Carbonate ion $\left(\mathrm{CO}_3^{2-}\right)=3$ bond pair +1 lone pair $\Rightarrow$ trigonal planar

Due to resonance all C$$-$$O bond length are equal.

All the similar bonds in a molecule do not have the same bond enthalpies. e.g., in $\mathrm{H}_2 \mathrm{O}(\mathrm{H}-\mathrm{O}-\mathrm{H})$ molecule after the breaking of first $\mathrm{O}-\mathrm{H}$ bond, the second $\mathrm{O}-\mathrm{H}$ bond undergoes some change because of changed chemical environment.

Therefore, in polyatomic molecules the term mean or average bond enthalpy is used. It is obtained by dividing total bond dissociation enthalpy by the number of bonds broken.

$$\begin{aligned} \text { e.g., } \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{H}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) ; \\ \Delta_{\mathrm{a}} H_1^{\circ}=502 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{OH}(\mathrm{g}) \longrightarrow \mathrm{H}+\mathrm{O}(\mathrm{g}) ; \\ \Delta_{\mathrm{a}} H_2^{\circ} =427 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\\ \text { Average } \mathrm{O}-\mathrm{H} \text { bond enthalpy }=\frac{502+427}{2}=464.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$

The bod enthalpies of $\mathrm{O}-\mathrm{H}$ bond in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{H}_2 \mathrm{O}$ are different because of the different chemical (electronic) environment around oxygen atom.

A. $\rightarrow$ (3)

B. $\rightarrow$ (1)

C. $\rightarrow(5)$

D. $\rightarrow(4)$

$$\text { A. } \begin{aligned} \mathrm{SF}_4 & =\text { number of } b p(4)+\text { number of } l p(1) \\ & =s p^3 d \text { hybridisation } \end{aligned}$$

B. $\mathrm{IF}_5=$ number of $b p(5)+$ number of $l p(1)$ $=s p^3 d^2$ hybridisation

$$\begin{aligned} \text { C. } \mathrm{NO}_2^{+} & =\text {number of } b p(2)+\text { number of } l p(0) \\ & =\text { sphybridisation }\end{aligned}$$

D. $\mathrm{NH}_4^{+}=$number of $b p(4)+$ number of $l p(0)$ $=s p^3$ hybridisation.

A. $\rightarrow(5)$

B. $\rightarrow$ (1)

C. $\rightarrow(2)$

$\mathrm{D} \rightarrow(3)$

A. $\mathrm{H}_3 \mathrm{O}^{+}=3 b p+1 / p$ pyramidal shape

B. $\mathrm{HC} \equiv \mathrm{CH} \Rightarrow$ linear as sphybridised shape

C. $\mathrm{ClO}_2^{-}=2 b p+2 / p \Rightarrow$ angular shape

D. $\mathrm{NH}_4^{+}=4 b p+0 / p \Rightarrow$ tetrahedral shape

A. $\rightarrow$ (3)

B. $\rightarrow$ (4)

C. $\rightarrow(1)$

D. $\rightarrow(2)$

$$ \begin{gathered} \text { A. } N O(7+8=15)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi \star 2 p_x^1 \\ \text { Bond order }=\frac{1}{2}\left(N_b-N_a\right)=\frac{10-5}{2}=2.5 \end{gathered}$$

B. $\mathrm{CO}(6+8=14)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2$

Bond order $=\frac{10-4}{2}=3$

C. $\mathrm{O}_2^{-}(8+8+1=17)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma * 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi * 2 p_x^2 \approx \pi * 2 p_y^1$

Bond order $=\frac{10-7}{2}=1.5$

D. $\mathrm{O}_2(8+8=16)=\sigma 1 s^2, \sigma^{\star} 1 s^2, \sigma 2 s^2, \sigma^{\star} 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \approx \pi 2 p_y^2, \pi^{\star} 2 p_x^1 \approx \pi^{\star} 2 p_y^1$

Bond order $=\frac{10-6}{2}=2$