Question

If one of the electrons in the $H{e}_2$ molecule is taken to the next excited state, then the bond order in $H{e}_2$:

• A. increases by 1
• B. decreases by 1
• C. increases by 0.5
• D. no change

Answer 1

The correct option is A increases by 1

$H{e}_2$ (4 electrons) : $\left(\sigma 1s^2\right)(\sigma \ast 1s{)}^2$
Bond order $=\frac{2-2}{2}=0$
Electron is taken to next excited state that is $\left(\sigma 2s\right)$, then the electronic configuration is:
$\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s{)}^2\right(\sigma 2s{)}^1$
number of electrons in bonding molecular orbital = 3
and in anti-bonding molecular orbital = 1
Bond order $=(3-1)/2=1$
Thus, the bond order increases by 1 .