IIT-JAM s-Block Chemistry — Trends, Anomalies and Named Compounds
The s-block looks like the easiest inorganic unit and is therefore the one students revise last and least. That is a mistake: JAM's inorganic questions are usually about reasoning — why lithium has the most negative electrode potential despite the highest ionisation energy, why barium sulphate is insoluble while magnesium sulphate is not, why the Solvay process works for sodium but not for potassium. Every one of those has a one-line explanation built on ionic size and enthalpy, and once you hold that framework the whole block becomes predictable. This guide gives the framework, the exceptions, and three worked calculations.
The framework — everything follows from size and charge
Group 1 elements have the configuration [noble gas]ns1 and form M+; group 2 are ns2 and form M2+. Going down either group the ion gets bigger, so its charge density (polarising power) falls. Two enthalpies compete in almost every question:
Hydration enthalpy — energy released when gaseous ions are surrounded by water. Depends mainly on the cation: roughly proportional to q / r.
Enthalpy of solution ≈ hydration enthalpy − lattice enthalpy
Which of the two falls faster as the cation grows decides the solubility trend, and that is why group 2 hydroxides become more soluble down the group while group 2 sulphates become less soluble. With a small anion such as OH−, the cation radius dominates the lattice term, so lattice enthalpy falls quickly and solubility rises. With a large anion such as SO42−, the anion already controls the lattice term, so lattice enthalpy barely changes while hydration enthalpy keeps falling — and solubility drops. Mg(OH)2 is sparingly soluble but Ba(OH)2 dissolves; MgSO4 dissolves freely but BaSO4 is famously insoluble.
The trends, side by side
| Property | Down group 1 | Down group 2 | Group 1 vs group 2 at the same period |
|---|---|---|---|
| Ionic radius (pm, 6-coordinate) | Li+ 76 → Cs+ 167 | Mg2+ 72 → Ba2+ 135 | M2+ is smaller than M+ |
| Ionisation energy | Decreases | Decreases | Group 2 higher (smaller, higher nuclear charge) |
| Hydration enthalpy | Becomes less negative | Becomes less negative | Group 2 much larger in magnitude (charge 2+) |
| Reactivity with water | Increases | Increases (Be does not react) | Group 1 far more reactive |
| Thermal stability of carbonates | All stable except Li2CO3 | Increases down the group | Group 1 carbonates much more stable |
| Flame colour | Li crimson, Na golden yellow, K lilac, Rb red-violet, Cs blue | Ca brick red, Sr crimson, Ba apple green | Be and Mg give none |
Be and Mg give no flame colour because their ionisation energies are too high for the small energy of a bunsen flame to promote an electron — a favourite one-line reasoning question.
Worked example 1 — polarising power and carbonate stability
Using 6-coordinate ionic radii Mg2+ 72 pm, Ca2+ 100 pm, Sr2+ 118 pm, Ba2+ 135 pm, compute charge/radius and explain the order of carbonate decomposition temperatures.
Mg2+: 2 / 72 = 0.0278 pm−1
Ca2+: 2 / 100 = 0.0200 pm−1
Sr2+: 2 / 118 = 0.0170 pm−1
Ba2+: 2 / 135 = 0.0148 pm−1
Polarising power falls by almost a factor of two from Mg2+ to Ba2+. A strongly polarising cation distorts the carbonate ion, weakens a C–O bond and helps it shed CO2. So the decomposition temperature increases in the order MgCO3 < CaCO3 < SrCO3 < BaCO3.
The same argument explains why Li2CO3 decomposes on heating while Na2CO3 and the rest of group 1 do not — Li+ is by far the most polarising alkali metal cation.
Worked example 2 — lime from limestone
How much quicklime is obtained by complete decomposition of 100.0 g of pure CaCO3? CaCO3 → CaO + CO2
M(CaCO3) = 40.078 + 12.011 + 3 × 15.999 = 40.078 + 12.011 + 47.997 = 100.086 g mol−1
M(CaO) = 40.078 + 15.999 = 56.077 g mol−1
moles of CaCO3 = 100.0 / 100.086 = 0.99914 mol
mass of CaO = 0.99914 × 56.077 = 56.03 g
Check by difference: CO2 lost = 100.0 − 56.03 = 43.97 g, and directly 0.99914 × 44.009 = 43.97 g. The two agree, so the working is sound.
Worked example 3 — gypsum to plaster of Paris
Find the percentage mass loss when gypsum is heated to about 393 K.
CaSO4·2H2O → CaSO4·½H2O + 1½H2O
M(CaSO4·2H2O) = 40.078 + 32.06 + 4 × 15.999 + 2 × 18.015
= 40.078 + 32.06 + 63.996 + 36.030 = 172.164 g mol−1
M(CaSO4·½H2O) = 40.078 + 32.06 + 63.996 + 9.0075 = 145.142 g mol−1
Mass lost = 172.164 − 145.142 = 27.022 g mol−1, which is exactly 1.5 × 18.015 — the stoichiometry checks itself.
Percentage loss = (27.022 / 172.164) × 100 = 15.70%, so 100 g of gypsum gives 84.30 g of plaster of Paris.
