Question

Arrange the following molecules/species in the order of their increasing bond orders.
(I) $O_2$ (II) ${O_2}^{-}$ (III) $O_2^{2-}$ (IV) $O_2$

• A. $III<II<I<IV$
• B. $IV<III<II<I$
• C. $III<II<IV<I$
• D. $II<III<I<IV$

Answer 1

The correct option is A $III<II<I<IV$

For $O_2$ molecular orbital configuration is:

${(\sigma }_{1s}{)}^2<{\left({\sigma }_{1s}^{\ast }\right)}^2<{(\sigma }_{2s}{)}^2<{\left({\sigma }_{2s}^{\ast }\right)}^2<{(\sigma }_{{2p}_z}{)}^2<({\Pi }_{{2p}_x}{)}^2={(\Pi }_{{2p}_y}{)}^2<{\left({\Pi }_{2p_x}^{\ast }\right)}^1={\left({\Pi }_{2p_y}^{\ast }\right)}^1$

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

For $O_2^{-}$ molecular orbital configuration is:

${(\sigma }_{1s}{)}^2<{\left({\sigma }_{1s}^{\ast }\right)}^2<{(\sigma }_{2s}{)}^2<{\left({\sigma }_{2s}^{\ast }\right)}^2<{(\sigma }_{{2p}_z}{)}^2<({\Pi }_{{2p}_x}{)}^2={(\Pi }_{{2p}_y}{)}^2<{\left({\Pi }_{2p_x}^{\ast }\right)}^2={\left({\Pi }_{2p_y}^{\ast }\right)}^1$

Bond order for $O_2:\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-7)=1.5$

For $O_2^{2-}$ molecular orbital configuration is:

${(\sigma }_{1s}{)}^2<{\left({\sigma }_{1s}^{\ast }\right)}^2<{(\sigma }_{2s}{)}^2<{\left({\sigma }_{2s}^{\ast }\right)}^2<{(\sigma }_{{2p}_z}{)}^2<({\Pi }_{{2p}_x}{)}^2={(\Pi }_{{2p}_y}{)}^2<{\left({\Pi }_{2p_x}^{\ast }\right)}^2={\left({\Pi }_{2p_y}^{\ast }\right)}^2$

Bond order for $O_2:\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-8)=1$

For $O_2^{+}$ molecular orbital configuration is:

${(\sigma }_{1s}{)}^2<{\left({\sigma }_{1s}^{\ast }\right)}^2<{(\sigma }_{2s}{)}^2<{\left({\sigma }_{2s}^{\ast }\right)}^2<{(\sigma }_{{2p}_z}{)}^2<({\Pi }_{{2p}_x}{)}^2={(\Pi }_{{2p}_y}{)}^2<{\left({\Pi }_{2p_x}^{\ast }\right)}^1={\left({\Pi }_{2p_y}^{\ast }\right)}^0$

Bond order for $O_2:\frac{1}{2}(N_b-N_a)=\frac{1}{2}(10-7)=2.5$

Order for increasing bond orders is:$O_2^{2-}$ < $O_2^{-}$ < $O_2$ < $O_2^{+}$

So, (A) is the correct option.