$$\eqalign{ & N{O^ - }\left( {16} \right) - B.O. - 2 \cr & {O_2}\left( {16} \right) - B.O. - 2 \cr & N{O^ + }\left( {14} \right) - B.O. - 3 \cr & NO\left( {15} \right) - B.O. - 2.5 \cr} $$
Higher the bond order lower is the bond length. Hence $$N{O^ + }$$  will have smallest bond.

$$O_2^ - \left( {17{e^ - }} \right) - K\,K\,\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_x^2,\left\{ {\pi 2p_y^2 = \pi 2p_z^2,\left\{ {{\pi ^*}2p_y^2 = {\pi ^*}2p_z^1} \right.} \right.$$
Thus, $$\,O_2^ - \,$$  has one unpaired electron; hence it is paramagnetic. Other species have no unpaired electron.
All of them have 14 electrons.

For the formation of an ionic bond, the electronegativities of two atoms should differ by more than 1.9.

According to the postulate of $$VSEPR$$   theory, a lone pair occupies more space than a bonding pair, since it lies closer to the central atom. This means that the repulsion between the different electron pairs follow the order.
$$lp - lp > lp - bp > bp - bp$$

\[As=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},\underbrace{3{{s}^{1}}3p_{x}^{1}\,\,3p_{y}^{1}\,\,3p_{z}^{1}\,\,3{{d}^{1}}}_{s{{p}^{3}}d\,\,\text{hybridisation}}\]
Due to this hybridisation, geometry of the $$As{F_5}$$  molecule is trigonal bipyramidal and the hybrid orbitals used by $$As - {\text{atoms}}$$    are $$s,{p_x},{p_y},{p_z}$$   and $${d_{xy}}.$$

$$M.O.{\text{ configuration of}}\,{O_2}:$$
$$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$      $$\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)$$    $$\left( {{\pi ^ * }2p_x^1 = {\pi ^ * }2p_y^1} \right)$$

The percentage of $$s$$ - character in $$s{p^3},s{p^2}$$  and $$sp$$  is 25%, 33% and 50% respectively. Order of size of orbitals is $$sp < s{p^2} < s{p^3}.$$

For $$NO$$
Total no. of electrons $$= 15$$
$$B.O = 2.5$$
$$Mag.$$  Behaviour = Paramagnetic
For $$N{O^ + }$$
Total no. of electrons $$= 14$$
$$B.O = 3$$
$$Mag.$$  Behaviour = Diamagnetic

The bond length follows the order
$$O_2^ + < {O_2} < O_2^ - < O_2^{2 - }$$
According to this the possible values are
$$1.12\mathop {\text{A}}\limits^{\text{o}} ,1.21\mathop {\text{A}}\limits^{\text{o}} ,1.30\mathop {\text{A}}\limits^{\text{o}} ,1.49\mathop {\text{A}}\limits^{\text{o}} $$

From amongst given species $$P{H_3},$$  $$N{H_3}$$  and $$Sb{H_3}$$  are all $$s{p^3}$$ hybridised. Their central atom has both bond pair as well as lone pair of electrons. The lone pair occupy the fourth orbital. $$CH_3^ + $$  has only three pairs of electrons so it is $$s{p^2}$$ hybridised.