Question

Assuming 2s-2p mixing is not operative the paramagnetic species among the following is:
A) $\text{ B}{\text{e}}_{\text{2 }}$
B) ${\text{N}}_{\text{2 }}$
C) ${\text{C}}_{\text{2 }}$
D) ${\text{B}}_{\text{2 }}$

Answer 1

Hint: The magnetic property of a molecule can be explained based on the molecular orbital theory. The molecule which does not contain the unpaired electron is known as the paramagnetic. The molecule which has all-electron paid-up does not contribute towards the magnetic property. It is diamagnetic in nature. To solve such a problem write down the MOT diagram of molecules.

Complete step by step solution:
The MOT diagram is drawn in such a way that we do not consider the s and p orbital mixing. Therefore the order of orbitals is not changed.
Let’s first draw the MOT of all molecules given in the problem.
A) $\text{ B}{\text{e}}_{\text{2 }}$ molecule: Beryllium atom has two electrons in the first shell and two-electrons in the second shell. The electronic configuration of the beryllium atom is as shown below,
$\text{ Be = 1}{\text{s}}^{\text{2}}\text{ 2}{\text{s}}^{\text{2}}$
The $\text{ B}{\text{e}}_{\text{2 }}$molecule contains a total of 8 electrons in it. These 8 electrons are arranged in the molecular orbitals.
The MOT is as shown below,

First of all, we can write the molecular orbital configuration of $\text{ B}{\text{e}}_{\text{2}}$ . The molecular orbital configuration of $\text{ B}{\text{e}}_{\text{2}}$ the molecule is as follows:
$$$$$$\sigma \text{ 1}{\text{s}}^{\text{2}}\text{,}{\sigma }^{\text{*}}\text{1}{\text{s}}^{\text{2}}\text{, }\sigma \text{ 2}{\text{s}}^{\text{2}}\text{, }{\sigma }^{\text{*}}\text{2}{\text{s}}^{\text{2}}\text{ OR KK , }\sigma \text{ 2}{\text{s}}^{\text{2}}\text{, }{\sigma }^{\text{*}}\text{2}{\text{s}}^{\text{2}}$$
Since, here all molecular orbitals contain the two electrons. These are all paired electrons. Thus $\text{ B}{\text{e}}_{\text{2}}$ the molecule is diamagnetic.
B) ${\text{N}}_{\text{2 }}$ molecule : The nitrogen atom has three electrons in the 2p shell and 4 electrons in the innermost shell. The electronic configuration of nitrogen atom is as shown below,
$\text{ N = 1}{\text{s}}^{\text{2}}\text{ 2}{\text{s}}^{\text{2}}\text{ 2}{{\text{p}}^{\text{1}}}_{\text{x}}\text{= 2}{{\text{p}}^{\text{1}}}_{\text{y}}=\text{2}{{\text{p}}^{\text{1}}}_z$
The ${\text{N}}_{\text{2 }}$molecule contains a total of 14 electrons in it. These 14 electrons are arranged in the molecular orbitals. The MOT is as shown below,

We can write the molecular orbital configuration of ${\text{N}}_{\text{2 }}$ . The molecular orbital configuration of $\text{ B}{\text{e}}_{\text{2}}$ the molecule is as follows:
$\sigma \text{ 1}{\text{s}}^{\text{2}}\text{,}{\sigma }^{\text{*}}\text{1}{\text{s}}^{\text{2}}\text{, }\sigma \text{ 2}{\text{s}}^{\text{2}}\text{, }{\sigma }^{\text{*}}\text{2}{\text{s}}^{\text{2}}\text{ , }\pi {\text{ 2p}}_{\text{y}}^{\text{2}}\text{ = }\pi {\text{ 2p}}_{\text{x}}^{\text{2}}\text{ , }\sigma {\text{ 2p}}_{\text{z}}^{\text{2}}$
Since, here all molecular orbitals contain the two electrons. These are all paired electrons. Thus ${\text{N}}_{\text{2 }}$ the molecule is diamagnetic.
C) ${\text{C}}_{\text{2 }}$ Molecule: The carbon atom has two electrons in the 2p shell and 4 electrons in the innermost shell. The electronic configuration of nitrogen atom is as shown below,$\text{ C = 1}{\text{s}}^{\text{2}}\text{ 2}{\text{s}}^{\text{2}}\text{ 2}{{\text{p}}^{\text{1}}}_{\text{x}}\text{= 2}{{\text{p}}^{\text{1}}}_{\text{y}}=\text{2}{{\text{p}}^0}_z$
The ${\text{C}}_{\text{2 }}$molecule contains a total of 12 electrons in it. These 12 electrons are arranged in the molecular orbitals.
The MOT is as shown below,

we can write the molecular orbital configuration of ${\text{C}}_{\text{2 }}$ . The molecular orbital configuration of ${\text{C}}_{\text{2}}$ the molecule is as follows:
$\sigma \text{ 1}{\text{s}}^{\text{2}}\text{,}{\sigma }^{\text{*}}\text{1}{\text{s}}^{\text{2}}\text{, }\sigma \text{ 2}{\text{s}}^{\text{2}}\text{, }{\sigma }^{\text{*}}\text{2}{\text{s}}^{\text{2}}\text{ , }\sigma {\text{ 2p}}_{\text{z}}^{\text{2}},\pi {\text{ 2p}}_{\text{y}}^1\text{ = }\pi {\text{ 2p}}_{\text{x}}^1$
Since the pi molecular orbitals contain unpaired electrons. Thus ${\text{C}}_{\text{2 }}$ the molecule is paramagnetic.
D) ${\text{B}}_{\text{2 }}$molecule: The boron atom has one electron in the 2p shell and 4 electrons in the innermost shell. The electronic configuration of nitrogen atom is as shown below,
$\text{ B = 1}{\text{s}}^{\text{2}}\text{ 2}{\text{s}}^{\text{2}}\text{ 2}{{\text{p}}^{\text{1}}}_{\text{x}}\text{= 2}{{\text{p}}^0}_{\text{y}}=\text{2}{{\text{p}}^0}_z$
The ${\text{B}}_{\text{2 }}$molecule contains a total of 10 electrons in it. These 10 electrons are arranged in the molecular orbitals.
The MOT is as shown below,

We can write the molecular orbital configuration of ${\text{B}}_{\text{2 }}$ . The molecular orbital configuration of ${\text{B}}_{\text{2 }}$ the molecule is as follows:$\sigma \text{ 1}{\text{s}}^{\text{2}}\text{,}{\sigma }^{\text{*}}\text{1}{\text{s}}^{\text{2}}\text{, }\sigma \text{ 2}{\text{s}}^{\text{2}}\text{, }{\sigma }^{\text{*}}\text{2}{\text{s}}^{\text{2}}\text{ , }\sigma {\text{ 2p}}_{\text{z}}^{\text{2}}$
Since the molecular orbitals theory does not contain the unpaired electron. Thus ${\text{B}}_{\text{2 }}$ the molecule is diamagnetic.
Thus, from here we know that only ${\text{C}}_{\text{2 }}$ molecule is paramagnetic.

Hence, (C) is the correct option.

Note: Note that, in atoms like $\text{ Li }$ , $\text{ Be }$ , $\text{ C }$ , $\text{ N }$ the energy difference is very small between the $\text{ 2s }$ and $\text{ 2p }$ orbitals. These are close to each other and thus p and s orbitals combine and lead to change in the expected order of the orbital energies. Here, we are not considering the mixing of s and p orbitals thus the carbon molecule is paramagnetic otherwise if we consider the mixing then the carbon molecule is diamagnetic.