In which of the following $m o l e c u l e / i o n$ no unpaired electron is present?

N_{2}^{+} \Rightarrow 13 e^{-}

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O_{2} \Rightarrow 16 e^{-}

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O_{2}^{2} \Rightarrow 18 e^{-}

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B_{2} \Rightarrow 10 e^{-}

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Solution

The correct option is C $O_{2}^{2} \Rightarrow 18 e^{-}$
$O_{2}$ has 8+8=16 electron so $O_{2}^{2} -$ will have 16+2=18 electron.
We know that for 18 electron the electronic configuration can be written as
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$ there we can see that all orbitals $s$ and $p$ are fully filled $s has 2 electrons and all p has 6 electrons so fully filled.$
So correct answer is C.

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In which of the following molecule/ion no unpaired electron is present?

In which of the following $m o l e c u l e / i o n$ no unpaired electron is present?