JAM Inorganic — Periodic Properties Questions
Periodic properties are examined at IIT-JAM level in a quite different way from school: instead of "state the trend", you are asked to calculate an effective nuclear charge, to explain an anomaly, or to order a set of isoelectronic species. The underlying quantity behind almost all of it is Zeff, and Slater's rules give you a way to put a number on it. This guide works those rules carefully, then applies them to the exceptions that actually get asked.
Effective nuclear charge and Slater's rules
Write the configuration in Slater's groupings, in this order and with these brackets:
Then, for the electron of interest:
| Electron being screened | Contribution to S |
|---|---|
| Any electron in a group to the right | 0 |
| Each other electron in the same group | 0.35 (but 0.30 within 1s) |
| For an ns/np electron: each electron in shell n − 1 | 0.85 |
| For an ns/np electron: each electron in shell n − 2 or lower | 1.00 |
| For an nd/nf electron: every electron in any group to the left | 1.00 |
Worked example 1 — a 3p electron in chlorine
Cl, Z = 17: (1s²)(2s²2p⁶)(3s²3p⁵)
Same group (3s,3p) has 7 electrons, so 6 others: 6 × 0.35 = 2.10
Shell n − 1 = (2s,2p), 8 electrons: 8 × 0.85 = 6.80
Shell n − 2 = (1s), 2 electrons: 2 × 1.00 = 2.00
S = 2.10 + 6.80 + 2.00 = 10.90 → Zeff = 17 − 10.90 = 6.10
Worked example 2 — why potassium loses its 4s electron so easily
K, Z = 19: (1s²)(2s²2p⁶)(3s²3p⁶)(4s¹)
Same group (4s,4p): no other electron → 0
Shell n − 1 = n = 3, i.e. 3s²3p⁶ = 8 electrons: 8 × 0.85 = 6.80
Shells n − 2 and lower = 10 electrons: 10 × 1.00 = 10.00
S = 16.80 → Zeff = 19 − 16.80 = 2.20
Only 2.2 units of positive charge are felt by that lone 4s electron against a nucleus of charge 19 — which is exactly why K has a low first ionisation energy and reactive metallic chemistry.
Worked example 3 — the 4s / 3d question, settled by numbers
Fe, Z = 26: (1s²)(2s²2p⁶)(3s²3p⁶)(3d⁶)(4s²)
For a 3d electron — same group: 5 others × 0.35 = 1.75; everything to the left (18 electrons) × 1.00 = 18.00. S = 19.75 → Zeff(3d) = 26 − 19.75 = 6.25
For a 4s electron — same group: 1 other × 0.35 = 0.35; shell n − 1 = n = 3, which is 3s²3p⁶3d⁶ = 14 electrons × 0.85 = 11.90; shells below = 10 × 1.00 = 10.00. S = 22.25 → Zeff(4s) = 26 − 22.25 = 3.75
The 3d electrons feel almost twice the effective charge that the 4s electrons do, which is why transition metals ionise from 4s first even though 4s fills first. That apparent contradiction is a standard JAM question, and this calculation is the answer.
Slater's rules are an empirical fit, not a derivation. They work well for main-group s and p electrons, become rough across the d block, and are superseded in accurate work by self-consistent-field values. Say so if a question asks about their reliability.
Why atoms shrink across a period
Li (Z = 3), 2s electron: S = 2 × 0.85 = 1.70 → Zeff = 1.30
F (Z = 9), 2p electron: S = (6 × 0.35) + (2 × 0.85) = 2.10 + 1.70 = 3.80 →
Zeff = 5.20
Across period 2 the added electrons go into the same shell and screen each other only 0.35 each, while the nuclear charge rises by a full unit per element. Zeff nearly quadruples, so the valence shell is pulled in and the atoms get smaller — the opposite of what "more electrons" naively suggests.
The anomalies, with their real explanations
| Anomaly | Explanation |
|---|---|
| IE₁: B (801) < Be (899 kJ mol⁻¹) | B loses a 2p electron, which is higher in energy and better screened than Be's paired 2s |
| IE₁: O (1314) < N (1402 kJ mol⁻¹) | O's 2p⁴ has one doubly occupied orbital; removing an electron relieves that pairing repulsion. N has a stable half-filled 2p³ |
| Electron affinity: Cl (−349) more exothermic than F (−328 kJ mol⁻¹) | F's 2p shell is so compact that the incoming electron suffers large electron–electron repulsion. Same reason S beats O |
| IE₂ of Na is enormous | It removes an electron from the complete neon core, not from the valence shell |
| Ga radius ≈ Al radius | The intervening 3d electrons screen poorly, so Zeff for Ga's 4p electron is unusually high (the "d-block contraction") |
| Zr (160 pm) ≈ Hf (159 pm) | Lanthanide contraction — 14 poorly shielding 4f electrons intervene, so the 5d elements are no larger than their 4d congeners |
The lanthanide contraction is worth understanding rather than memorising. It makes Zr/Hf, Nb/Ta and Mo/W chemically almost inseparable, raises the densities of the third-row transition metals, and contributes to the nobility of gold and platinum. Any of those can be the actual question.
The inert pair effect explains why the heavier p-block elements favour an oxidation state two below the group maximum — Tl(I) over Tl(III), Pb(II) over Pb(IV), Bi(III) over Bi(V). The ns² pair becomes progressively harder to use for bonding because of poor d and f shielding and, for the heaviest elements, relativistic contraction of the 6s orbital.
Isoelectronic series — ordering by nuclear charge alone
When several species have the same number of electrons, the electron–electron repulsion is identical and only Z differs. Radius therefore falls monotonically as Z rises.
This is the fastest question type in the whole topic once you spot that the species are isoelectronic — count electrons first, then simply rank by atomic number.
Electronegativity scales
Pauling's scale is thermochemical and relative; Mulliken's is absolute and defined per atom:
For chlorine, IE₁ = 1251 kJ mol⁻¹ and EA = 349 kJ mol⁻¹.
In electronvolts (divide by 96.485): IE = 12.97 eV, EA = 3.61 eV.
χM = (12.97 + 3.61)/2 = 8.29 eV
The absolute number is not comparable to a Pauling value without a conversion, and different sources use different conversions — so compare Mulliken with Mulliken, and quote the scale you are using. The ordering the two scales give is essentially the same, which is what the trends questions rely on.
Allred–Rochow, based on the electrostatic force Zeffe²/r², is the third scale in the syllabus and links directly back to the Slater calculations above.
Common mistakes
- Screening the 4s electron with only 3s and 3p. For a 4s electron the whole n = 3 shell, 3d included, contributes 0.85 each.
- Giving a d electron 0.85 for the shell below. For nd and nf, every electron to the left contributes a full 1.00.
- Counting the electron itself in its own group. Use "one fewer" than the group population.
- Reading electron affinity signs inconsistently. As an energy change it is negative for an exothermic attachment; as "electron affinity" tabulated positive it means the energy released. State which convention you are using.
- Explaining the N/O ionisation anomaly by "half-filled shell stability" alone. The operative point is the pairing repulsion in O's doubly occupied 2p orbital, which is relieved on ionisation.
- Comparing radii across different coordination numbers or bond types. Ionic, covalent, metallic and van der Waals radii are different quantities and are not interchangeable in a comparison.
Look the real numbers up while you practise. Slater calculations, isoelectronic orderings and Mulliken electronegativities all become much faster once you have the atomic number, configuration, ionisation energy and electron affinity in front of you for every element.
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