Question

Which of the following is diamagnetic?
[A] ${{H}_{2}}$
[B] ${{H}_{2}}^{-}$
[C] ${{H}_{2}}^{+}$
[D] $H{{e}_{2}}^{+}$

Answer 1

HINT: To solve this, firstly write down the electronic configurations of the given ions of the elements following the molecular orbital theory. For a substance to be diamagnetic, it should not have unpaired electrons. Use this to answer the question.

COMPLETE STEP BY STEP SOLUTION: To answer this question, firstly we have to understand the meaning of the term diamagnetic.
A substance which is not attracted by a magnetic field due to the absence of unpaired electrons are known as diamagnetic substances. When two electrons are paired with each other in an orbital, their total spin is zero and they repel magnetic fields.
For an element / atom / substance to own diamagnetic property, it should not have unpaired electrons.
Now, let us write down the electronic configuration of the given ions and find out if their electrons are paired or not.
Firstly we have ${{H}_{2}}$. We know that one atom of hydrogen has 1 electron. Therefore, for di-hydrogen we will have 2 electrons. So, we can write its electronic configuration as ${{\left( {{\sigma }_{1s}} \right)}^{2}}$ . We can see it has no unpaired electrons. Therefore, it is diamagnetic.
Then we have ${{H}_{2}}^{-}$. We have 2 electrons for the di-hydrogen and one additional electron as depicted by the negative charge. Therefore, configuration will be ${{\left( {{\sigma }_{1s}} \right)}^{2}}{{\left( {{\sigma }_{2s}} \right)}^{1}}$ . We can see it has 1 unpaired electron. Therefore, it is not diamagnetic.
Then we have ${{H}_{2}}^{+}$.We have 2 electrons for the di-hydrogen and one electron will be subtracted as depicted by the positive charge. Therefore, configuration will be ${{\left( {{\sigma }_{1s}} \right)}^{1}}$ . We can see it has 1 unpaired electron. Therefore, it is not diamagnetic.
And lastly, we have $H{{e}_{2}}^{+}$. We know that the atomic number of helium is 2 i.e. for every helium atom we have 2 electrons so for di-helium we have 4 electrons but 1 electron will be less due to the positive charge. Therefore, configuration will be ${{\left( {{\sigma }_{1s}} \right)}^{2}}{{\left( {{\sigma }_{2s}} \right)}^{1}}$ . We can see it has 1 unpaired electron. Therefore, it is not diamagnetic.
We can understand from the above discussion di-hydrogen has no unpaired electron.

Therefore, the correct answer is option [A] ${{H}_{2}}$.

NOTE: Presence of paired electrons makes an element diamagnetic similarly presence of unpaired electrons makes it paramagnetic. The substances which are attracted by a magnetic field are paramagnetic. Here, ${{H}_{2}}^{-}$, ${{H}_{2}}^{+}$ and $H{{e}_{2}}^{+}$ is paramagnetic.