Group 2 Elements and Their Reactions

AS · 11 min

The Group 2 metals, magnesium to barium, are the alkaline earth metals. They are a clean example of a group trend: the same reactions, the same formula types, and a reactivity that rises steadily down the group. This note covers their physical trends and their reactions with oxygen, water and dilute hydrochloric and sulfuric acids, with the equations and observations Cambridge expects, and explains the reactivity trend using ionisation energy. The compounds (oxides, hydroxides, carbonates, nitrates and sulfates) follow in the next note.

Electronic structure and physical trends

Every Group 2 atom has two electrons in its outer s sub-shell ( $ns^2$ ). In all their compounds the elements lose both to form $\ce{M^2+}$ ions, with oxidation number $+2$ .

element	electron configuration	atomic radius / nm	first IE / kJ mol⁻¹	second IE / kJ mol⁻¹ (approximate)
Mg	$[\ce{Ne}]3s^2$	0.160	736	1450
Ca	$[\ce{Ar}]4s^2$	0.197	590	1150
Sr	$[\ce{Kr}]5s^2$	0.215	548	1060
Ba	$[\ce{Xe}]6s^2$	0.224	502	970

Key result

Trends down Group 2

Atomic radius increases: each element has one more occupied shell.
Ionisation energies decrease: the outer electrons are further from the nucleus and more shielded by inner shells. This outweighs the increase in nuclear charge, so less energy is needed to remove them.
Reactivity increases: the metals lose their two outer electrons more easily, forming $\ce{M^2+}$ .
Ionic radius of $\ce{M^2+}$ increases for the same reason as atomic radius.

Melting points generally fall from calcium to barium, because the metallic bonding gets weaker as the ions get larger (the delocalised electrons are further from the ion centres). Magnesium does not fit the pattern neatly because it has a different metallic crystal structure; you will not be asked to explain this.

Reactions with oxygen

All the Group 2 metals burn in oxygen (or air) when heated to form white ionic oxides, $\ce{MO}$ .

\ce{2M(s) + O2(g) -> 2MO(s)}

metal	observation when burned	equation
Mg	brilliant white flame, white solid	$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$
Ca	brick-red (orange-red) flame, white solid	$\ce{2Ca(s) + O2(g) -> 2CaO(s)}$
Sr	crimson-red flame, white solid	$\ce{2Sr(s) + O2(g) -> 2SrO(s)}$
Ba	apple-green flame, white solid	$\ce{2Ba(s) + O2(g) -> 2BaO(s)}$

The metals react more readily down the group; calcium, strontium and barium tarnish quickly in air at room temperature and are stored under oil.

Tip

Extension: in excess oxygen, strontium and barium also form some peroxide, for example $\ce{Ba + O2 -> BaO2}$ (barium peroxide, containing the $\ce{O2^2-}$ ion). The syllabus only expects the oxides $\ce{MO}$ .

Reactions with water

Reactivity with water increases clearly down the group. The general equation with cold water is

\ce{M(s) + 2H2O(l) -> M(OH)2(aq) + H2(g)}

metal	reaction with cold water	product
Mg	very slow; a few bubbles over a long time	$\ce{Mg(OH)2}$ , almost insoluble; pH about 9–10
Ca	steady fizzing; metal sinks then rises with bubbles; white suspension forms	$\ce{Ca(OH)2}$ , slightly soluble (cloudy); pH about 11–12
Sr	rapid fizzing	$\ce{Sr(OH)2}$ , more soluble; pH about 13
Ba	vigorous fizzing	$\ce{Ba(OH)2}$ , soluble; pH about 13–14

With steam, magnesium reacts rapidly, burning with a white flame to form the oxide:

\ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)}

Two trends combine here. The metal is more reactive further down (faster fizzing), and the hydroxide formed is more soluble further down (higher pH). Both are worth stating when describing these reactions.

Watch out

Write $\ce{Ca(OH)2}$ as $\ce{Ca(OH)2(s)}$ or $\ce{Ca(OH)2(aq)}$ depending on context: with a lot of calcium it forms a white suspension because calcium hydroxide is only slightly soluble. "Calcium dissolves to form a colourless solution" is not what is seen; describe the cloudiness.

Reactions with dilute acids

Dilute hydrochloric acid

All the metals react with dilute hydrochloric acid, fizzing to give hydrogen and a colourless solution of the chloride (all Group 2 chlorides are soluble). The reaction becomes more vigorous down the group.

