Question

draw the molecular orbital diagram of N2 and calculate the bond order

Answer 1

BOND ORDER =

½(BONDINGelectron-ANTIBONDING electron)

= ½(8 - 2)

= 3

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Answer 2

Answer :

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

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

As there are 7 electrons present in nitrogen.

The number of electrons present in $N_2$ molecule = 2(7) = 14

The molecular orbital configuration of $N_2$ molecule will be,

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

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

The bonding order of $N_2$ = $\frac{1}{2}\times (10-4)=3$

The bond order of $N_2$ is, 3

The molecular orbital diagram of $N_2$ are shown below.