JAM Atomic Structure — Quantum Numbers and Electron Configuration
Atomic structure is the unit where IIT-JAM candidates most often feel confident and score badly. The reason is that the topic looks like a set of rules to memorise, when in fact it is a set of rules plus a set of consequences that the exam asks about far more often than the rules themselves. Counting nodes, assigning a valid set of quantum numbers, computing an effective nuclear charge, and writing the correct configuration for a transition-metal ion are all consequences. This article works through each.
The four quantum numbers and their allowed values
| Symbol | Name | Allowed values | What it determines |
|---|---|---|---|
| n | Principal | 1, 2, 3, … | Shell, size, and (for hydrogen-like atoms) the energy |
| l | Azimuthal / orbital angular momentum | 0 to n − 1 | Subshell and orbital shape; l = 0,1,2,3 → s,p,d,f |
| ml | Magnetic | −l to +l, i.e. (2l + 1) values | Spatial orientation of the orbital |
| ms | Spin magnetic | +½ or −½ | Electron spin orientation |
Two counting results follow immediately and are asked directly: a shell of principal quantum number n contains n² orbitals and can hold 2n² electrons. For n = 4 that is 16 orbitals (one 4s, three 4p, five 4d, seven 4f) and 32 electrons.
The orbital angular momentum magnitude is a separate quantity from l itself, and JAM exploits the confusion:
For a 3d electron, l = 2, so |L| = √6 ħ = 2.449 ħ — not 2ħ.
Node counting
Worked example 1. Count the nodes in 3p, 3d and 4d orbitals.
3p: n = 3, l = 1 → radial = 3 − 1 − 1 = 1, angular = 1, total = 2 ✔ (= n − 1)
3d: n = 3, l = 2 → radial = 3 − 2 − 1 = 0, angular = 2, total = 2 ✔
4d: n = 4, l = 2 → radial = 4 − 2 − 1 = 1, angular = 2, total = 3 ✔
That 3d has zero radial nodes is worth remembering, because the same is true of every orbital where l = n − 1: 1s, 2p, 3d, 4f. These are the orbitals whose radial distribution function has a single maximum.
Hydrogen-like systems
For any one-electron species — H, He⁺, Li²⁺, Be³⁺ — the energy depends only on n and the nuclear charge Z:
Worked example 2 — Balmer transition. Find the wavelength emitted when a hydrogen atom relaxes from n = 3 to n = 2.
ΔE = 13.6 (1/2² − 1/3²) = 13.6 (0.2500 − 0.1111) = 13.6 × 0.13889 = 1.889 eV
Using λ(nm) = 1239.84 / E(eV):
λ = 1239.84 / 1.889 = 656.4 nm
This is the red H-α line — the fact that it comes out at exactly the wavelength observed in the visible spectrum of hydrogen is the historical check on the whole model.
Worked example 3 — ionisation energy of Li²⁺. Li²⁺ has Z = 3 and one electron in n = 1.
E₁ = −13.6 × 3² / 1² = −13.6 × 9 = −122.4 eV
Ionisation energy = 0 − (−122.4) = 122.4 eV
Nine times that of hydrogen, because the energy scales as Z². The radius scales the other way: r₁ = 0.529 / 3 = 0.176 Å.
The de Broglie relation
An electron accelerated through 100 V therefore has λ = 12.27 / 10 = 1.227 Å, comparable to interatomic spacings — which is exactly why electron diffraction works as a structural technique. Pair this with the Heisenberg uncertainty principle, Δx · Δp ≥ ħ/2, which JAM usually asks as a numerical estimate of the minimum uncertainty in velocity.
Electron configuration and the ion trap
Fill orbitals in order of increasing (n + l), and for equal (n + l) take the lower n first — the Madelung rule. Obey the Pauli exclusion principle (no two electrons in an atom share all four quantum numbers) and Hund's rule of maximum multiplicity.
Then learn the exceptions and, more importantly, the reason: half-filled and completely filled d subshells gain exchange-energy stabilisation.
The trap that catches most candidates is the cation. Although 4s fills before 3d, electrons are removed from 4s first, because once the 3d orbitals are occupied they lie below 4s in energy:
Fe³⁺ being d⁵ — half filled — is precisely why it is the more stable oxidation state of iron in aqueous solution, and this connects atomic structure directly to the inorganic chemistry syllabus.
Slater's rules for effective nuclear charge
Zeff = Z − S, where the screening constant S is built from the grouping (1s)(2s,2p)(3s,3p)(3d)(4s,4p)… with these contributions for an s or p electron: 0.35 from each other electron in the same group (0.30 if the electron of interest is in 1s), 0.85 from each electron in the shell one lower, and 1.00 from every electron in shells further in. For a d or f electron, everything to its left contributes 1.00.
Worked example 4. Find Zeff for a 2p electron in oxygen (Z = 8) and for the 3s electron in sodium (Z = 11).
Oxygen, (1s²)(2s² 2p⁴). For one of the 2p electrons, the same group holds five
other electrons:
S = (5 × 0.35) + (2 × 0.85) = 1.75 + 1.70 = 3.45
Zeff = 8 − 3.45 = 4.55
Sodium, (1s²)(2s² 2p⁶)(3s¹). The 3s electron has no companions in its group:
S = (0 × 0.35) + (8 × 0.85) + (2 × 1.00) = 6.80 + 2.00 = 8.80
Zeff = 11 − 8.80 = 2.20
The contrast explains the periodic trend directly: oxygen's outer electron feels more than twice the effective nuclear pull that sodium's does, which is why the atomic radius drops across a period and jumps at the start of the next one.
The traps that cost atomic-structure marks
- Removing 3d electrons before 4s when writing a cation. Fill 4s first, empty 4s first. This is the highest-frequency error in the entire unit.
- Writing orbital angular momentum as lħ instead of √(l(l+1))ħ. The two agree at l = 0 and nowhere else.
- Confusing radial nodes with total nodes. A 3s orbital has two radial nodes; a 3d orbital has none. Both have two nodes in total.
- Applying the Bohr formula to a many-electron atom. En = −13.6 Z²/n² holds for one-electron species only. For anything else the energy depends on l as well as n.
- Assigning ml outside the range −l to +l, for example claiming n = 2, l = 2. Since l can only go up to n − 1, that set is simply not allowed — "which of these sets is impossible" is a stock question.
- Treating the Cr and Cu exceptions as arbitrary. If you know the exchange-energy reasoning you can also answer the Mo, Ag and Au versions rather than memorising a list.
Check every quantum-number set instantly. The Quantum Numbers calculator validates any combination of n, l, ml and ms, reports the subshell, the node count and the orbital capacity — exactly the checks a JAM question asks you to make under time pressure.
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