Chemical Bonding & Molecular Structure

$\begin{array}{cccc}\mathrm{NO}_2^{+} & \mathrm{NO}_2 & \mathrm{NO}_3^{-} & \mathrm{NO}_2^{-} \\ s p & s p^2 & s p^2 & s p^2\end{array}$

$\mathrm{NO}_2^{-}<\mathrm{NO}_3^{-}<\mathrm{NO}_2<\mathrm{NO}_2^{+}$bond angle order.

$\therefore $ Order of dipole moment : $\mathrm{BF}_3=\mathrm{NH}_4^{+}<\mathrm{NF}_3<\mathrm{NH}_3$

See-saw → 1

T-shape → 2

Trigonal pyramidal → zero

Square pyramidal → 3

$(P) \rightarrow(1) ;(Q) \rightarrow(2) ;(R) \rightarrow(5) ;(S) \rightarrow(3)$

Option A

A charge–charge interaction falls off as

$U_{qq}\propto\frac1r$

while a charge–dipole interaction goes as

$U_{q\mu}\propto\frac1{r^2}\,. $

So the charge–dipole potential actually decays faster.

⇒ A is false.

Option B

A fixed dipole–dipole energy is

$U_{\mu\mu}\propto\frac1{r^3}\,, $

but once both molecules tumble freely, the thermal (Keesom) average gives

$\langle U\rangle_{\rm Keesom}\propto -\frac{\mu^4}{kT\,(4\pi\varepsilon_0)^2\,r^6}\,. $

That falls off as $1/r^6$, not $1/r^3$.

⇒ B is false.

Option C

The permanent–induced dipole (Debye) interaction is

$U_{\rm Debye}=-\tfrac12\,\alpha\,E^2\propto -\frac{\alpha\,\mu^2}{(4\pi\varepsilon_0)^2\,r^6}\,, $

and when you average over random orientations (assuming the interaction is weak compared with $kT$), no $T$ appears in the result.

⇒ C is true.

Option D

Even totally nonpolar species experience instantaneous‐induced (London) dispersion forces that attract as

$U_{\rm disp}\propto -\frac1{r^6}\,. $

⇒ D is true.

Answer: C and D.

Molecular orbital (MO) energy levels for homonuclear diatomic molecules: For Molecular orbital theory, all the valence atomic orbitals are included. The total number of atomic orbitals must always equal the total number of molecular orbitals. Bonding molecular orbitals are more stable than the atomic orbitals and increase the electron density between the nuclei Antibonding molecular orbitals are. less stable than the atomic orbitals and decrease the electron density between the nuclei. (A) Bond older can be calculated using the formula:
Bond order $=\frac{1}{2}\left(N_b-N_a\right)$
$N_b$ - Number of bonding electrons
$\mathrm{N_a}^{\text { }}$ - Number of antibonding electrons
For $\mathrm{Ne}_2$, total number of electrons $=20 \quad\left\{\begin{array}{l}\text { Number of electrons } \\ \text { for } \mathrm{Ne}=10\end{array}\right.$

The total electrons are filled in the molecular orbitals as,${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * \sigma _{2{p_z^2}}^ *$$ \begin{aligned} N_b=10, & N_a=10 \\ \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-10) \\ & =\frac{1}{2}(0) \\ & =0 \end{aligned} $Bond order of $\mathrm{Ne}_2$ is zero. It is correct. B) $H O M O$ of $F_2$ The number of electrons in $F$ is 9 . So, the total number of electrons in $F_2$ molecule is $18(9+9)$ The 18 electrons are filled in the molecular orbitals as,${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $
Homo is the highest occupied molecular orbital. For $F_2$, the last electron is occupied in the $\pi^*$ orbital. So, the highest occupied molecular orbital of $F_2$ is $\pi$ - type, not $\sigma$-type. It is not correct. c) Bond energy is directly proportional to bond order. The strength of a chemical bond is directly proportional to the amount of energy required to break it. To compare the bond energy of $\mathrm{O}_2^{+}$and $\mathrm{O}_2$, calculate the bond older. The molecule with higher bond order has higher bond energy$\mathrm{O}_2$ Number of elections of oxygen $(0)=8$ number of elections of $\mathrm{O}_2$ molecule $=16$ The 16 elections are filled in the molecular orbitals as: ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * $

