What is the number of unpaired electron(s) in | vedclass

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

Question

(A) $N_2$ $(14 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $HOMO$ is $\sigma 2p_z$,whic

Vedclass · Accepted Answer

$0, 1, 2, 1$

Vedclass · Answer

(A) $N_2$ $(14 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $HOMO$ is $\sigma 2p_z$,which has $0$ unpaired electrons.$N_2^{+}$ $(13 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. $HOMO$ is $\sigma 2p_z$,which has $1$ unpaired electron.$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $HOMO$ is $\pi^* 2p$,which has $2$ unpaired electrons.$O_2^{+}$ $(15 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $HOMO$ is $\pi^* 2p$,which has $1$ unpaired electron.Thus,the number of unpaired electrons is $0, 1, 2, 1$ respectively.

What is the number of unpaired electron$(s)$ in the highest occupied molecular orbital $(HOMO)$ of the following species: $N_2$,$N_2^{+}$,$O_2$,$O_2^{+}$?

Similar Questions

Write the significance of a plus $(+)$ and a minus $(-)$ sign shown in representing the orbitals.

In which of the following pairs do the two molecules have identical bond order?

The correct statement with regard to $H_{2}^{+}$ and $H_{2}^{-}$ is

$N_2$ and $O_2$ are converted into monocations,$N_2^+$ and $O_2^+$ respectively. Which of the following statements is wrong?

The correct molecular orbital diagram for $F_2$ molecule in the ground state is

Vedclass Test Series

Exam Paper Generator

Online Exam Module