JEE Main 2023 · 31 January · evening · Single correct · +4 / −1

Arrange the following orbitals in decreasing order of energy.
A. n=3, l=0, m=0
B. n=4, l=0, m=0
C. n=3, l=1, m=0
D. n=3, l=2, m=1
The correct option for the order is :

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

Answer (d) D>B>C>A

Solution

Chemistry ma’am · 3 steps

Turn each set of quantum numbers into an orbital name, then rank them by the (n + l) rule.

  1. Name the orbitals. l = 0 is s, l = 1 is p and l = 2 is d. So A is 3s, B is 4s, C is 3p and D is 3d. The value of m only picks one orbital within a subshell, so it does not change the energy.
  2. Aufbau energy rule. A larger (n + l) means higher energy. If two orbitals have the same (n + l), the one with the larger n has the higher energy.
    LabelOrbitaln + lRank
    D3d3 + 2 = 5highest
    B4s4 + 0 = 4, n = 4second
    C3p3 + 1 = 4, n = 3third
    A3s3 + 0 = 3lowest
  3. So. 4s and 3p tie at 4, and 4s has the larger n, so it is higher. The order is D > B > C > A, which is option (d).

The idea to keep. For orbital energies, add n + l first. When there is a tie, the larger n wins. Ignore m: every orbital in a subshell has the same energy.

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

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Asked about 170 times in JEE Main since 2020. This question is from 2023.