$ \begin{aligned} & N_b=10, N_a=6 \end{aligned} $ $ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-6) \\ & =\frac{1}{2} \cdot(4) \\ & =2 \end{aligned} $ $\mathrm{O}_2^{+}$ Total number of elections for $0_2$ molecule $=16$ For $\mathrm{O}_2^{+}$, one electron is removed from $\mathrm{O}_2$. So, the total number of electrons $=16-1=15$ The 15 electrons are filled in the molecular orbitals as, ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * $$ \begin{aligned} N_b=10, & N_a=5 \\ \text { Bond order } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(10-5) \\ & =\frac{1}{2}(5) \\ & =2.5 \end{aligned} $Bond order of $\mathrm{O}_2^{+}$is higher than the bond order of $\mathrm{O}_2$. Bond order and bond energy are directly proportional so, the higher the bond oder, the higher the bond energy. Therefore, $\mathrm{O}_2{ }^{+}$molecule which has higher bond order and also has higher bond energy.So, bond energy of $\mathrm{O}_2^{+}$is higher than the bond energy of $\mathrm{O}_2$.$ \text { It is not correct: } $ D) Bond length of $Li_2$ and $B_2$ Bond length and bond older are inversely proportional to each other so; higher the bond order, shorter the bond length. To get the bond length of $L_2$ and $B_2$, first calculate the bond order of $Li_2$ and $B_2$. $\mathrm{Li}_2$ Number of electrons of $\mathrm{Li}=3$ : total number of electrons in $\mathrm{Li}_2$ molecule $=6$ The total electrons are filled in the molecular orbitals as Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$$ N_b=4, N_a=2 $$ \begin{aligned} \text { Bond loeder } & =\frac{1}{2}\left(N_b-N_a\right) \\ & =\frac{1}{2}(4-2) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $ $B_2$ Number of electrons of Balon $(B)=5$ Total number of electrons of $B_2$ molecule $=10$ The total electrons are filled in the molecular orbitals as ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$ N_b=6 \quad N_a=4 $$ \begin{aligned} \text { Bond older } & =\frac{1}{2}\left(N_b+N_a\right) \\ & =\frac{1}{2}(6-4) \\ & =\frac{1}{2}(2) \\ & =1 \end{aligned} $Bond order is equal for $Li_{2}$ and $B_2$ so, atomic size is consider Comparing the size of lithium and boron, lithium is a larger atom than boron. Bond length generally increases with the size of atom so, bond length of $\mathrm{Li}_2$ is larger than the bond length of $\mathrm{B_2}$ Size of $Li>B$ So, Bond length of $Li_2>B_2$ This statement is correct: Answer: The incorrect statements are (B) and (C). (B) and (C).

$\mathrm{XeF}_2 \Rightarrow 2$ sigma bonds and 3 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation , geometry will be trigonal bipyramidal.

$\mathrm{P}-5$

$\mathrm{XeF}_4 \Rightarrow 4$ sigma bonds and 2 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=6, \mathrm{sp}^3 \mathrm{~d}^2$ hybridisation , geometry will be octahedral.

Q-3

$\mathrm{XeO}_3 \Rightarrow 3$ sigma bonds and 1 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=4, \mathrm{sp}^3$ hybridisation , geometry will be tetrahedral.

R-2

$\mathrm{XeO}_3 \mathrm{F}_2 \Rightarrow 5$ sigma bonds and 0 lone pairs on $\mathrm{Xe}$, number of hybrid orbitals $=5, \mathrm{sp}^3 \mathrm{~d}$ hybridisation , geometry will be trigonal bipyramidal.

S-4

The octet rule states that atoms tend to combine in such a way that they each have eight electrons in their valence shell, giving them the same electronic configuration as a noble gas. Let's examine each option to determine which molecules follow the octet rule.

