JEE Main 2024 — Chemistry Physical Chemistry

2024mcqeasy

Compare the energies of following sets of quantum numbers for multielectron system. (A) $\mathrm{n}=4,1=1$ (B) $\mathrm{n}=4, \mathrm{l}=2$ (C) $\mathrm{n}=3,1=1$ (D) $\mathrm{n}=3,1=2$ (E) $\mathrm{n}=4,1=0$ Choose the correct answer from the options given below :

Official previous-year question

Held on 9 Apr 2024 · Verified 6 Jul 2026.

Options

A
$\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }$
B
$\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })$
C
$\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)$
D
$\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }$

Did you get this right?

Solution

Energy level can be determined by comparing ( $\mathrm{n}+\ell$ ) values (A) $\mathrm{n}=4, \ell=1 \Rightarrow(\mathrm{n}+\ell)=5$ (B) $\mathrm{n}=4, \ell=2 \Rightarrow(\mathrm{n}+\ell)=6$ (C) $\mathrm{n}=3, \ell=1 \Rightarrow(\mathrm{n}+\ell)=4$ (D) $\mathrm{n}=3, \ell=2 \Rightarrow(\mathrm{n}+\ell)=5$ (E) $\mathrm{n}=4, \ell=0 \Rightarrow(\mathrm{n}+\ell)=4$

For same value of $(n+\ell)$ , orbital having higher value of $n$ , will have more energy. (B) $>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

Options

$\text { (B) } > \text { (A) } > \text { (C) } > \text { (E) } > \text { (D) }$

$\text { (E) } < (\text { C }) < \text { (D) } < (\text { A }) < (\text { B })$

$\text { (E) } > (\text { C }) > (\text { A }) > (\text { D }) > (B)$

$\text { (C) } < \text { (E) } < \text { (D) } < \text { (A) } < \text { (B) }$

Solution

For same value of $(n+\ell)$ , orbital having higher value of $n$ , will have more energy. (B) $>(\mathrm{A})>(\mathrm{D})>(\mathrm{E})>(\mathrm{C})$

JEE Main 2024 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2024 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry