Question

Question 1) By drawing molecular orbital diagrams for B2, C2, N2, O2, and F2, predict which of these homonuclear diatomic molecules are magnetic.

Question 2) Based on the molecular orbital diagram for NO, which of the following electronic configurations and statements are most correct?

Answer 1

Concepts and reason

The concept used to solve this problem is based on the molecular orbital theory.

Firstly the molecular orbital diagram is drawn then bond order is calculated using the bonding and antibonding electrons. After that, check for its magnetic behaviour.

Fundamentals

Bond Order:

It tells the number of chemical bonds between the atoms.

The bond order can be calculated by using the formula given below.

${\rm{Bond order }} = \frac{{{\rm{ }}\left[ {{N_{\rm{b}}}{\rm{ }} - {N_{\rm{a}}}{\rm{ }}} \right]}}{{\rm{2}}}$ …… (1)

Here, ${N_{\rm{b}}}$ is a number of electrons in bonding molecular orbital and ${N_{\rm{a}}}$ number of electrons in an antibonding molecular orbital.

(A)

The total number of electrons of ${{\rm{B}}_{\rm{2}}}$ is $10$ . The molecular orbital configuration of ${{\rm{B}}_{\rm{2}}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^1 = {\rm{\pi 2}}{{\rm{p}}_X}^1$

The total number of electrons of ${{\rm{O}}_{\rm{2}}}$ is $16$ . The molecular orbital configuration of ${{\rm{O}}_{\rm{2}}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1 = {{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1$

The total number of electrons of ${{\rm{C}}_{\rm{2}}}$ is $12$ . The molecular orbital configuration of ${{\rm{C}}_{\rm{2}}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2$

The total number of electrons of ${{\rm{N}}_{\rm{2}}}$ is $14$ . The molecular orbital configuration of ${{\rm{N}}_{\rm{2}}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2{\rm{,\sigma 2}}{{\rm{p}}_z}^{\rm{2}}$

The total number of electrons of ${{\rm{F}}_{\rm{2}}}$ is $18$ . The molecular orbital configuration of ${{\rm{F}}_{\rm{2}}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^2 = {{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^2$

(B)

The total number of electrons of ${\rm{NO}}$ is $15$ . The molecular orbital configuration of ${\rm{NO}}$ of increasing order of energy is shown below.

${\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}},{{\rm{\sigma }}^*}{\rm{1}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}_z}^{\rm{2}},{\rm{\pi 2}}{{\rm{p}}_X}^2 = {\rm{\pi 2}}{{\rm{p}}_X}^2,{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}_X}^1$

There is an unpaired electron present in the orbital of ${\rm{NO}}$ . Therefore, it is paramagnetic in nature.

Ans: Part A

The homonuclear diatommic molecule ${{\bf{B}}_{\bf{2}}}$ and ${{\bf{O}}_{\bf{2}}}$ are magnetic.

Part B

The configuration ${\rm{\sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{,}}{{\rm{\sigma }}^*}{\rm{2}}{{\rm{s}}^{\rm{2}}},{\rm{\sigma 2}}{{\rm{p}}^{\rm{2}}},{\rm{\pi 2}}{{\rm{p}}^4},{{\rm{\pi }}^*}{\rm{2}}{{\rm{p}}^1}$ magnetic is the most correct.