Question

Which of the following molecules is expected to exhibit diamagnetic behaviour?
A. ${{\text{N}}_{2}}$
B. ${{\text{O}}_{2}}$
C. ${{\text{S}}_{2}}$
D. ${{\text{C}}_{2}}$

Answer 1

For this problem, we have to write the electronic configuration of the given molecule based on the Molecular orbital theory. After which we can calculate the total number of unpaired and paired electrons.

Complete answer:
- In the given question, we have to identify the molecule which will show diamagnetic behaviour.
- As we know that those elements which show diamagnetic behaviour have all the electrons in a paired form.
- If there is a presence of even one unpaired electron then the element will be considered as a paramagnetic.
- Now, the Molecular Orbital Theory, considers that the atomic orbitals combine to form the molecular orbital and the filling of the orbitals take place by bonding and antibonding interaction.
- So, in a nitrogen molecule, we know that the total atomic number of the molecule will be 7 + 7 = 14 so the electronic configuration will be:
$\text{KK (}\sigma \text{ 2s}{{\text{)}}^{2}}\text{ (}{{\sigma }^{*}}\text{ 2s}{{\text{)}}^{2}}\text{ (}\sigma \text{ 2px}{{\text{)}}^{2}}\text{ (}\pi \text{ 2py}{{\text{)}}^{2}}\text{ (}\pi \text{ 2pz}{{\text{)}}^{2}}$
- As we can see that every electron is paired due to which nitrogen will show diamagnetic behaviour.

- Now, in an oxygen molecule, we know that the total atomic number of the molecule is 8 + 8 = 16. So, the electronic configuration will be:
$\text{KK (}\sigma \text{ 2s}{{\text{)}}^{2}}\text{ (}{{\sigma }^{*}}\text{ 2s}{{\text{)}}^{2}}\text{ (}\sigma \text{ 2px}{{\text{)}}^{2}}\text{ (}\pi \text{ 2py}{{\text{)}}^{2}}\text{ (}\pi \text{ 2pz}{{\text{)}}^{2}}\text{ (}{{\pi }^{*}}\text{ 2py}{{\text{)}}^{1}}\text{ (}{{\pi }^{*}}\text{ 2pz}{{\text{)}}^{2}}$
- Now, here we can see that the electron of antibonding, 2py is unpaired due to which it will show the paramagnetic behaviour.

- Now in the Sulphur molecule we know that the total atomic number of the molecule is 16 + 16 = 32. So, the electronic configuration will be:
$\text{LL (}\sigma \text{ 3s}{{\text{)}}^{2}}\text{ (}{{\sigma }^{*}}\text{ 3s}{{\text{)}}^{2}}\text{ (}\sigma \text{ 3px}{{\text{)}}^{2}}\text{ (}\pi \text{ 3py}{{\text{)}}^{2}}\text{ (}\pi \text{ 3pz}{{\text{)}}^{2}}\text{ (}{{\pi }^{*}}\text{ 3py}{{\text{)}}^{1}}\text{ (}{{\pi }^{*}}\text{ 3pz}{{\text{)}}^{1}}$

- Now, here we can see that the electron of antibonding, 3py and 3pz are unpaired due to which it will show the paramagnetic behaviour.

- Similarly, in a carbon molecule, we know that the total atomic number of the molecule is 6 + 6 = 12. So, the electronic configuration will be:
$\text{KK (}\sigma \text{ 2s}{{\text{)}}^{2}}\text{ (}{{\sigma }^{*}}\text{ 2s}{{\text{)}}^{2}}\text{ (}\pi \text{ 2py}{{\text{)}}^{2}}\text{ (}\pi \text{ 2pz}{{\text{)}}^{2}}$

- Now, here we can see that the electron of antibonding, 3py and 3pz are unpaired due to which it will show the paramagnetic behaviour.
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So, the correct answer is “Option A and B”.

Note: Molecular orbital theory helps predict the arrangement of the electrons in a molecule. The bonding orbitals are considered as stable due to the presence of electrons between the nuclei.