Question

In the molecular orbital diagram for the molecule ion, ${N_2}^{+}$, the number of electrons in the ${\sigma }_{2p}$ molecular orbital is:
A. 1
B. 2
C. 0
D. 3

Answer 1

Hint: The Molecular Orbital Theory is often abbreviated as MOT. As we know that valence bond theory fails to explain how certain molecules contain two or more equivalent bonds whose bond orders lie between that of a single bond and double bond and for this we use MOT.

Complete step-by-step answer: The electronic configuration of Nitrogen represented as N is represented as $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2{p_x}^2\pi 2{p_y}^2\sigma 2{p_z}^1$ but molecular orbital configuration is written in different manner which can be explained as $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2{p_z}^2\pi 2{p_x}^2\pi 2{p_y}^2$ here we have to write the molecular orbital configuration for ${N_2}^{+}$, this corresponds to the loss of one electron only 13 electrons are there to fill in the orbital which can be defined as:
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2{p_x}^2\pi 2{p_y}^2\sigma 2{p_z}^1$, here two type of electrons can be seen which are in $\sigma$and $\pi$ are known as bonding molecular orbital whereas when electrons are present in ${\sigma }^{\ast }$ orbital are known as anti-bonding molecular orbital.
Now from the electronic configuration we can see that the number of electrons in the ${\sigma }_{2p}$ molecular orbital is 1.

Thus, option A is the correct answer.

Note: Probability of finding the electrons is more in the case of bonding molecular orbitals as compared to anti-bonding orbitals as in case of anti-bonding molecular orbitals there is a node present between two nuclei where the electron density is zero so probability of finding electron will low.