Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds

1. Overview

This section covers the Group 2 metals magnesium, calcium, strontium and barium. They react in similar ways, so one general equation, with M standing for the metal, covers most types of reaction. What changes down the group is how readily they react, how well their hydroxides and sulfates dissolve, and how easily their carbonates and nitrates decompose. These patterns let you predict how an unfamiliar Group 2 compound will behave.

Key Definitions

• Group 2 element: an element in Group 2 of the Periodic Table, whose atoms have two outer electrons, both in an s orbital (ns²); magnesium to barium lose these two electrons to form M²⁺ ions in their compounds.
• Reducing agent: a substance that gives electrons to another substance and is itself oxidised; the Group 2 metals act as reducing agents in all their reactions on this page.
• Basic oxide: an oxide that reacts with an acid, but not with an alkali, to form a salt and water; the oxides of magnesium to barium are basic.
• Alkali: a base that dissolves in water to give a solution containing hydroxide ions, OH⁻(aq).
• Thermal decomposition: the breaking down of a compound into two or more simpler substances by heating alone, with no other reactant.
• Thermal stability: how well a compound resists thermal decomposition; a more thermally stable compound has to be heated to a higher temperature before it decomposes.
• Solubility: the maximum amount of a solute that dissolves in a given amount of solvent at a stated temperature, giving a saturated solution.
• Precipitate: an insoluble solid that forms in a solution during a reaction, for example when two solutions are mixed.
• Limewater: a saturated aqueous solution of calcium hydroxide; carbon dioxide gives a white precipitate of calcium carbonate with it.

Content

Group 2 at a glance

Each atom has two electrons in its outer s sub-shell: Mg [Ne]3s², Ca [Ar]4s², Sr [Kr]5s², Ba [Xe]6s². In every reaction of the metals on this page, the atom loses both outer electrons:

M → M²⁺ + 2e⁻

So the metal is oxidised and acts as a reducing agent, and its compounds contain the M²⁺ ion. Formulae follow from the charges: MO, M(OH)₂, MCO₃, M(NO₃)₂, MSO₄ and MCl₂.

Property	Mg to Ba	Why
Atomic radius	increases	one more occupied shell each step
M²⁺ ion radius	increases	same reason
1st + 2nd ionisation energy	decreases	outer electrons further out, more shielded
Electronegativity	decreases	larger atoms attract bonding electrons less
Reactivity	increases	outer electrons lost more easily
Melting point	no simple trend	do not predict it from position

The Data Booklet values show the ionisation energy trend. Adding the first and second ionisation energies gives 2186 kJ mol⁻¹ for Mg (736 + 1450), 1740 for Ca, 1608 for Sr and 1468 for Ba. The nuclear charge rises down the group, but the greater distance and extra shielding outweigh it. The booklet's electronegativities fall from 1.3 (Mg) to 1.0 (Ca); it gives none for Sr and Ba.

Reactions of the metals with oxygen

All four metals burn when heated in air or oxygen, forming a white solid oxide:

2M(s) + O₂(g) → 2MO(s); for example, 2Ca(s) + O₂(g) → 2CaO(s)

Metal	Flame when it burns
magnesium	brilliant white
calcium	brick-red (orange-red)
strontium	red (crimson)
barium	pale green

Burning magnesium in air also forms some magnesium nitride, Mg₃N₂, and barium in plenty of oxygen also forms some barium peroxide, BaO₂, but the equation to learn is the one for the oxide, MO.

Reactions of the metals with water

Magnesium reacts only very slowly with cold water: a few bubbles of hydrogen form over a long time.

Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g)

Heated magnesium reacts quickly with steam, burning with a bright white glow. The product is white magnesium oxide, not the hydroxide:

Mg(s) + H₂O(g) → MgO(s) + H₂(g)

Calcium, strontium and barium react with cold water, fizzing steadily as hydrogen forms, and the reaction gets more vigorous down the group. The metal disappears and the solution formed is alkaline:

M(s) + 2H₂O(l) → M(OH)₂(aq) + H₂(g)

With calcium, a white precipitate forms and the mixture looks cloudy, because calcium hydroxide is only slightly soluble. Barium hydroxide is much more soluble, so far less solid appears.

