Question

Which of the following are diamagnetic?
A. ${{C}_{2}}$
B. $O_{2}^{2-}$
C. $L{{i}_{2}}$
D. $N_{2}^{+}$

Answer 1

We should know the molecular electronic configuration to know about the paramagnetic and diamagnetic behavior of the molecules. If the molecules contain unpaired electrons then the molecule is paramagnetic in nature.

Complete answer:
- In the question it is given to find the diamagnetic molecules among the given options.
- Coming to option A, ${{C}_{2}}$.
- The electronic configuration of carbon atoms is $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$ .
- The molecular electronic configuration of ${{C}_{2}}$ molecule is $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}{{(\pi 2{{p}_{y}})}^{2}}={{(\pi 2{{p}_{z}})}^{2}}$ .
- We can see that all the electrons in the molecular electronic configuration of ${{C}_{2}}$ are paired, then ${{C}_{2}}$ is diamagnetic in nature.
- Coming to option B, $O_{2}^{2-}$ .
- The electronic configuration of oxygen atom is $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$.
- The molecular electronic configuration of $O_{2}^{2-}$ is $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}{{(\pi 2{{p}_{y}})}^{2}}={{(\pi 2{{p}_{z}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{2}}={{({{\pi }^{*}}2{{p}_{z}})}^{2}}$
- We can see that all the electrons in the molecular electronic configuration of $O_{2}^{2-}$ are paired, then $O_{2}^{2-}$ is diamagnetic in nature.
- Coming to option C, $L{{i}_{2}}$.
- The electronic configuration of lithium atoms is $1{{s}^{2}}2{{s}^{1}}$.
- The molecular electronic configuration of $L{{i}_{2}}$ is $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}$ .
- We can see that all the electrons in the molecular electronic configuration of $L{{i}_{2}}$ are paired, then $L{{i}_{2}}$ is diamagnetic in nature.
- Coming to option D, $N_{2}^{+}$ .
- The electronic configuration of nitrogen atom is $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$ .
- The molecular electronic configuration of $N_{2}^{+}$ is $sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{2}}\sigma 2{{s}^{2}}{{\sigma }^{*}}2{{s}^{2}}{{(\pi 2{{p}_{y}})}^{2}}={{(\pi 2{{p}_{z}})}^{2}}{{(\sigma 2{{p}_{x}})}^{1}}$
- We can see that one electron in the molecular electronic configuration of $N_{2}^{+}$ in $(\sigma 2{{p}_{x}})$ is not paired, then $N_{2}^{+}$ is paramagnetic in nature.

Therefore the molecules in options A, B, and C show diamagnetic behavior.

Note:
If the electrons are not paired means unpaired electrons are present in the electronic configuration of any molecule then the molecule will be paramagnetic in nature and they are colored also in nature.