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(B) The bond order $(BO)$ is calculated using the formula: $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding molecular orbi

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$II > I > III$

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(B) The bond order $(BO)$ is calculated using the formula: $BO = \frac{N_b - N_a}{2}$,where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.For $NO^+$ ($14$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $BO = \frac{10 - 4}{2} = 3.0$.For $NO$ ($15$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{10 - 5}{2} = 2.5$.For $NO^-$ ($16$ electrons): The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $BO = \frac{10 - 6}{2} = 2.0$.Comparing the bond orders: $3.0 (II) > 2.5 (I) > 2.0 (III)$.Thus,the decreasing order is $II > I > III$.

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:
$(I)$ $NO$
$(II)$ $NO^+$
$(III)$ $NO^-$

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Number of molecules having bond order $2$ from the following molecules is: $C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$

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

Find out the bond order in $He_2, He_2^{+}, O_2$ and $O_2^{+}$ respectively.

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