Question

The bond orders of $H{e}_2^{+}$ and $H{e}_2$ are respectively

• A. $\frac{1}{2},0$
• B. $0,\frac{1}{2}$
• C. 0,1
• D. 1,0

Answer 1

Answer: (A)

$H{e}_2$ molecular orbital has electronic configuration as $\sigma l{s}^2,{\sigma }^{\ast }l{s}^2$

Bond order $=\frac{\text{(electrons in bonding molecular orbital) - <br/> <br/> (electrons in antibonding molecular orbital)}}{2}$

For $H{e}_2$ molecule bond order $=\frac{2-2}{2}=0$

Bond order is zero so, $H{e}_2$ molecule does not exist.

In case of $H{e}_2^{+}$ electronic configuration of molecular orbital is $\sigma l{s}^2{\sigma }^{\ast }1s^1$.

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