JEE Main 2024 · 31 January · evening · Single correct · +4 / −1

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

  1. (a) n=3, l=0, m=1, s=+1/2
  2. (b) n=4, l=0, m=0, s=+1/2
  3. (c) n=2, l=0, m=0, s=+1/2
  4. (d) n=4, l=2, m=-1, s=+1/2
Answer and solution

Answer (b) n=4, l=0, m=0, s=+1/2

Solution

Chemistry ma’am · 3 steps

Write potassium's configuration first. The last electron sits in 4s, and the four numbers follow from that.

  1. Configuration. Potassium has 19 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. The 4s fills before 3d because 4s is lower in energy, so the outermost electron is the single 4s electron.
  2. Read the numbers. For a 4s electron, n = 4. An s orbital has l = 0, and l = 0 allows only m = 0. A lone electron in an orbital is taken as spin up, s = +1/2.
    OptionOrbital it describesVerdict
    (a) n=3, l=0, m=1none, because m must be 0 when l=0not allowed
    (b) n=4, l=0, m=04soutermost electron of K
    (c) n=2, l=0, m=02sinner shell
    (d) n=4, l=2, m=-14dempty in K
  3. So. The outermost electron of potassium has n = 4, l = 0, m = 0, s = +1/2, which is option (b).

The idea to keep. To get quantum numbers, write the configuration and find the last orbital: n is the shell, l comes from the letter (s = 0, p = 1, d = 2), and m runs only from -l to +l.

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 2024.