Question
Find the four quantum numbers (n, l, m_l, m_s) for the 25th electron of an atom. Use the Aufbau principle, Pauli exclusion principle, and Hund’s rule to justify your answer.
This is a direct NCERT Class 11 question and a standard 2-marker in CBSE boards.
Solution — Step by Step
The element with atomic number 25 is Manganese (Mn). We need to fill 25 electrons in orbitals following the Aufbau order.
Filling orbitals in order of increasing (n + l) value:
Let’s count the running total: 2 + 2 + 6 + 2 + 6 + 2 = 20 electrons through 4s². The remaining 5 electrons go into 3d.
Electrons 21 through 25 fill the five 3d orbitals. By Hund’s rule, each orbital gets one electron before any pairing begins — all with spin up (+½).
| Electron # | Orbital | m_l |
|---|---|---|
| 21 | 3d | −2 |
| 22 | 3d | −1 |
| 23 | 3d | 0 |
| 24 | 3d | +1 |
| 25 | 3d | +2 |
The 25th electron lands in the last available 3d orbital: m_l = +2.
For the 25th electron:
| Quantum Number | Value | Meaning |
|---|---|---|
| n | 3 | 3rd shell |
| l | 2 | d subshell (l = 2 for d) |
| m_l | +2 | last 3d orbital |
| m_s | +½ | spin up (first pass, no pairing) |
Final answer: n = 3, l = 2, m_l = +2, m_s = +½
Why This Works
The Aufbau principle tells us which subshell to fill next. We use the (n + l) rule: lower sum fills first. For 3d, n + l = 3 + 2 = 5. For 4s, n + l = 4 + 0 = 4 — so 4s fills before 3d. That’s why electrons 19 and 20 go into 4s, not 3d.
Once we’re inside the 3d subshell, Hund’s rule takes over. The five 3d orbitals have m_l values: −2, −1, 0, +1, +2. Each gets exactly one electron (all spin up) before any pairing. This happens because electrons repel each other — spreading out minimises electron–electron repulsion and gives a lower energy state.
The Pauli exclusion principle is our guardrail throughout: no two electrons in the same atom can share all four quantum numbers. Since electrons 21–25 each occupy a different m_l value, they’re all distinguishable and no rule is violated.
Alternative Method
Box diagram shortcut — draw five boxes for 3d and fill arrows left to right. The 5th arrow (25th electron) lands in the rightmost box. That box corresponds to m_l = +2. This visual approach is faster in exams and less error-prone than counting through a table.
Draw the 3d orbital diagram:
3d: [↑] [↑] [↑] [↑] [↑]
ml: -2 -1 0 +1 +2The 5th electron (our 25th) fills the m_l = +2 box with spin up. Same answer, faster execution.
Common Mistake
Wrong subshell order: Many students write the configuration as ...3d⁵ 4s² in energy order and mistakenly assign the 25th electron to 4s. Remember — we fill in Aufbau order (4s before 3d), so when counting the 25th electron by fill sequence, 4s² is already done at electron 20. The 25th goes to 3d, not back to 4s.
Similarly, some students assign m_l = −2 to the 25th electron, forgetting that electrons 21–24 already occupied m_l = −2, −1, 0, +1 in that order.