3.1.3.1 - Ionic Bonding

What ionic bonding really is

Many common ionic compounds form when electrons are transferred from metal atoms to non-metal atoms. Ionic compounds can also contain compound ions, such as ammonium, so a metal is not always present. That transfer creates charged particles called ions. A sodium atom, for example, loses one electron to form $\ce{Na^+}$, while a chlorine atom gains one electron to form $\ce{Cl^-}$.

Keep two ideas separate here. Electron transfer is how the ions are formed, but it is not itself the bond. The bond is the attraction between the positive and negative ions after those ions exist.

In sodium chloride, the ions do not sit as one isolated $\ce{Na^+}$ next to one isolated $\ce{Cl^-}$. They arrange in a giant three-dimensional ionic lattice with alternating positive and negative ions. Each positive ion is attracted to negative ions around it, and each negative ion is attracted to positive ions around it. That repeated electrostatic attraction is why ionic compounds are stable.

The diagram below separates the two linked ideas in ionic bonding: electron transfer forms the ions, and the giant lattice holds those ions together by repeated electrostatic attraction.

[DIAGRAM: asset_name: 1.3.1 - Ionic Bonding - Diagram 1; asset_slug: 1.3.1 - Ionic Bonding - Diagram 1; recommended_method: deterministic_code; description: Outer-electron transfer from sodium to chlorine forms Na+ and Cl−; dots and crosses identify electron origins.]

This also explains why the formula of an ionic compound shows the simplest ratio of ions in the lattice, not a small molecule. $\ce{NaCl}$ means the ratio is 1 sodium ion to 1 chloride ion throughout the lattice.

Predicting charges on simple ions

A simple ion is an ion formed from a single atom. For the main-group elements in this part of the course, the position of the element in the Periodic Table helps you predict the charge of the ion it forms.

The pattern comes from how many electrons an atom needs to lose or gain to reach a full outer shell:

• Group 1 elements lose 1 electron to form $1+$ ions, for example $\ce{Na^+}$ and $\ce{K^+}$.
• Group 2 elements lose 2 electrons to form $2+$ ions, for example $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$.
• Group 3 (13) elements such as aluminium lose 3 electrons to form $3+$ ions, for example $\ce{Al^{3+}}$.
• Group 5 (15) elements such as nitrogen can gain 3 electrons to form $3-$ ions, for example $\ce{N^{3-}}$.
• Group 6 (16) elements commonly gain 2 electrons to form $2-$ ions, for example $\ce{O^{2-}}$ and $\ce{S^{2-}}$.
• Group 7 (17) elements gain 1 electron to form $1-$ ions, for example $\ce{Cl^-}$ and $\ce{F^-}$.

You can think of the charge as a balance between protons and electrons. Magnesium has 12 protons. When it loses 2 electrons, it still has 12 protons but now only 10 electrons, so the ion has an overall $2+$ charge.

For this spec point, these charge predictions work best for the common main-group simple ions. Group 0 elements do not normally form ions, and Group 4 does not usually form simple ions in this model.

When you write an ion, include both the symbol and the charge.

The compound ions you must know

Not all ions are single atoms. Some are groups of covalently bonded atoms that carry an overall charge.

For this lesson, you need the formulae of these common compound ions:

• sulfate, $\ce{SO4^{2-}}$
• hydroxide, $\ce{OH^-}$
• nitrate, $\ce{NO3^-}$
• carbonate, $\ce{CO3^{2-}}$
• ammonium, $\ce{NH4^+}$

The charge belongs to the whole ion, not to one atom inside it. For example, $\ce{NO3^-}$ is one nitrate ion with a $1-$ charge overall. The three oxygen atoms are part of the ion; they are not three separate oxide ions.

Ammonium is especially important because it is a positive compound ion. The others in this list are negative compound ions. Once you know these five ions securely, building ionic formulae becomes much quicker and more reliable.

Worked example: calcium nitrate contains $\ce{Ca^{2+}}$ ions and $\ce{NO3^-}$ ions. One calcium ion has charge $+2$, so it needs two nitrate ions with total charge $-2$. The formula is $\ce{Ca(NO3)2}$.

Learn these as whole units. If you recognise the ion quickly, you are much less likely to invent incorrect formulae such as $\ce{NO6}$ or $\ce{SO8}$.

Constructing formulae for ionic compounds

To construct the formula of an ionic compound, use charges to find the smallest whole-number ratio of ions that gives no overall charge.

This four-step method is reliable:

1. Write the ions with their charges.
2. Choose the ratio that makes the total positive charge equal the total negative charge.
3. Write the formula without the charges.
4. Use brackets if you need more than one of a compound ion.

For magnesium chloride:

• ions: $\ce{Mg^{2+}}$ and $\ce{Cl^-}$
• charge balance: one $\ce{Mg^{2+}}$ needs two $\ce{Cl^-}$
• formula: $\ce{MgCl2}$

For aluminium oxide:

• ions: $\ce{Al^{3+}}$ and $\ce{O^{2-}}$
• charge balance: two $\ce{Al^{3+}}$ give $+6$, and three $\ce{O^{2-}}$ give $-6$
• formula: $\ce{Al2O3}$

The diagram below shows that the formula $\ce{Al2O3}$ comes directly from balancing the total positive and negative charges in the smallest whole-number ratio.

[DIAGRAM: asset_name: 1.3.1 - Ionic Bonding - Diagram 2; asset_slug: 1.3.1 - Ionic Bonding - Diagram 2; recommended_method: retained_png; description: Two Al3+ ions balance three O2− ions to form the formula Al2O3.]

For ammonium sulfate:

• ions: $\ce{NH4^+}$ and $\ce{SO4^{2-}}$
• charge balance: two $\ce{NH4^+}$ ions give $+2$, which balances one $\ce{SO4^{2-}}$
• formula: $\ce{(NH4)2SO4}$

The brackets matter because there are two ammonium ions. Without brackets, $\ce{NH42SO4}$ would mean something different and would be chemically wrong.

Two easy mistakes are worth avoiding:

• Do not leave charges in the final formula. Write $\ce{MgCl2}$, not $\mathrm{Mg^{2+}Cl^{-}_{2}}$.
• Do not change the inside of a compound ion. Hydroxide stays $\ce{OH}$, so write calcium hydroxide as $\ce{Ca(OH)2}$ so the repeated hydroxide ion is explicit. $\ce{CaO2H2}$ has the same elemental composition but obscures the compound-ion grouping.

If the charges already balance in a 1:1 ratio, keep the formula simple. Calcium oxide is $\ce{CaO}$, not $\ce{Ca2O2}$.

The big idea to remember is that an ionic formula is a charge-balance statement. If the total positive and negative charges cancel in the smallest whole-number ratio, your formula is correct.