Ionic Bonding: Electron Transfer, Charges, and Formula Writing

What an Ionic Bond Is (and What It Is Not)

An ionic bond is the attraction between oppositely charged ions formed when electrons are transferred from one atom to another. The key idea is that the bond itself is not a shared pair of electrons (that would be covalent bonding). Instead, an ionic bond is an electrostatic attraction: positive and negative charges pull toward each other.

In real substances, ionic bonding usually creates a large, repeating three-dimensional structure (often called an ionic lattice), not a single isolated “pair” of atoms. When you write a formula like NaCl, you are not describing one sodium atom stuck to one chlorine atom as a single molecule. You are describing the simplest whole-number ratio of ions in the solid: 1 Na⁺ for every 1 Cl⁻ in the crystal.

Electron Transfer: Who Gives and Who Takes?

Ionic bonding begins with electron transfer. One atom loses one or more electrons (becoming a positive ion), and another atom gains them (becoming a negative ion). In beginner chemistry problems, you can often predict the direction of transfer by recognizing that metals tend to form positive ions and nonmetals tend to form negative ions. The “reason” is that this transfer leads to more stable electron arrangements, but in this chapter we will focus on how to use the idea to determine charges and write formulas correctly.

Step-by-step: Predicting the Ions Formed

When you are given two elements and asked to form an ionic compound, use this practical sequence:

• Step 1: Identify which element forms the cation. This is typically the metal (or a known positive polyatomic ion if present).
• Step 2: Identify which element forms the anion. This is typically the nonmetal (or a known negative polyatomic ion if present).
• Step 3: Assign the most common ion charge. Many main-group elements have predictable charges; some metals can have multiple charges and must be specified.
• Step 4: Ensure total charge is zero. Adjust the number of each ion so the sum of charges cancels.
• Step 5: Write the empirical formula. Use the smallest whole-number ratio of ions.

Example: magnesium and chlorine. Magnesium forms Mg²⁺. Chlorine forms Cl⁻. To balance charge, you need two chloride ions for each magnesium ion, giving MgCl₂.

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Charges: Reading and Using Ion Charges Correctly

Writing ionic formulas is mostly a charge-balancing skill. The compound must be electrically neutral overall, meaning the total positive charge equals the total negative charge.

Common Charges You Will Use Often

In formula-writing problems, you will repeatedly use a small set of common charges:

• Group 1 metals: +1 (e.g., Li⁺, Na⁺, K⁺)
• Group 2 metals: +2 (e.g., Mg²⁺, Ca²⁺)
• Aluminum: typically +3 (Al³⁺)
• Halogens (Group 17): −1 (e.g., F⁻, Cl⁻, Br⁻)
• Oxygen family (Group 16): −2 (e.g., O²⁻, S²⁻)
• Nitrogen family (Group 15): −3 (e.g., N³⁻, P³⁻)

Transition metals and a few main-group metals can form more than one charge. In those cases, the charge is usually given in the compound name (for example, iron(III) means Fe³⁺).

Charge Balance as an Equation

A helpful way to think is: (number of cations × cation charge) + (number of anions × anion charge) = 0.

Example: aluminum oxide. Aluminum is Al³⁺, oxide is O²⁻. Find the smallest numbers that make the total zero:

• 2 Al³⁺ gives +6
• 3 O²⁻ gives −6

So the formula is Al₂O₃.

Formula Writing with Monatomic Ions (Single-Atom Ions)

Monatomic ions are ions made from a single element. Many beginner ionic compounds are built from one metal ion and one nonmetal ion. The main task is to balance charges and reduce to the simplest ratio.

The “Criss-Cross” Method (with Understanding)

A popular shortcut is the criss-cross method: take the magnitude (ignore the sign) of each ion’s charge and use it as the subscript of the other ion. This works because it often produces a neutral combination. However, you must still reduce subscripts if they share a common factor.

Example: calcium and nitrogen. Ca²⁺ and N³⁻. Criss-cross gives Ca₃N₂. Check: 3×(+2)=+6 and 2×(−3)=−6, total 0.

Example where reduction is needed: magnesium and oxygen. Mg²⁺ and O²⁻. Criss-cross gives Mg₂O₂, but reduce to MgO.

