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

• A. $I > IV > II > III$
• B. $III > II > IV > I$
• C. $III > IV > II > I$
• D. $III > IV > I > II$

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.

Answer 1

The Correct Option is B

Step 1: Know the rule for energy order.
The energy of an electron in a subshell is decided by the sum $(n + l)$. A bigger $(n + l)$ means more energy. This is called the Bohr-Bury rule.

Step 2: Know the tie-breaker.
If two subshells have the same $(n + l)$ value, then the one with the bigger $n$ has more energy. This small extra rule settles ties.

Step 3: Name each subshell.
The value of $l$ tells us the type of subshell: $l = 0$ is s, $l = 1$ is p, and $l = 2$ is d. So electron I ($n=3, l=1$) is 3p, II ($n=4, l=1$) is 4p, III ($n=4, l=2$) is 4d, and IV ($n=3, l=2$) is 3d.

Step 4: Add up $(n + l)$ for each.
For I: $3 + 1 = 4$. For II: $4 + 1 = 5$. For III: $4 + 2 = 6$. For IV: $3 + 2 = 5$.

Step 5: Compare the sums.
III has the highest sum of 6, so it has the most energy. I has the lowest sum of 4, so it has the least energy. II and IV both give 5, so we need the tie-breaker.

Step 6: Break the tie and write the order.
For II and IV the sum is equal, so the bigger $n$ wins. II has $n = 4$ and IV has $n = 3$, so II is higher than IV. Putting it all together from highest to lowest gives III, then II, then IV, then I. \[ \boxed{III > II > IV > I} \]