Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $$(B \lt Al), (Al \lt Ga), (Ga \lt In)$$ and $$(In \lt Tl)$$. Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius ($$M^{3+}$$) than the other one. The atomic number of the element (X) is

The five members of group 13 are arranged in the order of increasing atomic number as
$$B\;(Z = 5),\;Al\;(13),\;Ga\;(31),\;In\;(49),\;Tl\;(81).$$

In any group the atomic radius normally increases downwards because a new electron shell is added at every step. However, in the case of gallium the presence of ten 3d-electrons offers poor shielding to the valence electrons. The higher effective nuclear charge produced by this poor shielding makes the atomic radius of gallium slightly smaller than that of aluminium. This anomaly is called d-block contraction.

Using this idea we examine the four relations supplied in the question:

$$\text{(i)}\;B \lt Al,\qquad \text{(ii)}\;Al \lt Ga,\qquad \text{(iii)}\;Ga \lt In,\qquad \text{(iv)}\;In \lt Tl.$$

Relations (i), (iii) and (iv) are true, but relation (ii) is wrong because the actual order is
$$Ga \lt Al.$$

Thus the incorrect pair is $$Al \text{ and } Ga.$$

Next we compare the ionic (M3+) radii of these two elements. Standard ionic-radius data (coordination number 6) give
$$r_{Al^{3+}} \approx 0.50\;\text{Å},\qquad r_{Ga^{3+}} \approx 0.62\;\text{Å}.$$

Therefore gallium forms the larger $$M^{3+}$$ ion. Hence the required element X is gallium, whose atomic number is $$31.$$

Option A is correct.