JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
The electron gain enthalpy (in KJ/mol) of fluorine, chlorine, bromine and iodine, respectively are:
- (a) -296, -325, -333 and -349
- (b) 349, -333, -325 and -296
- (c) -333, -349, -325 and -296
- (d) -333, -325, -349 and -296
Answer and solution
Answer (c) -333, -349, -325 and -296
Solution
Chemistry ma’am · 3 steps
Chlorine, not fluorine, has the most negative value. Spot that and the options sort themselves out.
- Sign first. A halogen atom is one electron short of a noble gas configuration, so adding an electron releases energy. All four values must be negative. That rules out (b), which gives fluorine +349.
- Why fluorine dips. Going down the group, the atom gets bigger and the added electron is held less tightly, so the value becomes less negative from Cl to I. Fluorine breaks this trend. Its 2p subshell is very small, and the incoming electron is strongly repelled by the electrons already there, so less energy is released than for chlorine.
Element Electron gain enthalpy (kJ mol⁻¹) Reason F -333 Small 2p shell, strong electron–electron repulsion Cl -349 Most negative: larger 3p shell, less repulsion Br -325 Bigger atom, weaker hold on the added electron I -296 Biggest atom, weakest hold - So. The order is F -333, Cl -349, Br -325, I -296, which is option (c).
The idea to keep. In the halogens, Cl has the most negative electron gain enthalpy. F loses out because its tiny 2p shell repels the incoming electron. After Cl, the values become less negative down the group.
Answer from the NTA final answer key. Solution by PrepWiser.