JEE Chemistry 2023 Question with Solution

Given: The set contains $\mathrm{O^{2-}},\ \mathrm{F^-},\ \mathrm{Al},\ \mathrm{Mg^{2+}},\ \mathrm{Na^+},\ \mathrm{O},\ \mathrm{Mg},\ \mathrm{Al^{3+}},\ \mathrm{F}$.

Find: The total number of isoelectronic species.

Isoelectronic species are atoms or ions that have the same number of electrons.

Calculate the number of electrons in each species:

$$\begin{aligned} \mathrm{O^{2-}} &: 8 + 2 = 10 \\ \mathrm{F^-} &: 9 + 1 = 10 \\ \mathrm{Na^+} &: 11 - 1 = 10 \\ \mathrm{Mg^{2+}} &: 12 - 2 = 10 \\ \mathrm{Al^{3+}} &: 13 - 3 = 10 \\ \mathrm{O} &: 8 \\ \mathrm{F} &: 9 \\ \mathrm{Mg} &: 12 \\ \mathrm{Al} &: 13 \end{aligned}$$

Thus, the species having $10$ electrons are:

$$\mathrm{O^{2-}},\ \mathrm{F^-},\ \mathrm{Na^+},\ \mathrm{Mg^{2+}},\ \mathrm{Al^{3+}}$$

Therefore, the total number of isoelectronic species is $5$.

Given: You need to identify species with the same number of electrons.

Find: How many species belong to one isoelectronic group.

Use the rule: add electrons for a negative charge and subtract electrons for a positive charge. Checking only the charged species:

$$\mathrm{O^{2-}} = 10,\ \mathrm{F^-} = 10,\ \mathrm{Na^+} = 10,\ \mathrm{Mg^{2+}} = 10,\ \mathrm{Al^{3+}} = 10$$

These five form the isoelectronic set. Therefore, the answer is $5$.