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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 $II > I > III$.

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

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