the resonance structures for
1. Ozone molecule
2. $\mathrm{N}_2 \mathrm{O}$
Answer:
1. Ozone molecule, $\mathrm{O}_3$

$\text { 2. Nitrous oxide, } \mathrm{N}_2 \mathrm{O}$

Question 5.
Of the two molecules $\mathrm{OCS}$ and $\mathrm{CS}_2$ which one has higher dipole moment value? Why?
Answer:
$\mathrm{OCS}$ and $\mathrm{CS}_2$

The dipole moment $\mu_{\mathrm{OCS}}=0.7149 \pm 0.0003$ Debye.
$\mathrm{CS}_2$
$
\mathrm{S}=\mathrm{C}=\mathrm{S}
$
In $\mathrm{CS}_2$, the bond dipoles of $2 \mathrm{C}=\mathrm{S}$ have same values and bond dipoles cancel each other so dipole moment of $\mathrm{CS}_2$ is zero. Among $\mathrm{OCS}$ and $\mathrm{CS}_2$, OCS has a higher dipole moment because in OCS oxygen is more electronegative than sulphur and $\mathrm{C}=\mathrm{S}$ and $\mathrm{C}=\mathrm{O}$ bonds in $\mathrm{OCS}$ molecules do not cancel each other.
On the other hand $\mathrm{CS}_2$, due to linear structure, the bond dipole of two $\mathrm{C}=\mathrm{S}$ bonds cancel each other. On the other hand, $\mathrm{CS}_2$ due to linear structure, the bond dipoles of two $\mathrm{C}=\mathrm{S}$ bonds cancel each other and the recultant dipo'c moment value is zero. So OCS has a higher dipole moment than $\mathrm{CS}_2$.
Question 6.
Arrange the following in the decreasing order of Bond angle
1. $\mathrm{CH}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$
2. $\mathrm{C}_2 \mathrm{H}_2, \mathrm{BF}_3, \mathrm{CCl}_4$
Answer:
1. $\mathrm{CH}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$.
$\mathrm{NH}_3=107^{\circ}$, Water $=104.5^{\circ}, \mathrm{CH}_4=109.5^{\circ}$
Decreasing order of bond angle: $\mathrm{H}_2 \mathrm{O}<\mathrm{NH}_3<\mathrm{CH}_4$
2. $\mathrm{C}_2 \mathrm{H}_2, \mathrm{BF}_3, \mathrm{CCl}_4$ :
$\mathrm{C}_2 \mathrm{H}_2=1800, \mathrm{BF}_3=120^{\circ}, \mathrm{CCl}_4=109.5^{\circ}$
Decreasing order of bond angle: $\mathrm{CCl}_4<\mathrm{BF}_3<\mathrm{C}_2 \mathrm{H}_2$

Question 7.
Bond angle in $\mathrm{PH}_4^{+}$is higher than in $\mathrm{PH}_3$. Why?
Answer:
Phosphorous in both $\mathrm{PH}_3$ and $\mathrm{PH}_4{ }^{+}$is $\mathrm{sp}^3$ hybridised. Due to the absence of lone pair - bond pair repulsion and presence of four identical bond pair - bond pair interactions, $\mathrm{PH}_4{ }^{+}$assumes tetrahedral geometry with a bond angle of $109^{\circ} 28$
But $\mathrm{PH}_3$ has three bond pairs and one lone pair around $\mathrm{P}$. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from $109^{\circ} 28^{\prime}$ to $93.6^{\circ}$. As a result $\mathrm{PH}_3$ is pyramidal.
$\mathrm{PH}_3$ - Pyramidal with bond angle of $93.6^{\circ} . \mathrm{PH}_4^{+}$Tetrahedral with bond angle of $109^{\circ} 28$.
Question 8.
Explain the bond formation in $\mathrm{SF}_4$ and $\mathrm{CCl}_4$ using hybridisation concept.
Answer:
In $\mathrm{SF}_4$, the central atom is $\mathrm{sp}^3 \mathrm{~d}$ hybridised.
1. The molecule $\mathrm{SF}^3$ 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 $3 \mathrm{~d}$ 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 $3 \mathrm{~s}$ orbital three $3 \mathrm{p}$ orbitais and one $3 \mathrm{~d}$ orbital. So the central atom is $\mathrm{sp}^3 \mathrm{~d}$ hybridised.

$\mathrm{CCl}_4$
1. It is not necessary to invoke hybridisation especially in $\mathrm{CCl}_4$. 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 $1 \mathrm{~s}^2 \mathrm{~s}^2 2 \mathrm{p}^2$.
3. $\mathrm{CCl}_4$ 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 $2 \mathrm{~s}$ orbital and three $2 \mathrm{p}$ orbitals of carbon atoms are mixed
to produce four identical orbitals, a process known as $\mathrm{sp}^3$ hybridisation.

Question 9.
The observed bond length of $\mathrm{N}_2^{+}$is larger than $\mathrm{N}_2$ while the bond length in $\mathrm{NO}^{+}$is less than in $\mathrm{NO}$. Why?

Answer:
(a)
(1) By molecular orbital theory, the bond order of both $\mathrm{N}_2^{+}$is 2.5 whereas $\mathrm{N}_2$ is 3 .
2. $\mathrm{N}_2$ has $5 \mathrm{e}^{-}$in the antibonding molecular orbital whereas $\mathrm{N}_2^{+}$has $4 \mathrm{e}^{-}$in the antibonding molecular orbital. So $\mathrm{N}_2{ }^{+}$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 $\mathrm{N}_2$ has more bond length than $\mathrm{N}_2$
Bond order in
$
\begin{aligned}
\mathrm{N}_2 & =\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{10-4}{2}=\frac{6}{2}=3 \\
\mathrm{~N}_2^{+} & =\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{9-4}{2}=\frac{5}{2}=2.5
\end{aligned}
$
Bond order in
So, $\mathrm{N}_2$ is more stable than $\mathrm{N}_2^{+}$. But bond length $\mathrm{N}_2^{+}$is greater than $\mathrm{N}_2$.
(b) $\mathrm{NO}^{+} \& \mathrm{NO}$
Bond order of $\mathrm{NO}=2.5$
Bond order of $\mathrm{NO}^{+}=3$
Due to lesser bond order in NO, the bond length is greater than $\mathrm{NO}^{+} \mathrm{So}, \mathrm{NO}^{+}$bond length is shorter than NO bond length.
Question 10.
Draw the MO diagram for acetylide ion $\mathrm{C}_2{ }^{2-}$ and calculate its bond order.
Answer:
Acetylide ion, $\mathrm{C}_2{ }^{2-}$ in acetylene
$
\begin{array}{cl}
\mathrm{C} \longrightarrow & \mathrm{C}^{2-}+2 e^{-} \\
1 s^2 2 s^2 2 p^4 & 1 s^2 2 s^2 2 p^2
\end{array}
$
Electronic configuration of $\mathrm{C}^2$ ion is:
$\sigma 1 \mathrm{~s}^2 \sigma^* 1 \mathrm{~s}^2 \sigma 1 \mathrm{~s}^2 \pi 2 \mathrm{p}_{\mathrm{x}}^2 \pi 2 \mathrm{p}_{\mathrm{y}}^2$

Bond order
$
=\frac{\mathrm{N}_b-\mathrm{N}_a}{2}=\frac{8-4}{2}=2
$