The correct orders among the following are
Atomic radius : B<Al<Ga<In<Tl
Electronegativity : Al<Ga<In<Tl<B
Density : Tl<In<Ga<Al<B
1st Ionisation Energy : In<Al<Ga<Tl<B
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 4 steps

In group 13, Ga and Tl break the smooth trends, so check each order against the actual data.

1. Poor shielding. Ga comes right after the ten 3d electrons, and Tl comes after the 4f and 5d electrons. These inner d and f electrons shield the nucleus poorly. So the outer electrons in Ga and Tl feel a larger effective nuclear charge than the usual trend predicts.

   Statement	Order given	Actual values (NCERT)	Verdict
   A: Atomic radius	B < Al < Ga < In < Tl	B 85, Al 143, Ga 135, In 167, Tl 170 pm	Wrong: Ga < Al
   B: Electronegativity	Al < Ga < In < Tl < B	Al 1.5, Ga 1.6, In 1.7, Tl 1.8, B 2.0	Correct
   C: Density	Tl < In < Ga < Al < B	B 2.35, Al 2.70, Ga 5.90, In 7.31, Tl 11.85 g cm⁻³	Wrong: the order is reversed
   D: 1st ionisation energy	In < Al < Ga < Tl < B	In 558, Al 577, Ga 579, Tl 589, B 801 kJ mol⁻¹	Correct

2. A and C fail. Poor shielding by the 3d electrons pulls Ga's outer shell in, so Ga is smaller than Al. Density increases down the group, because the atomic mass rises much faster than the atomic volume. So the order in C is upside down.
3. D holds. The same poor shielding makes Ga's ionisation energy slightly higher than Al's. The lanthanoid contraction makes Tl's higher than In's. B is small, so its value is far above the rest.
4. So. Only statements B and D are correct, which is option (a).

The idea to keep. In group 13, expect Ga and Tl to break the trends. Poor d and f shielding makes Ga smaller than Al and raises the ionisation energies of Ga and Tl. Density rises steadily down the group.

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