The correct stability order for N_2 and its g | vedclass

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

Question

(A) Stability is directly proportional to bond order. Bond order $= \frac{1}{2} 	imes (N_B - N_{AB})$. $1$. $N_2$ (valence electrons $= 10$): $(\sigm

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Stability is directly proportional to bond order. Bond order $= \frac{1}{2} 	imes (N_B - N_{AB})$. $1$. $N_2$ (valence electrons $= 10$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2$. Bond order $= \frac{1}{2} 	imes (8 - 2) = 3$. $2$. $N_2^+$ (valence electrons $= 9$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^1$. Bond order $= \frac{1}{2} 	imes (7 - 2) = 2.5$. $3$. $N_2^-$ (valence electrons $= 11$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^* 2p)^1$. Bond order $= \frac{1}{2} 	imes (8 - 3) = 2.5$. $4$. $N_2^{2-}$ (valence electrons $= 12$): $(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p)^4(\sigma 2p)^2(\pi^* 2p)^2$. Bond order $= \frac{1}{2} 	imes (8 - 4) = 2$. Although $N_2^+$ and $N_2^-$ have the same bond order,$N_2^-$ has an electron in an antibonding orbital,which decreases its stability compared to $N_2^+$. Thus,the stability order is $N_2 > N_2^+ > N_2^- > N_2^{2-}$. Therefore,the correct option is $A$.

The correct stability order for $N_2$ and its given ions is :-

Similar Questions

The molecular orbital configuration of a diatomic molecule is
$\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_x^2 \pi 2p_y^2 \pi 2p_z^2$
Its bond order is

The bond order of $O_2^+$ is the same as in

Use the molecular orbital energy level diagram to show that $N_{2}$ would be expected to have a triple bond,$F_{2}$ a single bond and $Ne_{2}$ no bond.

Among the following,the species with the highest bond order is

Which of the following species is paramagnetic?

Vedclass Test Series

Exam Paper Generator

Online Exam Module