JEE Chemistry 2025 Question with Solution

Given: The pairs are $(\mathrm{B}<\mathrm{Al})$, $(\mathrm{Al}<\mathrm{Ga})$, $(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ in terms of atomic radius.

Find: The incorrect pair, then the element in that pair having larger ionic radius in the $\mathrm{M}^{3+}$ state, and finally its atomic number.

As we move down group $13$, atomic radius generally increases. However, due to poor shielding by $d$-electrons, an anomaly appears between aluminum and gallium.

From the given pairs:

1. $(\mathrm{B}<\mathrm{Al})$ is correct.
2. $(\mathrm{Al}<\mathrm{Ga})$ is incorrect.
3. $(\mathrm{Ga}<\mathrm{In})$ is correct.
4. $(\mathrm{In}<\mathrm{Tl})$ is correct.

So, the incorrect pair is:

Now compare the ionic radii of their $3+$ ions:

Hence, the element $X$ is gallium.

The atomic number of gallium is $31$. Therefore, the correct option is A.

Given: A comparison of atomic radii in group $13$.

Find: The atomic number of the element with larger ionic radius in the incorrect pair.

The only exceptional comparison in group $13$ is between aluminum and gallium because of $d$-block contraction. Therefore:

is the incorrect pair.

For the ions, the solution gives:

So the required element is gallium, whose atomic number is $31$. The correct option is A.