Step-by-step Practice Examples

Example 1: potassium sulfide

• Identify ions: K⁺ and S²⁻
• Balance charges: need 2 K⁺ to balance 1 S²⁻
• Formula: K₂S

Example 2: aluminum bromide

• Ions: Al³⁺ and Br⁻
• Balance: need 3 bromides for each aluminum
• Formula: AlBr₃

Example 3: barium phosphide

• Ions: Ba²⁺ and P³⁻
• Balance: LCM of 2 and 3 is 6, so 3 Ba²⁺ (+6) and 2 P³⁻ (−6)
• Formula: Ba₃P₂

Polyatomic Ions: Treat Them as a Unit

Many ionic compounds contain polyatomic ions, which are groups of atoms that carry a net charge and behave as a single ion in formulas. When writing formulas, you do not break them apart. You balance the charge of the entire polyatomic ion against the other ion.

Common Polyatomic Ions You Will See Often

• Ammonium: NH₄⁺
• Hydroxide: OH⁻
• Nitrate: NO₃⁻
• Carbonate: CO₃²⁻
• Sulfate: SO₄²⁻
• Phosphate: PO₄³⁻
• Acetate: C₂H₃O₂⁻ (also written CH₃COO⁻)

You will often need parentheses in the formula when more than one polyatomic ion is required. Parentheses show that the subscript applies to the whole ion group.

Step-by-step: Writing Formulas with Polyatomic Ions

Example 1: calcium nitrate

• Ions: Ca²⁺ and NO₃⁻
• Balance: need 2 nitrates to balance +2
• Formula: Ca(NO₃)₂

Example 2: aluminum sulfate

• Ions: Al³⁺ and SO₄²⁻
• Balance: LCM of 3 and 2 is 6, so 2 Al³⁺ and 3 sulfate ions
• Formula: Al₂(SO₄)₃

Example 3: ammonium phosphate

• Ions: NH₄⁺ and PO₄³⁻
• Balance: need 3 ammonium ions to balance −3
• Formula: (NH₄)₃PO₄

Example 4: magnesium hydroxide

• Ions: Mg²⁺ and OH⁻
• Balance: need 2 hydroxides
• Formula: Mg(OH)₂

Variable-Charge Metals: Using Roman Numerals to Get the Right Formula

Some metals can form more than one cation charge. When that happens, the compound name typically includes a Roman numeral that tells you the charge on the metal ion. Your job is to use that charge to balance the anion.

Step-by-step: From Name to Formula

Example 1: iron(III) chloride

• Iron(III) means Fe³⁺
• Chloride is Cl⁻
• Balance: need 3 chlorides
• Formula: FeCl₃

Example 2: iron(II) chloride

• Iron(II) means Fe²⁺
• Chloride is Cl⁻
• Balance: need 2 chlorides
• Formula: FeCl₂

Example 3: copper(I) oxide

• Copper(I) means Cu⁺
• Oxide is O²⁻
• Balance: need 2 Cu⁺ for 1 O²⁻
• Formula: Cu₂O

Example 4: copper(II) oxide

• Copper(II) means Cu²⁺
• Oxide is O²⁻
• Balance: 1:1
• Formula: CuO

From Formula to Charges: Finding the Metal Charge in a Compound

Sometimes you are given a formula and asked to determine the charge on a variable-charge metal. You can do this by using the known charge of the anion and the fact that the total charge must be zero.

Step-by-step: Solve for the Unknown Charge

Example 1: Determine the charge on iron in Fe₂O₃

• Oxide is O²⁻
• There are 3 oxygens: total negative charge = 3 × (−2) = −6
• The compound is neutral, so total positive charge must be +6
• There are 2 iron atoms, so each iron must be +3
• Iron is Fe³⁺, so the name would be iron(III) oxide

Example 2: Determine the charge on copper in Cu(NO₃)₂

• Nitrate is NO₃⁻
• There are 2 nitrates: total negative charge = 2 × (−1) = −2
• Total positive charge must be +2
• There is 1 copper, so copper is +2
• Name: copper(II) nitrate

Writing Names vs. Writing Formulas: Avoiding Common Mix-ups

Even if naming rules were covered elsewhere, formula writing has a few recurring traps that are worth addressing directly because they cause incorrect subscripts and incorrect charge balance.

Trap 1: Forgetting to Make the Total Charge Zero

Students sometimes write the charges as subscripts without balancing. For example, writing AlO because aluminum is +3 and oxygen is −2. The correct approach is to find a combination that cancels: Al₂O₃.

Trap 2: Not Using Parentheses with Polyatomic Ions

If you need more than one polyatomic ion, you must use parentheses. For calcium nitrate, Ca(NO₃)₂ is correct. Writing CaNO₃₂ is incorrect because it suggests only the oxygen count is doubled, not the entire nitrate group.

