Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. (B<Al),(Al<Ga),(Ga<In) and (In<Tl) Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius (M³⁺) than the other one. The atomic number of the element (X) is

Solution

Chemistry ma’am · 3 steps

Gallium is the odd one out: smaller than Al as an atom, but its 3+ ion is bigger than Al³⁺.

1. Check each pair. Atomic radius generally grows down group 13, but the 3d electrons in Ga shield the nuclear charge poorly. So Ga's outer electrons are pulled in harder.

   Pair	Atomic radii (pm)	Given relation
   B, Al	85 and 143	B < Al, correct
   Al, Ga	143 and 135	Al < Ga, incorrect
   Ga, In	135 and 167	Ga < In, correct
   In, Tl	167 and 170	In < Tl, correct

2. Compare the ions. Al³⁺ has the configuration [Ne], with its outermost electrons in n = 2. Ga³⁺ is [Ar] 3d¹⁰, with its outermost electrons in n = 3. One more shell makes Ga³⁺ (62 pm) larger than Al³⁺ (53.5 pm).
3. So. The incorrect pair is (Al, Ga), and X is gallium, atomic number 31, option (c).

The idea to keep. Ga breaks the group 13 size trend for atoms because of poor 3d shielding. Its M³⁺ ion still follows the normal trend, because it has an extra shell compared with Al³⁺.

Answer from the NTA final answer key. Solution by PrepWiser.