Question

The magnetic behavior of Li₂O, Na₂O₂ and KO₂, respectively, are

• A. Paramagnetic, paramagnetic and diamagnetic
• B. Diamagnetic, diamagnetic and paramagnetic
• C. Paramagnetic, diamagnetic and paramagnetic
• D. Diamagnetic, paramagnetic and diamagnetic

Hint:

For magnetic properties of oxides:
   O\(^{2-}\) and O\(_2^{2-}\) ions are diamagnetic due to paired electrons.
   O\(_2^-\) ions are paramagnetic due to the presence of unpaired electrons.

Answer 1

The Correct Option is B

To determine the magnetic behavior of Li₂O, Na₂O₂, and KO₂, we need to understand their electron configurations and the nature of their chemical bonds.

1. Li₂O: Lithium oxide (Li₂O) is an ionic compound with lithium and oxygen. Lithium has a +1 oxidation state, while oxygen is in a -2 oxidation state. The oxygen in Li₂O has a closed-shell configuration, which means all the electrons are paired. Therefore, Li₂O is diamagnetic.
2. Na₂O₂: This compound is known as sodium peroxide. In Na₂O₂, sodium has a +1 oxidation state, and the peroxide ion (O₂^2-) involves a pair of oxygen atoms with a -1 oxidation state each. In the peroxide ion, the oxygen atoms share electrons to form a bond, resulting in paired electrons in the molecular orbital. As all electrons are paired, Na₂O₂ is diamagnetic.
3. KO₂: Potassium superoxide (KO₂) contains the superoxide anion (O₂^-), in which there is one unpaired electron. The presence of an unpaired electron means that KO₂ exhibits paramagnetism, allowing it to be attracted by a magnetic field.

Given this analysis, the correct answer is: Diamagnetic, diamagnetic, and paramagnetic.

Hence, the correct option is Diamagnetic, diamagnetic and paramagnetic.