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Chemistry Question on Chemical bonding and molecular structure

Which one of the following molecules is paramagnetic?

A

F2F_{2}

B

B2B_{2}

C

Li2Li_{2}

D

N2N_{2}

Answer

B2B_{2}

Explanation

Solution

Molecules in which unpaired electrons are present, are paramagnetic while that have absence of unpaired electrons, are diamagnetic.
F2(18)=σ1s2,σ∗ 1s2,σ2s2σ∗ 2s2,σ2pz2,{{F}_{2}}(18)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}}\overset{*}{\sigma}\,2{{s}^{2}},\sigma 2p_{z}^{2},
π2px2≈π2py2,π∗ 2px2≈π∗ 2py2\pi 2p_{x}^{2}\approx \pi 2p_{y}^{2},\overset{*}{\pi }\,2p_{x}^{2}\approx \overset{*}{\pi}\,2p_{y}^{2}
(No unpaired electron, so diamagnetic)
B2(10)=σ1s2,σ∗ 1s2σ2s2,σ∗ 2s2,π2px1≈π2py1{{B}_{2}}(10)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}}\sigma 2{{s}^{2}},\overset{*}{\sigma }\,2{{s}^{2}},\pi 2p_{x}^{1}\approx \pi 2p_{y}^{1}
(Two unpaired electrons, paramagnetic)
Li2(6)=σ1s2,σ∗ 1s2,σ2s2L{{i}_{2}}(6)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}}
(No unpaired electrons, diamagnetic)
N2(14)=σ1s2,σ∗ 1s2,σ2s2,σ∗ 2s2,π2px2{{N}_{2}}(14)=\sigma 1{{s}^{2}},\overset{*}{\sigma }\,1{{s}^{2}},\sigma 2{{s}^{2}},\overset{*}{\sigma }\,2{{s}^{2}},\pi 2p_{x}^{2}
≈π2py2,σ2pz2\approx \pi 2p_{y}^{2},\sigma 2p_{z}^{2}
(No unpaired electrons, diamagnetic)