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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 $B_2$.

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

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

Similar Questions

The lowest $O-O$ bond length in the following molecules is:

Consider the given figure showing the formation of $H^{+}_2$ ion depending on internuclear distance versus potential energy of the system.
Which is the correct statement?

Match the following:

Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

Arrange the following species in the decreasing order of their bond orders:
$(I)$ $NO$
$(II)$ $NO^+$
$(III)$ $NO^-$

Which of the following species has a bond order of $1.5$?

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