The total number of species from the followin | vedclass

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

Question

(C) To determine the number of unpaired electrons,we use Molecular Orbital Theory $(MOT)$: $N_2$: $(14 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^2,

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) To determine the number of unpaired electrons,we use Molecular Orbital Theory $(MOT)$: $N_2$: $(14 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$ ($0$ unpaired $e^-$) $O_2$: $(16 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$ ($2$ unpaired $e^-$) $C_2^{-}$: $(13 \ e^-) \rightarrow \dots, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^1$ ($1$ unpaired $e^-$) $O_2^{-}$: $(17 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^1$ ($1$ unpaired $e^-$) $O_2^{2-}$: $(18 \ e^-) \rightarrow \dots, \pi^* 2p_x^2 = \pi^* 2p_y^2$ ($0$ unpaired $e^-$) $H_2^{+}$: $(1 \ e^-) \rightarrow \sigma 1s^1$ ($1$ unpaired $e^-$) $CN^{-}$: $(14 \ e^-) \rightarrow \dots, \sigma 2p_z^2$ ($0$ unpaired $e^-$) $He_2^{+}$: $(3 \ e^-) \rightarrow \sigma 1s^2, \sigma^* 1s^1$ ($1$ unpaired $e^-$) Species with one unpaired electron are $C_2^{-}, O_2^{-}, H_2^{+}, He_2^{+}$. Total count = $4$.

The total number of species from the following in which one unpaired electron is present,is . . . . . . $N_2, O_2, C_2^{-}, O_2^{-}, O_2^{2-}, H_2^{+}, CN^{-}, He_2^{+}$

Similar Questions

Which of the following species has a bond order of $1.5$?

$A$ pair of isoelectronic species having bond order of $1$ is

In which of the following ionization processes has the bond energy increased and the magnetic behaviour changed from paramagnetic to diamagnetic?

Which of the following is paramagnetic?

Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module