Respuesta :
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 7 electrons present in nitrogen and 9 electrons in fluorine.
(a) The number of electrons present in [tex]NF[/tex] molecule = 7 + 9 = 16
The molecular orbital configuration of [tex]NF[/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 formula of bonding order = [tex]\frac{1}{2}\times (\text{Number of bonding electrons}-\text{Number of anti-bonding electrons})[/tex]
The bonding order of [tex]NF[/tex] = [tex]\frac{1}{2}\times (10-6)=2[/tex]
(b) The number of electrons present in [tex]NF^+[/tex] molecule = 7 + 9 - 1 = 15
The molecular orbital configuration of [tex]NF^+[/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}^*)^0],(\sigma_{2p_z}^*)^0[/tex]
The bonding order of [tex]NF^+[/tex] = [tex]\frac{1}{2}\times (10-5)=2.5[/tex]
(c) The number of electrons present in [tex]NF^-[/tex] molecule = 7 + 9 + 1 = 17
The molecular orbital configuration of [tex]NF^-[/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}^*)^2=(\pi_{2p_y}^*)^1],(\sigma_{2p_z}^*)^0[/tex]
The bonding order of [tex]NF^-[/tex] = [tex]\frac{1}{2}\times (10-7)=1.5[/tex]
