Valence Electrons and Ionic Compounds: Study Guide (Chemistry 1.8)

Why valence electrons control chemistry

When two atoms meet, their outer electrons interact first. The inner (core) electrons are held tightly and take no part in ordinary reactions. So the chemistry of an element is set by its valence electrons: the electrons in the outermost shell, beyond the noble-gas core.

For main-group elements you can read the number of valence electrons straight from the periodic table:

• Groups 1 and 2: the number of valence electrons equals the group number (Na has 1, Ca has 2).
• Groups 13 to 18: subtract 10 from the group number (Al in group 13 has 3, O in group 16 has 6, Cl in group 17 has 7).
• Helium is the exception in group 18: it has only 2.

Whether two elements bond, and how, comes down to a tug of war: each nucleus attracts its own valence electrons and those of the other atom. Judge it with Coulomb’s law from Topic 1.7: the attraction grows with charge and falls with distance. For a valence electron, the charge that matters is the effective nuclear charge left after shielding by the core electrons.

That gives three broad cases:

The two elements	What the nuclei do	Likely result
A metal with weakly held valence electrons, and a nonmetal that strongly attracts extra electrons	The nonmetal’s nucleus pulls electrons away from the metal	Electrons transfer; ions form; an ionic compound results (NaCl)
Two nonmetals that both hold electrons strongly	Neither nucleus can take electrons from the other	Electrons are shared in covalent bonds (H₂O, Cl₂)
A noble gas with any partner	Its full valence shell is held very tightly, and an extra electron would have to start a new shell further out	Little or no tendency to bond

This topic focuses on the first case. You will study bond types in detail in Topic 2.1.

Reactivity and periodicity

Reactivity is something you can see: how fast and how violently an element reacts. Trends in the attraction between nucleus and valence electrons produce trends in reactivity.

Down a group of metals: reactivity increases

Group 1 metals react by losing their single valence electron. The easier that electron is to remove, the more readily the metal reacts. Going down the group, the valence electron is in a shell further from the nucleus, and the extra core electrons shield it. The effective nuclear charge it feels stays about the same, but the distance grows, so the Coulomb attraction is weaker.

Metal	Valence electron	First ionization energy (kJ mol⁻¹)	What you see with cold water
Li	2s¹	520	Fizzes steadily
Na	3s¹	496	Melts into a ball and darts across the surface
K	4s¹	419	Reacts violently; the hydrogen produced ignites with a lilac flame

The macroscopic observation (more violent reaction) links to the particle-level cause (a valence electron that is further from the nucleus and easier to remove). Group 2 metals follow the same trend, but each is less reactive than the group 1 metal beside it, because it must lose two electrons and its nucleus has a larger effective charge.

Down a group of nonmetals: reactivity decreases

The halogens (group 17) react by gaining an electron. Going down the group the atoms get larger, so an incoming electron sits further from the nucleus and is more shielded from it. The attraction for that extra electron is weaker, and the halogen is less reactive. Fluorine is the most reactive halogen; iodine is the least reactive of the common ones.

So the direction of the trend depends on whether the element loses or gains electrons.

Same group, same kind of compound

Elements in one column have the same number of valence electrons arranged in the same way (ns¹ for group 1, ns²np⁵ for group 17, and so on). They therefore tend to form analogous compounds: compounds with the same type of formula.

Group	Example compounds with oxygen and chlorine
1	Li₂O, Na₂O, K₂O; LiCl, NaCl, KCl
2	MgO, CaO, SrO; MgCl₂, CaCl₂, BaCl₂
16 (with hydrogen)	H₂O, H₂S

So if you know one compound, you can predict the formula for any element in the same group: from Na₂O, predict K₂O and Rb₂O; from H₂O, predict H₂Se. The formula type is shared, but properties change gradually down the group.

Predicting ion charges from position

The pattern in Figure 1 follows from the electron counts:

• Metals in groups 1, 2 and 13 lose all their valence electrons. Na (1s²2s²2p⁶3s¹) loses one electron to become Na⁺ (1s²2s²2p⁶), which has the same arrangement as neon. Mg becomes Mg²⁺ and Al becomes Al³⁺, both also with the neon arrangement.
• Nonmetals in groups 15, 16 and 17 gain electrons until the valence shell holds 8. The charge is the group number minus 18: N³⁻, O²⁻, F⁻; P³⁻, S²⁻, Cl⁻.
• Group 14 elements such as carbon and silicon would need to lose or gain four electrons. In this course they usually share electrons instead.

The ions formed (Na⁺, Mg²⁺, Al³⁺, N³⁻, O²⁻, F⁻) all have 10 electrons: the arrangement of neon.

Why the charge stops where it does

“Atoms want a full shell” is a slogan, not an explanation. The real reason is energy and Coulomb’s law. Removing the valence electrons of a metal costs a moderate amount of energy. Removing the next electron means taking one from the core: a shell closer to the nucleus, with far less shielding. That costs much more energy, more than the formation of the ionic solid can repay. You can see this directly in successive ionization energies.

On the nonmetal side, the same logic applies in reverse. An oxygen atom can accept two electrons into its 2p subshell. A third extra electron would have to start the 3s shell, much further from the nucleus, and would be strongly repelled by the electrons already there. So O²⁻ is the typical oxide ion.

(Background, beyond what you need to calculate here: adding the second electron to O⁻ actually costs energy in the gas phase. The strong attraction between the doubly charged ions in a solid such as MgO more than repays it. You will look at the attraction between ions in solids again in Topic 2.3.)

