Chemical Bonding and Molecular Structure - Result Question 122
####126. The ground state electronic configuration of valence shell electrons in nitrogen molecule $(N_2)$ is written as $KK \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_x^{2}, \pi 2 p_y^{2} \sigma 2 p_z^{2}$ Bond order in nitrogen molecule is
[1995]
(a) 0
(b) 1
(c) 2
(d) 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 $N_b=8$ and $N_a=2$.
$\therefore \quad$ Bond order $=\frac{1}{2}(N_b-N_a)=\frac{1}{2}(8-2)=3$.
