Which of the following have identical bond order ?

O_{2}^{2 +}

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N_{2}

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C_{2}^{2 +}

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C_{2}^{2 -}

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Solution

The correct options are
A $O_{2}^{2 +}$
B $N_{2}$
D $C_{2}^{2 -}$
Bond order for $O_{2}^{2 +}$ is:
$(16 - 2 = 14) O_{2}^{2 +} = σ 1 s^{2} σ^{} 1 s^{2} σ 2 s^{2} σ^{} 2 s^{2} σ 2 P_{z}^{2} π 2 P_{x}^{2} = π 2 P_{y}^{2}$
Bond order $= \frac{10 - 4}{2} = \frac{6}{2} = 3$

Bond order for $N_{2}$ is
$(7 + 7) N_{2} = 14$
$σ 1 s^{2} σ^{} 1 s^{2} σ 2 s^{2} σ^{} 2 s^{2} π 2 P_{x}^{2} = π 2 P_{y}^{2} σ 2 P_{z}^{2}$
Bond order $= \frac{N_{b} - N_{a}}{2}$
$= \frac{10 - 4}{2} = \frac{6}{2} = 3$

Bond order for $C_{2}^{2 +}$ is:
$(6 + 6 - 2) C_{2}^{2 +} = 10$ $σ 1 s^{2} σ^{} 1 s^{2} σ 2 s^{2} σ^{} 2 s^{2} π 2 P_{x}^{1} = π 2 P_{y}^{1}$
Bond order $= \frac{N_{b} - N_{a}}{2}$
$= \frac{6 - 4}{2} = \frac{2}{2} = 1$

Bond order for $C_{2}^{2 -}$ is:
$(6 + 6 + 2) C_{2}^{2 -} = 14$
$σ 1 s^{2} σ^{} 1 s^{2} σ 2 s^{2} σ^{} 2 s^{2} π 2 P_{x}^{2} = π 2 P_{y}^{2} σ 2 P_{z}^{2}$
Bond order $= \frac{N_{b} - N_{a}}{2}$
$= \frac{10 - 4}{2} = 3$
Hence option A, B and D are correct.