Draw the resonating structure of
(a) ozone molecule
(b) nitrate ion
(a) The resonating structure of ozone molecule may be written as
(b) The resonating structure of nitrate ion $\left(\mathrm{NO}_3^{-}\right)$ is
Predict the shapes of the following molecules on the basis of hybridisation.
$$\mathrm{BCl}_3, \mathrm{CH}_4, \mathrm{CO}_2, \mathrm{NH}_3$$
In $\mathrm{BCl}_3$, the geometry is trigonal planar is due to $s p^2$ hybridisation.
The shape of $\mathrm{CH}_4$ is tetrahedral due to $s p^3$ hybridisation.
CO$$_2$$ show linear shape because of sp hybridisation.
The geometry of $\mathrm{NH}_3$ is pyramidal shape and has $s p^3$ hybridisation.
All the $\mathrm{C}-0$ bonds in carbonate ion $\left(\mathrm{CO}_3^{2-}\right)$ are equal in length. Explain.
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.
What is meant by the term average bond enthalpy? Why is there difference in bond enthalpy of 0-H bond in ethanol $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)$ and water?
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.
Match the species in Column I with the type of hybrid orbitals in Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | SF$$_4$$ | 1. | sp$$^3$$d$$^2$$ |
| B. | IF$$_5$$ | 2. | d$$^2$$sp$$^3$$ |
| C. | NO$$_2^+$$ | 3. | sp$$^3d$$ |
| D. | NH$$_4^+$$ | 4. | sp$$^3$$ |
| 5. | sp |
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.