Question:
Quantum number sets of four electrons I, II, III, IV are given below.
I. $n=3$, $l=1$, $m_{l}=-1$, $m_{s}=+\frac{1}{2}$
II. $n=4$, $l=1$, $m_{l}=0$, $m_{s}=+\frac{1}{2}$
III. $n=4$, $l=2$, $m_{l}=-2$, $m_{s}=+\frac{1}{2}$
IV. $n=3$, $l=2$, $m_{l}=-1$, $m_{s}=-\frac{1}{2}$ The correct order of the energy of these electrons is
Quantum number sets of four electrons I, II, III, IV are given below.
I. $n=3$, $l=1$, $m_{l}=-1$, $m_{s}=+\frac{1}{2}$
II. $n=4$, $l=1$, $m_{l}=0$, $m_{s}=+\frac{1}{2}$
III. $n=4$, $l=2$, $m_{l}=-2$, $m_{s}=+\frac{1}{2}$
IV. $n=3$, $l=2$, $m_{l}=-1$, $m_{s}=-\frac{1}{2}$ The correct order of the energy of these electrons is
Show Hint
Always sum $n + l$ first to evaluate relative subshell stability. If tied, the electron configuration with the larger $n$ wins the higher energy slot.
Updated On: Jun 3, 2026
- $I > IV > II > III$
- $III > II > IV > I$
- $III > IV > II > I$
- $III > IV > I > II$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
According to the Bohr-Bury $(n+l)$ rule, the subshell with a higher value of $(n+l)$ has higher energy. If two subshells have the same $(n+l)$ value, the one with the larger primary quantum number ($n$) possesses higher energy.
Step 2: Meaning
We identify the subshells using the angular momentum quantum number $l$: $l=0$ is s, $l=1$ is p, and $l=2$ is d.
Step 3: Analysis
Let us evaluate each electron set: * **I:** $n=3, l=1 \rightarrow 3p \rightarrow n+l = 3+1 = 4$ * **II:** $n=4, l=1 \rightarrow 4p \rightarrow n+l = 4+1 = 5$ * **III:** $n=4, l=2 \rightarrow 4d \rightarrow n+l = 4+2 = 6$ * **IV:** $n=3, l=2 \rightarrow 3d \rightarrow n+l = 3+2 = 5$ Comparing values: III has the highest value (6). For II and IV, both have $(n+l) = 5$, but II ($n=4$) has greater energy than IV ($n=3$). Electron I has the lowest value (4).
Step 4: Conclusion
Sorting them from highest to lowest energy yields $III > II > IV > I$.
Final Answer: (B)
According to the Bohr-Bury $(n+l)$ rule, the subshell with a higher value of $(n+l)$ has higher energy. If two subshells have the same $(n+l)$ value, the one with the larger primary quantum number ($n$) possesses higher energy.
Step 2: Meaning
We identify the subshells using the angular momentum quantum number $l$: $l=0$ is s, $l=1$ is p, and $l=2$ is d.
Step 3: Analysis
Let us evaluate each electron set: * **I:** $n=3, l=1 \rightarrow 3p \rightarrow n+l = 3+1 = 4$ * **II:** $n=4, l=1 \rightarrow 4p \rightarrow n+l = 4+1 = 5$ * **III:** $n=4, l=2 \rightarrow 4d \rightarrow n+l = 4+2 = 6$ * **IV:** $n=3, l=2 \rightarrow 3d \rightarrow n+l = 3+2 = 5$ Comparing values: III has the highest value (6). For II and IV, both have $(n+l) = 5$, but II ($n=4$) has greater energy than IV ($n=3$). Electron I has the lowest value (4).
Step 4: Conclusion
Sorting them from highest to lowest energy yields $III > II > IV > I$.
Final Answer: (B)
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