Among the following species,identify the pair | vedclass

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(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$

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$CN^{-}$ and $NO^{+}$

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(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.$1$. For $CN^{-}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $B.O. = \frac{10 - 4}{2} = 3$.$2$. For $O_2^-$ ($17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. $B.O. = \frac{10 - 7}{2} = 1.5$.$3$. For $NO^{+}$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{10 - 4}{2} = 3$.$4$. For $CN^{+}$ ($12$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. $B.O. = \frac{8 - 4}{2} = 2$.Thus,$CN^{-}$ and $NO^{+}$ both have a bond order of $3$.

Among the following species,identify the pair having the same bond order: $CN^{-}, O_2^-, NO^{+}, CN^{+}$

Similar Questions

Given below are two statements:
Statement $I$: $A$ hypothetical diatomic molecule with bond order zero is quite stable.
Statement $II$: As bond order increases,the bond length increases.
In the light of the above statements,choose the most appropriate answer from the options given below.

Give the number of electrons in the species $H_2^+$,$H_2$,and $O_2^+$.

Consider the following species: $CN^{+}$,$CN^{-}$,$NO$,and $CN$. Which one of these will have the highest bond order?

Which one of the following species is diamagnetic in nature?

The correct order of bond length is :-

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