Question

Which one of the following oxides is expected to exhibit paramagnetic behavior?
(A)- $C{O_2}$
(B)- $Si{O_2}$
(C)- $S{O_2}$
(D)- $Cl{O_2}$

Answer 1

In chemistry, paramagnetic is a type of behavior of compounds in which the central atom has an odd number of electrons in the valence shell. This type of question can be solved by finding out the number of electrons in the central atom of the given compound.

Complete step by step answer:
Let's talk about each option and find out which one exhibits paramagnetic behavior.
- $C{O_2}$ : Carbon dioxide is a simple covalent molecule that most people have heard about. Carbon dioxide has the formula $C{O_2}$ and at the center of this molecule is a carbon atom joined by two pairs of double bonds to the oxygen atoms in the manner $O = C = O$. At room temperature, carbon dioxide is a colorless gas and it is also a linear molecule with a bond angle of 180°. Carbon has 4 electrons in its valence shell which is bonded by two of the given oxygens. So, in result it doesn’t have any valence electron left. Therefore, it won't show paramagnetic behavior.
- $Si{O_2}$ : Silicon also comes from the carbon family as it lies below carbon in the periodic table so it also has 4 electrons so it has the similar structure as carbon dioxide. Therefore, it also doesn’t exhibit paramagnetic character.
- $S{O_2}$ : We know that the shape which minimizes the repulsions of electronics pairs is adopted by the molecule to form the structure which minimizes their energy, increasing stability. The molecular shape of $S{O_2}$ is the same as the molecular geometry of Carbon Dioxide i.e. linear. Sulphur has six electrons and oxygen has four of them among which one electron is used for each bond. So, there are a total number of ten electrons in five pairs. To make bonds, four pairs are needed, so one pair remains alone which acts as a lone pair but there is not any unpaired electron. Therefore, it also won't exhibit paramagnetic character.
- $Cl{O_2}$ : In this structure, chlorine has 7 valence electrons and after making bonds with two oxygen it will be left with 3 electrons in the valence shell. So, it will exhibit paramagnetic character.

Therefore, the correct answer of the above question is option (D)- $Cl{O_2}$.

Note: Always remember the difference between paramagnetic and diamagnetic character. Paramagnetic compounds have unpaired electrons in their valence shell while in diamagnetic compounds the electrons all have paired spins,