JEE Main 2022 · 29 July · evening · Single correct · +4 / −1

Given below are the quantum numbers for 4 electrons.
A. n=3,l=2, m₁=1, mₛ=+1 / 2
B. n=4,l=1, m₁=0, mₛ=+1 / 2
C. n=4,l=2, m₁=-2, mₛ=-1 / 2
D. n=3,l=1, m₁=-1, mₛ=+1 / 2
The correct order of increasing energy is :

  1. (a) D < B < A < C
  2. (b) D < A < B < C
  3. (c) B < D < A < C
  4. (d) B < D < C < A
Answer and solution

Answer (b) D < A < B < C

Solution

Chemistry ma’am · 3 steps

Name each orbital, then rank by n + l; when two tie, the one with smaller n comes first.

  1. What decides energy. For a multi-electron atom, orbital energy depends on n and l only. The values of mₗ and mₛ do not change it, so ignore them here.
  2. Ranking rule. A lower n + l means lower energy. If two orbitals have the same n + l, the one with the smaller n has lower energy.
    ElectronOrbitaln + lRank
    D3p3 + 1 = 4lowest
    A3d3 + 2 = 5, n = 3second
    B4p4 + 1 = 5, n = 4third
    C4d4 + 2 = 6highest
  3. So. A and B tie at n + l = 5, and 3d has the smaller n, so it sits below 4p. The order is D < A < B < C, which is option (b).

The idea to keep. Only n and l set orbital energy. Rank by n + l first, and break a tie by giving the lower energy to the smaller n.

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

Study Atomic Structure

Asked about 170 times in JEE Main since 2020. This question is from 2022.