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Question

Question: Which molecule/ion out of the following does not contain unpaired electrons?...

Which molecule/ion out of the following does not contain unpaired electrons?

A

N2+N_{2}^{+}

B

O2O_{2}

C

O22O_{2}^{2 -}

D

B2B_{2}

Answer

O22O_{2}^{2 -}

Explanation

Solution

: N2+(13)=σ1s2,σ1s2,σ2s2,π2px2=π2Py2σ2Pz1N_{2}^{+}(13) = \sigma 1s^{2},\sigma*1s^{2},\sigma 2s^{2},\pi 2p_{x}^{2} = \pi 2P_{y}^{2}\sigma 2P_{z}^{1}

O2(16)=σ15s2,σ1s2,σ2s2,σ2s2,σ2pz2O_{2}(16) = \sigma 15s^{2},\sigma*1s^{2},\sigma 2s^{2},\sigma*2s^{2},\sigma 2p_{z}^{2}

π2px2=π2py2,π2px1=π2Py1\pi 2p_{x}^{2} = \pi 2p_{y}^{2},\pi*2p_{x}^{1} = \pi*2P_{y}^{1}

O22(18)=σ1s2,σ1s2,σ2s2,σ2s2,σ2pz2,π2px2O_{2}^{2 -}(18) = \sigma 1s^{2},\sigma*1s^{2},\sigma 2s^{2},\sigma*2s^{2},\sigma 2p_{z}^{2},\pi 2p_{x}^{2}

=π2py2,π2px2=π2Py2= \pi 2p_{y}^{2},\pi*2p_{x}^{2} = \pi*2P_{y}^{2}

B2(10)=σ1s2,σ1s2,σ2s2,σ2s2,π2px1=π2py1B_{2}(10) = \sigma 1s^{2},\sigma*1s^{2},\sigma 2s^{2},\sigma*2s^{2},\pi 2p_{x}^{1} = \pi 2p_{y}^{1}

Thus, O22O_{2}^{2 -}does not contain unpaired electrons.