Question

The bond length of $H_2^{+},H_2^{-}$ and $H_2$ are in the following order

• A. $H_2^{+}>H_2>H_2^{-}$
• B. $H_2>H_2^{+}>H_2^{-}$
• C. $H_2^{-}>H_2>H_2^{+}$
• D. $H_2^{-}>H_2^{+}>H_2$

Answer 1

Answer: (D)

$H_2^{+}(2-1=1)=\sigma 1s^1$

$BO=\frac{1-0}{2}=0.5$

$H_2^{-}(2+1=3)=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^1$

$BO=\frac{2-1}{2}=0.5$

$H_2(2)=\sigma 1s^2$

$BO=\frac{2-0}{2}=1$

$\because$ The order of bond order is

$H_2>H_2^{+}>H_2^{-}$

Thus, the order of bond length is

$H_2^{-}>H_2^{+}>H_2$