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

Pentagonal bipyramidal shape.

Out of $$S{O_2},CO_3^{2 - },SO_4^{2 - }$$    and $$C{O_2},C{O_2}$$   have $$sp$$  hybridisation, thus have linear geometry.

$${N_2}\left( {7 + 7 = 14} \right);\sigma 1{s^2},\sigma {1^*}{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$         $$\left\{ {_{{\pi ^*}2p_z^2}^{{\pi ^*}2p_y^2},\,\,\sigma 2p_x^2} \right.$$

$${F_2}\left( {9 + 9 = 18} \right);\sigma 1{s^2},\sigma {1^*}{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$         $$\left\{ {_{\pi 2p_z^2}^{\pi 2p_y^2},\left\{ {_{{\pi ^*}2p_z^2}^{{\pi ^*}2p_y^2}} \right.} \right.$$

$$O_2^ - \left( {8 + 8 + 1 = 17} \right);\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$          $$\left\{ {_{\pi 2p_z^2}^{\pi 2p_y^2},\left\{ {_{{\pi ^*}2p_z^1}^{{\pi ^*}2p_y^2}} \right.} \right.$$

$$O_2^{2 - }\left( {8 + 8 + 2 = 18} \right);\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$          $$\left\{ {_{\pi 2p_z^2}^{\pi 2p_y^2},\left\{ {_{{\pi ^*}2p_z^2}^{{\pi ^*}2p_y^2}} \right.} \right.$$
∴ $$O_2^ - $$  is the only species having unpaired electron.

In $$Br{F_3},$$  both bond pairs as well as lone pairs of electrons are present. Due to the presence of lone pairs of electrons $$(lp)$$  in the valence shell, the bond angle is contracted and the molecule takes the $$T$$ - shape. This is due to greater repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pairs.

$$\eqalign{ & Cl{F_3} \to {\text{Hybridisation}} \cr & = 3 + \frac{1}{2}\left[ {7 - 3} \right] = 5\left( {s{p^3}d} \right) \cr & XeO{F_2} \to {\text{Hybridisation}} \cr & = 3 + \frac{1}{2}\left[ {8 - 4} \right] = 5\left( {s{p^3}d} \right) \cr & XeF_3^ + \to {\text{Hybridisation}} \cr & = 3 + \frac{1}{2}\left[ {8 - 3 - 1} \right] \cr & = 5\left( {s{p^3}d} \right) \cr} $$
All molecules have $$s{p^3}d$$  hybridization and 2 lone pairs. Hence all have identical ( $$T$$ - shape).

TIPS/Formulae :
Hybridisation = $$\frac{1}{2}\left( {V + M - C + A} \right)$$
where; $$V$$ = no. of electron in valence shell of central atom
$$M=$$  no. of monovalent atoms, $$C=$$  charge on cation,
$$A=$$  charge on anion
$$\left( {\text{i}} \right)C{O_2},H = \frac{1}{2}\left( {4 + 0 - 0 + 0} \right) = 2$$
∴ $$sp$$  hybridisation
$$\left( {{\text{ii}}} \right)S{O_2},H = \frac{1}{2}\left( {6 + 0 - 0 + 0} \right) = 3$$
∴ $$s{p^2}$$ hybridisation
$$\left( {{\text{iii}}} \right)N{O_2}$$   has $$V$$ shaped structure.
$$\left( {{\text{iv}}} \right)Cl{O_2}$$  has $$V$$ shaped structure.
∴ $$C{O_2}$$ having $$sp$$  hybridation has linear shape.

$$Cis$$  1, 2 - dichloro benzene will have some

Hydrogen bonding $$ \propto $$ electronegativity $$ \propto $$  $$\frac{1}{{{\text{Size of atom to which }}H{\text{ is covalently bonded }}}}$$
Since, $$F$$  is most electronegative and has smaller size, $$HF$$  shows maximum strength of hydrogen bond.

$$\ddot{\underset{\dot{}}{N}}=\underset{¨}{\ddot{O}}$$
For $$NO,$$  the octet rule is not followed due to the presence of odd electrons on $$N.$$