Question

Given below are the quantum numbers for $4$ electrons.

A. $\text{n}=3,l=2,{\text{m}}_1=1,{\text{m}}_{\text{s}}=+1/2$

B. $\text{n}=4,l=1,{\text{m}}_1=0,{\text{m}}_{\text{s}}=+1/2$

C. $\text{n}=4,l=2,{\text{m}}_1=-2,{\text{m}}_{\text{s}}=-1/2$

D. $\text{n}=3,l=1,{\text{m}}_1=-1,{\text{m}}_{\text{s}}=+1/2$

The correct order of increasing energy is :

• A. D $<$ A $<$ B $<$ C
• B. D $<$ B $<$ A $<$ C
• C. B $<$ D $<$ A $<$ C
• D. B $<$ D $<$ C $<$ A

Answer 1

The correct option is A D $<$ A $<$ B $<$ C

Energy order of subshells is decided by $(n+\ell )$ rule.
The higher the value of n + l, the more the energy.
If any 2 electrons have same value of n + l, then the electron with higher value of n will have greater energy.

$\begin{array}{c}\text{A}\Rightarrow 3\text{d}\Rightarrow \text{n}+1=3+2=5\\ \text{B}\Rightarrow 4\text{p}\Rightarrow \text{n}+\ell =4+1=5\\ \text{C}\Rightarrow 4\text{d}\Rightarrow \text{n}+\ell =4+2=6\\ \text{D}\Rightarrow 3\text{p}\Rightarrow (\text{n}+\ell )=3+1=4\\ \\ \text{So order will be:}\\ \text{D}<\text{A}<\text{B}<\text{C}\end{array}$