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(C) The molecular orbital configurations and bond orders $(BO)$ are calculated as follows:$He_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^1$; $BO = \frac{2-1}{

Vedclass · Accepted Answer

$B_2$

Vedclass · Answer

(C) The molecular orbital configurations and bond orders $(BO)$ are calculated as follows:$He_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^1$; $BO = \frac{2-1}{2} = 0.5$. It is paramagnetic due to one unpaired electron.$Li_2^{+}: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^1$; $BO = \frac{3-2}{2} = 0.5$. It is paramagnetic.$B_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^1 (\pi 2p_y)^1$; $BO = \frac{6-4}{2} = 1.0$. It is paramagnetic due to two unpaired electrons.$C_2: (\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2$; $BO = \frac{8-4}{2} = 2.0$. It is diamagnetic.Thus,the molecule with a bond order of $1.0$ and paramagnetic nature is $B_2$.

The molecule/ion having paramagnetic nature and a bond order of $1.0$ is

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