Question

Which of the following are arranged in an increasing order of their bond strengths?

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

Answer 1

The correct option is B $O_2^{2-}<O_2^{-}<O_2<O_2^{+}$

Bond strength $\propto$ bond order

$O_2\to \sigma 1s^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\sigma 2p_z^2<\pi 2p_x^2=\pi 2p_y^2<{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$

Bond order of $O_2=\frac{10-6}{2}=2$

$O_2^{+}\Rightarrow BO=\frac{10-5}{2}=2.5$

$O_2^{-}\Rightarrow BO=\frac{10-7}{2}=1.5$

$O_2^{2-}\Rightarrow BO=\frac{10-8}{2}=1$

So, the correct order is: $O_2^{2-}<O_2^{-}<O_2<O_2^{+}$