Reactions of the metals with dilute acids

With dilute hydrochloric acid, every metal fizzes and dissolves, giving a colourless solution of the chloride. The reaction is faster down the group.

M(s) + 2HCl(aq) → MCl₂(aq) + H₂(g); ionic equation: M(s) + 2H⁺(aq) → M²⁺(aq) + H₂(g)

With dilute sulfuric acid, the result depends on whether the sulfate dissolves:

• Magnesium reacts like it does with hydrochloric acid, because magnesium sulfate is soluble: Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g).
• Calcium, strontium and barium fizz at first, but their sulfates are only slightly soluble or insoluble. A layer of solid sulfate forms on the metal and keeps the acid away, so the reaction slows and may stop before all the metal has reacted: M(s) + H₂SO₄(aq) → MSO₄(s) + H₂(g).

Worked example 1: identifying a metal from the hydrogen it gives

Question: 0.360 g of a Group 2 metal, M, reacts completely with an excess of dilute hydrochloric acid. The hydrogen collected occupies 98.6 cm³ at room conditions. (a) Identify M. (b) Describe what you would see if a piece of M were added to dilute sulfuric acid instead, and write the equation. ($A_{\text{r}}$: Mg 24.3, Ca 40.1, Sr 87.6, Ba 137.3; $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ at room conditions)

Step 1: use the equation. M(s) + 2HCl(aq) → MCl₂(aq) + H₂(g), so 1 mol of M gives 1 mol of H₂.

Step 2: amount of hydrogen. 24.0 dm³ = 24 000 cm³. $$n(\text{H}_2) = \frac{98.6}{24\,000} = 4.108 \times 10^{-3}\ \text{mol}$$

Step 3: amount and $A_{\text{r}}$ of the metal. $n(\text{M}) = 4.108 \times 10^{-3}$ mol, so $$\text{molar mass of M} = \frac{0.360}{4.108 \times 10^{-3}} = 87.6\ \text{g mol}^{-1}$$

Step 4: match. 87.6 is the $A_{\text{r}}$ of strontium.

Step 5: sulfuric acid. Fizzing starts, but a white solid, strontium sulfate, forms on the metal, so the fizzing slows and may stop before the metal is used up. Sr(s) + H₂SO₄(aq) → SrSO₄(s) + H₂(g).

Answer: (a) $A_{\text{r}}$ = 87.6 (3 s.f.), so M is strontium. (b) Brief fizzing that slows or stops as insoluble SrSO₄ coats the metal.

Oxides, hydroxides and carbonates with water

Oxides react with water to form the hydroxide, MO(s) + H₂O(l) → M(OH)₂, with state symbol (s) or (aq) depending on how much dissolves. No gas is given off.

• Magnesium oxide reacts only slightly. The white solid seems unchanged, and the mixture is only weakly alkaline.
• Calcium oxide reacts vigorously with a little water, releasing a lot of heat, and crumbles to a white powder: CaO(s) + H₂O(l) → Ca(OH)₂(s).
• Strontium oxide and barium oxide react readily and dissolve, giving strongly alkaline solutions: BaO(s) + H₂O(l) → Ba(OH)₂(aq).

Hydroxides dissolve only to a limited extent (magnesium hydroxide hardly at all), releasing hydroxide ions:

M(OH)₂(s) ⇌ M²⁺(aq) + 2OH⁻(aq)

The more hydroxide dissolves, the higher the concentration of OH⁻(aq) and the higher the pH. As typical values, a saturated solution of magnesium hydroxide has a pH of about 10 and one of barium hydroxide a pH above 13, with calcium and strontium hydroxides in between.

Carbonates of all four metals are insoluble in water (their solubilities are very low), so no reaction is seen.

Oxides, hydroxides and carbonates with dilute acids

All three types of compound react with acids as bases, forming a salt and water. Carbonates also give carbon dioxide.

With dilute hydrochloric acid:

• Oxide: MO(s) + 2HCl(aq) → MCl₂(aq) + H₂O(l)
• Hydroxide: M(OH)₂(s) + 2HCl(aq) → MCl₂(aq) + 2H₂O(l)
• Carbonate: MCO₃(s) + 2HCl(aq) → MCl₂(aq) + H₂O(l) + CO₂(g)

With dilute sulfuric acid:

• Oxide: MO(s) + H₂SO₄(aq) → MSO₄ + H₂O(l)
• Hydroxide: M(OH)₂(s) + H₂SO₄(aq) → MSO₄ + 2H₂O(l)
• Carbonate: MCO₃(s) + H₂SO₄(aq) → MSO₄ + H₂O(l) + CO₂(g)

Give MSO₄ the state symbol (aq) for magnesium and (s) for calcium, strontium and barium.

