Question

How many unpaired electrons in the F22+ ion are based on molecular orbital theory? The order of the molecular orbitals are (σ2s)(σ*2s)(σ2p)(π2p)(π*2p)(σ*2p).

Answer 1

Answer : The number of unpaired electrons in the $F_2^{2+}$ = 2

The bond order of $F_2^{2+}$ is, 2

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

$(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)$

As there are 9 electrons present in fluorine.

The number of electrons present in $F_2^{2+}$ molecule = 2(9) = 18 - 2 = 16

The molecular orbital configuration of $F_2^{2+}$ molecule will be,

$(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],[(\pi_{2p_x}^*)^1=(\pi_{2p_y}^*)^1],(\sigma_{2p_z}^*)^0$

The number of unpaired electrons in the $F_2^{2+}$ = 2

The formula of bond order = $\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})$

The bonding order of $F_2^{2+}$ = $\frac{1}{2}\times (10-6)=2$

The bond order of $F_2^{2+}$ is, 2