Heating further, above about 473 K, drives off the remaining water to give dead-burnt anhydrous CaSO4, which no longer sets with water. That temperature detail is exactly the kind of thing JAM turns into a single-line question.
The anomalies you must be able to explain
- Lithium has the most negative E°(M+/M) in group 1 (about −3.04 V) even though it has the highest ionisation energy. Electrode potential is about the whole process in solution: sublimation, ionisation and hydration. Li+ is tiny, so its hydration enthalpy is by far the most negative in the group (roughly −519 kJ mol−1 against about −264 for Cs+), and that term more than compensates. This is the single most-asked "explain" question in the s-block.
- Products of burning in air differ down group 1. Li gives mainly the normal oxide Li2O, Na gives the peroxide Na2O2, and K, Rb and Cs give superoxides MO2. Larger, less polarising cations stabilise the larger, more diffuse anions.
- Lithium alone forms a nitride, Li3N, by direct combination with N2 — and so does magnesium, which is the diagonal partner.
- Beryllium is essentially covalent. BeCl2 is a polymeric chain in the solid with bridging chlorides, dimeric in the vapour at moderate temperature and linear monomeric only at high temperature. BeO and Be(OH)2 are amphoteric while the rest of group 2 give basic oxides. Be also has a maximum covalency of 4 because it has no d orbitals available.
- Beryllium does not react with water, magnesium reacts only with steam, while calcium onwards react with cold water — a clean reactivity gradient worth memorising as a set.
Solutions in liquid ammonia
Alkali metals dissolve in liquid ammonia to give a deep blue solution containing solvated cations and ammoniated electrons. Dilute solutions are blue, paramagnetic and good electrical conductors — the colour is due to the solvated electron. As concentration rises the solution turns bronze, becomes metallic in conductivity and diamagnetic, because the electrons pair up in clusters. On standing, and rapidly in the presence of a transition metal catalyst, the solution decomposes:
These solutions are powerful reducing agents and are the reagent behind the Birch reduction in organic chemistry — a nice cross-link that JAM sometimes exploits by asking an inorganic fact inside an organic question.
Diagonal relationships
| Pair | Shared behaviour |
|---|---|
| Li and Mg | Both form nitrides directly with N2; both carbonates decompose to the oxide on heating; LiF and MgF2 are sparingly soluble; LiCl and MgCl2 are deliquescent and dissolve in ethanol; neither forms a stable superoxide |
| Be and Al | Both oxides and hydroxides are amphoteric; both chlorides are covalent Lewis acids that bridge or dimerise; both are passivated by concentrated nitric acid; both form fluoro-complexes, [BeF4]2− and [AlF6]3−; both carbides give methane on hydrolysis |
The cause is the same in both cases: moving one place right increases charge and moving one place down increases size, and the two changes roughly cancel to leave a similar charge density.
Industrial compounds worth knowing by name
- NaOH — chlor-alkali electrolysis of brine, with H2 at the cathode and Cl2 at the anode.
- Na2CO3 (Solvay process) — brine, ammonia and CO2 give NaHCO3, which is filtered off and calcined. The process fails for potassium because KHCO3 is too soluble to precipitate. That "why not potassium" question appears again and again.
- CaO, Ca(OH)2, CaCO3 — the lime cycle, used in mortar, steelmaking and water treatment.
- Plaster of Paris, CaSO4·½H2O, which rehydrates back to gypsum and sets.
- Bleaching powder, from Cl2 on dry slaked lime, whose active chlorine content is the usual numerical question.
- Crown ethers and cryptands, which complex alkali metal cations selectively by cavity size — 18-crown-6 fits K+ well — and so make alkali metal salts soluble in organic solvents.
Common mistakes that cost marks
- "Smaller ion means more soluble." It depends on the anion. Compare hydroxides with sulphates before answering — the trends run in opposite directions.
- Explaining electrode potential with ionisation energy alone. E° is a solution-phase quantity; hydration enthalpy must be in the answer.
- Saying all alkali metals give normal oxides. Only lithium mainly does.
- Claiming Be and Mg give flame colours. They do not.
- Treating BeCl2 as ionic. It is covalent, and its structure changes with temperature.
- Forgetting Li2CO3's instability when asked to compare carbonate stability across group 1.
- Confusing the diagonal pairs. Li goes with Mg; Be goes with Al. Write both on your formula sheet.
How to prepare this unit
Build one page: two trend tables (group 1 and group 2), one list of anomalies with the one-line reason next to each, one list of named compounds with their process, and the two diagonal pairs. Revising a trend table is faster and more reliable than re-reading paragraphs. For the syllabus itself and the current paper pattern, use the official IIT-JAM notification for your year — not any coaching summary, including this one.
See the trends instead of memorising them. The interactive periodic table gives you atomic and ionic data, electronic configurations and group position for all 118 elements, so you can read a trend down group 1 or group 2 directly instead of trusting a half-remembered list.
Open the Interactive Periodic Table →Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches — classroom at the Gurugram coaching centre and live online classes for students across India. Details at abcchemistry.in.