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s-Block Elements · TOPIC LESSON

Alkali and Alkaline-Earth Compounds

Explain s-block trends, industrial salts, thermal and solution chemistry, and the distinct biological roles of Na, K, Mg and Ca.

JEE Main · JEE Advanced · NEET UG55 min study guide10 indexed sections

BY THE END, YOU CAN

  • Use ion charge, size and hydration to explain trends and exceptions.
  • Balance carbonate, nitrate, peroxide and industrial-salt reactions.
  • Apply solubility and hydrate stoichiometry with explicit assumptions.
  • Explain sodium–potassium gradients and pump stoichiometry.
  • Relate magnesium and calcium coordination to biological functions.

BEFORE YOU START

  • Ionic bonding, oxidation states and periodic trends.
  • Mole ratios, acid–base chemistry and Ksp.
  • Basic cell membranes and ATP.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Charge, size and the first-member exceptions

Group 1 metals have an ns¹ outer configuration and usually form M⁺; group 2 metals have ns² and usually form M²⁺. Their compounds are largely ionic, but small highly polarizing ions, especially Be²⁺, introduce substantial covalent character. Removing a further electron after the noble-gas configuration requires a large additional ionization energy, explaining the preferred charges.

Atomic size generally increases down each group while first ionization energy decreases. Yet an aqueous reducing-power ranking cannot be read from ionization energy alone: atomization, ionization and hydration all enter the overall free-energy change. Lithium is a particularly strong aqueous reductant because hydration of its small ion is strongly favourable.

The metals are electropositive and normally occur as compounds rather than uncombined deposits. Aqueous electrolysis does not simply deposit sodium from brine: water is reduced instead. Industrial sodium production uses molten salts. This distinction follows competition between possible electrode reactions, not whether sodium ions are present.

Flame colours arise from electronic excitation and emission, not from the bulk colour of the salt. Common observations include lithium crimson, sodium yellow, potassium lilac, calcium brick-red, strontium crimson and barium green. Be and Mg do not give the standard characteristic visible flame colours under the usual test. These are analytical facts, not instructions for handling reactive metals.

02 / UNDERSTAND & APPLY

2. Water and oxygen products depend on the metal

The group 1 water balance is 2M+2H₂O → 2MOH+H₂. It is strongly exothermic and can be violent; equations in this lesson are for interpretation only. Group 2 behaviour differs: Be is protected by its oxide layer, Mg reacts slowly with cold water and reacts with steam to give MgO and H₂, while Ca, Sr and Ba react with cold water to form hydroxides and hydrogen.

Oxygen compounds must be identified by the anion. Oxide contains O²⁻, peroxide contains O₂²⁻ with an O–O bond, and superoxide contains O₂⁻. Representative dominant products in excess oxygen are Li₂O, Na₂O₂ and KO₂, with heavier alkali metals also favouring superoxides. Conditions matter; these are characteristic products rather than a claim that each metal has only one oxygen compound.

Larger cations can stabilize the larger peroxide and superoxide anions more effectively in the relevant solids. Oxygen oxidation states are −2 in oxide, −1 in peroxide and average −1/2 in superoxide. The last is an average over two equivalent oxygen atoms, not half an electron literally confined to each atom.

Peroxide with water can give hydrogen peroxide and hydroxide: Na₂O₂+2H₂O → 2NaOH+H₂O₂. Superoxide undergoes disproportionation; a useful overall balance is 2KO₂+2H₂O → 2KOH+H₂O₂+O₂. Subsequent peroxide decomposition can alter the final observed mixture.

ORIGINAL WORKED EXAMPLE

Distinguish three oxygen salts

Assign oxygen oxidation states in Li₂O, Na₂O₂ and KO₂. Which contains an O–O unit?

  1. Each alkali metal is +1. Charge balance gives O=−2 in Li₂O.
  2. Two Na⁺ require the O₂ unit to be −2, giving each O an average −1. One K⁺ requires O₂⁻, giving average −1/2.
  3. Peroxide and superoxide contain O–O units; oxide does not.

Answer −2, −1 and −1/2 respectively; Na₂O₂ and KO₂ contain O–O units.

ORIGINAL PRACTICE · 1

Balance calcium reacting with cold water.

Reveal answer & reasoning

Ca+2H₂O → Ca(OH)₂+H₂.

Calcium loses two electrons, while two water hydrogens are reduced to hydrogen gas; the remaining hydroxide groups balance Ca²⁺.

