JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
Given below are two statements :
Statement I : The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga.
Statement II : The d orbitals in Ga are completely filled.
In the light of the above statements, choose the most appropriate answer from the options given below
- (a) Statement I is correct but statement II is incorrect
- (b) Statement I is incorrect but statement II is correct
- (c) Both the statements I and II are correct
- (d) Both the statements I and II are incorrect
Answer and solution
Answer (c) Both the statements I and II are correct
Solution
Chemistry ma’am · 4 steps
Put the three IE values side by side, then ask what sits between Al and Ga: ten 3d electrons.
- Look at the numbers. The first ionisation enthalpies are B 801, Al 577 and Ga 579 kJ mol⁻¹. Going from B to Al, the value falls by 224 kJ mol⁻¹. Going from Al to Ga, it hardly changes; it even rises by 2 kJ mol⁻¹.
Element Configuration Δᵢ H₁ (kJ mol⁻¹) Change from the one above B [He] 2s² 2p¹ 801 none, first in group Al [Ne] 3s² 3p¹ 577 large drop of 224 Ga [Ar] 3d¹⁰ 4s² 4p¹ 579 almost no change - Statement I. The drop from B to Al is large and the change from Al to Ga is tiny. So the statement is correct.
- Statement II. Ga comes right after the first transition series, so its 3d subshell is completely filled (3d¹⁰). These d electrons shield the outer 4p electron poorly. The extra nuclear charge is felt strongly, and this cancels the effect of the larger size. The statement is correct, and it also explains Statement I.
- So. Both statements are correct, which is option (c).
The idea to keep. Down a group, IE falls as atoms grow. But when a filled d (or f) subshell comes in between, its poor shielding lets the nucleus hold on harder, and the drop stalls or even reverses.
Answer from the NTA final answer key. Solution by PrepWiser.