Concept: The last (valence) electron in a Halogen occupies a \( p \)-orbital.
Formula: None required.
Solution: - Chlorine has an atomic number \( Z=17 \).
- The full electronic configuration is \( 1s^2, 2s^2, 2p^6, 3s^2, 3p^5 \).
- The valence subshell is \( 3p \). Expanding this using Hund's Rule gives: \( 3p_x^2, 3p_y^2, 3p_z^1 \).
- The single unpaired electron is in the \( 3p \) orbital.
- For a \( 3p \) orbital, the principal quantum number \( n = 3 \).
- The azimuthal quantum number for a \( p \)-orbital is \( l = 1 \).
- The magnetic quantum number \( m \) can be \( -1, 0, \) or \( +1 \). Based on the provided options, \( (3, 1, 1) \) is the only valid set that correctly describes a \( 3p \) orbital.
Why other options are incorrect: Options A and B point to the second shell (\( n=2 \)), which is fully filled in Chlorine. Option D points to a \( 3s \) orbital, which is fully paired (\( 3s^2 \)).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.