03 / UNDERSTAND & APPLY

3. Why lithium and beryllium need their own row

Lithium differs from its heavier group 1 neighbours: it forms a stable nitride directly with nitrogen, Li₂CO₃ decomposes on heating, and LiNO₃ gives oxide rather than nitrite on strong heating. These exceptions relate to the small ion and strong interactions with anions; they should be learned alongside the trend, not appended as unrelated trivia.

Lithium and magnesium show a diagonal resemblance: both form nitrides, their carbonates decompose to oxides and CO₂ on heating, and their nitrates produce oxides, NO₂ and O₂. Similar charge density and polarization help rationalize these selected similarities, but they do not make Li and Mg chemically identical or give them the same ionic charge.

Beryllium and aluminium show another useful diagonal resemblance. Their oxides/hydroxides are amphoteric, their halides have substantial covalent character, and protective oxide layers influence their reactions. Be(OH)₂ reacts with acid and with excess strong base; Mg(OH)₂ is treated as basic in the usual school comparison.

Aqueous amphoterism can be represented by Be(OH)₂+2H⁺ → Be²⁺+2H₂O and Be(OH)₂+2OH⁻ → [Be(OH)₄]²⁻. The product in strongly alkaline solution is a hydroxo complex, not bare BeO₂²⁻ floating without hydration. Beryllium compounds are toxic; their inclusion is conceptual.

ORIGINAL PRACTICE · 1

Why is “all group 2 hydroxides are purely basic” too broad?

Reveal answer & reasoning

Be(OH)₂ is amphoteric.

It reacts with both acids and excess strong base; the small Be²⁺ ion makes the first member atypical.

04 / UNDERSTAND & APPLY

4. Carbonates and nitrates: balance the exception first

Group 2 carbonates decompose on sufficiently strong heating: MCO₃ → MO+CO₂. Their thermal stability generally increases down the group. Small highly charged cations strongly polarize the carbonate electron distribution and stabilize the oxide product relative to carbonate; the complete energetic explanation compares both solid lattices, not only one bond in an isolated ion.

Most group 1 carbonates are relatively resistant to ordinary heating, while Li₂CO₃ → Li₂O+CO₂ is the key exception. The decomposition temperature depends on conditions including CO₂ pressure; a single group arrow does not supply a universal measured temperature.

Group 1 nitrates other than lithium characteristically give nitrite and oxygen: 2NaNO₃ → 2NaNO₂+O₂. Lithium nitrate behaves like group 2 nitrates: 4LiNO₃ → 2Li₂O+4NO₂+O₂. The group 2 balance is 2M(NO₃)₂ → 2MO+4NO₂+O₂.

Carbonate decomposition is not redox: C remains +4. Nitrate decomposition involves changes in oxidation state; in the nitrite route N changes from +5 to +3 while some oxygen is oxidized to O₂. Confusing NO₂ with NO₂⁻ changes both the chemistry and the atom balance.

ORIGINAL WORKED EXAMPLE

Mass loss from limestone

A pure 10.0 g CaCO₃ sample decomposes completely. Using CaCO₃=100.0, CaO=56.0 and CO₂=44.0 g mol⁻¹, calculate residue and gas masses.

  1. n(CaCO₃)=10.0/100.0=0.100 mol.
  2. The 1:1:1 balance gives 0.100 mol CaO and 0.100 mol CO₂.
  3. Residue=0.100×56.0=5.60 g; gas=0.100×44.0=4.40 g. Their sum is 10.0 g.

Answer 5.60 g CaO remains and 4.40 g CO₂ is released.

ORIGINAL PRACTICE · 1

What gases and amounts arise from complete decomposition of 0.20 mol Mg(NO₃)₂?

Reveal answer & reasoning

0.40 mol NO₂ and 0.10 mol O₂.

Divide 2Mg(NO₃)₂ → 2MgO+4NO₂+O₂ by two to obtain the per-mole ratios.

05 / UNDERSTAND & APPLY

5. Solubility is a competition of free energies

Dissolving an ionic solid requires disrupting its lattice and solvating the released ions. The balance also includes entropy: ΔG°=ΔH°−TΔS°. A stronger hydration enthalpy alone cannot prove greater solubility, because the original solid may also have a more stable lattice. The solid phase, hydration state and temperature must be specified for quantitative work.