Option A: $\mathrm{CO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{NO}$ and $\mathrm{HCl}$

1. $\mathrm{CO}_2$: Carbon dioxide follows the octet rule. Carbon forms double bonds with two oxygen atoms, allowing it to have 8 electrons in its valence shell.

2. $\mathrm{C}_2 \mathrm{H}_4$: Ethylene follows the octet rule. Each carbon forms a double bond with the other carbon and single bonds with two hydrogen atoms, resulting in 8 electrons in the valence shell of the carbons.

3. $\mathrm{NO}$: Nitric oxide does not follow the octet rule. It has an odd number of electrons, making it a free radical.

4. $\mathrm{HCl}$: Hydrogen chloride follows the octet rule. Chlorine has 8 electrons in its valence shell when bonded with hydrogen.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{C}_2 \mathrm{H}_4$, and $\mathrm{HCl}$ follow the octet rule here.

Option B: $\mathrm{NO}_2, \mathrm{O}_3, \mathrm{HCl}$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{NO}_2$: Nitrogen dioxide does not follow the octet rule. It has an odd number of electrons.

2. $\mathrm{O}_3$: Ozone follows the octet rule. It has a resonance structure that allows each oxygen atom to have 8 electrons in its valence shell.

3. $\mathrm{HCl}$: As mentioned, hydrogen chloride follows the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule. All atoms achieve stable configurations through bonding.

Thus, three molecules $\mathrm{O}_3$, $\mathrm{HCl}$, and $\mathrm{H}_2 \mathrm{SO}_4$ follow the octet rule in this option.

Option C: $\mathrm{BCl}_3, \mathrm{NO}, \mathrm{NO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$

1. $\mathrm{BCl}_3$: Boron trichloride does not follow the octet rule. Boron has only 6 electrons in its valence shell.

2. $\mathrm{NO}$: As mentioned, nitric oxide does not follow the octet rule.

3. $\mathrm{NO}_2$: As mentioned, nitrogen dioxide does not follow the octet rule.

4. $\mathrm{H}_2 \mathrm{SO}_4$: Sulfuric acid follows the octet rule.

Thus, only one molecule, $\mathrm{H}_2 \mathrm{SO}_4$, follows the octet rule in this option.

Option D: $\mathrm{CO}_2, \mathrm{BCl}_3, \mathrm{O}_3$ and $\mathrm{C}_2 \mathrm{H}_4$

1. $\mathrm{CO}_2$: As mentioned, carbon dioxide follows the octet rule.

2. $\mathrm{BCl}_3$: As mentioned, boron trichloride does not follow the octet rule.

3. $\mathrm{O}_3$: As mentioned, ozone follows the octet rule.

4. $\mathrm{C}_2 \mathrm{H}_4$: As mentioned, ethylene follows the octet rule.

Thus, three molecules $\mathrm{CO}_2$, $\mathrm{O}_3$, and $\mathrm{C}_2 \mathrm{H}_4$ follow the octet rule in this option.

Conclusion: The correct options are Option A and Option D, both having at least three molecules that follow the octet rule.

(1) Upto 14 electrons, molecular orbital configuration is ->


N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(2) After 14 electrons to 20 electrons molecular orbital configuration is ->



Here N_a = 10

and N_b = 10

Now from question, in F atom 9 electrons present, so in F₂, 9 $ \times $ 2 = 18 electrons present.

Molecular orbital configuration of F₂(18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

The molecules which gives permanent dipole moment are polar in nature.





Thus, options (a, c) are correct.

Option (A) : According to molecular orbital theory, the arrangement of the electrons in the molecular orbitals is as follows :

For $C_2^{2 - }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

There is no unpaired electron and thus it is diamagnetic.

Option (B) : For $O_2^{2 + }$, the total number of electrons is 14. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^0}^ * \,\, = \pi _{2p_y^0}^ * $

Thus the bond order = (10 $-$ 4)/2 = 3;

Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

So O$_2$ has 2 unpaired electrons. for O₂, bond order = 2.

As bond order is inversely proportional to bond length, thus the bond length of $O_2^{2 + }$ is less than the bond length of O₂.

