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Chemistry · Structure of Atom

JEE Main 2024 — 31 January, Shift 2 — Question 69

The four quantum numbers for the electron in the outermost orbital of potassium (Z=19Z=19) are:

  1. Option A:

    n=4,l=2, m=−1,s=+12\mathrm{n}=4, l=2, \mathrm{~m}=-1, s=+\frac{1}{2}

  2. Option B:

    n=4,l=0, m=0,s=+12\mathrm{n}=4, l=0, \mathrm{~m}=0, s=+\frac{1}{2}

    Correct
  3. Option C:

    n=3,l=0, m=1,s=+12\mathrm{n}=3, l=0, \mathrm{~m}=1, s=+\frac{1}{2}

  4. Option D:

    n=2,l=0, m=0,s=+12\mathrm{n}=2, l=0, \mathrm{~m}=0, s=+\frac{1}{2}

Answer: B

Step-by-step solution

Potassium (Z=19Z=19)

Electronic configuration: [Ar] 4s1[Ar]\,4s^1

The outermost electron is in 4s orbital.

For an s orbital, we have

ℓ=0\ell = 0 mℓ=0m_\ell = 0

Since it is a single valence electron, thus s=+12s = +\frac{1}{2}

Therefore the quantum numbers are n=4, ℓ=0, mℓ=0, s=+12n = 4,\ \ell = 0,\ m_\ell = 0,\ s = +\frac{1}{2}.

Hence, the given option B is correct.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
31 January, Shift 2
Subject
Chemistry
Chapter
Structure of Atom
Topic
Quantum Numbers and Sommerfeld
The four quantum numbers for the electron in the outermost orbital of… | JEE Main 2024 PYQ with Solution · DhiX AI