Cation size is decreasing in the order : $$N{a^ + } > M{g^{2 + }} > A{l^{3 + }}$$
$$A{l^{3 + }}$$  has maximum polarisation effect and $$N{a^ + }$$  has minimum polarisation effect.
The covalent nature is in the order : $$AlC{l_3} > MgC{l_2} > NaCl$$

$$s{p^3}{d^2}$$  hybridisation
Number of $${90^ \circ }$$ angle between bonds $$ = 12$$

Key Idea For $$s{p^2}$$  hybridisation, there must be $$3\sigma - {\text{bonds}}\,\,{\text{or}}\,2\sigma - {\text{bonds}}\,$$     along with a lone pair of electrons.
$$\left( {\text{i}} \right)NO_2^ - \Rightarrow 2\sigma + 1\,lp = 3,$$      $${\text{i}}{\text{.e}}{\text{.}}\,s{p^2}\,\,{\text{hybridisation}}$$
$$\left( {{\text{ii}}} \right)N{H_3} \Rightarrow 3\sigma + 1\,lp = 4,$$      $${\text{i}}{\text{.e}}{\text{.}}\,s{p^3}\,\,{\text{hybridisation}}$$
$$\left( {{\text{iii}}} \right)B{F_3} \Rightarrow 3\sigma + 0\,lp = 3,$$      $${\text{i}}{\text{.e}}{\text{.}}\,s{p^2}\,\,{\text{hybridisation}}$$
$$\left( {{\text{iv}}} \right)NH_2^ - \Rightarrow 2\sigma + 2\,lp = 4,$$      $${\text{i}}{\text{.e}}{\text{.}}\,s{p^3}\,\,{\text{hybridisation}}$$
$$\left( {\text{v}} \right){H_2}O \Rightarrow 2\sigma + 2\,lp = 4,$$      $${\text{i}}{\text{.e}}{\text{.}}\,s{p^3}\,\,{\text{hybridisation}}$$
Thus, among the given pairs, only $$B{F_3}$$  and $$NO_2^ - $$  have $$s{p^2}$$  hybridisation.

Formation of $$B{F_3}$$
Ground state of $$B:$$ 
Excited state of $$B:$$ 

No explanation is given for this question. Let's discuss the answer together.

$$HF$$  molecules have linear polymeric structure due to hydrogen bonding.
$$H - F{\text{ - - - }}H - F\mathop {{\text{ - - - }}}\limits_{\mathop \uparrow \limits_{H - bond} } H - F{\text{ - - - }}H - F$$

Bond order of $$O_2^ + = \frac{{10 - 5}}{2} = 2.5$$
Bond order of $$O_2^ - = \frac{{10 - 7}}{2} = 1.5$$
Bond order of $$O_2^{2 - } = \frac{{10 - 8}}{2} = 1$$
Bond order of $${O_2} = \frac{{10 - 6}}{2} = 2$$
$$\because $$ Maximum bond order = minimum bond length.
∴ Bond length is minimum for $$O_2^ + $$

$$XeO{F_2};\,s{p^3}d \Rightarrow T{\text{ - shaped}}$$

Number of electrons in each species are
$$C{N^ - } = 6 + 7 + 1 = 14,\,CO = 6 + 8 = 14\,$$        $$N{O^ + } = 7 + 8 - 1 = 14$$
Each of the species has $$14$$  electrons which are distributed in $$\,MOs$$  as below
$$\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},$$     $$\left\{ {\pi 2p_y^2 = \pi 2p_z^2,{o^ - }2p_x^2} \right.$$
Bond order $$\, = \frac{{10 - 4}}{2} = 3$$

$$\eqalign{ & {\text{Hybridisation :}} \cr & 1.\,\,\,\,:\ddot S{F_2} \Rightarrow \frac{1}{2}\left( {6 + 2} \right) = 4 = s{p^3} \cr & 2.\,\,\,:S{F_4} \Rightarrow \frac{1}{2}\left( {6 + 4} \right) = 5 = s{p^3}d \cr & 3.\,\,\,S{F_6} \Rightarrow \frac{1}{2}\left( {6 + 6} \right) = 6 = s{p^3}{d^2} \cr} $$