Option (C) : For $N_2^ + $, the total number of electrons is 13. The molecular orbital configuration is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Bond order of $N_2^ + $ = (9 $-$ 4)/2 = 2.5

   $N_2^{ - }$ has 15 electrons.

Moleculer orbital configuration of $N_2^{ - }$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^0}^ *$

$\therefore\,\,\,\,$N_a = 5

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right] = 2.5$

Thus, the bond orders are the same.

Option (D) : As some energy is released during the formation of $He_2^ + $ from two isolated He atoms, thus it has lesser energy as compared to two isolated He atoms.

Structure of each of the compound is represented as :

(A) BrF₅



Structure : Square pyramidal

(i) One lone pair

(ii) Four bond pair

(B) ClF₃



Structure : T-Shaped

(i) Two lone pair

(ii) Three bond pair

(C) XeF₄



Structure : Square planar

(i) Two lone pair

(ii) Four bond pair

(D) SF₄



Structure : See-saw

(i) One lone pair

(ii) Four bond pair

Write the molecular orbital electronic configuration keeping in mind that there is no $2 s-2 p$ mixing, then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic.

Note: Since there is no mixing of $2 s$ and $2 p$ orbitals of each of the atom in a diatomic molecule, the energy of bonding molecular orbital $2 p z$, i.e., $2 p$ is less than that of bonding molecular orbital of $2 p x$ and $2 p y$, i.e., $\left(\pi_{2 p x}^*\right.$ and $\left.\pi_{2 p y}^*\right)$.

$ \therefore $ The order of molecular orbital energy levels, i.e., bonding and anti-bonding energy levels is as follows:

$ \begin{aligned} \left(\sigma_{1 s}\right)\left(\sigma_{1 s}^*\right)\left(\sigma_{2 s}\right)\left(\sigma_{2 s}^*\right)\left(\sigma_{2 p z}\right)\left(\pi_{2 p x}\right. & \left.\equiv \pi_{2 p y}\right) \left(\pi_{2 p x}^*\right. \left.\equiv \pi_{2 p y}^*\right)\left(\sigma_{2 p z}^*\right) \end{aligned} $

(A) Total number (no.) of electrons in $\mathrm{Be}_2=4+4=$ 8. The electronic configuration of $\mathrm{Be}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2 $

All the electrons in $\mathrm{Be}_2$ are paired up, $\mathrm{Be}_2$ is diamagnetic.

(B) Total number (no.) of electrons in $\mathrm{B}_2=5+5=$ 10. The electronic configuration of $B_2$ is:

$\left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2$

All the electrons in $B_2$ are paired up, $B_2$ is diamagnetic.

(C) Total no. of electrons in $\mathrm{C}_2=6+6=12$ The electronic configuration of $\mathrm{C}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 1} \equiv \pi_{2 p y}^{* 1}\right) $

There are two unpaired electrons in doubly degenerate.

$\pi_{2 p x}$ and $\pi_{2 p y} \mathrm{C}_2$ is paramagnetic.

(D) Total no. of electrons in $\mathrm{N}_2=7+7=14$

The electronic configuration of $\mathrm{N}_2$ is:

$ \left(\sigma_{1 s}\right)^2\left(\sigma_{1 s}^*\right)^2\left(\sigma_{2 s}\right)^2\left(\sigma_{2 s}^*\right)^2\left(\sigma_{2 p z}\right)^2\left(\pi_{2 p x}^{* 2} \equiv \pi_{2 p y}^{* 2}\right) $

All the electrons in $\mathrm{N}_2$ are paired up; $\mathrm{N}_2$ is diamagnetic.

(a) Ice floats in water due to the low density of ice as compare to water which is due to open cage like structure (formed by intermolecular H-bonding).

(b) Basic strength of RNH₂ > R₃N. It is also explained by hydrogen bonding.


(c)
(d) Dimerisation of acetic acid in benzene is due to intermolecular hydrogen bonding.

