Topic 8 · page 10 of 29

Reactions of Group 1 and 2 metals

The metals burn in oxygen and chlorine to give oxides and chlorides, and react with water to give a hydroxide and hydrogen. In every case the metal is oxidised, and the vigour increases down the group.

13 min read8.13

Read first: Ionisation energy and reactivity down Groups 1 and 2

With oxygen and chlorine 8.13

Group 1 metals form 1+ ions and Group 2 metals form 2+ ions. The formula of the product follows from that, so you can write each equation from the group number alone.

With oxygen With chlorine
Group 1 (Li, Na, K) 4 M(s)+OX2(g)→2 MX2O(s)\ce{4M(s) + O2(g) -> 2M2O(s)} 2 M(s)+ClX2(g)→2 MCl(s)\ce{2M(s) + Cl2(g) -> 2MCl(s)}
Group 2 (Mg, Ca, Sr, Ba) 2 M(s)+OX2(g)→2 MO(s)\ce{2M(s) + O2(g) -> 2MO(s)} M(s)+ClX2(g)→MClX2(s)\ce{M(s) + Cl2(g) -> MCl2(s)}

All the products are white ionic solids: oxides containing M⁺ or M²⁺ and O²⁻, and chlorides containing M⁺ or M²⁺ and Cl⁻.

Worked exampleWriting the equations from the group number

Write the equations for lithium with oxygen, and for calcium with chlorine.

  1. Lithium is Group 1, so Li⁺. The oxide has O²⁻, so two Li⁺ to one O²⁻: Li₂O. Balance: 4 Li+OX2→2 LiX2O\ce{4Li + O2 -> 2Li2O}
  2. Calcium is Group 2, so Ca²⁺. The chloride has Cl⁻, so two Cl⁻ to one Ca²⁺: CaCl₂. Balance: Ca+ClX2→CaClX2\ce{Ca + Cl2 -> CaCl2}

In terms of oxidation numbers, the metal goes from 0 to +1 or +2, so it is oxidised and is the reducing agent. Oxygen goes from 0 to −2 and chlorine from 0 to −1, so they are reduced and are the oxidising agents.

The reactions are more vigorous down each group, because the metals lose electrons more easily. The metals also burn with flames of characteristic colours, which are explained on page 15: lithium red, sodium yellow, potassium lilac, magnesium bright white, calcium orange-red, strontium red, barium green.

With water 8.13

A Group 1 metal reacts with cold water to give the hydroxide and hydrogen. The solution is alkaline, because of the OH⁻ ions.

2 M(s)+2 HX2O(l)→2 MOH(aq)+HX2(g) \ce{2M(s) + 2H2O(l) -> 2MOH(aq) + H2(g)}

Group 2 metals form the hydroxide (and hydrogen) in the same way.

M(s)+2 HX2O(l)→M(OH)X2(aq)+HX2(g) \ce{M(s) + 2H2O(l) -> M(OH)2(aq) + H2(g)}

In both the metal is oxidised (0 → +1 or +2) and hydrogen in the water is reduced (+1 → 0). Oxygen stays −2. The hydrogen can be identified with a lighted splint: it burns with a squeaky pop.

Metal What you see with cold water
Lithium Floats, fizzes steadily, gradually disappears. Does not melt.
Sodium Floats, melts into a ball, fizzes vigorously and darts across the surface, disappears.
Potassium Melts, fizzes violently, the hydrogen burns with a lilac flame; sometimes pops or explodes.
Magnesium Very slow: a few bubbles of hydrogen over many minutes or hours.
Calcium Sinks and fizzes steadily. The solution warms and turns cloudy white: calcium hydroxide is only slightly soluble.
Strontium Sinks, fizzes faster than calcium. Cloudy solution.
Barium Sinks, fizzes vigorously. The solution stays clearer: barium hydroxide is soluble.

Magnesium needs steam to react quickly. It burns with a bright white flame, forming the oxide not the hydroxide.

Mg(s)+HX2O(g)→MgO(s)+HX2(g) \ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)}

Check yourself

  1. Write balanced equations with state symbols for (a) sodium with oxygen, (b) magnesium with chlorine, (c) potassium with water.
  2. State the change in oxidation number of the metal and of the non-metal element in your equation (a).
  3. Describe what you see when a small piece of sodium is dropped on to water, and give the equation.
  4. How could you show that the solution formed when calcium reacts with water is alkaline?
  5. Explain why calcium reacts with water more vigorously than magnesium.
  6. Write the equation for magnesium with steam, and state why this is a different type of product from magnesium with cold water.

Answers cover these until you have tried the questions

  1. (a) 4 Na(s)+OX2(g)→2 NaX2O(s)\ce{4Na(s) + O2(g) -> 2Na2O(s)} (b) Mg(s)+ClX2(g)→MgClX2(s)\ce{Mg(s) + Cl2(g) -> MgCl2(s)} (c) 2 K(s)+2 HX2O(l)→2 KOH(aq)+HX2(g)\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}
  2. Na: 0 → +1 (oxidised). O: 0 → −2 (reduced).
  3. The sodium floats, melts into a ball and fizzes vigorously, moving around the surface, and disappears. 2 Na(s)+2 HX2O(l)→2 NaOH(aq)+HX2(g)\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}
  4. Add a few drops of universal indicator: it turns purple or blue. (Or test with damp red litmus paper, which turns blue.) The OH⁻ ions make the solution alkaline.
  5. Calcium's outer electrons are further from the nucleus and more shielded, so its ionisation energies are lower. It loses electrons more easily, so it is oxidised more readily and reacts faster.
  6. Mg(s)+HX2O(g)→MgO(s)+HX2(g)\ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)}. With steam at high temperature the magnesium hydroxide would decompose, so the product is the oxide; cold water gives the hydroxide, which is almost insoluble, and the reaction is very slow.