Give the order of relative stability for N_2, | vedclass

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Question

(B) The stability of a molecule is directly proportional to its bond order.Bond order calculations:$N_2 (14 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma

Vedclass · Accepted Answer

$N_2 > N_2^+ = N_2^- > N_2^{2+}$

Vedclass · Answer

(B) The stability of a molecule is directly proportional to its bond order.Bond order calculations:$N_2 (14 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3.0$.$N_2^+ (13 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$.$N_2^- (15 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$.$N_2^{2+} (12 \ e^-): \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(8-4)/2 = 2.0$.Comparing bond orders: $N_2 (3.0) > N_2^+ (2.5) = N_2^- (2.5) > N_2^{2+} (2.0)$.Thus,the stability order is $N_2 > N_2^+ = N_2^- > N_2^{2+}$.

Give the order of relative stability for $N_2, N_2^+, N_2^-,$ and $N_2^{2+}$.

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