What you see:

• With hydrochloric acid, the white solid dissolves to give a colourless solution. Only the carbonates fizz; the carbon dioxide gives a white precipitate with limewater: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l).
• With sulfuric acid, the magnesium compounds behave in the same way. For calcium, strontium and barium, the slightly soluble or insoluble sulfate coats the solid, so a carbonate's fizzing soon slows or stops. Barium hydroxide solution gives a white precipitate: Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2H₂O(l).

Thermal decomposition of carbonates and nitrates

Carbonates break down into the oxide and carbon dioxide:

MCO₃(s) → MO(s) + CO₂(g)

The white solid stays white, its mass falls, and the gas gives a white precipitate with limewater.

Nitrates break down into the oxide, nitrogen dioxide and oxygen:

2M(NO₃)₂(s) → 2MO(s) + 4NO₂(g) + O₂(g)

You see brown fumes of nitrogen dioxide, an acidic gas that turns damp blue litmus paper red; the oxygen is colourless. Hydrated nitrate crystals give off water first, which condenses on the cool part of the tube. To check the balancing, count oxygen atoms: 12 on the left, 2 + 8 + 2 = 12 on the right.

The trend: thermal stability increases down the group for both carbonates and nitrates. Of the four carbonates, magnesium carbonate decomposes at the lowest temperature and barium carbonate needs the highest; the nitrates follow the same order.

Comparing them in the laboratory. Heat equal amounts, in moles, of each carbonate in identical tubes with the same flame, and time how long the limewater takes to give a white precipitate (for nitrates, until brown gas appears). A shorter time means a less thermally stable compound. In a Bunsen flame, strontium and barium carbonates may not give a precipitate at all in the time available; record "no white precipitate" and treat it as a sign of high thermal stability.

Checking the decomposition is complete. Heat, cool and weigh, and repeat until two masses agree closely (heating to constant mass). If dilute acid added to the cooled residue fizzes, some carbonate is still there.

At AS Level you describe this trend. Explaining it, through the size of the M²⁺ ion and its effect on the large anion, is A Level work (topic 27.1).

Worked example 2: heating barium nitrate

Question: A student heats 6.25 g of barium nitrate, Ba(NO₃)₂, in a fume cupboard until it has fully decomposed. (a) Write the equation, with state symbols. (b) Calculate the mass of solid left. (c) Calculate the volume of oxygen formed, at room conditions. (d) Calculate the percentage loss in mass. (e) Would the same amount of magnesium nitrate decompose at a higher or a lower temperature? ($A_{\text{r}}$: Ba 137.3, N 14.0, O 16.0; $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ at room conditions)

Step 1: equation. 2Ba(NO₃)₂(s) → 2BaO(s) + 4NO₂(g) + O₂(g)

Step 2: amount of barium nitrate. $M_{\text{r}}$ of Ba(NO₃)₂ = 137.3 + 2(14.0 + 3 × 16.0) = 261.3; $M_{\text{r}}$ of BaO = 137.3 + 16.0 = 153.3. $$n = \frac{6.25}{261.3} = 0.023919\ \text{mol}$$

Step 3: mass of residue. 2 mol of Ba(NO₃)₂ gives 2 mol of BaO, so $n(\text{BaO}) = 0.023919$ mol. $$\text{mass} = 0.023919 \times 153.3 = 3.667\ \text{g}$$

Step 4: volume of oxygen. 2 mol of Ba(NO₃)₂ gives 1 mol of O₂, so $n(\text{O}_2) = 0.023919 \div 2 = 0.011960$ mol. $$V = 0.011960 \times 24.0 = 0.2870\ \text{dm}^3 = 287\ \text{cm}^3$$

Step 5: percentage loss in mass. $$\frac{6.25 - 3.667}{6.25} \times 100 = 41.3\%$$

As a check, the 2.583 g lost should equal the mass of the gases. With $n(\text{NO}_2) = 2 \times 0.023919 = 0.04784$ mol: $0.04784 \times 46.0 + 0.011960 \times 32.0 = 2.583$ g.