The electronic configuration of Xe is 5s² 5p⁶; and that of Xe in excited state is 5s² 5p⁵ 5d¹. The hybridisation is sp³d and geometry is see-saw. The actual shape is trigonal bipyramidal but due to the preaence of lone pair it gets distorted to see-saw structure.

$\mathrm{OSF}_2$ has $s p^3$ hybridisation. According to VSEPR theory, we can conclude that the shape of $\mathrm{OSF}_2$ is a trigonal pyramid.

Sulphur has a non-bonding pair of electrons and three sulphur-oxygen ( $\mathrm{S}-\mathrm{O})$ bond pairs. $\mathrm{SO}_2$ is $s p^2$ hybridised with three bond pairs (two sigma and one pi) and one lone pair. It is replace trigonal by bent.

$\mathrm{BrF}_3$ is $s p^3 d$ hybridized with three bond pair and two lone pairs. It is T shaped.

$\mathrm{SiO}_3^{2-}$ is $s p^2$ hybridised with four bond pairs (three sigma and one pi) and no lone pair. It has trigonal planar shape.

Molecular orbital configuration of B₂ (10 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here in B₂, 2 unpaired electrons present.

Since the Hund's rule is violated, two electrons are placed in $\pi 2 p_x$ molecular orbital, so

$ \mathrm{B}_2(10): {\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} $

Thus, bond order $=\frac{6-4}{2}=1$.

As there are no unpaired electrons, the nature is diamagnetic.

(A) ${B_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^1 = \pi 2p_y^1$

It is paramagnetic due to two unpaired electrons. The bond order is

${{{N_b} - {N_a}} \over 2} = {{6 - 4} \over 2} = 1$

The gain of electron increases bond order, so reduction is possible.

(B) ${N_2} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\pi 2p_x^2 = \pi 2p_y^2\sigma 2p_z^2$

There are no unpaired electrons. Bond order is

${{{N_b} - {N_a}} \over 2} = {{10 - 4} \over 2} = {6 \over 2} = 3$

Mixing of 2s and 2p orbitals is possible because of similar energies.

(C) $O_2^ - \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^2\,\pi * 2p_y^1$

The molecule is paramagnetic due to presence of unpaired electron. Bond order is less than 2.

${{{N_b} - {N_a}} \over 2} = {{10 - 7} \over 2} = {3 \over 2}$

Loss of electron increases the bond order so oxidation is possible.

(D) $O_2^{} \to \sigma 1{s^2}\,\sigma * 1{s^2}\,\sigma 2{s^2}\,\sigma * 2{s^2}\sigma 2p_z^2\pi 2p_x^2 = \pi 2p_y^2\,\pi * 2p_x^1\,\pi * 2p_y^1$

The molecule is paramagnetic due to presence of unpaired electrons. Bond order is

${{{N_b} - {N_a}} \over 2} = {{10 - 6} \over 2} = {4 \over 2} = 2$

Loss of electron causes increase in bond order, so it undergoes oxidation.

Species having the same number of electrons will have the same bond order.

Number of electrons in CO are 14 (8 from oxygen and 6 from carbon).

NO$^-$ has 16 electrons (7 from nitrogen, 8 from oxygen, add 1 for negative charge).

NO$^+$ has 14 electrons (7 from nitrogen, 8 from oxygen, subtract 1 for positive charge).

CN$^-$ has 14 electrons (7 from nitrogen, 6 from carbon, add 1 for negative charge).

N$_2$ has 14 electrons (7 from each nitrogen).

$\begin{array}{ll}\text { (a) } \mathrm{Na}_2 \mathrm{O}_2 & \rightarrow \mathrm{Na}^{+} \quad 8 \mathrm{e}^{-} \text {all shells are filled. } \\ \mathrm{O}_2^{2-} & \left.\rightarrow \quad \begin{array}{lll}1 s^2 & 2 s^2 & 2 p^5 \\ 1 s^2 & 2 s^2 & 2 p^5\end{array}\right] 10 e^{-}\end{array}$

(c) $\mathrm{N}_2 \mathrm{O}$

$ : \mathrm{N}=\mathrm{N} \rightarrow \ddot{\mathrm{O}}: $

all electrons are paired.

