Which of the following is paramagnetic?

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Question

(D) To determine the magnetic nature,we calculate the total number of electrons and fill the molecular orbitals:$NO^{+} : 14 	ext{ electrons} \Righta

Vedclass · Accepted Answer

$B_2$

Vedclass · Answer

(D) To determine the magnetic nature,we calculate the total number of electrons and fill the molecular orbitals:$NO^{+} : 14 	ext{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, (\sigma 2p_z)^2$ (All paired,diamagnetic)$CO : 14 	ext{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, (\sigma 2p_z)^2$ (All paired,diamagnetic)$O_2^{2-} : 18 	ext{ electrons}$ $\Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p_z)^2, (\pi 2p_x)^2 = (\pi 2p_y)^2, \pi^*(2p_x)^2 = \pi^*(2p_y)^2$ (All paired,diamagnetic)$B_2 : 10 	ext{ electrons} \Rightarrow KK, \sigma(2s)^2, \sigma^*(2s)^2, \pi(2p_x)^1 = \pi(2p_y)^1$ (Contains unpaired electrons,paramagnetic)Therefore,$B_2$ is paramagnetic.

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