$$\eqalign{ & O_2^ + \left( {15} \right) = KK\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr & \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^1 = \pi 2p_z^0} \right.} \right.\, \cr & {\text{Bond}}\,{\text{order}}\, = \frac{1}{2}\left( {8 - 3} \right) = \frac{5}{2} = 2.5 \cr & {O_2}\left( {16} \right) = KK\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr & \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1} \right.} \right.\, \cr & {\text{Bond}}\,{\text{order}}\, = \frac{1}{2}\left( {8 - 4} \right) = 2 \cr & O_2^ - \left( {17} \right) = KK\sigma 2{s^2},\,{\sigma ^*}2{s^2},\sigma 2p_x^2, \cr & \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^1} \right.} \right. \cr & {\text{Bond}}\,\,{\text{order}}\, = \frac{1}{2}\left( {8 - 5} \right) - 1.5 \cr & O_2^{2 - }\left( {18} \right) = KK\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_x^2, \cr & \left\{ {\pi 2p_y^2 = \pi 2p_z^2,\,\,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^2} \right.} \right. \cr & {\text{Bond}}\,\,{\text{order}} = \frac{1}{2}\left( {8 - 6} \right) = 1 \cr} $$
NOTE : As we know that as the bond order decreases, stability also decreases and hence the bond strength also decreases. Hence the correct order of their increasing bond strength is
$$O_2^{2 - } < \,O_2^ - < {O_2} < O_2^ + $$

All have tetrahedral structure.

Molecular electronic configuration of
$$CO:\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\sigma 2p_x^2} \right.$$
Therefore, bond order $$ = \frac{{{N_b} - {N_a}}}{2} = \frac{{10 - 4}}{2} = 3$$
$$N{O^ + }:\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$       $$\left\{ {\pi 2p_y^2 = \pi p_z^2} \right.$$
Bond order $$ = \frac{{10 - 4}}{2} = 3$$
$$C{N^ - } = \sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$       $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,\sigma 2p_x^2$$
Bond order $$ = \frac{{10 - 4}}{2} = 3$$
$${N_2}:\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$       $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,\sigma 2p_x^2$$
Bond order $$\, = \frac{{10 - 4}}{2} = 3$$
$$N{O^ - }:\,\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2p_x^2,$$       $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2} \right.,$$    $$\left\{ {{\pi ^*}2p_y^1 = {\pi ^*}2p_z^1} \right.$$
Bond order $$\, = \frac{{10 - 6}}{2} = 2$$
∴ $$N{O^ - }$$  has different bond order from that in $$CO.$$

Due to high $$EN$$  of $$F$$ than $$N, H$$  - bonding is stronger in $$HF$$  than $$N{H_3}$$  thus $$b.pt.$$  of $$HF > N{H_3}.$$

$${H_3}N \to B{F_3}$$   where both $$N,B$$  are attaining tetrahedral geomerty.

TIPS/Formulae : Hydrogen bonding is formed in those compounds in which $$F$$ or $$O$$ or $$N$$ atoms are attached to hydrogen atom.
$$\because $$  $$HCl$$  does not have $$F$$ or $$N$$ or $$O$$
$$\therefore $$  It does not form hydrogen bond.

In $$Cl{F_3}$$  all bonds are not equal due to its trigonal-bipyramidal ($$s{p^3}d$$  hybridisation) geometry

$$B{F_3}$$  and $$Al{F_3}$$  show trigonal symmetric structure due to $$s{p^2}$$  hybridisation.

$$N{F_3}$$  shows pyramidal geometry due to $$s{p^3}$$  hybridisation.

$$H$$ -bond is weakest bond because its bond dissociation energy is very low as compared to other given bonds $$\left( {10\,kJ\,mo{l^{ - 1}}} \right).$$

Compounds involved in chelation become non-polar. Consequently such compounds are soluble in nonpolar solvents like ether, benzene etc. and are only sparingly soluble in water whereas meta and para isomers are more soluble in water & less soluble in non-polar solvents.

$$B{F_3}$$
  $$ \Rightarrow 3\sigma $$  - bonds, i.e. $$s{p^2}$$  hybridisation
Planar structure
$$NO_2^ - $$
  $$ \Rightarrow 2\sigma $$  - bonds $$+1$$  lone pair of electrons, i.e. $$s{p^2}$$  hybridisation
$$NH_2^ - $$
  $$ \Rightarrow 2\sigma $$  - bonds $$+2$$  lone pairs, i.e. $$s{p^3}$$  hybridisation

$$ \Rightarrow 2\sigma $$  - bonds $$+2$$  lone pairs, i.e. $$s{p^3}$$  hybridisation, Thus, in $$B{F_3}$$  and $$NO_2^ - ,$$  central atom is $$s{p^2}$$  hybridised, while $$N{H_2},N{H_3}$$   and $${H_2}O$$  are $$s{p^3}$$  hybridised.