(d) $\mathrm{KO}_2 \rightarrow \mathrm{~K}^{+} \mathrm{O}_2^{-}$

$ \left.\mathrm{O}_2^{-}=\begin{array}{ccc} 1 s^2 & 2 s^2 & 2 p^4 \\ 1 s^2 & 2 s^2 & 2 p^5 \end{array}\right] 9 e^{-} $

Statement 1: Boron always forms covalent bonds.

→ Boron has a small size and high ionization energy. It does not easily lose electrons to form $ B^{3+} $ ions, since removing three electrons requires a lot of energy.

→ Therefore, boron compounds (like BF₃, BCl₃, BH₃, etc.) are covalent in nature.

✅ Statement 1 is True.

Statement 2: The small size of $ B^{3+} $ favors formation of covalent bond.

→ According to Fajan’s rules, small cations with high charge density have strong polarizing power, which distorts the electron cloud of the anion, leading to covalency.

→ $ B^{3+} $ is extremely small and highly charged, therefore its compounds are predominantly covalent.

✅ Statement 2 is True.

Furthermore, Statement 2 gives the correct reason why boron compounds are covalent — due to the small size and high charge of $ B^{3+} $.

✅ Correct Option: A

Answer: Option A — Both statements are true, and Statement 2 is the correct explanation for Statement 1.

To determine which species has the maximum number of lone pairs of electrons on the central atom, we'll examine the electron configuration of each molecule's central atom:

Option A: $[ClO_3]^-$

The central atom is chlorine (Cl).

Chlorine has 7 valence electrons. In the $[ClO_3]^-$ ion, chlorine forms three bonds with oxygen atoms and there is an additional negative charge (one extra electron). Thus, chlorine has:

$7 + 1 - 3 = 5$ valence electrons left, which form 2.5 lone pairs. However, due to the requirement to minimize formal charges, typically chlorine in $[ClO_3]^-$ is involved in resonance structures reducing the average lone pairs per resonance structure to 1.

So, $[ClO_3]^-$ has 1 lone pair on the central atom.

Option B: $XeF_4$

The central atom is xenon (Xe).

Xenon has 8 valence electrons. In $XeF_4$, xenon forms four bonds with four fluorine atoms. Thus, xenon has:

$8 - 4 = 4$ valence electrons left, which form 2 lone pairs.

So, $XeF_4$ has 2 lone pairs on the central atom.

Option C: $SF_4$

The central atom is sulfur (S).

Sulfur has 6 valence electrons. In $SF_4$, sulfur forms four bonds with four fluorine atoms. Thus, sulfur has:

$6 - 4 = 2$ valence electrons left, which form 1 lone pair.

So, $SF_4$ has 1 lone pair on the central atom.

Option D: $[I_3]^-$

The central atom is iodine (I).

Iodine has 7 valence electrons. In the $[I_3]^-$ ion, iodines form a linear structure where the central iodine forms two bonds with two other iodine atoms and there is an additional negative charge (one extra electron). Thus, the central iodine has:

$7 + 1 - 2 = 6$ valence electrons left, which form 3 lone pairs.

So, $[I_3]^-$ has 3 lone pairs on the central atom.

Hence, the species with the maximum number of lone pairs of electrons on the central atom is:

Option D: $[I_3]^-$

To determine which molecular species have unpaired electrons, let's examine the electron configurations of each species using molecular orbital (MO) theory. The presence of unpaired electrons can be deduced from the electron filling in the molecular orbitals.

Option A: N₂

For nitrogen diatomic (N₂), each nitrogen atom has 7 electrons, so the total number of electrons is 14. According to the molecular orbital theory, the electrons fill the molecular orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$.

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2$

So, N₂ has no unpaired electrons.

Option B: F₂

For fluorine diatomic (F₂), each fluorine atom has 9 electrons, resulting in a total of 18 electrons. They fill the orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals also results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, F₂ has no unpaired electrons.

