Lesson 1.7 · Unit 1 · CED topic 1.8

Valence electrons and ionic compounds

Sodium is a soft metal that reacts violently with water; chlorine is a poisonous green gas. Put them together and you get table salt, a stable white solid you sprinkle on food. The change is explained entirely by valence electrons: sodium has one it can give up easily, and chlorine has a vacancy for one it holds strongly. This lesson uses periodic trends to predict which elements bond, what charges they take, and what formulas their ionic compounds have.

Learning objective · 1.8.A

Explain the relationship between trends in the reactivity of elements and periodicity.

Required content · EK 1.8.A.1 to 1.8.A.3

The likelihood that two elements will form a chemical bond is determined by the interactions between the valence electrons and nuclei of elements.
Elements in the same column of the periodic table tend to form analogous compounds.
Typical charges of atoms in ionic compounds are governed by the number of valence electrons and predicted by their location on the periodic table.

Valence electrons and the periodic table

For the main-group elements (groups 1, 2 and 13 to 18), the number of valence electrons is set by the column: group 1 has 1, group 2 has 2, group 13 has 3, and so on up to 8 for the noble gases (helium has 2). Because elements in the same column have the same valence configuration (all of group 1 is \(ns^1\), all of group 17 is \(ns^2 np^5\)), they behave alike and form analogous compounds. If you know that NaCl, KCl and LiCl exist, you can predict RbCl; if you know H₂O, you can predict H₂S.

Why ions form: ionization energy and electron affinity

Atoms with low ionization energies (metals, at the left of the table) lose their valence electrons fairly easily. Atoms with high electron affinity (nonmetals, at the right) gain electrons. When a metal meets a nonmetal, electrons move from the metal to the nonmetal. The result is a positive cation and a negative anion, held together by the Coulomb attraction between opposite charges: an ionic bond.

Figure 1.7b Formation of an ionic compound. The valence electron of sodium moves to chlorine; each ion ends with a noble-gas electron configuration, and the oppositely charged ions attract each other (Coulomb’s law).

Notice that removing electrons costs energy, so the transfer by itself is not favorable (sodium’s ionization energy is 496 kJ/mol while chlorine releases only 349 kJ/mol on gaining an electron). The energy that makes the compound stable is the large attraction of the ions in the solid lattice, which you will compare in Unit 2. What the periodic trends tell you now is which atoms are good donors and acceptors.

Typical charges

Main-group atoms tend to form ions with the electron configuration of the nearest noble gas, because the next electron to remove (or add) would be from or to a very different energy level. This is why successive ionization energies jump after the valence electrons, as you saw in the last lesson. The typical charge follows from the group:

Figure 1.7a Typical ion charges of main-group elements follow from the valence electrons: metals lose electrons to reach the electron configuration of the previous noble gas, and nonmetals gain electrons to reach that of the next noble gas.

Group	Valence e⁻	Typical ion	Examples
1	1	1+	Li⁺, Na⁺, K⁺
2	2	2+	Mg²⁺, Ca²⁺, Ba²⁺
13	3	3+	Al³⁺
15	5	3−	N³⁻, P³⁻
16	6	2−	O²⁻, S²⁻
17	7	1−	F⁻, Cl⁻, Br⁻

The transition metals (groups 3 to 12) do not follow a single rule, because their valence electrons are in both the s and d subshells, which are close in energy. They commonly form several charges: iron makes Fe²⁺ and Fe³⁺, copper makes Cu⁺ and Cu²⁺. Their ionic compounds are named with a Roman numeral, as in iron(III) oxide. The cation always has the same charge-balance rule as any other.

Formulas from charges

An ionic compound is electrically neutral overall, so the total positive charge equals the total negative charge. Find the smallest numbers of each ion that balance.

Worked example · writing formulas

Calcium and chlorine. Ca²⁺ and Cl⁻: one calcium needs two chlorides, so CaCl₂.

Aluminum and oxygen. Al³⁺ and O²⁻: the least common multiple of 3 and 2 is 6, so two aluminum (+6) and three oxygen (−6): Al₂O₃.

Magnesium and nitrogen. Mg²⁺ and N³⁻: three magnesium (+6) and two nitrogen (−6): Mg₃N₂.

Check. The formula units come out as the empirical formulas of Lesson 1.3. Check that the charges sum to zero: \(2(+3) + 3(-2) = 0\).

Reactivity trends

The reactivity of a metal depends on how easily it loses electrons. Going down group 1, ionization energy falls (Li 520, Na 496, K 419 kJ/mol), so reactivity with water rises: lithium fizzes steadily, sodium melts into a ball and darts about, potassium bursts into flame. For nonmetals, reactivity depends on how strongly an atom attracts electrons. Going down group 17 the electron-attracting ability falls, so fluorine is the most reactive halogen, then chlorine, bromine and iodine. Both statements are Coulomb’s law again: farther from the nucleus and more shielded means weaker control over electrons.

Worked example · predicting a reaction

Predict the formula of the compound formed by strontium and iodine, and compare its likely reactivity with that of calcium and iodine.

Strontium is in group 2 (Sr²⁺) and iodine in group 17 (I⁻), so the compound is SrI₂. Strontium is in the same group as calcium, so it forms an analogous compound, CaI₂. Strontium is below calcium, with a lower first and second ionization energy, so it loses electrons slightly more easily and reacts a little more vigorously.

Exam focus

The group number gives the valence electrons and, for main-group elements, the typical charge.
Ions are neutral overall. Always check that the formula’s charges balance.
Reasoning from trends, not from memory. “Sodium reacts more vigorously than lithium because its valence electron is farther away and more shielded, so less energy is needed to remove it” earns the point; “because it is bigger” does not.
Transition metals vary. Do not give them one fixed charge; use the name or the problem to decide.
Use analogies. If you are asked about an unfamiliar compound, find a member of the same group you do know.

Practice

Write the formula of the compound formed from barium and bromine.
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Barium is in group 2 (Ba²⁺) and bromine in group 17 (Br⁻). Two bromides balance one barium: BaBr₂.
Write the formula of aluminum sulfide and state the charge on each ion.
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Al³⁺ and S²⁻. Two Al (+6) balance three S (−6): Al₂S₃.
An element in period 3 has first through fourth ionization energies (kJ/mol) of 496, 4562, 6910 and 9543. What is the formula of its compound with chlorine?
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The jump after the first ionization energy shows one valence electron (group 1, sodium), so the ion is Na⁺ and the compound is NaCl.
Explain why potassium is more reactive than lithium toward water.
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Potassium’s valence electron is in the fourth shell, farther from the nucleus and shielded by more core electrons than lithium’s valence electron in the second shell. By Coulomb’s law the attraction to the nucleus is weaker, so the first ionization energy is lower (419 against 520 kJ/mol) and potassium loses its electron more easily.
Why do sodium and potassium form analogous compounds, such as NaCl and KCl?
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Both are in group 1 and have one valence electron (ns¹), so both form 1+ ions and combine with chlorine in a 1 : 1 ratio.
Predict the charge of the most common ion of selenium (group 16) and write the formula of its compound with sodium.
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Selenium has 6 valence electrons and typically gains 2 to form Se²⁻. Two Na⁺ balance one Se²⁻, so the formula is Na₂Se.