VKC Notes · Inorganic chemistry · Geo-Scientist P-I, IFoS P-I
14.1 Periodic Properties: Slater’s Rules, Radii, Ionisation and Electron-Gain Enthalpy, Electronegativity
★★ 2021–2025: 2 parts, 20 marks asked directly (Geo-Scientist Paper I 2025, 5 marks; IFoS Paper I 2022, 15 marks). Few direct questions, but these ideas are the reason behind answers on lanthanide contraction, the 5d series, lattice energies and bond polarity in every paper.
Syllabus line. No Mains syllabus has a separate line for this. It sits inside Geo-Scientist Paper I unit 9 (“d-block elements: general comparison of 3d, 4d and 5d elements … f-block elements: … ionization enthalpies, … variation in atomic and ionic (3+) radii”) and the chemical bonding unit of IFoS and Civil Services Paper I (“Ionic bond, characteristics of ionic compounds, … lattice energy, Born-Haber cycle; … polarities of bonds in molecules”).
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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes
14.1.1 Shielding, penetration and Slater’s rules ★★★
- Each electron is shielded from the nucleus by the others, so it feels an effective nuclear charge Zeff = Z − σ. Orbitals that penetrate the core feel more of Z: s > p > d > f for the same n. That is why 2s lies below 2p and 4s fills before 3d in K and Ca.SN
(14.1.1)
- Write the configuration in Slater’s groups: (1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p)…; electrons to the right of the chosen group contribute nothing.SN
- For an ns or np electron: each other electron in the same group 0.35 (0.30 within 1s); each electron in shell n − 1: 0.85; each electron in n − 2 or lower: 1.00.SN
- For an nd or nf electron: same group 0.35; every electron in groups to the left: 1.00.SN
| Electron | σ | Zeff | Explains |
|---|---|---|---|
| N 2p | 4 × 0.35 + 2 × 0.85 = 3.10 | 3.90 | +0.65 per step across a period |
| Cl 3p | 6 × 0.35 + 8 × 0.85 + 2 × 1.00 = 10.90 | 6.10 | high I₁ and χ of Cl |
| K 4s / 3d | 16.80 / 18.00 | 2.20 / 1.00 | 19th electron enters 4s |
| Fe 4s / 3d | 22.25 / 19.75 | 3.75 / 6.25 | 4s lost first: Fe²⁺ [Ar]3d⁶ |
| Zn 4s / 3d | 25.65 / 21.15 | 4.35 / 8.85 | 3d is core by Zn |
14.1.2 Atomic and ionic radii; the lanthanide contraction ★★★
- Metallic radius: half the nearest-neighbour distance in the metal; covalent radius: half the A–A distance in a molecule; ionic radius: the cation–anion distance shared out, taking O²⁻ = 140 pm. Radii fall across a period (Zeff rises, same shell) and rise down a group (new shell): Li 157 → F 71 pm; Li 157 → K 235 pm.SN
- Cations are smaller and anions larger than the atom. In an isoelectronic series the radius falls as Z rises (Fig. 14.1.1).SN
Draw it in 60 s: six circles shrinking left to right; write N3− 146, O2− 140, F− 133, Na+ 102, Mg2+ 72, Al3+ 53 and Z = 7 to 13 under them.
- Lanthanide contraction: across La–Lu the added 4f electrons shield each other poorly, so Zeff on the outer electrons keeps rising. The 5d elements that follow are no bigger than their 4d partners: Zr 160 ≈ Hf 159 pm; Mo 140 ≈ W 141 pm. Zr and Hf are therefore chemically almost identical and hard to separate (the contraction itself is in 15.1).SN
- The same effect after the 3d series (d-block contraction): Ga (140 pm) is hardly bigger than Al (143 pm). In heavy atoms, relativistic contraction of s and p orbitals adds to it.SN
Draw it in 60 s: left: three curves dipping in the middle; 4d and 5d almost on top of each other, 3d below. Right: three rising lines; 5d on top from group 5 on, ending with Hg.
14.1.3 Ionisation enthalpy: trends, anomalies, jumps ★★★ asked 2 times
(14.1.2)
- I₁ is lowest at the lower left (Cs) and highest at the upper right (He, F, Ne). It rises with Zeff and falls with size and shielding; s electrons are harder to remove than p electrons of the same shell.SN
Draw it in 60 s: saw-tooth: tall peaks at He, Ne, Ar; deep minima at Li, Na, K; small dips after Be, N, Mg, P: red dots at B, O, Al, S.
- Anomalies: Be 899 > B 801 and Mg 737 > Al 577 (the np¹ electron penetrates less than ns² and is shielded by it); N 1402 > O 1314 and P 1011 > S 1000 (in O the fourth 2p electron is paired and repelled by its partner; N is half-filled) kJ mol⁻¹.SN
- Successive energies jump when a closed shell is broken: Na 495, 4562, 6911; Mg 737, 1476, 7732; Al 577, 1816, 2744, 11 574 kJ mol⁻¹. The position of the first large jump gives the group.SN
Solved PYQs Geo 2025 Q1(j) 5m IFoS 2022 Q2(b) 15m
- Why is I₁ of the 5d series higher than 3d and 4d? From 4d to 5d, Z rises by 32, fourteen of them across the 4f block. The diffuse 4f electrons shield poorly, so Zeff on the 6s electron rises sharply, while the lanthanide contraction keeps the size equal to the 4d atom (Hf ≈ Zr, Ta ≈ Nb). Relativistic contraction of 6s adds to it (largest for Au and Hg). Values (kJ mol⁻¹): Ni 737, Pd 805, Pt 870; Cu 745, Ag 731, Au 890; Zn 906, Cd 868, Hg 1007. Group 4 is the exception: Hf 642 is a little below Zr 660 and Ti 658.SN
- Second ionisation energy of Na from a Born–Haber cycle (Fig. 14.1.4). For NaCl: −381 = (S + I₁) + 121 − 349 − 757, so S + I₁ = 604. For NaCl₂: +2530 = (S + I₁) + I₂ + 242 − 2 × 349 − 2180 = I₂ − 2032, so I₂ = 4562 kJ mol⁻¹, as in the tables. Removing an electron from the closed 2p⁶ shell of Na⁺ costs so much that NaCl₂ has a large positive ΔfH: it does not exist.SN
Draw it in 60 s: a ladder: from Na(s) + ½Cl2 up +604 (S + I1), up +121 (½D), down −349 (Ea), down −757 (U) to NaCl(s) at −381; dashed line for ΔfH.
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