Arrange the molecules B_2, He_2, N_2 and C_2 | vedclass

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Question

(C) Bond order can be calculated by using the formula,$	ext{B.O.} = \frac{1}{2} [N_b - N_a]$.Molecular orbital electronic configuration of $B_2 = (\s

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$He_2 < B_2 < C_2 < N_2$

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(C) Bond order can be calculated by using the formula,$	ext{B.O.} = \frac{1}{2} [N_b - N_a]$.Molecular orbital electronic configuration of $B_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^1 = \pi 2p_y^1)$.Bond order of $B_2 = \frac{1}{2} [6 - 4] = 1$.Molecular orbital electronic configuration of $He_2 = (\sigma 1s)^2 (\sigma^* 1s)^2$.Bond order of $He_2 = \frac{1}{2} [2 - 2] = 0$.Molecular orbital electronic configuration of $N_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2) (\sigma 2p_z)^2$.Bond order of $N_2 = \frac{1}{2} [10 - 4] = 3$.Molecular orbital electronic configuration of $C_2 = (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x^2 = \pi 2p_y^2)$.Bond order of $C_2 = \frac{1}{2} [8 - 4] = 2$.Therefore,the increasing order of bond order values is $He_2 (0) < B_2 (1) < C_2 (2) < N_2 (3)$.

Arrange the molecules $B_2, He_2, N_2$ and $C_2$ in the increasing order of their bond order values.

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