Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

$$H = \frac{1}{2}\left( {V + M - C + A} \right)$$
where $$H = $$  No. of orbitals involved in hybridisation $$\left( {viz.2\,,3\,,\,4,\,5,\,6} \right)$$    and hence nature of hybridisation $$\left( {viz.\,s{p^2},\,s{p^3},\,s{p^3}d,\,s{p^3}{d^2}} \right)$$     can be ascertained.
$$V = $$  No. of electrons in valence shell of the central atom, $$M = $$  No. of monovalent atoms,$$C = $$  Charge on cation, $$A = $$  Charge on anion,
For $$ClO_2^ - $$  we have, $$H = \frac{1}{2}\left( {7 + 0 - 0 + 1} \right)$$
$$ \Rightarrow H = \frac{1}{2}\left( {7 + 1} \right) = 4\,or\,s{p^3}$$       hybridisation as $$4\,$$ orbitals are involved

$$\eqalign{ & NO_2^ - \to s{p^2} \cr & NO_3^ - \to s{p^2} \cr & NH_2^ + \to s{p^3} \cr & NH_4^ - \to s{p^3} \cr & SC{N^ + } \to sp \cr} $$
$$NO_2^ - $$  and $$NO_3^ - $$  both have the same hybridisation, i.e. $$s{p^2}.$$

Configuration of \[{C_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2 = \pi 2p_y^2\]
Configuration of \[C_2^ - = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2 = \pi 2p_y^2\sigma 2p_z^1\]
$${\text{Bond order}} = \frac{{{\text{No}}{\text{. of bounding }}{e^ - } - {\text{No}}{\text{. of antibonding }}{e^ - }}}{2}$$
\[{C_2}\] has $$s{\text{ - }}p$$  mixing and the HOMO is \[\pi 2{p_x} = \pi 2{p_y},\]   and LUMO is \[\sigma 2{p_z}.\]. So, the extra electron will occupy bounding molecular orbital and this will lead to an increase in bond order.
\[C_2^ - \] has more bond order than \[{C_2}.\]

NOTE : Isoelectronic species have same number of electrons.
Electrons in $$\,CO\, = 6 + 8 = 14$$
Electrons in $$C{N^ - } = 6 + 7 + 1 = 14$$
Electrons in $$O_2^ - = 8 + 8 + 1 = 17$$
Electrons in $$O_2^ + = 8 + 8 - 1 = 15$$
∴ $$CO$$  and $$C{N^ - }$$ are isoelectronic.

TIPS/Formulae:
$$4\sigma \,{\text{bonds}} - s{p^3}\,{\text{hybridisation}}$$
$$2\sigma \,{\text{and}}\,2\pi \,{\text{bonds}} - s{p^2}\,{\text{hybridisation}}\,$$
$$1\sigma \,{\text{and}}\,3\pi \,{\text{bonds}} - sp\,\,\,{\text{hybridisation}}$$
[For hybridization only $$\sigma {\text{ - bonds}}$$   are considered]

$$\left( {\text{a}} \right)3\sigma ,1\pi \,\,\left( {\text{b}} \right)3\sigma ,1\pi \,\,\left( {\text{c}} \right)4\sigma \,\,\left( {\text{d}} \right)3\sigma ,1\pi $$
$$\therefore {\left( {C{H_3}} \right)_3}COH$$    has $$4\sigma $$  bonds and thus it has $$s{p^3}$$  hybridisation.

Due to presence of lone pairs, the repulsion is more which changes the bond angle to $${104.5^ \circ }.$$

$$HF$$  have strongest hydrogen bond because the electronegativity of $$F$$ -atom is high and produce strong electrostatic force of attraction.

$$C{N^ - }$$   and $$CO$$  are isoelectronic because they have equal number of electrons.
In $$C{N^ - }$$   the number of electron $$ = 6 + 7 + 1 = 14$$
In $$CO$$   the number of electrons $$ = 6 + 8 = 14$$