Respuesta :
Answer : The number of unpaired electrons in the [tex]F_2^{2+}[/tex] = 2
The bond order of [tex]F_2^{2+}[/tex] is, 2
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration will be,
[tex](\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}^*)[/tex]
As there are 9 electrons present in fluorine.
The number of electrons present in [tex]F_2^{2+}[/tex] molecule = 2(9) = 18 - 2 = 16
The molecular orbital configuration of [tex]F_2^{2+}[/tex] molecule will be,
[tex](\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[/tex]
The number of unpaired electrons in the [tex]F_2^{2+}[/tex] = 2
The formula of bond order = [tex]\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})[/tex]
The bonding order of [tex]F_2^{2+}[/tex] = [tex]\frac{1}{2}\times (10-6)=2[/tex]
The bond order of [tex]F_2^{2+}[/tex] is, 2
