Question

Among the following species, the diamagnetic molecule is:
  (A) ${{O}_{2}}$
  (B) $NO$
  (C) ${{B}_{2}}$
  (D) $CO$

Answer 1

Hint: From the molecular orbital theory we can write the electronic configuration of molecular orbitals and we could identify the paramagnetic and diamagnetic species. Diamagnetic species have all the electrons paired and the paramagnetic species will have unpaired electrons in it.

Complete step by step solution:
- In the same way that the distribution of electrons in atoms is described using atomic orbitals, the Molecular orbital theory defines the distribution of electrons in molecules. From the MO theory we could find the electronic configuration of molecules, bond order of molecules, magnetic behaviour etc.
- Lets understand the concept of paramagnetism and diamagnetism. The diamagnetic species will have all its electrons paired up in any subshell of the orbital while the paramagnetic species will have unpaired electrons in the orbitals.
- Let’s write the electronic configuration or the molecular orbital configuration of all the given species and check whether the species contains unpaired electrons or not. The first one is oxygen molecule (${{O}_{2}}$) (16 electrons) and its configuration can be written as
\[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{z}^{2},\Pi 2p_{x}^{2},\Pi 2p_{y}^{2},{{\Pi }^{*}}2p_{x}^{1},{{\Pi }^{*}}2p_{y}^{1}\]
As we can see there are two unpaired electrons in ${{O}_{2}}$ and hence ${{O}_{2}}$ is paramagnetic.
 - The second one is nitric oxide ( $NO$) molecule (15 electrons) and its configuration can be written as follows
\[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{x}^{2},\Pi 2p_{y}^{2},\Pi 2p_{z}^{2},{{\Pi }^{*}}2p_{y}^{1}\] $NO$ also contains unpaired electrons and thus $NO$ is paramagnetic.
- The third one is ${{B}_{2}}$ molecule (10 electrons) and its configuration can be written as
\[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{x}^{1},\Pi 2p_{y}^{1}\]
${{B}_{2}}$ molecule also contains unpaired electrons and hence is paramagnetic.
- The next one is $CO$(14 electrons) and its configuration can be written as
\[\sigma 1{{s}^{2}},{{\sigma }^{*}}1{{s}^{2}},\sigma 2{{s}^{2}},{{\sigma }^{*}}2{{s}^{2}},\sigma 2p_{x}^{2},\Pi 2p_{y}^{2},\Pi 2p_{z}^{2}\]. All the electrons in a carbon monoxide molecule are paired and thus it’s diamagnetic in nature.


Therefore the answer is option (D). $CO$


Note: Keep in mind that if we have only one atom, then we are talking about atomic orbitals and its subshells, shells etc. are filled accordingly whereas in the case of molecules which consists of minimum two atoms we will be dealing with molecular orbitals. Also the paired or unpaired electrons are decided in the basis of filling up of orbitals