Given below are the quantum numbers for 4 electrons.
A. $\mathrm{n}=3, l=2, \mathrm{~m}_{\square}=1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
B. $\mathrm{n}=4, l=1, \mathrm{~m}_{\mathrm{a}}=0, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
C. $\mathrm{n}=4, l=2, \mathrm{~m}_{\mathrm{a}}=-2, \mathrm{~m}_{\mathrm{s}}=-1 / 2$
D. $\mathrm{n}=3, l=1, \mathrm{~m}_{\mathrm{a}}=-1, \mathrm{~m}_{\mathrm{s}}=+1 / 2$
The correct order of increasing energy is :
Select the correct option:
A
D < B < A < C
B
D < A < B < C
C
B < D < A < C
D
B < D < C < A
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Energy order of subshell decided by ( $n+\square$ ) rule.
$$
\begin{aligned}
& A \Rightarrow 3 d \Rightarrow n+l=5 \\
& B \Rightarrow 4 p \Rightarrow n+l=5 \\
& C \Rightarrow 4 d \Rightarrow n+l \Rightarrow 6
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{D} \Rightarrow 3 \mathrm{~s} \Rightarrow(\mathrm{n}+\mathrm{l})=4 \\
& \mathrm{D}<\mathrm{A}<\mathrm{B}<\mathrm{C}
\end{aligned}
$$
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