Number of molecules/species from the followin | vedclass

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

Question

(B) According to $M.O.T.$ (Molecular Orbital Theory):$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) According to $M.O.T.$ (Molecular Orbital Theory):$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Unpaired electrons $= 2$.$O_2^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Unpaired electrons $= 1$.$NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Unpaired electrons $= 1$.$CN^{-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Unpaired electrons $= 0$.$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Unpaired electrons $= 0$.Thus,the species with one unpaired electron are $O_2^{-}$ and $NO$. The total number is $2$.

Number of molecules/species from the following having one unpaired electron: $O_2, O_2^{-}, NO, CN^{-}, O_2^{2-}$

Similar Questions

How do you express the bond strength in terms of bond order?

Consider the given figure showing the formation of $H^{+}_2$ ion depending on internuclear distance versus potential energy of the system.
Which is the correct statement?

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

The correct order of energies of molecular orbitals of $N_2$ molecule is

Vedclass Test Series

Exam Paper Generator

Online Exam Module