Arrange the following orbitals in decreasing order of energy.
A. $\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0$
B. $\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0$
C. $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0$
D. $\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1$
The correct option for the order is :
Solution
(A) n = 3; l = 0; m = 0 ; 3s orbital
<br/><br/>(B) n = 4; l = 0; m = 0 ; 4s orbital
<br/><br/>(C) n = 3; l = 1; m = 0 ; 3p orbital
<br/><br/>(D) n = 3; l = 2; m = 0 ; 3d orbital
<br/><br/>As per Hund’s ruleo of energy is given by (n + l) value.
If value of (n + l) remains same then energy is given
by n only.
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Bohr's Model
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