\ce{M(s) + 2HCl(aq) -> MCl2(aq) + H2(g)}

The ionic equation is the same for all of them: $\ce{M(s) + 2H+(aq) -> M^2+(aq) + H2(g)}$ .

Dilute sulfuric acid

\ce{M + H2SO4 -> MSO4 + H2}

Here the solubility of the sulfate matters.

Magnesium reacts vigorously and completely: magnesium sulfate is soluble, giving a colourless solution. $\ce{Mg(s) + H2SO4(aq) -> MgSO4(aq) + H2(g)}$
Calcium reacts at first, but the fizzing slows and a white solid forms: calcium sulfate is only sparingly soluble and coats the metal.
Strontium and barium start to react but stop almost at once: the insoluble sulfate ( $\ce{SrSO4}$ , $\ce{BaSO4}$ ) forms a layer on the surface of the metal, preventing further contact with the acid. $\ce{Ba(s) + H2SO4(aq) -> BaSO4(s) + H2(g)}$

So with sulfuric acid, the observed reactivity appears to decrease down the group, even though the metals themselves are more reactive. This is a favourite exam trap.

Explaining the trend in reactivity

Key result

Why reactivity increases down Group 2

In every reaction the metal atom loses its two outer electrons: $\ce{M -> M^2+ + 2e-}$ .
Down the group the atomic radius increases and there is more shielding by inner shells.
The outer electrons are attracted less strongly by the nucleus (the increase in nuclear charge is outweighed).
So the sum of the first and second ionisation energies decreases, the electrons are lost more easily, and the metal reacts more readily.

Worked examples

Calcium and water

Describe what you would see when a small piece of calcium is added to water containing a few drops of universal indicator. Write an equation with state symbols.

Solution

Observations: effervescence (bubbles of gas), the calcium sinks and may rise with the bubbles, it gradually disappears, the mixture turns cloudy white (a suspension of calcium hydroxide), the indicator turns blue/purple, and the mixture gets warm.

\ce{Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g)}

(Calcium hydroxide may be shown as $\ce{(s)}$ when the suspension is mentioned.)

Explaining the reactivity trend

Explain why barium reacts more vigorously with water than magnesium does.

Solution

Both metals react by losing two electrons to form $\ce{M^2+}$ ions. A barium atom is much larger than a magnesium atom: its outer ( $6s$ ) electrons are further from the nucleus and shielded by more inner shells. The attraction between the nucleus and the outer electrons is weaker, despite the larger nuclear charge, so the first and second ionisation energies are lower. Barium loses its outer electrons more easily, so it reacts more vigorously.

Exam-style: the sulfuric acid trap

Equal pieces of magnesium and of barium are added to separate samples of dilute sulfuric acid. Magnesium reacts completely, but the barium stops reacting almost immediately. Explain these observations, with equations.

Solution

$\ce{Mg(s) + H2SO4(aq) -> MgSO4(aq) + H2(g)}$

$\ce{Ba(s) + H2SO4(aq) -> BaSO4(s) + H2(g)}$

Magnesium sulfate is soluble, so it dissolves as it forms and the acid can keep reaching the metal. Barium sulfate is insoluble: it forms a coating on the surface of the barium which stops the acid reaching the metal, so the reaction stops even though barium is the more reactive metal. The solubility of the Group 2 sulfates decreases down the group.

Exam-style: barium and water, with a titration

$1.37\ \text{g}$ of barium is added to water. When the reaction is complete, the solution is made up to $250\ \text{cm}^3$ .

(a) Calculate the volume of hydrogen produced at room conditions.

(b) A $25.0\ \text{cm}^3$ portion is titrated with $0.100\ \text{mol dm}^{-3}$ hydrochloric acid. Calculate the expected titre.

( $A_r$ : Ba 137.3)

Solution

(a) $n(\ce{Ba}) = \dfrac{1.37}{137.3} = 9.978 \times 10^{-3}\ \text{mol}$

$\ce{Ba + 2H2O -> Ba(OH)2 + H2}$ : $n(\ce{H2}) = 9.978 \times 10^{-3}\ \text{mol}$

$V = 9.978 \times 10^{-3} \times 24.0 = 0.239\ \text{dm}^3$ ( $239\ \text{cm}^3$ ).

