Chemical Bonding And Molecular Structure Ques 29
The ground state electronic configuration of valence shell electrons in nitrogen molecule $(N_2)$ is written as $KK \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_x^{2}, \pi 2 p_y^{4} \sigma 2 p_z^{2}$ Bond order in nitrogen molecule is
[1995]
0
1
2
3
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Solution:
- (d) In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding molecular orbital. So that $N_b=4$ and $N_a=1$.
$\therefore \quad$ Bond order $=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(8-2)=3$.
BETA