Trap 3: Reducing When You Should Not (and Not Reducing When You Should)

You should reduce subscripts to the simplest whole-number ratio for ionic formulas. Mg₂O₂ must become MgO. However, do not “reduce” inside a polyatomic ion. For example, you cannot simplify (NO₃)₂ to N₂O₆ in an ionic formula-writing context, because nitrate is treated as a unit and the standard formula keeps it as NO₃⁻.

Trap 4: Confusing Subscripts with Charges

Charges are not written as subscripts in the final neutral formula. You may use charges while working (like Ca²⁺ and Cl⁻), but the final compound formula is CaCl₂, not Ca²⁺Cl₂⁻.

A Reliable “Least Common Multiple” Method for Any Ionic Formula

If the criss-cross method feels like a trick, you can use a method that works the same way every time and makes it obvious why the subscripts are what they are.

Step-by-step LCM Method

• Step 1: Write the ion charges (including signs).
• Step 2: Find the least common multiple (LCM) of the charge magnitudes.
• Step 3: Determine how many of each ion are needed to reach that LCM in total positive and total negative charge.
• Step 4: Write subscripts and reduce if possible.

Example: aluminum phosphate

• Ions: Al³⁺ and PO₄³⁻
• LCM of 3 and 3 is 3
• Need 1 Al³⁺ and 1 PO₄³⁻
• Formula: AlPO₄

Example: sodium carbonate

• Ions: Na⁺ and CO₃²⁻
• LCM of 1 and 2 is 2
• Need 2 Na⁺ and 1 carbonate
• Formula: Na₂CO₃

Connecting Electron Transfer to the Final Formula

It can help to connect the “electron transfer” story to the “charge balance” math. The number of electrons lost by the metal must equal the number of electrons gained by the nonmetal (or by the polyatomic ion as a whole). That equality is another way of seeing why the total charge must be zero.

Example: magnesium chloride. Magnesium forms Mg²⁺, meaning it lost 2 electrons. Each chlorine becomes Cl⁻, meaning each gained 1 electron. You need two chlorines to accept the two electrons magnesium lost. That leads directly to MgCl₂.

Example: aluminum oxide. Each Al becomes Al³⁺ (lost 3 electrons). Each O becomes O²⁻ (gained 2 electrons). The smallest match is 2 aluminum atoms losing 6 electrons total and 3 oxygen atoms gaining 6 electrons total, giving Al₂O₃.

Practice Set with Worked Answers

Use these to check your process. For each, identify ions, balance charges, and write the simplest formula.

Set A: Monatomic Ions

1) lithium fluoride

• Li⁺ and F⁻ balance 1:1
• Formula: LiF

2) calcium iodide

• Ca²⁺ and I⁻
• Need 2 iodides
• Formula: CaI₂

3) aluminum nitride

• Al³⁺ and N³⁻
• Balance 1:1
• Formula: AlN

4) strontium phosphide

• Sr²⁺ and P³⁻
• LCM 6, so Sr₃P₂
• Formula: Sr₃P₂

Set B: Polyatomic Ions

5) potassium sulfate

• K⁺ and SO₄²⁻
• Need 2 potassium ions
• Formula: K₂SO₄

6) barium carbonate

• Ba²⁺ and CO₃²⁻
• Balance 1:1
• Formula: BaCO₃

7) ammonium sulfide

• NH₄⁺ and S²⁻
• Need 2 ammonium ions
• Formula: (NH₄)₂S

8) aluminum hydroxide

• Al³⁺ and OH⁻
• Need 3 hydroxides
• Formula: Al(OH)₃

Set C: Variable-Charge Metals

9) tin(II) chloride

• Sn²⁺ and Cl⁻
• Need 2 chlorides
• Formula: SnCl₂

10) tin(IV) chloride

• Sn⁴⁺ and Cl⁻
• Need 4 chlorides
• Formula: SnCl₄

11) cobalt(III) oxide

• Co³⁺ and O²⁻
• LCM 6, so Co₂O₃
• Formula: Co₂O₃

12) Determine the metal charge in MnO₂

• Each O is −2, so 2 oxygens total −4
• Manganese must be +4 to balance
• Name would use manganese(IV)

Compact Algorithm You Can Use on Any Problem

If you want a repeatable checklist for homework and tests, use this algorithm:

1. Write the cation and anion with their charges. (Include polyatomic ions as a single unit.)
2. Find the smallest whole numbers a and b so that: a(cation charge) + b(anion charge) = 0.
3. Write the formula with subscripts a and b.
4. If a polyatomic ion has a subscript > 1, use parentheses.
5. Reduce subscripts to the simplest ratio if possible (but do not alter the inside of a polyatomic ion).