Step 6: trend. Magnesium is above barium, and thermal stability increases down the group, so magnesium nitrate decomposes at a lower temperature.

Answer: (a) the equation in Step 1; (b) 3.67 g of barium oxide; (c) 287 cm³ of oxygen; (d) 41.3% (all 3 s.f.); (e) a lower temperature.

Solubility of the hydroxides and sulfates

The two trends run in opposite directions:

Metal	Hydroxide	Sulfate
Mg	very sparingly soluble	soluble
Ca	slightly soluble	slightly soluble
Sr	moderately soluble	very sparingly soluble
Ba	soluble	insoluble

• The solubility of the hydroxides increases down the group.
• The solubility of the sulfates decreases down the group.

They explain two common tests:

• Adding aqueous sodium hydroxide to a solution of Mg²⁺ ions gives a white precipitate of magnesium hydroxide: Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s). Ca²⁺(aq) gives a white precipitate unless the solution is very dilute, while Ba²⁺(aq) gives at most a faint one.
• Adding Ba²⁺(aq), as barium chloride or barium nitrate solution, to a solution of sulfate ions gives a white precipitate even when the solution is dilute: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). This is the test for sulfate ions (the precipitate does not dissolve in excess dilute strong acid). Ca²⁺(aq) gives a precipitate only when it is concentrated, and Mg²⁺(aq) gives none.

Explaining these solubility trends, using lattice energies and enthalpy changes of hydration, is A Level work (topic 27.1). At AS Level you state them.

Making predictions from the trends

Property, magnesium to barium	Trend
Reaction of the metal with oxygen, water and dilute acids	more vigorous
Reaction of the oxide with water	more vigorous; the solution formed is more alkaline
Solubility of the hydroxide, and pH of its saturated solution	increase
Solubility of the sulfate	decreases
Thermal stability of the carbonate and the nitrate	increases

To predict for strontium, place it between calcium and barium: strontium sulfate is less soluble than calcium sulfate but more soluble than barium sulfate. For an element below barium, extend each trend one step further; it still forms M²⁺ ions, so the formulae and equation types stay the same. Take care extrapolating upwards: beryllium, above magnesium, differs in some ways (its oxide and hydroxide are amphoteric).

Worked example 3: identifying three white solids

Question: Three unlabelled white powders are magnesium oxide, calcium carbonate and barium hydroxide. Using only water, Universal Indicator, dilute hydrochloric acid and limewater, describe how to identify each, with an equation for each reaction with the acid.

Step 1: water and indicator. Shake a little of each powder with water and add Universal Indicator. Barium hydroxide dissolves and the indicator turns purple (strongly alkaline). The other two do not visibly dissolve, and the indicator shows at most a weakly alkaline colour.

Step 2: acid on the other two. Add dilute hydrochloric acid. Calcium carbonate fizzes, and the gas gives a white precipitate with limewater (carbon dioxide). Magnesium oxide dissolves with no fizzing, giving a colourless solution.

Step 3: equations.

• CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)
• MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l)
• Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)

Answer: dissolves in water, strongly alkaline: barium hydroxide. Fizzes with acid: calcium carbonate. Dissolves in acid without fizzing: magnesium oxide.

Key Equations

General equations (M = Mg, Ca, Sr or Ba):

• Metal + oxygen: 2M(s) + O₂(g) → 2MO(s)
• Metal + cold water (Ca, Sr, Ba): M(s) + 2H₂O(l) → M(OH)₂(aq) + H₂(g)
• Magnesium + steam: Mg(s) + H₂O(g) → MgO(s) + H₂(g)
• Metal + dilute acid: M(s) + 2H⁺(aq) → M²⁺(aq) + H₂(g)
• Oxide + water: MO(s) + H₂O(l) → M(OH)₂
• Oxide + dilute acid: MO(s) + 2H⁺(aq) → M²⁺(aq) + H₂O(l)
• Hydroxide + dilute acid: M(OH)₂(s) + 2H⁺(aq) → M²⁺(aq) + 2H₂O(l)
• Carbonate + dilute acid: MCO₃(s) + 2H⁺(aq) → M²⁺(aq) + H₂O(l) + CO₂(g)
• Carbonate, heated: MCO₃(s) → MO(s) + CO₂(g)
• Nitrate, heated: 2M(NO₃)₂(s) → 2MO(s) + 4NO₂(g) + O₂(g)

Give M(OH)₂ the state symbol (s) or (aq) according to how much dissolves: (s) for magnesium, and for calcium when only a little water is used. With sulfuric acid, the ionic equations above apply only to magnesium. For calcium, strontium and barium the sulfate forms as a solid, so write the full equation with MSO₄(s).

