JEE Main 2026 — Jan 23, Shift 1 — Chemistry solution

JEE Main2026Jan 23, Shift 1Chemistry

Given,

(A) $n = 5, m_{l} = - 1$

(B) $n = 3, l = 2, m_{l} = - 1, m_{s} = + \frac{1}{2}$

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :

A
26 and 1
B
4 and 1
C
2 and 4
D
8 and 1

Solution

(A): $n = 5$ , $m_{l} = - 1$ . For each allowed $l$ (which must satisfy $∣ m_{l} ∣ \leq l$ , so $l = 1, 2, 3, 4$ ), $m_{l} = - 1$ designates exactly one orbital. Each orbital holds 2 electrons.

Number of orbitals with $m_{l} = - 1$ at $n = 5$ is 4 (one each in p, d, f, g sub-shells), so total $= 4 \times 2 = 8$ electrons.

(B): All four quantum numbers are fixed, so the Pauli principle allows exactly 1 electron.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →