Back to the on-screen lesson ·
Why a metal and a non-metal make a lattice of charged ions, and why the charges decide the formula.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say what charge a main-group atom takes on when it becomes an ion, and why — and then write the formula of the compound it makes with any other ion, by finding the ratio in which the two charges cancel. You will also be able to say what an ionic formula is a count of, which is a ratio in a lattice and not the atoms in a molecule, and why an ionic solid melts high and shatters rather than bending.
You can read a formula as a count of atoms, and you know that a subscript belongs to the symbol in front of it. You have met the periodic table as a grid whose group number is the number of electrons in the outer shell. What this lesson adds is the reason a formula has the subscripts it has — which is a question the course has so far answered with because it does.
| Term | What it means |
|---|---|
| Ion | An atom that has lost or gained electrons and so carries a charge. |
| Cation | A positive ion. |
| Anion | A negative ion. |
| Ionic bond | The attraction between a cation and an anion. |
| Lattice | The repeating three-dimensional arrangement the ions settle into. |
| Formula unit | The smallest whole-number ratio of ions in the lattice. |
An atom is neutral because it has as many electrons as protons. Move an electron and the balance goes.
Put a metal and a non-metal together and the electrons move across. What is left is a positive ion and a negative ion, and opposite charges attract — that attraction is the ionic bond.
Now the formula. A jar of solid has no charge on it, so the positive and negative charges in it must cancel exactly. Magnesium gives $2+$ and chloride takes $1-$, so one magnesium needs two chlorides: $\mathrm{MgCl_2}$. Aluminum gives $3+$ and oxide takes $2-$; six is the smallest number both go into, so two aluminum and three oxide: $\mathrm{Al_2O_3}$.
That is the whole method, and it is the reason the subscripts of every ionic formula in this course could have been predicted before you ever met the compound.
Another way: picture
Think of the charges as debts and credits that have to balance to zero. A $3+$ ion is a credit of three; a $2-$ ion is a debt of two. Two credits of three and three debts of two both come to six, and the account closes — which is $\mathrm{Al_2O_3}$. Any other pairing leaves the account open, and an ionic compound never leaves it open.
Another way: steps
To write the formula of an ionic compound:
Find the metal ion's charge. From its group for groups 1, 2 and 13; from the roman numeral in the name for a transition metal.
Find the non-metal ion's charge. Eight less its outer-shell count, as a negative charge: group 17 gives $1-$, 16 gives $2-$, 15 gives $3-$.
Find the cancelling number. The smallest number both charge sizes divide into. For $3+$ and $2-$ that is 6.
Divide to get the subscripts. The cancelling number over each charge.
Write the formula. Metal first, subscripts after each symbol, any subscript of 1 left out.
Check the work. Multiply each subscript by its ion's charge: the positive total and the negative total must be equal. Are the subscripts in their lowest terms? $\mathrm{Mg_2O_2}$ cancels too, but the formula unit is the smallest ratio, $\mathrm{MgO}$. Does every charge have its sign? A table of values that reads 2 and 1 for the charges, with no signs, has lost half its meaning.
Reading a main-group charge from the group is allowed because an atom reaches a full outer shell by the fewest moves. A group 2 metal sheds two electrons rather than gaining six; a group 16 non-metal gains two rather than shedding six. The group number says how many moves each way.
Requiring the charges to cancel is allowed because matter we can hold has no overall charge. Even a small excess of one kind of ion would give a sample an enormous electric charge, and nothing of the kind is ever seen.
Using the smallest cancelling number is allowed because a formula unit is defined as the smallest ratio. Any multiple describes the same lattice; the lattice does not contain separate groups of two magnesium and two oxide.
Writing the metal first is a convention, not chemistry, but every formula and name in this course follows it so that a formula can be read at a glance. The name follows the same order: the metal keeps its own name, and the non-metal takes the ending -ide, so chlorine becomes chloride, oxygen becomes oxide and nitrogen becomes nitride. Reading a name therefore tells you which ion to write first, and the next unit uses exactly that to go from a name to a formula, one ion at a time.
A crystal of sodium chloride is not a heap of $\mathrm{NaCl}$ particles. It is a single repeating arrangement in which every sodium ion has six chloride ions around it and every chloride ion has six sodium ions around it, on and on to the edge of the crystal. Pick one sodium ion and there is no honest way to say which chloride it is bonded to; it is attracted to all six, and to the ones beyond them a little as well.
So $\mathrm{NaCl}$ does not name a particle. It names a ratio: one sodium ion for every chloride ion, throughout. That is what a formula unit is.
This is not pedantry, and two things follow from it directly. An ionic solid melts very high, because to melt it you have to overcome the attraction of every ion to all of its neighbors at once — magnesium oxide, with $2+$ against $2-$, melts at $2852\,^\circ\mathrm{C}$, while sodium chloride, with $1+$ against $1-$, melts at $801\,^\circ\mathrm{C}$. And an ionic solid shatters rather than bends, because sliding one layer by a single ion brings like charges face to face and the crystal drives itself apart.
Every rule above is about the outer shell, so it works for the main groups and stops at the transition metals. Iron forms $\mathrm{Fe^{2+}}$ and $\mathrm{Fe^{3+}}$; copper forms $\mathrm{Cu^{2+}}$ and $\mathrm{Cu^+}$; lead forms $\mathrm{Pb^{2+}}$ and $\mathrm{Pb^{4+}}$. Nothing about the position of iron in the table chooses between two and three.
That is exactly why their names carry a roman numeral. Iron(II) sulfate and iron(III) sulfate are two different substances with two different formulas, and the numeral is not decoration: it is the charge, supplied because it could not be worked out.
A few transition metals are in practice reliable — zinc is always $2+$ and silver always $1+$ — and their names carry no numeral for that reason.
| Ion | Formula | Charge |
|---|---|---|
| iron(II) | $\mathrm{Fe^{2+}}$ | $2+$ |
| iron(III) | $\mathrm{Fe^{3+}}$ | $3+$ |
| copper(II) | $\mathrm{Cu^{2+}}$ | $2+$ |
| zinc | $\mathrm{Zn^{2+}}$ | $2+$ |
| silver | $\mathrm{Ag^+}$ | $1+$ |
| lead(II) | $\mathrm{Pb^{2+}}$ | $2+$ |
Minnesota's road crews spread hundreds of thousands of tons of de-icing salt every winter, and the choice between salts comes straight out of ionic formulas. Ordinary rock salt is sodium chloride, $\mathrm{NaCl}$: one $1+$ ion for one $1-$ ion, two ions per formula unit. Calcium chloride, $\mathrm{CaCl_2}$, is one $2+$ ion for two $1-$ ions, three ions per formula unit.
That difference matters on the road. Salt melts ice because dissolved particles get in the way of water freezing, and the effect depends on how many particles dissolve, not which. A formula unit of calcium chloride breaks into three ions where sodium chloride gives two, so calcium chloride lowers the freezing point further and keeps working down to about $-25$ degrees Fahrenheit, where rock salt stops being useful at around 15 degrees.
Calcium chloride also gives off heat as it dissolves, which helps it start melting at once. It costs several times as much per ton, so crews keep it for the coldest nights and spread rock salt the rest of the time. A charge of $2+$ on the calcium ion, set by its two outer electrons, is ultimately what a highway department is paying extra for.
The sodium fluoride in many American toothpastes is $\mathrm{NaF}$: one $\mathrm{Na^+}$ for one $\mathrm{F^-}$. Its formula follows from sodium's single outer electron and fluorine being one short of a full shell.
The subscript is taken from the group number. Magnesium is in group 2, so $\mathrm{MgCl_2}$ looks as though the 2 came from the group. It did not; it came from the charge on magnesium being $2+$ and the charge on chloride being $1-$. The coincidence breaks immediately: aluminum is in group 13 and $\mathrm{Al_2O_3}$ has a 2 on the aluminum, not a 13.
The two ions are said to share. An ionic bond is not a shared pair. The electrons have gone across and stayed there, and what is left is two charged particles pulling on each other. Sharing is the next lesson.
The formula is read as a molecule. $\mathrm{CaCl_2}$ is often described as one calcium joined to two chlorines, as though that little group floated about. There is no such group; there is a lattice, and the formula is the ratio in it.
The subscripts are left unreduced. $\mathrm{Mg_2O_2}$ cancels its charges, but the formula unit is the smallest ratio: $\mathrm{MgO}$.
A charge is written without its sign, or with the sign in front. The convention is the size first and the sign after: $\mathrm{Mg^{2+}}$, not $\mathrm{Mg^{+2}}$, and never a bare $\mathrm{Mg^2}$, which says something else entirely.
Find calcium's charge.
$\text{group } 2 \Rightarrow \mathrm{Ca^{2+}}$
A metal gives away its outer electrons.
Find fluoride's charge.
$\text{group } 17 \Rightarrow \mathrm{F^{-}}$
One short of a full shell.
Find the cancelling number.
$2 \text{ and } 1 \Rightarrow 2$
The smallest number both divide into.
Divide by each charge.
$2 \div 2 = 1; \ 2 \div 1 = 2$
These are the subscripts.
Write the formula.
$\mathrm{CaF_2}$
A subscript of 1 is not written.
Find aluminum's charge.
$\text{group } 13 \Rightarrow \mathrm{Al^{3+}}$
Three outer electrons given away.
Find oxide's charge.
$\text{group } 16 \Rightarrow \mathrm{O^{2-}}$
Two short of a full shell.
