Names read positive ion first, then negative ion

Definition

Ionic compound

A compound made of positive and negative ions, whose formula shows the smallest whole-number ratio of ions that gives no overall charge.

The positive ion is named first and the negative ion second, so $NaCl\text{NaCl}$ is sodium chloride.
The metal supplies the positive ion, and it keeps its own element name unchanged.
The negative ion's name carries an ending that identifies which ion it is.
A finished formula carries no charges, because the positive and negative charges inside it cancel.
Naming and formula writing are two directions of one step: names give the ions, and charges give the ratio.

Key Idea

Name order is fixed: positive ion first, negative ion second.
A formula is the smallest whole-number ratio of ions that leaves no overall charge.

The ending -ide marks an ion made from one element

An -ide ending shows a negative ion built from a single element.
Oxygen forms the oxide ion $O2−\text{O}^{2-}$ , so $MgO\text{MgO}$ is magnesium oxide.
The halogens form halide ions: fluoride $F−\text{F}^{-}$ , chloride $Cl−\text{Cl}^{-}$ , bromide $Br−\text{Br}^{-}$ and iodide $I−\text{I}^{-}$ .
$CaCl2\text{CaCl}_2$ is therefore calcium chloride, whatever the number of chloride ions in it.
Hydroxide, $OH−\text{OH}^{-}$ , is the exception, because it ends in -ide but contains two elements.
Oxide and hydroxide are different ions and are never interchangeable in a formula.

Common Mistake

$OH−\text{OH}^{-}$ is not $O2−\text{O}^{2-}$ , so sodium hydroxide is $NaOH\text{NaOH}$ while sodium oxide is $Na2O\text{Na}_2\text{O}$ .
The -ide ending does not change with the ratio, so $NaCl\text{NaCl}$ and $MgCl2\text{MgCl}_2$ are both named as chlorides.

The ending -ate marks an ion that also contains oxygen

Definition

Polyatomic ion

A group of two or more atoms covalently bonded together that carries an overall charge and behaves as a single ion.

An -ate ending shows a negative ion that contains oxygen alongside another element.
Nitrate is $NO3−\text{NO}_3^{-}$ , carbonate is $CO32−\text{CO}_3^{2-}$ and sulfate is $SO42−\text{SO}_4^{2-}$ .
The atoms inside one of these ions are held by covalent bonds and travel as a single unit.
The ion keeps both its name and its internal formula inside any compound.
$KNO3\text{KNO}_3$ is potassium nitrate, $CaCO3\text{CaCO}_3$ is calcium carbonate and $MgSO4\text{MgSO}_4$ is magnesium sulfate.

Example

$KNO3\text{KNO}_3$ : potassium nitrate, from $K+\text{K}^{+}$ and $NO3−\text{NO}_3^{-}$ in a $1 : 1$ ratio.
$CaCO3\text{CaCO}_3$ : calcium carbonate, from $Ca2+\text{Ca}^{2+}$ and $CO32−\text{CO}_3^{2-}$ in a $1 : 1$ ratio.
$Na2SO4\text{Na}_2\text{SO}_4$ : sodium sulfate, from two $Na+\text{Na}^{+}$ ions and one $SO42−\text{SO}_4^{2-}$ ion.

Balancing the charges gives the formula

Write the positive ion with its charge and the negative ion with its charge.
Find the smallest numbers of each ion that bring the total charge to zero.
With $Am+\text{A}^{m+}$ and $Bn−\text{B}^{n-}$ , taking $n$ of A and $m$ of B always balances, and the ratio is then simplified if both share a factor.
Write each count as a subscript, leaving out a subscript of one.
Put brackets round a polyatomic ion whenever more than one of it is needed.
$Al3+\text{Al}^{3+}$ with $O2−\text{O}^{2-}$ needs two and three, giving $Al2O3\text{Al}_2\text{O}_3$ .
$Ca2+\text{Ca}^{2+}$ with $OH−\text{OH}^{-}$ needs one and two, giving $Ca(OH)2\text{Ca(OH)}_2$ .
$Al3+\text{Al}^{3+}$ with $NO3−\text{NO}_3^{-}$ needs one and three, giving $Al(NO3)3\text{Al(NO}_3)_3$ .

