Question

The bond order is maximum in:

• A. $O_2$
• B. $O_2^{-}$
• C. $O_2^{+}$
• D. $O_2^{2-}$

Answer 1

The correct option is C $O_2^{+}$

According to MOT, bond order is given by:

B.O.=$\frac{1}{2}(N_B-N_A)$​ where $N_A$ and $N_B$​are the number of electrons in antibonding and bonding molecular orbital respectively,

$O_2\Rightarrow \sigma \left(1s{)}^2{\sigma }^{\ast }\right(1s{)}^2\sigma \left(2s{)}^2{\sigma }^{\ast }\right(2s{)}^2\sigma \left(2p_z{)}^2\pi \right(2p_x{)}^2\pi \left(2p_y{)}^2{\pi }^{\ast }\right(2p_x{)}^1{\pi }^{\ast }(2p_y{)}^1$

Bond order for $O_2$ ​= $\frac{1}{2}(10-6)$​ = 2

Bond order for $O_2^{+}$​ = $\frac{1}{2}(10-5)$ ​= 2.5

Bond order for $O_2^{-}$ ​ = $\frac{1}{2}(10-7)$ ​= 1.5

Bond order for $O_2^{2-}=\frac{1}{2}(10-8)$​ = 1