Find the cancelling number.
$3 \text{ and } 2 \Rightarrow 6$
The lowest common multiple.
Divide by each charge.
$6 \div 3 = 2; \ 6 \div 2 = 3$
Two aluminum ions, three oxide ions.
Write the formula.
$\mathrm{Al_2O_3}$
Metal first.
Check by multiplying back.
$2 \times 3 = 6; \ 3 \times 2 = 6$
Equal and opposite: neutral.
Read the iron ion's charge.
$\text{iron(III)} \Rightarrow \mathrm{Fe^{3+}}$
The roman numeral is the charge.
Find chloride's charge.
$\text{group } 17 \Rightarrow \mathrm{Cl^{-}}$
One short of a full shell.
Find the cancelling number.
$3 \text{ and } 1 \Rightarrow 3$
The smallest number both divide into.
Divide by each charge.
$3 \div 3 = 1; \ 3 \div 1 = 3$
One iron ion, three chloride ions.
Write the formula.
$\mathrm{FeCl_3}$
The 1 on iron is not written.
Check by multiplying back.
$1 \times 3 = 3; \ 3 \times 1 = 3$
Neutral.
Compare iron(II) chloride.
$\mathrm{Fe^{2+}} \Rightarrow \mathrm{FeCl_2}$
A different charge, a different substance.
Find both charges.
$\mathrm{Mg^{2+}}; \ \mathrm{N^{3-}}$
Group 15 non-metals are three short.
Find the cancelling number.
Divide and write the formula.
Match each ion to the charge it carries.
| a charge of $3-$ | a charge of $1+$ | a charge of $2-$ | a charge of $3+$ | |
|---|---|---|---|---|
| nitride ion | ||||
| potassium ion | ||||
| oxide ion | ||||
| aluminum ion |
Complete the worked solution: aluminum, in group 13, combines with a non-metal from group $17$. Find the charge on each ion, and how many of each ion one formula unit holds.
Find aluminum's charge.
$\text{aluminum gives up its outer electrons: } +$ a
Group 13 metals have three outer electrons.
Find the non-metal's charge.
$\text{eighteen} - (\text{group}) = -$ b
It takes what its outer shell is short of.
Find the aluminum ions in one formula unit.
$(\text{cancelling number}) \div (\text{aluminum's charge}) =$ c
The cancelling number is the smallest both charges divide.
Find the non-metal ions in one formula unit.
$(\text{cancelling number}) \div (\text{non-metal's charge}) =$ d
Then the charges cancel exactly.
One formula unit of sodium oxide holds $2$ sodium ions and $1$ oxide ions. Why that ratio, and not some other?
calcium oxide is made of $\mathrm{Ca^{2+}}$ ions and $\mathrm{O^{2-}}$ ions. Write its formula, and say how many ions altogether one formula unit holds. Write the formula the way a keyboard writes it, with the numbers on the line: sodium chloride is NaCl and magnesium chloride is MgCl2.
The formula of calcium oxide is a, and one formula unit of it holds b ions in all.
calcium chloride is the road salt Minnesota crews spread in winter. It is made of $\mathrm{Ca^{2+}}$ and $\mathrm{Cl^{-}}$ ions. How many ions does one formula unit hold in all?
The answer: a.
Three jars in a school stockroom have lost their formula labels; only the names and a note of the two ionic charges survive. Write the formula that belongs on each jar, with the numbers on the line as a keyboard writes them.
| charge on the metal ion | charge on the non-metal ion | formula | |
|---|---|---|---|
| magnesium chloride | 2 | -1 | |
| calcium fluoride | 2 | -1 | |
| barium oxide | 2 | -2 |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
calcium fluoride is built from calcium ions and fluoride ions. Work out the charge on each ion from where its element sits in the periodic table, then say how many of each ion one formula unit holds.
| value | |
|---|---|
| charge on the calcium ion | |
| charge on the fluoride ion | |
| calcium ions in one formula unit | |
| fluoride ions in one formula unit |
You can turn a group number into an ionic charge, and two charges into a formula. Say out loud why aluminum oxide is $\mathrm{Al_2O_3}$ and not $\mathrm{AlO}$, and what the word lattice adds that the word molecule would get wrong. Next: what happens instead when two non-metals meet, and neither of them will give an electron up.
15. Your turn: magnesium nitride, from magnesium in group 2 and nitrogen in group 15, step 2
$2 \text{ and } 3 \Rightarrow 6$
Six positive charges must meet six negative ones.
15. Your turn: magnesium nitride, from magnesium in group 2 and nitrogen in group 15, step 3
$6 \div 2 = 3; \ 6 \div 3 = 2 \Rightarrow \mathrm{Mg_3N_2}$
Three magnesium ions, two nitride ions.