Question

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

• A. Increased by $1$
• B. Decreased by $1$
• C. Increased by $0.5$
• D. Not changed

Answer 1

The correct option is A Increased by $1$

$B{e}_2$ ($8$ electrons) : $\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s{)}^2\right(\sigma 2s^2\left)\right({\sigma }^{\ast }2s{)}^2$
Bond order $=\frac{4-4}{2}=0$
Electron is taken to next excited state that is $\left(\pi 2p_x\right)$, then the electronic configuration is:
$\left(\sigma 1s^2\right)\left({\sigma }^{\ast }1s{)}^2\right(\sigma 2s^2\left)\right({\sigma }^{\ast }2s{)}^1\left(\pi 2p_x{)}^1\right(\pi 2p_y{)}^0$
Number of electrons in bonding molecular orbital $=5$
and in anti-bonding molecular orbital $=3$
Bond order $=(5-3)/2=1$
Thus, the bond order increases by $1$ .