Boron hydrides are electron deficient compound because boron atom has three half-filled orbitals in excited state.

Electronic configuration of O₂

$$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1$$

Thus the incoming electron will enter in $$\pi ^*2p_x$$ to form $$O_2^+$$

O₂^- $$ \to $$ 1

B₂ $$ \to $$ 1

O₂⁺ $$ \to $$ 2.5

NO⁺ $$ \to $$ 3

NO $$ \to $$ 2.5

CO $$ \to $$ 3

N₂ $$ \to $$ 3

O₂ $$ \to $$ 2

Diatomic species	Bond order
NO	2.5
$$O_2^-$$	1.5
$$C_2^{2-}$$	3.0
$$He_2^+$$	0.5

The increasing order: $$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$$