JEE Main 2020 · 8 January · morning · Single correct · +4 / −1
The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are :
- (a) 496, 577, 786, 737
- (b) 496, 737, 577, 786
- (c) 786, 737, 577, 496
- (d) 496, 577, 737, 786
Answer and solution
Answer (b) 496, 737, 577, 786
Solution
Chemistry ma’am · 3 steps
Ionisation energy rises across period 3, but Al dips below Mg: its 3p electron is easier to pull out.
- General trend. Across a period the nuclear charge grows and the atom shrinks, so ionisation energy generally rises. That puts Na lowest and Si highest of these four.
- The exception. Mg (3s²) has a filled 3s subshell, which is stable and penetrates close to the nucleus. Al (3s² 3p¹) loses a 3p electron, which is higher in energy and shielded by the 3s pair, so it comes off more easily. Hence IE of Al is less than IE of Mg.
Element Electron removed from First IE (kJ mol⁻¹) Na 3s (single electron) 496 Mg 3s (filled pair) 737 Al 3p (single electron) 577 Si 3p 786 - So. The order is Na < Al < Mg < Si, and the values in the order Na, Mg, Al, Si are 496, 737, 577, 786. That is option (b).
The idea to keep. Across a period IE rises, with two dips: group 13 falls below group 2 (a p electron leaves), and group 16 falls below group 15 (a paired electron leaves).
Answer from the NTA final answer key. Solution by PrepWiser.