(b) $n(\ce{Ba(OH)2})$ in $25.0\ \text{cm}^3 = 9.978 \times 10^{-4}\ \text{mol}$ .

$\ce{Ba(OH)2 + 2HCl -> BaCl2 + 2H2O}$ : $n(\ce{HCl}) = 2 \times 9.978 \times 10^{-4} = 1.996 \times 10^{-3}\ \text{mol}$

V = \frac{1.996 \times 10^{-3}}{0.100} = 0.01996\ \text{dm}^3 = 19.96\ \text{cm}^3

Exam-hard: predicting the chemistry of radium

Radium, Ra, is below barium in Group 2. Predict, with reasons: (a) its reaction with cold water, with an equation; (b) its reaction with dilute sulfuric acid; (c) the relative sizes of its first ionisation energy and that of barium.

Solution

(a) Radium should react very vigorously with cold water, more so than barium, because its outer ( $7s$ ) electrons are even further from the nucleus and more shielded, so they are lost more easily. $\ce{Ra(s) + 2H2O(l) -> Ra(OH)2(aq) + H2(g)}$ . The hydroxide should be very soluble (solubility of hydroxides increases down the group), so the solution would be strongly alkaline.

(b) It would start to react but stop almost at once, because radium sulfate would be insoluble (even less soluble than $\ce{BaSO4}$ ) and would coat the metal. $\ce{Ra(s) + H2SO4(aq) -> RaSO4(s) + H2(g)}$

(c) Radium's first ionisation energy should be lower than barium's (larger atom, more shielding). (In reality it is very slightly higher because of relativistic effects in very heavy atoms, but the expected A Level answer is "lower", with the reason.)

Watch out

The reaction of Mg with cold water gives $\ce{Mg(OH)2}$ ; with steam it gives $\ce{MgO}$ .
"More reactive because the atom has more electrons" is not an explanation. Talk about distance from the nucleus, shielding and ionisation energy.
Do not say barium is unreactive with sulfuric acid. It reacts, but the insoluble sulfate layer stops the reaction.
Group 2 metals form $\ce{M^2+}$ only: formulas are $\ce{MO}$ , $\ce{M(OH)2}$ , $\ce{MCl2}$ , $\ce{MSO4}$ , $\ce{MCO3}$ , $\ce{M(NO3)2}$ .

Exam tip

Questions often give the trend for two metals and ask you to predict a third (usually Sr). State the prediction and the trend it comes from.
Observation marks: "effervescence" or "fizzing" (not "gas produced", which is a conclusion), "metal disappears", "white precipitate / suspension", "heat released".
Write ionic equations when asked: $\ce{Ca(s) + 2H+(aq) -> Ca^2+(aq) + H2(g)}$ .
The explanation for reactivity carries about three marks: larger radius (or more shells), more shielding, lower ionisation energy so electrons lost more easily.

Practical skills

To compare the reactivity of Group 2 metals with water or acid fairly, use equal masses (or equal surface areas) of metal, the same volume and concentration of acid, and the same starting temperature. The dependent variable can be the volume of hydrogen collected in a gas syringe (or over water in an inverted measuring cylinder) after a fixed time, or the time for the metal to disappear. Sources of error: oxide coatings on the metals (clean magnesium ribbon with emery paper first), gas lost before the bung is inserted, and the heat released raising the temperature. With sulfuric acid, the insoluble sulfate coating makes rate comparisons for Ca, Sr and Ba meaningless; use hydrochloric acid instead.

Summary

Group 2 atoms are $ns^2$ ; they form $\ce{M^2+}$ (oxidation number $+2$ ).
Down the group: atomic radius increases, ionisation energies decrease, reactivity increases.
With oxygen: $\ce{2M + O2 -> 2MO}$ (Mg white flame, Ca brick red, Sr crimson, Ba apple green).
With water: $\ce{M + 2H2O -> M(OH)2 + H2}$ , faster down the group; Mg is very slow with cold water but burns in steam to give $\ce{MgO}$ .
With $\ce{HCl}$ : $\ce{M + 2HCl -> MCl2 + H2}$ , more vigorous down the group.
With $\ce{H2SO4}$ : Mg reacts fully; Ca, Sr, Ba stop as insoluble sulfate coats the metal.
Reactivity increases because the two outer electrons are further from the nucleus and more shielded, so are lost more easily.

