Which of the following is the correct relatio | vedclass

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(C) The bond order of $N_2$ is $3.0$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sig

Vedclass · Accepted Answer

Bond dissociation energy of $N_2 > $ Bond dissociation energy of $N_2^+$

Vedclass · Answer

(C) The bond order of $N_2$ is $3.0$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$).The bond order of $N_2^+$ is $2.5$ (electronic configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$).Since bond dissociation energy is directly proportional to bond order,the bond dissociation energy of $N_2$ is greater than that of $N_2^+$.

Which of the following is the correct relationship between the bond dissociation energy of $N_2$ and $N_2^+$?

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