Question

Use molecular orbital theory to complete this table BY FILLING FLANKS, 0,1,2,3, or 4NF = (?1s) ___(?1s*) ___(?2s) ___(?2s*) ___(? 2p)___ (?2p) ___(? 2p*) Bonding order=NF+ =(?1s) ___(?1s*) ___(?2s)___ (?2s*) ___(? 2p) ___(?2p) ___(? 2p*) Bonding order=NF- = (?1s)___ (?1s*)___ (?2s) ___(?2s*)___ (? 2p)___ (?2p) ___(? 2p*) Bonding order=

Answer 1

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 7 electrons present in nitrogen and 9 electrons in fluorine.

(a) The number of electrons present in $NF$ molecule = 7 + 9 = 16

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

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

(b) The number of electrons present in $NF^+$ molecule = 7 + 9 - 1 = 15

The molecular orbital configuration of $NF^+$ molecule will be,

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

The bonding order of $NF^+$ = $\frac{1}{2}\times (10-5)=2.5$

(c) The number of electrons present in $NF^-$ molecule = 7 + 9 + 1 = 17

The molecular orbital configuration of $NF^-$ molecule will be,

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

The bonding order of $NF^-$ = $\frac{1}{2}\times (10-7)=1.5$