Practice

Question

Write the electron configuration of strontium using a noble-gas core, and state its oxidation number in its compounds.
Write equations, with state symbols, for the reactions of strontium with oxygen and with water.
Describe two differences between the reaction of magnesium with cold water and the reaction of barium with cold water.
Write the ionic equation for the reaction of calcium with dilute hydrochloric acid.
Explain why the first ionisation energy decreases from magnesium to barium.
Calculate the volume of hydrogen, at room conditions, formed when $0.243\ \text{g}$ of magnesium reacts with excess dilute hydrochloric acid. ( $A_r$ : Mg 24.3)
A student says, "Calcium is less reactive than magnesium, because calcium stops reacting with sulfuric acid but magnesium does not." Explain why this conclusion is wrong.
Calculate the mass of barium sulfate formed when $0.0100\ \text{mol}$ of barium reacts with excess dilute sulfuric acid, assuming the reaction goes to completion. ( $A_r$ : Ba 137.3, S 32.1, O 16.0)
Strontium is heated in steam. Predict the products and write an equation. Explain why, unlike magnesium, strontium does not need steam to react with water at a useful rate.
A $0.200\ \text{g}$ sample of a Group 2 metal M reacts with excess water, producing $120\ \text{cm}^3$ of hydrogen at room conditions. Identify M and predict two observations you would make during the reaction.

Answers

$[\ce{Kr}]5s^2$ ; oxidation number $+2$ .
$\ce{2Sr(s) + O2(g) -> 2SrO(s)}$ ; $\ce{Sr(s) + 2H2O(l) -> Sr(OH)2(aq) + H2(g)}$ .
Any two: barium fizzes vigorously while magnesium gives only a few bubbles very slowly; barium disappears quickly while magnesium barely changes; the solution with barium is strongly alkaline (pH 13–14) whereas with magnesium it is only weakly alkaline (pH 9–10), because $\ce{Ba(OH)2}$ is much more soluble than $\ce{Mg(OH)2}$ ; the barium reaction releases more heat.
$\ce{Ca(s) + 2H+(aq) -> Ca^2+(aq) + H2(g)}$ .
Down the group the atoms have more occupied shells, so the outer electron is further from the nucleus and more shielded by inner electrons. The increased nuclear charge is outweighed, so the attraction for the outer electron is weaker and less energy is needed to remove it.
$n(\ce{Mg}) = 0.243 / 24.3 = 0.0100\ \text{mol} = n(\ce{H2})$ ; $V = 0.0100 \times 24.0 = 0.240\ \text{dm}^3 = 240\ \text{cm}^3$ .
Calcium is more reactive than magnesium (lower ionisation energies). It stops reacting with sulfuric acid because calcium sulfate is only sparingly soluble and forms a coating on the metal that keeps the acid away. Magnesium sulfate is soluble, so magnesium keeps reacting. The observation reflects sulfate solubility, not metal reactivity; hydrochloric acid would show calcium reacting faster.
$\ce{Ba + H2SO4 -> BaSO4 + H2}$ : $n(\ce{BaSO4}) = 0.0100\ \text{mol}$ ; $M = 137.3 + 32.1 + 64.0 = 233.4\ \text{g mol}^{-1}$ ; mass $= 2.33\ \text{g}$ .
$\ce{Sr(s) + H2O(g) -> SrO(s) + H2(g)}$ : strontium oxide and hydrogen. Strontium is lower in the group than magnesium, so its outer electrons are further from the nucleus and more shielded, its ionisation energies are lower and it loses electrons more readily. It therefore reacts rapidly even with cold water, forming $\ce{Sr(OH)2}$ and hydrogen.
$n(\ce{H2}) = 0.120 / 24.0 = 0.00500\ \text{mol}$ . $\ce{M + 2H2O -> M(OH)2 + H2}$ is $1 : 1$ , so $n(\ce{M}) = 0.00500\ \text{mol}$ and $A_r = 0.200 / 0.00500 = 40.0$ : calcium. Observations: steady effervescence; the metal gradually disappears; a white suspension (cloudy mixture) of calcium hydroxide forms; the mixture becomes warm.

Electronic structure and physical trends

Reactions with oxygen

Reactions with water

Reactions with dilute acids

Dilute hydrochloric acid

Dilute sulfuric acid

Explaining the trend in reactivity

Worked examples

Practice

Console