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The s-block looks like the easiest chapter in inorganic chemistry: two groups, one or two valence electrons, smooth trends. Then a question asks why lithium, with the highest ionisation energy in its group, is the strongest reducing agent in water, and the smooth trends stop helping. This post sorts the s-block into trends, exceptions and a short list of compounds, with worked numbers and the mistakes students make most often.
Which exams still test the s-block in detail?
The JEE (Advanced) 2026 information brochure has a full "s-Block Elements" topic: reactivity of alkali and alkaline earth metals towards air, water, dihydrogen, halogens and acids, their solutions in liquid ammonia, their oxides, hydroxides, halides and oxoacid salts, the anomalous behaviour of lithium and beryllium, and named compounds of sodium and calcium. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus have no s-block unit; they cover s-block elements only within periodic trends. The current NCERT Class 11 Part I contents and Part II contents also have no separate s-block chapter. If you are preparing for JEE Advanced, you need this chapter from another standard textbook; check the current brochure first.
What are the main trends down groups 1 and 2?
Going down a group, atoms get larger and the outer electron is held less tightly.
Ionisation energy falls. PubChem's periodic table data give first ionisation energies of 5.39 eV for lithium and 3.89 eV for caesium; in group 2, beryllium is 9.32 eV and barium 5.21 eV.
Group 2 is higher than group 1 in the same period. Magnesium (7.65 eV) is well above sodium (5.14 eV), because magnesium's nuclear charge is higher and its 3s electrons are paired in a full subshell.
Electronegativity falls, from 0.98 for lithium to 0.79 for caesium on the Pauling scale (same PubChem data).
Flame colours come from electrons excited by heat and falling back: lithium crimson, sodium yellow, potassium lilac, calcium brick red, strontium crimson, barium apple green. Beryllium and magnesium give no colour, because their electrons are held too tightly for a flame to excite.
Why is lithium the strongest reducing agent in water?
The NCERT electrochemistry chapter's table of standard electrode potentials gives −3.05 V for Li⁺/Li, more negative than potassium (−2.93 V) or sodium (−2.71 V). That looks like it contradicts the ionisation energies. It does not, because the electrode potential covers the whole change from solid metal to ion in water: atomising the metal, removing the electron, and hydrating the ion. Li⁺ is so small that water molecules cluster tightly around it, and the energy released by that hydration more than makes up for the harder ionisation.
The common mistake: answering "reducing power" questions with ionisation energy alone. In the gas phase, caesium loses its electron most easily. In water, lithium wins.
What makes lithium and beryllium anomalous?
Small size and high charge density give the first member of each group partly covalent behaviour.
Lithium burns in air to give mainly the oxide Li₂O; sodium gives mainly the peroxide Na₂O₂, and potassium, rubidium and caesium give superoxides MO₂.
Lithium reacts directly with nitrogen to form Li₃N, as magnesium forms Mg₃N₂. This likeness is the diagonal relationship.
Li₂CO₃ decomposes on heating to Li₂O and CO₂, as MgCO₃ does, while the carbonates of the heavier alkali metals are stable to ordinary heating.
Be(OH)₂ is amphoteric, dissolving in both acid and alkali, like Al(OH)₃; the other group 2 hydroxides are basic.
Which solubility trends flip in group 2?
This is the most tested pair of trends.
Hydroxides get more soluble down the group. Mg(OH)₂ is sparingly soluble; Ba(OH)₂ dissolves much more.
Sulphates get less soluble down the group. MgSO₄ dissolves readily; BaSO₄ is so insoluble that it is the precipitate used to test for sulphate ions.
Carbonates get more stable to heat down the group. MgCO₃ decomposes at a lower temperature than CaCO₃, which decomposes more easily than BaCO₃. A small cation distorts the large carbonate ion and weakens its bonds.
Worked example 1: hydrogen from sodium and water
How much hydrogen does 0.23 g of sodium give with excess water, at 273.15 K and 1 bar?
Equation: 2Na + 2H₂O → 2NaOH + H₂
Moles of sodium: 0.23 / 23 = 0.010 mol.
Moles of hydrogen: half of that, 0.0050 mol.
The NIST molar volume of an ideal gas at 273.15 K and 100 kPa is 22.71 L, so the volume is 0.0050 × 22.71 = 0.114 L, about 114 mL.
The common mistake: taking one mole of hydrogen per mole of sodium. Each sodium atom gives one electron, and each H₂ needs two.
Worked example 2: gypsum to plaster of Paris
Heating gypsum, CaSO₄·2H₂O, gives plaster of Paris, CaSO₄·½H₂O. What percentage of its mass does gypsum lose?
Molar mass of CaSO₄ = 40.08 + 32.06 + 4 × 16.00 = 136.14 g/mol. Water is 18.02 g/mol.
Gypsum = 136.14 + 2 × 18.02 = 172.18 g/mol.
Water lost per formula unit = 1.5 × 18.02 = 27.03 g.
Percentage lost = 27.03 / 172.18 × 100 = 15.7%.
A practice routine for the s-block
Make a one-page table of the trends above, with the reason for each in five words or fewer.
List the lithium and beryllium anomalies and pair each with its diagonal partner.
Do ten numericals on the reactions of s-block metals and their compounds with water, acids and heat.
Test yourself from memory every few days instead of rereading. Our guide to learning inorganic chemistry with retrieval shows how, and periodic trends and their exceptions covers the wider table.
If a trend will not stick because the idea behind it is unclear, the Fermi Concept Library covers every concept in chemistry, each with its own lesson, so you can revise hydration energy or lattice energy on its own before coming back.
This week, write the three group 2 solubility and stability trends from memory, then check them against this post.
