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JEE Main 2024 · ChemistryMultiple choiceSingle correctEasyApplication

JEE Main 31 January 2024, Shift 2, Chemistry Q62

Question 62 of 90 in this shift, Chemistry question 2 of 30, Section A.

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

(1) $n=2,\ l=0,\ m=0,\ s=+\frac{1}{2}$
(2) $n=3,\ l=0,\ m=1,\ s=+\frac{1}{2}$
(3) $n=4,\ l=0,\ m=0,\ s=+\frac{1}{2}$ Official answer
(4) $n=4,\ l=2,\ m=-1,\ s=+\frac{1}{2}$

Official answer

Option 3

NTA final key.

Chapter: Structure of Atom
Topic: Quantum Numbers
Idea tested: Quantum Numbers of the Differentiating Electron

Same idea in other shifts

Asked 2× in all

29 Jan 2024, Shift 1 · Q61The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is :EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.