Option C: $O_2^-$

The dioxygen anion ($O_2^-$) has one extra electron compared to neutral O₂, which normally has 16 electrons. Hence, $O_2^-$ has 17 electrons. These electrons fill the orbitals in the following order: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals, the last electron will be placed in one of the degenerate $\pi_{2p_x}^{*}$ or $\pi_{2p_y}^{*}$ orbitals, resulting in one unpaired electron:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$

So, $O_2^-$ has one unpaired electron.

Option D: $O_2^{2-}$

The peroxide ion ($O_2^{2-}$) has two extra electrons compared to neutral O₂, giving it a total of 18 electrons. These electrons fill the orbitals as follows: $\sigma_{1s}$, $\sigma_{1s}^*$, $\sigma_{2s}$, $\sigma_{2s}^*$, $\pi_{2p_x}$ = $\pi_{2p_y}$, $\sigma_{2p_z}$, $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$, $\sigma_{2p_z}^*$.

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, $O_2^{2-}$ has no unpaired electrons.

Based on the molecular orbital theory, the molecular species that has unpaired electron(s) is Option C: $O_2^-$.

To answer the question about the common features among the species CN^-, CO, and NO⁺, let's first understand the electronic configurations and bond orders of each species.

1. Electronic Configurations:

- CN^-: The cyanide ion (CN^-) has an electronic configuration similar to N₂ because CN has 10 electrons (from C and N); adding one more electron for the negative charge brings it to 11 electrons.

- CO: Carbon monoxide (CO) has a total of 10 valence electrons (4 from Carbon and 6 from Oxygen).

- NO⁺: The nitrosonium ion (NO⁺) has a total of 10 electrons as well, considering the positive charge removes one electron from the original 11 valence electrons of NO.

Thus, all three species CN^-, CO, and NO⁺ are isoelectronic, meaning they all have the same number of electrons (10 valence electrons).

2. Bond Order: The bond order is the number of chemical bonds between a pair of atoms. It is calculated as half the difference between the number of bonding electrons and the number of antibonding electrons. All three species have a bond order of 3, as can be seen from their molecular orbital diagrams.

3. Ligand Field Strength:

CO and CN^- are generally considered strong field ligands due to their ability to partake in back-donation, where electrons from the metal center can be donated back into the empty π* orbitals of the ligand. NO⁺ is also a reasonably strong field ligand due to its ability to accept electrons.

Conclusion:

Based on the above analysis, the correct option is:

Option A: bond order three and isoelectronic

This option accurately describes the common features among CN^-, CO, and NO⁺.

To determine the order of increasing C-O bond lengths for CO, $CO_3^{2-}$ and CO₂, it's essential to understand the types of bonding in these molecules and ions.

1. Carbon Monoxide (CO):

Carbon monoxide has a triple bond between carbon and oxygen. Triple bonds are stronger and shorter compared to single and double bonds. Thus, CO has a very short bond length.

2. Carbon Dioxide (CO₂):

In CO₂, carbon is double-bonded to each oxygen atom. Double bonds are longer than triple bonds but shorter than single bonds. Therefore, the C-O bond length in CO₂ is longer than in CO but shorter than in a molecule with single bonds.

3. Carbonate Ion ($CO_3^{2-}$):

In the carbonate ion, the bond order for each C-O bond is 1.33. This is because the carbonate ion has a resonance structure where one carbon-oxygen bond is a double bond and the others are single bonds. Due to the resonance, the effective bond order between carbon and oxygen is between a single and a double bond, making its bond length longer than that in CO₂ and shorter than a pure single bond.

Therefore, the increasing order of C-O bond length is:

CO < CO₂ < $CO_3^{2-}$

The correct option is:

Option D

CO < CO₂ < $CO_3^{2-}$

$\mathrm{H}_2 \mathrm{~S}$ has $s p^3$ hybridised sulphur, therefore, angular in shape with non-zero dipole moment.

(Non-linear, polar molecule)

The correct answer is Option C: If assertion is correct but reason is incorrect.