Within the familiar Mg–Ba group 2 series, hydroxides become more soluble down the group, whereas sulfates become less soluble. MgSO₄ is readily soluble while BaSO₄ is very sparingly soluble. Group 2 carbonates are generally sparingly soluble; avoid converting that useful statement into an exact monotonic numerical trend for every member and hydrate.

For M(OH)₂(s) ⇌ M²⁺+2OH⁻, a dilute ideal solution in pure water has [M²⁺]=s and [OH⁻]≈2s when its contribution greatly exceeds water autoionization, so Ksp≈4s³. Added hydroxide changes the common-ion condition and invalidates the simple pure-water substitution.

A sparingly soluble hydroxide can still give a strongly alkaline saturated solution. Solubility measures how much solid dissolves; base strength concerns the behaviour of the dissolved species. Precipitation is predicted by comparing the ionic reaction quotient Q with Ksp, not merely by seeing ions with familiar names.

ORIGINAL WORKED EXAMPLE

A pure-water hydroxide model

For a hypothetical M(OH)₂ with Ksp=4.0×10⁻¹² at a stated temperature, estimate its molar solubility in pure water under the dilute ideal model.

  1. Set Ksp=[M²⁺][OH⁻]²=s(2s)²=4s³.
  2. s³=1.0×10⁻¹² and s=1.0×10⁻⁴ mol L⁻¹.
  3. [OH⁻]≈2.0×10⁻⁴ mol L⁻¹, much greater than 10⁻⁷ mol L⁻¹ at 25 °C, so ignoring water autoionization is consistent.

Answer s≈1.0×10⁻⁴ mol L⁻¹; these are illustrative data, not a tabulated real-salt constant.

ORIGINAL PRACTICE · 1

Would the formula Ksp=4s³ apply unchanged in a solution already containing 0.10 mol L⁻¹ NaOH?

Reveal answer & reasoning

No.

The initial hydroxide concentration contributes to [OH⁻]; for small dissolution, Ksp≈s(0.10)² rather than 4s³.

06 / UNDERSTAND & APPLY

6. Sodium hydroxide, carbonate and hydrogen carbonate

Chlor-alkali electrolysis of aqueous brine has the overall balance 2NaCl+2H₂O → 2NaOH+Cl₂+H₂, with product separation essential in industrial equipment. At the cathode water is reduced; sodium metal is not deposited from the aqueous solution. Sodium hydroxide is a strong base whose solutions absorb CO₂ to form carbonate or hydrogen carbonate depending on proportions.

In the Solvay process, ammonia assists precipitation of sparingly soluble NaHCO₃ from carbonated brine: NaCl+NH₃+CO₂+H₂O → NaHCO₃+NH₄Cl. Heating the hydrogen carbonate gives 2NaHCO₃ → Na₂CO₃+CO₂+H₂O. Ammonia is recovered using lime: 2NH₄Cl+Ca(OH)₂ → 2NH₃+CaCl₂+2H₂O.

With limestone supplying CO₂ and lime, the simplified net process is 2NaCl+CaCO₃ → Na₂CO₃+CaCl₂. The intermediate recycle explains why ammonia appears in the process but not in that net equation. Potassium carbonate is not obtained by a simple identical substitution because KHCO₃ is more soluble and does not precipitate as required.

Washing soda is Na₂CO₃·10H₂O; baking soda is NaHCO₃. Water of crystallization changes formula mass without changing the carbonate ion charge. Carbonate solution is alkaline because CO₃²⁻+H₂O ⇌ HCO₃⁻+OH⁻. Hydrogen carbonate is amphiprotic and reacts with acid to release CO₂.

ORIGINAL WORKED EXAMPLE

Heating baking soda

What mass of Na₂CO₃ remains after complete heating of 8.40 g NaHCO₃? Use formula masses 84.0 and 106.0 g mol⁻¹.

  1. Initial amount is 8.40/84.0=0.100 mol NaHCO₃.
  2. Two moles NaHCO₃ give one mole Na₂CO₃, so residue amount is 0.0500 mol.
  3. Mass=0.0500×106.0=5.30 g; the lost mass is water and carbon dioxide.

Answer 5.30 g anhydrous Na₂CO₃.

ORIGINAL PRACTICE · 1

How many moles of carbonate ions are in 28.6 g Na₂CO₃·10H₂O if its formula mass is 286 g mol⁻¹?

Reveal answer & reasoning

0.100 mol.

