Question:

Which of the following set of quantum numbers is not applicable for an electron in an atom?

Show Hint

For \(n=1\), only one orbital exists: \(l=0\) (1s orbital). No p, d, or f orbitals are possible.
Updated On: Apr 15, 2026
  • \(n=1, l=1, m=1, s=+\frac{1}{2}\)
  • \(n=1, l=0, m=0, s=+\frac{1}{2}\)
  • \(n=2, l=0, m=0, s=+\frac{1}{2}\)
  • \(n=2, l=0, m=0, s=-\frac{1}{2}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Concept: Allowed Quantum Numbers
• Principal quantum number: \( n = 1,2,3,\dots \)
• Azimuthal quantum number: \( l = 0 \text{ to } (n-1) \)
• Magnetic quantum number: \( m = -l \text{ to } +l \)
• Spin quantum number: \( s = \pm \frac{1}{2} \)

Step 1: Check option (A)
\[ n = 1 \Rightarrow l = 0 \text{ only} \] But given: \[ l = 1 \quad \text{(not allowed)} \] Hence, option (A) is not possible.

Step 2: Check remaining options
(B), (C), (D) all satisfy: \[ l \leq n-1,\quad m \in [-l, +l],\quad s = \pm \tfrac{1}{2} \] So they are valid. Final Answer: A
Was this answer helpful?
0
0