Transition metals

For transition metals, position is a much weaker guide. Many form more than one ion: iron forms Fe²⁺ and Fe³⁺, and copper forms Cu⁺ and Cu²⁺. Exam questions normally give you the charge, often as a Roman numeral in the name, such as iron(III) oxide for Fe₂O₃.

Writing ionic formulas

An ionic compound has no overall charge. Its formula shows the smallest whole-number ratio of ions (one formula unit), not a molecule. A reliable method:

1. Find each ion’s charge from the periodic table.
2. Find the lowest common multiple of the two charge sizes.
3. Use enough of each ion to reach that multiple.
4. Check: (number of cations × cation charge) + (number of anions × anion charge) = 0.

Worked example 1: formulas from position

Question. Predict the formula of the ionic compound formed by (a) aluminium and oxygen, (b) calcium and nitrogen.

(a) Aluminium and oxygen

1. Al is in group 13 with 3 valence electrons, so it forms Al³⁺. O is in group 16 with 6 valence electrons, so it forms O²⁻.
2. The lowest common multiple of 3 and 2 is 6.
3. Two Al³⁺ give +6; three O²⁻ give −6.
4. Check: 2(+3) + 3(−2) = 0.

Answer. Al₂O₃.

(b) Calcium and nitrogen

1. Ca is in group 2, so Ca²⁺. N is in group 15 with 5 valence electrons, so it gains 3 to form N³⁻.
2. The lowest common multiple of 2 and 3 is 6.
3. Three Ca²⁺ give +6; two N³⁻ give −6.
4. Check: 3(+2) + 2(−3) = 0.

Answer. Ca₃N₂ (calcium nitride).

Interpretation. The subscripts are not the charges. In Ca₃N₂, calcium is still 2+; the “3” says there are three calcium ions per formula unit.

Worked example 2: from ionization energies to formulas

Question. The first three ionization energies of magnesium are 738, 1451 and 7733 kJ mol⁻¹ (Figure 2). (a) Use the data to explain the typical charge of a magnesium ion. (b) Write the formula of magnesium chloride. (c) Predict the formula of strontium bromide, and justify the prediction.

(a) Compare each ionization energy with the one before:

• IE₂ ÷ IE₁ = 1451 ÷ 738 = 1.97. Doubling is expected: the second electron leaves an ion that is already positive.
• IE₃ ÷ IE₂ = 7733 ÷ 1451 = 5.33. This much bigger jump (an increase of 6282 kJ mol⁻¹) means the third electron comes from a new, inner shell.

So magnesium has 2 valence electrons (3s²). Losing them costs a moderate amount of energy; losing a third, core electron costs far more. The typical ion is Mg²⁺, which has the neon arrangement.

(b) Cl is in group 17 and forms Cl⁻. One Mg²⁺ balances two Cl⁻: MgCl₂. Check: 1(+2) + 2(−1) = 0.

(c) Strontium is in group 2, like magnesium, so it also has 2 valence electrons and forms Sr²⁺. Bromine is in group 17, like chlorine, so it forms Br⁻. The compound is analogous to MgCl₂: SrBr₂.

Worked example 3: connecting reactivity you can see to valence electrons

Question. Sodium reacts with water: 2Na + 2H₂O → 2NaOH + H₂. (a) Write the equation for the reaction of potassium with water. (b) A teacher adds equal small pieces of sodium and potassium to separate troughs of water. Predict which reacts more vigorously and explain at the particle level.

(a) Potassium is in group 1, like sodium, so it forms the analogous hydroxide KOH:

2K + 2H₂O → 2KOH + H₂

Check the balance: 2 K, 4 H and 2 O on each side.

(b) Potassium reacts more vigorously. In both reactions the metal atom loses its single valence electron to form M⁺. Potassium’s valence electron is in the fourth shell (4s¹), further from the nucleus than sodium’s 3s¹ electron, and it is shielded by more core electrons. By Coulomb’s law the attraction between the nucleus and that electron is weaker, so it is removed more easily (first ionization energies: K 419 kJ mol⁻¹, Na 496 kJ mol⁻¹). The easier the electron is to lose, the faster and more violent the reaction.

Interpretation. A macroscopic observation (how violent the reaction looks) is explained with a particle-level model (distance and shielding of the valence electron).

Common misconceptions

• “Atoms want to have a full shell.” Atoms do not want anything. Ions with a noble-gas arrangement are typical because removing or adding one more electron would involve a different shell and cost much more energy.
• “The subscripts show the charges.” In Al₂O₃ the ions are Al³⁺ and O²⁻. The subscripts give the ratio of ions that makes the total charge zero.
• Not reducing the ratio. Mg²⁺ and O²⁻ give MgO, not Mg₂O₂. An ionic formula is always the simplest ratio.
• “Reactivity always increases down a group.” True for metals, which lose electrons. For halogens, which gain electrons, reactivity decreases down the group.
• “Nonmetal charge equals the group number.” Group 16 elements form −2 ions, not +6 or −6. For groups 15 to 17 the charge is the group number minus 18.
• “Analogous compounds have identical properties.” They share a formula type, but properties such as melting point and solubility change gradually down a group.
• Treating an ionic formula as a molecule. NaCl describes a ratio of ions in a large lattice. That is why chemists speak of formula units (Topic 1.1).
• Using position to predict transition metal charges. Use the charge or Roman numeral given in the question.

Where this leads

This is the last topic of Unit 1. It builds on Topic 1.7, Periodic Trends, and it opens Unit 2, Compound Structure and Properties, where you will classify bonds as ionic, covalent or metallic (Topic 2.1) and see how ion charge and size control the structure and properties of ionic solids (Topic 2.3). Try the practice questions now, then use the revision notes and the checklist to consolidate.