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According to Molecular Orbital Theory,
($A$) $\mathrm{C}_2^{2-}$ is expected to be diamagnetic
($B$) $\mathrm{O}_2{ }^{2+}$ is expected to have a longer bond length than $\mathrm{O}_2$
($C$) $\mathrm{N}_2^{+}$and $\mathrm{N}_2^{-}$have the same bond order
($D$) $\mathrm{He}_2^{+}$has the same energy as two isolated He atoms
$A,D$
$A,C$
$A,B$
$A,D,B$
Solution
The correct options are
$A C _2^{2-}$ is expected to be diamagnetic
$CN _2^{+}$and $N _2^{-}$have the same bond order
Total number of electrons in $C _2^{2-}$ is 14 and is diamagnetic according to MOT:
$\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2} \pi_{2 p_y^2} \sigma_{2 p_x^2}^2$
$O _2^{2+}$ is having a bond order of 3 while $O _2$ has a bond order of 2 and thus $O _2$ is expected to have a longer bond length than $O _2^{2+}$.
Bond order of $N _2^{+}=$bond order of $N _2^{-}=2.5$
Bond order of $He _2^{+}$is $0.5$ and there will be some energy change during the formation of $He _2^{+}$from two isolated He atoms.
