Eduqas A-Level Chemistry

Formulae of ions and ionic compounds

An ionic compound is electrically neutral, so its formula has the smallest whole-number ratio of ions whose charges cancel. The charge of the common ions has to be known, and the ones below cover most of the course.

Group 1 metals form $+1$ ions, Group 2 metals form $+2$ ions, aluminium forms $+3$, the halogens form $-1$ ions and oxygen and sulfur form $-2$ ions. A roman numeral in a name gives the oxidation number of the metal, so iron(III) chloride contains $\ce{Fe^3+}$ and is $\ce{FeCl3}$.

To write a formula, find the lowest common multiple of the two charges. Aluminium sulfate combines $\ce{Al^3+}$ with $\ce{SO4^2-}$; the lowest common multiple of 3 and 2 is 6, so two aluminium ions balance three sulfate ions and the formula is $\ce{Al2(SO4)3}$. Brackets go round a polyatomic ion when more than one is needed, as in $\ce{Ca(OH)2}$ and $\ce{(NH4)2SO4}$, but not when only one is present.

Oxidation numbers

The oxidation number of an atom is the charge it would have if all its bonds were ionic. It is written with the sign first, as $+2$ or $-1$, and in a name as a roman numeral. The rules are applied in this order.

1. An atom in an uncombined element has oxidation number 0, including in molecules such as $\ce{O2}$, $\ce{P4}$ and $\ce{S8}$.
2. In a simple ion the oxidation number equals the charge.
3. The oxidation numbers in a neutral compound add up to 0; in a polyatomic ion they add up to the charge on the ion.
4. Group 1 metals are $+1$ and Group 2 metals $+2$ in compounds; fluorine is $-1$.
5. Hydrogen is $+1$ except in metal hydrides such as $\ce{NaH}$, where it is $-1$.
6. Oxygen is $-2$ except in peroxides such as $\ce{H2O2}$, where it is $-1$, and in $\ce{OF2}$, where it is $+2$.
7. Chlorine, bromine and iodine are $-1$ except when combined with oxygen or a more electronegative halogen.

Some worked values follow.

• In $\ce{MnO4-}$ the four oxygens contribute $-8$ and the ion has charge $-1$, so manganese is $+7$.
• In $\ce{Cr2O7^2-}$ the seven oxygens contribute $-14$, so the two chromiums together are $+12$ and each is $+6$.
• In $\ce{ClO-}$ chlorine is $+1$, and in $\ce{ClO3-}$ it is $+5$, so the ions are named chlorate(I) and chlorate(V).
• In $\ce{S2O3^2-}$ the three oxygens contribute $-6$, so the two sulfurs together are $+4$, an average of $+2$ each. In $\ce{S4O6^2-}$ the six oxygens contribute $-12$ and the four sulfurs together are $+10$, an average of $+2.5$.

Oxidation numbers give a way to recognise redox. An element that increases in oxidation number has been oxidised and one that decreases has been reduced, and in a balanced redox equation the total increase equals the total decrease. The same idea also shows disproportionation, where one element in one species is both oxidised and reduced. In $\ce{Cl2 + 2NaOH -> NaCl + NaClO + H2O}$ chlorine goes from 0 to $-1$ in the chloride and from 0 to $+1$ in the chlorate(I).

Balanced equations and state symbols

A balanced equation has the same number of each kind of atom on both sides and, where ions are involved, the same total charge. Balance by changing the numbers in front of formulae, never the formulae themselves. State symbols follow each formula: $\ce{(s)}$ solid, $\ce{(l)}$ liquid, $\ce{(g)}$ gas and $\ce{(aq)}$ dissolved in water.

Balancing starts with the atom that appears in the fewest places, here aluminium, then sulfate as a unit, then hydrogen, and oxygen last when it appears in several substances. If a fraction is needed, such as $\tfrac12\ce{O2}$ in an enthalpy equation, it is allowed when the equation is defined for one mole of a named substance.

Ionic equations

An ionic equation shows only the particles that change. Strong acids, soluble ionic compounds and strong bases are written as separate ions when dissolved, and everything else (solids, water, gases, weak acids) is written in full. Ions that appear unchanged on both sides are spectator ions and are cancelled.

Barium chloride solution and sodium sulfate solution in full are

Written as ions this is $\ce{Ba^2+ + 2Cl- + 2Na+ + SO4^2- -> BaSO4(s) + 2Na+ + 2Cl-}$, and after cancelling the sodium and chloride spectators

The charges are $+2-2=0$ on the left and 0 on the right. A redox ionic equation needs the electrons to balance as well as the atoms, which is done with half-equations, and the same check on charge applies.

Here both sides carry a total charge of $+2$, and zinc rises from 0 to $+2$ while copper falls from $+2$ to 0.

Neutralisation and carbonate reactions

The ionic equation for any strong acid with a strong base is $\ce{H+(aq) + OH-(aq) -> H2O(l)}$. An acid with a carbonate gives a salt, water and carbon dioxide, and the ionic form is $\ce{CO3^2-(aq) + 2H+(aq) -> H2O(l) + CO2(g)}$; with a solid carbonate the carbonate is written as $\ce{CO3^2-(s)}$ within the compound, for example $\ce{CaCO3(s) + 2H+(aq) -> Ca^2+(aq) + H2O(l) + CO2(g)}$.