Each hydrated formula unit contains one carbonate; the ten waters contribute to the mass but not the carbonate count.

07 / UNDERSTAND & APPLY

7. Lime, gypsum and plaster are different chemical systems

Quicklime is CaO; slaked lime is Ca(OH)₂. Hydration CaO+H₂O → Ca(OH)₂ releases heat. Limewater becomes cloudy when CO₂ precipitates CaCO₃: Ca(OH)₂+CO₂ → CaCO₃+H₂O. With sufficient excess CO₂, the precipitate can dissolve as aqueous hydrogen carbonate: CaCO₃+CO₂+H₂O → Ca²⁺+2HCO₃⁻.

Gypsum is CaSO₄·2H₂O. Controlled partial dehydration produces plaster of Paris, conventionally CaSO₄·½H₂O. Its setting with water reforms gypsum crystals that interlock into a solid mass. The half-water formula is a composition ratio: it does not require half of an individual water molecule to exist.

The idealized setting balance is 2(CaSO₄·½H₂O)+3H₂O → 2(CaSO₄·2H₂O). Excessive heating can produce forms of anhydrous calcium sulfate with different hydration behaviour, so all dehydrated gypsum should not be treated as identical plaster. Portland cement is a multi-component calcium-silicate/aluminate material; its setting is hydration chemistry, not merely drying wet powder.

These industrial compounds demonstrate why the same calcium ion does not imply the same properties. Anion identity, water content, crystal structure and reaction conditions all matter. None of these equations is a recommendation to handle caustic quicklime outside suitable laboratory or industrial controls.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Lime chemistry and gypsum plaster hydrationTop panel shows calcium carbonate losing carbon dioxide on heating to calcium oxide, followed by hydration to calcium hydroxide. Carbonation returns hydroxide to carbonate. Bottom panel shows gypsum calcium sulfate dihydrate becoming hemihydrate by partial dehydration and returning to dihydrate during setting with one and a half additional waters per calcium sulfate.Two calcium-material cyclesCarbonate → oxide → hydroxideCaCO₃CaOCa(OH)₂heat−CO₂+H₂OCa(OH)₂ + CO₂ → CaCO₃ + H₂OGypsum and plaster: water-content changeCaSO₄·2H₂OCaSO₄·½H₂Opartial dehydrationsetting: add 1½ H₂O per CaSO₄Interlocking gypsum crystals make the set plaster solid.Formula ratios only; neither panel depicts crystal geometry.
Reaction and composition map. Arrows summarize balanced equations taught in the text; heat and hydration conditions distinguish the pathways.

ORIGINAL PRACTICE · 1

How much water is consumed in the ideal setting of 0.20 mol CaSO₄·½H₂O?

Reveal answer & reasoning

0.30 mol, or 5.4 g using 18.0 g mol⁻¹.

Each formula amount needs 1.5 mol additional water to reach the dihydrate composition.

08 / UNDERSTAND & APPLY

8. Sodium and potassium: gradients, not interchangeable salts

In typical animal cells, Na⁺ is more concentrated in extracellular fluid and K⁺ more concentrated in the cytoplasm. These distributions contribute to osmotic balance, membrane electrical behaviour and transport of other solutes. They are concentrations of dissolved ions, not deposits of elemental sodium or potassium in the body.

The Na⁺/K⁺-ATPase couples one ATP hydrolysis cycle to export of three Na⁺ and import of two K⁺. It maintains unequal distributions against their electrochemical tendencies and is electrogenic, transferring one net positive charge outward per cycle. Selective ion channels and other transporters also contribute to membrane potential; the pump alone is not the whole action potential.

Electrochemical driving force combines concentration and electrical terms. A high intracellular K⁺ concentration tends to drive K⁺ outward through an open selective channel, while a negative interior attracts it inward. Which tendency wins depends on the membrane voltage and concentration ratio. One should not say that every positive ion always moves into a cell.

