Give the order of relative stability for N_2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $N_2^{2+}$.

Similar Questions

Among the following species,the diamagnetic molecule is

Which of the following pairs of species are paramagnetic?

"Isoelectronic molecules and ions have identical bond orders." Explain by examples.

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

When two atomic orbitals combine,they form:

Vedclass Test Series

Exam Paper Generator

Online Exam Module