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Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order ?
A
$C_2^{2-} \lt He_2^+ \lt NO \lt O_2^-$
B
$He_2^+ \lt O_2^- \lt NO \lt C_2^{2-}$
C
$O_2^- \lt NO \lt C_2^{2-} \lt He_2^+$
D
$NO \lt C_2^{2-} \lt O_2^- \lt He_2^+$
Detailed Solution
Bond order $= \dfrac{1}{2}(N_b - N_a)$, where $N_b$ and $N_a$ are the numbers of electrons in bonding and antibonding molecular orbitals.
$He_2^+$ (3 electrons): $\sigma 1s^2\,\sigma^* 1s^1$; bond order $= \dfrac{1}{2}(2 - 1) = 0.5$
$O_2^-$ (17 electrons): one more antibonding electron than $O_2$ (bond order 2); bond order $= \dfrac{1}{2}(10 - 7) = 1.5$
$NO$ (15 electrons): bond order $= \dfrac{1}{2}(10 - 5) = 2.5$
$C_2^{2-}$ (14 electrons, isoelectronic with $N_2$): bond order $= \dfrac{1}{2}(10 - 4) = 3.0$
Bond orders: 0.5, 1.5, 2.5 and 3.0 respectively.
Increasing order of bond order: $He_2^+ \lt O_2^- \lt NO \lt C_2^{2-}$
$He_2^+$ (3 electrons): $\sigma 1s^2\,\sigma^* 1s^1$; bond order $= \dfrac{1}{2}(2 - 1) = 0.5$
$O_2^-$ (17 electrons): one more antibonding electron than $O_2$ (bond order 2); bond order $= \dfrac{1}{2}(10 - 7) = 1.5$
$NO$ (15 electrons): bond order $= \dfrac{1}{2}(10 - 5) = 2.5$
$C_2^{2-}$ (14 electrons, isoelectronic with $N_2$): bond order $= \dfrac{1}{2}(10 - 4) = 3.0$
Bond orders: 0.5, 1.5, 2.5 and 3.0 respectively.
Increasing order of bond order: $He_2^+ \lt O_2^- \lt NO \lt C_2^{2-}$