Ion selectivity depends on the geometry and energetics of hydrated ions interacting with proteins. Sharing a +1 charge does not make Na⁺ and K⁺ biologically interchangeable. In plants, K⁺ also participates in osmotic regulation including stomatal function. These are biological chemistry concepts, not dietary or treatment recommendations.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Sodium and potassium transport across a cell membraneA membrane separates extracellular fluid above from cytoplasm below. One complete Na/K ATPase cycle exports three sodium ions and imports two potassium ions while hydrolyzing one ATP. Sodium is typically more concentrated outside and potassium inside animal cells. The net charge transfer is one positive charge outward.Na⁺/K⁺ pump: one complete cycleEXTRACELLULAR: higher Na⁺, lower K⁺membraneNa⁺/K⁺-ATPase3 Na⁺ out2 K⁺ inCYTOPLASM: higher K⁺, lower Na⁺1 ATP hydrolyzed per cycleNet: one positive charge transferred outwardOther channels and transporters also affect membrane voltage.
Stoichiometric transport schematic, not a molecular protein structure. Concentration labels describe typical animal cells; other transport proteins are omitted.

ORIGINAL WORKED EXAMPLE

Count ions and charge per pump cycle

A model membrane performs 200 complete Na⁺/K⁺ pump cycles. Ignore all other ion movements. Count Na⁺ exported, K⁺ imported and net positive charges exported.

  1. Each cycle exports 3 Na⁺, so 200×3=600 Na⁺ move out.
  2. Each imports 2 K⁺, so 200×2=400 K⁺ move in.
  3. Net outward positive-charge count is 600−400=200 elementary charges; 200 ATP molecules are hydrolyzed in the ideal cycle model.

Answer 600 Na⁺ outward, 400 K⁺ inward and net 200 positive charges outward.

ORIGINAL PRACTICE · 1

Why does a concentration gradient alone not completely predict K⁺ movement through a membrane channel?

Reveal answer & reasoning

The electrical potential difference also acts on the charged ion.

The relevant driving force is electrochemical; electrical and concentration contributions may oppose one another.

09 / UNDERSTAND & APPLY

9. Magnesium and calcium: coordination, structure and signals

Mg²⁺ coordinates oxygen-rich ligands and helps many enzymes function. ATP in many enzyme reactions is recognized as a magnesium-associated species: coordination modifies charge distribution and organizes the phosphate groups. Magnesium is also the central metal in chlorophyll pigments, linking coordination chemistry with light absorption in photosynthesis.

Ca²⁺ has both structural and regulatory roles. Bone and teeth contain calcium phosphate mineral, commonly represented by hydroxyapatite Ca₁₀(PO₄)₆(OH)₂, within a biological composite. Bone is not simply a block of pure calcium carbonate, and elemental calcium is not the structural material.

Free cytosolic Ca²⁺ is normally maintained far below extracellular and storage-compartment levels. Transient local increases act as signals, including in muscle contraction and secretion. Pumps, channels and binding proteins control these changes. Total calcium in a tissue and free cytosolic calcium concentration answer different questions.

The biological roles of Na⁺ and K⁺ described in the preceding section complete the s-block picture: monovalent ions support osmotic/electrical gradients, while divalent ions are important in coordination, mineral structures and signaling. This is a useful organizing comparison, not an absolute separation; all can have multiple functions. Beryllium and barium should not be inferred to be safe nutrients merely because they share a group with Mg and Ca.

ORIGINAL PRACTICE · 1

Check charge neutrality of Ca₁₀(PO₄)₆(OH)₂.

Reveal answer & reasoning

10(+2)+6(−3)+2(−1)=0.

Calcium contributes +20, phosphate −18 and hydroxide −2; the formula is electrically neutral.

10 / UNDERSTAND & APPLY

10. Revision: identify the species and the conditions

Start a salt question with the cation charge, anion identity and hydration state. Then distinguish the process: dissolution, heating, reaction with acid/base, electrolysis or a biological transport cycle. The same formula can participate in different reactions under different conditions.

Keep a compact exception set: lithium nitrate/carbonate versus the heavier alkali salts, amphoteric beryllium compounds, oxide/peroxide/superoxide distinctions, and aqueous versus molten electrolysis. For solubility, compare free energies and use the specified equilibrium model rather than memorizing an unsupported universal arrow.

In biological questions, distinguish ion distribution from pump direction and free-ion concentration from total elemental content. The worked examples and practice are original teaching cases; they are not presented as authentic past examination questions.

KEEP THE BIG IDEAS

Quick revision

  • Lithium and beryllium are important first-member exceptions.
  • Oxide, peroxide and superoxide have different anions and oxygen states.
  • Dissolution depends on free energy, not hydration enthalpy alone.
  • Hydrated salts require the complete formula mass.
  • Na/K transport and Mg/Ca coordination have specific biological roles.
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