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JEE Main 202227 JulyEvening Shift

Exam: JEE Main
Year: 2022
Date: 27 July
Shift: Evening Shift
Subject: Chemistry
Chapter: Structure of Atom

JEE Main 2022 · 27 July · Evening Shift · Chemistry · Structure of Atom

The correct decreasing order of energy, for the orbitals having,...

The correct decreasing order of energy, for the orbitals having, following set of quantum numbers:

(A) $n = 3, 1 = 0, m = 0$

(B) $n = 4, l = 0, m = 0$

(C) $n = 3, l = 1, m = 0$

(D) $n = 3, 1 = 2, m = 1$

Options

A $\left(D\right) > \left(B\right) > \left(C\right) > \left(A\right)$
B $\left(B\right) > \left(D\right) > \left(C\right) > \left(A\right)$
C $\left(C\right) > \left(B\right) > \left(D\right) > \left(A\right)$
D $\left(B\right) > \left(C\right) > \left(D\right) > \left(A\right)$

Correct answer

A. $\left(D\right) > \left(B\right) > \left(C\right) > \left(A\right)$

Step-by-step solution

(A) $n + l = 3 + 0 = 3⇒ 3s$
(B)~ $n + l = 4 + 0 = 4⇒ 4s$
(C) $n + l = 3 + 1 = 4⇒ 3p$
(D)~ $n + l = 3 + 2 = 5⇒ 3d$
Higher the~ $n + l$ value, higher the energy of the orbital and if two orbitals have same $n + l$ ~value, then the orbital with higher $n$ value has higher~energy.
Thus:& $D > B > C > A$ .

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