Common Mistake

$CaOH2\text{CaOH}_2$ is not $Ca(OH)2\text{Ca(OH)}_2$ , because without brackets the subscript applies to hydrogen alone.
Charges stay in the working, and never appear on the finished formula.

Checking a finished formula

Multiply each ion's charge by its subscript and add the results, which must come to zero.
$Al2(SO4)3\text{Al}_2(\text{SO}_4)_3$ gives $\times +3) + (3 \times -2) = 0$ .
Confirm that each polyatomic ion still carries its own internal formula.
Confirm that the ratio cannot be simplified any further.
Confirm that no charges have been left on the finished formula.

Exam technique

Crossing the charge numbers over gives a starting ratio, but a shared factor still has to be cancelled, so $Mg2+\text{Mg}^{2+}$ with $O2−\text{O}^{2-}$ gives $MgO\text{MgO}$ rather than $Mg2O2\text{Mg}_2\text{O}_2$ .
Writing both ions with their charges before anything else turns the formula into a single balancing step.
The charge check at the end costs a moment and catches a wrong subscript.

Self review

What does an -ide ending tell you about a negative ion?
Why does hydroxide end in -ide even though it contains two elements?
What is the formula of the compound formed from $Al3+\text{Al}^{3+}$ and $O2−\text{O}^{2-}$ ?
When are brackets needed around a polyatomic ion?
How do you check that an ionic formula carries no overall charge?

Names read positive ion first, then negative ion

Definition

Ionic compound

A compound made of positive and negative ions, whose formula shows the smallest whole-number ratio of ions that gives no overall charge.

The positive ion is named first and the negative ion second, so $NaCl\text{NaCl}$ is sodium chloride.
The metal supplies the positive ion, and it keeps its own element name unchanged.
The negative ion's name carries an ending that identifies which ion it is.
A finished formula carries no charges, because the positive and negative charges inside it cancel.
Naming and formula writing are two directions of one step: names give the ions, and charges give the ratio.

Key Idea

Name order is fixed: positive ion first, negative ion second.
A formula is the smallest whole-number ratio of ions that leaves no overall charge.

The ending -ide marks an ion made from one element

An -ide ending shows a negative ion built from a single element.
Oxygen forms the oxide ion $O2−\text{O}^{2-}$ , so $MgO\text{MgO}$ is magnesium oxide.
The halogens form halide ions: fluoride $F−\text{F}^{-}$ , chloride $Cl−\text{Cl}^{-}$ , bromide $Br−\text{Br}^{-}$ and iodide $I−\text{I}^{-}$ .
$CaCl2\text{CaCl}_2$ is therefore calcium chloride, whatever the number of chloride ions in it.
Hydroxide, $OH−\text{OH}^{-}$ , is the exception, because it ends in -ide but contains two elements.
Oxide and hydroxide are different ions and are never interchangeable in a formula.

Common Mistake

$OH−\text{OH}^{-}$ is not $O2−\text{O}^{2-}$ , so sodium hydroxide is $NaOH\text{NaOH}$ while sodium oxide is $Na2O\text{Na}_2\text{O}$ .
The -ide ending does not change with the ratio, so $NaCl\text{NaCl}$ and $MgCl2\text{MgCl}_2$ are both named as chlorides.

The ending -ate marks an ion that also contains oxygen

Definition

Polyatomic ion

A group of two or more atoms covalently bonded together that carries an overall charge and behaves as a single ion.

An -ate ending shows a negative ion that contains oxygen alongside another element.
Nitrate is $NO3−\text{NO}_3^{-}$ , carbonate is $CO32−\text{CO}_3^{2-}$ and sulfate is $SO42−\text{SO}_4^{2-}$ .
The atoms inside one of these ions are held by covalent bonds and travel as a single unit.
The ion keeps both its name and its internal formula inside any compound.
$KNO3\text{KNO}_3$ is potassium nitrate, $CaCO3\text{CaCO}_3$ is calcium carbonate and $MgSO4\text{MgSO}_4$ is magnesium sulfate.

