Question
An electron in a 3p orbital — write down all four quantum numbers for it, and explain what each one tells us about the electron’s state.
Solution — Step by Step
The “3” in “3p” directly gives us n = 3. The principal quantum number tells us which shell the electron is in. Higher n means more energy and greater distance from the nucleus.
The letter “p” determines l. The subshell-to-l mapping is fixed: s=0, p=1, d=2, f=3. So for a 3p electron, l = 1. This tells us the shape of the orbital — p orbitals are dumbbell-shaped.
Once l is fixed, m_l ranges from −l to +l in integer steps. For l = 1, that means m_l = −1, 0, or +1. Each value corresponds to one of the three p orbitals: p_x, p_y, p_z. The question asks for “all possible values,” so we list all three.
This one doesn’t depend on n, l, or m_l at all. Every electron has m_s = +½ or −½ — spin-up or spin-down. No exceptions.
So a complete set of quantum numbers for one electron in the 3p subshell is:
The 3p subshell can hold 6 electrons total (3 orbitals × 2 spins each).
Why This Works
The four quantum numbers together form a complete address for any electron in an atom — this is the content of the Pauli Exclusion Principle: no two electrons in the same atom can have the same set of all four quantum numbers. Think of it like a seat number: row (n), section (l), seat (m_l), and whether you’re in the upper or lower bunk (m_s).
The hierarchy matters. n determines the shell → l can be 0 to (n−1) → m_l can be −l to +l → m_s is always ±½. You can’t have l ≥ n, which is why 2d doesn’t exist (n=2, but d requires l=2, and l must be < n).
This framework directly predicts how many electrons each shell holds: shell n has n² orbitals, so 2n² electrons. Shell 3 has 9 orbitals → 18 electrons. You’ll see this number come up constantly in JEE questions about electronic configurations.
Alternative Method — Using the Address Analogy for Exam Questions
When the exam gives you a set of quantum numbers and asks “which orbital?”, work backwards:
| Given | What it means |
|---|---|
| n = 4, l = 2 | 4d orbital |
| n = 3, l = 0 | 3s orbital |
| n = 5, l = 3 | 5f orbital |
Just reverse the s/p/d/f → 0/1/2/3 mapping. This is faster in MCQs than deriving from scratch.
For NEET and JEE Main, memorise the l-to-subshell table as “s=0, p=1, d=2, f=3.” You’ll use it in almost every question on atomic structure. A quick mnemonic: Sharp=0, Principal=1, Diffuse=2, Fundamental=3 (old spectroscopic names — they’re the actual origin of the letters).
Common Mistake
The most common error in board exams: writing m_l values as 1, 2, 3 instead of −1, 0, +1. The range is −l to +l, centred on zero. For l=1, there are three values: −1, 0, and +1 — not 1, 2, 3. This mistake loses full marks on a 3-mark NCERT-type question because none of the values are correct.
A related slip: confusing m_l (magnetic) with m_s (spin). m_l depends on l and can be multiple values; m_s is always just ±½, independent of everything else.