Question

Which of the following has fractional bond order?

• A. $O_2^{2+}$
• B. $O_2^{2-}$
• C. $F_2$
• D. $H_2^{-}$

Answer 1

The correct option is D $H_2^{-}$

$O_2^{2+}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2(\pi 2p_x^2=\pi 2p_y^2)$
(16-2=14)
Bond order $=\frac{10-4}{2}=\frac{6}{2}=3$

The bond order of $O_2^{2-}$ is
$O_2^{2-}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2(\pi 2p_x^2=\pi 2p_y^2)({\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^2)$
$\left(18e^{-}\right)$
$=\frac{10-8}{2}=\frac{2}{2}=1$

The bond order of $F_2^{2-}$ is
$\underset{\left(18\right)}{F_2}=\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2(\pi 2p_x^2=\pi 2p_y^2)({\pi }^{\ast }2p_x^2={\pi }^{\ast }2p_y^2)$

Bond order $=\frac{10-8}{2}=\frac{2}{2}=1$

Bond order for $H_2^{-}$ is:
$\underset{H_2^{-}}{\left(3\right)}\sigma 1s^2{\sigma }^{\ast }1s^1$
$BondOrder=\frac{2-1}{2}=\frac{1}{2}=0.5$
Hence, option (d) is correct.