Example

$KNO3\text{KNO}_3$ : potassium nitrate, from $K+\text{K}^{+}$ and $NO3−\text{NO}_3^{-}$ in a $1 : 1$ ratio.
$CaCO3\text{CaCO}_3$ : calcium carbonate, from $Ca2+\text{Ca}^{2+}$ and $CO32−\text{CO}_3^{2-}$ in a $1 : 1$ ratio.
$Na2SO4\text{Na}_2\text{SO}_4$ : sodium sulfate, from two $Na+\text{Na}^{+}$ ions and one $SO42−\text{SO}_4^{2-}$ ion.

Balancing the charges gives the formula

Write the positive ion with its charge and the negative ion with its charge.
Find the smallest numbers of each ion that bring the total charge to zero.
With $Am+\text{A}^{m+}$ and $Bn−\text{B}^{n-}$ , taking $n$ of A and $m$ of B always balances, and the ratio is then simplified if both share a factor.
Write each count as a subscript, leaving out a subscript of one.
Put brackets round a polyatomic ion whenever more than one of it is needed.
$Al3+\text{Al}^{3+}$ with $O2−\text{O}^{2-}$ needs two and three, giving $Al2O3\text{Al}_2\text{O}_3$ .
$Ca2+\text{Ca}^{2+}$ with $OH−\text{OH}^{-}$ needs one and two, giving $Ca(OH)2\text{Ca(OH)}_2$ .
$Al3+\text{Al}^{3+}$ with $NO3−\text{NO}_3^{-}$ needs one and three, giving $Al(NO3)3\text{Al(NO}_3)_3$ .

Common Mistake

$CaOH2\text{CaOH}_2$ is not $Ca(OH)2\text{Ca(OH)}_2$ , because without brackets the subscript applies to hydrogen alone.
Charges stay in the working, and never appear on the finished formula.

Checking a finished formula

Multiply each ion's charge by its subscript and add the results, which must come to zero.
$Al2(SO4)3\text{Al}_2(\text{SO}_4)_3$ gives $\times +3) + (3 \times -2) = 0$ .
Confirm that each polyatomic ion still carries its own internal formula.
Confirm that the ratio cannot be simplified any further.
Confirm that no charges have been left on the finished formula.

Exam technique

Crossing the charge numbers over gives a starting ratio, but a shared factor still has to be cancelled, so $Mg2+\text{Mg}^{2+}$ with $O2−\text{O}^{2-}$ gives $MgO\text{MgO}$ rather than $Mg2O2\text{Mg}_2\text{O}_2$ .
Writing both ions with their charges before anything else turns the formula into a single balancing step.
The charge check at the end costs a moment and catches a wrong subscript.

Self review

What does an -ide ending tell you about a negative ion?
Why does hydroxide end in -ide even though it contains two elements?
What is the formula of the compound formed from $Al3+\text{Al}^{3+}$ and $O2−\text{O}^{2-}$ ?
When are brackets needed around a polyatomic ion?
How do you check that an ionic formula carries no overall charge?

2.3.2 Naming and deducing formulae of ionic compounds

Names read positive ion first, then negative ion

The ending -ide marks an ion made from one element

The ending -ate marks an ion that also contains oxygen

Balancing the charges gives the formula

Checking a finished formula

Review 2.3.2 Naming and deducing formulae of ionic compounds by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.3.2 Naming and deducing formulae of ionic compounds

Names read positive ion first, then negative ion

The ending -ide marks an ion made from one element

The ending -ate marks an ion that also contains oxygen

Balancing the charges gives the formula

Checking a finished formula

Review 2.3.2 Naming and deducing formulae of ionic compounds by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.3.2 Naming and deducing formulae of ionic compounds

Names read positive ion first, then negative ion

The ending -ide marks an ion made from one element

The ending -ate marks an ion that also contains oxygen

Balancing the charges gives the formula

Checking a finished formula

Review 2.3.2 Naming and deducing formulae of ionic compounds by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.3.2 Naming and deducing formulae of ionic compounds Revision Guide

2.3.2 Naming and deducing formulae of ionic compounds

Names read positive ion first, then negative ion

The ending -ide marks an ion made from one element

The ending -ate marks an ion that also contains oxygen

Balancing the charges gives the formula

Checking a finished formula

Review 2.3.2 Naming and deducing formulae of ionic compounds by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.3.2 Naming and deducing formulae of ionic compounds Revision Guide