Explanation:

• Assertion: LiCl (Lithium Chloride) is indeed predominantly a covalent compound. This is due to the high polarizing power of the small lithium ion (Li⁺), which distorts the electron cloud of the larger chloride ion (Cl⁻), resulting in a significant degree of electron sharing and hence covalency.


• Reason: The reason provided is incorrect. The electronegativity difference between lithium (Li) and chlorine (Cl) is substantial. It is this difference in electronegativity that creates the polarity in the Li-Cl bond, making it an ionic bond with some covalent character.

Key Point: The high polarizing power of the Li⁺ ion, due to its small size and high charge density, is the main reason for the covalent character of LiCl, not the small electronegativity difference.

The critical temperature of a substance is the temperature above which it cannot be liquefied, regardless of the pressure applied. It depends on the intermolecular forces within the substance. To understand why the critical temperature of water (H₂O) is higher than that of oxygen (O₂), we need to examine the characteristics and interactions of these molecules.

Let’s break down the options:

Option A: Fewer electrons than O₂

This option is incorrect. The number of electrons in a molecule does not directly determine the critical temperature. While electron count can affect molecular interactions to an extent, it is not the primary reason for the difference in critical temperatures between H₂O and O₂.

Option B: Two covalent bonds

Although H₂O does have two covalent bonds, this feature alone does not explain its higher critical temperature compared to O₂. The nature of the bonds plays a more crucial role, particularly the type of intermolecular forces these bonds enable.

Option C: V-shape

H₂O indeed has a V-shaped (or bent) molecular geometry, while O₂ is linear. This V-shape contributes to some properties of water, like its polarity, but it is not the most direct reason for the higher critical temperature.

Option D: Dipole moment

This is the correct answer. H₂O has a significant dipole moment due to its molecular geometry and the difference in electronegativity between hydrogen and oxygen atoms. This results in strong hydrogen bonding between water molecules. Hydrogen bonds are a type of dipole-dipole interaction, which are much stronger than the van der Waals forces (or London dispersion forces) experienced by O₂ molecules. These strong intermolecular hydrogen bonds in water require more energy (higher temperature) to break, resulting in a higher critical temperature.

Therefore, the primary reason the critical temperature of water is higher than that of oxygen is due to the strong dipole moment leading to significant hydrogen bonding in water. The correct answer is:

Option D: dipole moment

To determine which of the given compounds has sp² hybridization, we need to understand the structure of each molecule and the hybridization of the central atom in each case.

Option A: CO₂

In carbon dioxide (CO₂), the carbon atom is the central atom. Carbon forms two double bonds with two oxygen atoms.

In this configuration, carbon has two regions of electron density (each double bond counts as one region of electron density). According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, this would result in a linear structure with a bond angle of 180 degrees. The hybridization of the carbon atom, in this case, is sp, because it needs two hybrid orbitals to form the two double bonds.

So, CO₂ does not have sp² hybridization.

Option B: SO₂

In sulfur dioxide (SO₂), the sulfur atom is the central atom. Sulfur forms two double bonds with two oxygen atoms.

However, sulfur also has a lone pair of electrons. This results in three regions of electron density around the sulfur atom (two double bonds and one lone pair). According to the VSEPR theory, this configuration leads to a bent molecular shape.

With three regions of electron density, the hybridization of sulfur is sp². Therefore, SO₂ has sp² hybridization.

Option C: N₂O

In nitrous oxide (N₂O), one nitrogen atom is central. The molecule can be arranged as N=N=O, where the central nitrogen forms a double bond with each of the terminal nitrogen and oxygen atoms.

This gives the central nitrogen two regions of electron density, resulting in linear geometry. The hybridization of the central nitrogen atom here is sp, not sp².

So, N₂O does not have sp² hybridization.

Option D: CO

In carbon monoxide (CO), the carbon atom forms a triple bond with the oxygen atom. This involves one sigma bond and two pi bonds, with one lone pair on each atom.

This gives each atom two regions of electron density, resulting in linear geometry. The hybridization of carbon in CO is also sp.

So, CO does not have sp² hybridization.

Based on the analysis of each option, the correct answer is:

Option B: SO₂