Amount from a mass ($m$ in g; molar mass in g mol⁻¹, the same number as $A_{\text{r}}$ or $M_{\text{r}}$): $$n = \frac{m}{\text{molar mass}}$$

Amount of a gas at room conditions ($V$ in dm³; $V_{\text{m}}$ = 24.0 dm³ mol⁻¹ at room conditions). If a question gives a temperature and pressure instead, use $pV = nRT$ (topic 4.1). $$n = \frac{V}{V_{\text{m}}}$$

Common Mistakes to Avoid

• Mistake: giving the steam reaction when asked about magnesium and cold water, or the reverse. Fix: cold water gives the hydroxide, Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g), very slowly; steam gives the oxide, Mg(s) + H₂O(g) → MgO(s) + H₂(g).
• Mistake: an unbalanced nitrate equation, or one with no oxygen. Fix: write 2M(NO₃)₂ → 2MO + 4NO₂ + O₂ (ratio 2 : 2 : 4 : 1) and count oxygen atoms: 12 on each side. In Worked example 2, 0.023919 mol of Ba(NO₃)₂ gives 0.04784 mol of NO₂ and 0.011960 mol of O₂.
• Mistake: describing fizzing when an oxide or hydroxide reacts with water or an acid. Fix: only the metals give a gas with water (hydrogen); with dilute acids, only the metals (hydrogen) and the carbonates (carbon dioxide) do. SrO(s) + H₂O(l) → Sr(OH)₂(aq) gives a solution and heat, but no bubbles.
• Mistake: mixing up the two solubility trends. Fix: hydroxides become more soluble down the group, sulfates less soluble. So NaOH(aq) precipitates Mg(OH)₂ from Mg²⁺(aq), while Ba²⁺(aq) is the ion that precipitates sulfate, as BaSO₄.
• Mistake: wrong formulae such as MgOH, CaCO₂ or Mg(CO₃)₂, or BaSO₄(aq). Fix: build each formula from the charges: M²⁺ needs two OH⁻, one CO₃²⁻ or two NO₃⁻, so Mg(OH)₂, CaCO₃ and Mg(NO₃)₂. Give insoluble products the state symbol (s): Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
• Mistake: saying compounds lower in the group decompose more easily. Fix: thermal stability increases down the group. In the same flame, magnesium carbonate gives carbon dioxide sooner than barium carbonate.
• Mistake: vague practical descriptions, or assuming a decomposition is complete. Fix: write "a white precipitate forms" for carbon dioxide and limewater, not "it goes milky". Heat to constant mass, and treat fizzing when acid is added to the residue as a sign that decomposition was incomplete.

Exam Tips

• Describe a reaction by saying what you would see (fizzing, flame colour, a solid dissolving, a white precipitate, brown fumes) and naming the products; add a balanced equation with state symbols if asked.
• State a trend with its direction and the compounds it applies to: "the solubility of the hydroxides increases from Mg(OH)₂ to Ba(OH)₂", not just "it increases".
• Predict by placing the element in the group, applying the trend, and comparing it with a named neighbour: "strontium sulfate is less soluble than calcium sulfate".
• Distinguish two substances by naming one reagent and giving the result for each: with dilute sulfuric acid, strontium chloride solution gives a white precipitate of SrSO₄, but magnesium chloride solution shows no visible change.
• Calculate with the equation's mole ratio written down first. Give answers with units, to as many significant figures as the data justify (usually 3), and identify an element by the $A_{\text{r}}$ closest to your result.
• Keep to AS depth: describe the thermal stability and solubility trends. The explanations (ion size, lattice and hydration energies) belong to A Level topic 27.1.