# Valence electrons and ionic compounds

> AP Chemistry · AP Chemistry CED Unit 1
> Source: https://www.owlsprep.com/study/ap-chemistry-u1-valence-electrons-and-ionic-compounds/

This module covers valence electron identification, Lewis dot notation, ionic charge prediction, neutral ionic formula writing, and lattice energy trends per Coulomb's law, a foundational topic for AP Chemistry Unit 1 tested across MCQ and FRQ.

**Prerequisites:** Atomic electron configuration and principal quantum number rules; Periodic table group and period organization

## Learning objectives

- Identify valence electrons for main group elements using periodic table and electron configuration
- Draw Lewis dot structures for neutral main group atoms
- Predict charges of stable main group ions using the octet rule
- Write correct neutral formulas for ionic compounds including polyatomic ions
- Predict and justify lattice energy trends using Coulomb's law

## Identifying Valence Electrons and Lewis Dot Notation

Valence electrons are electrons occupying the highest principal energy level $n$ of an atom, and are the electrons that participate in chemical bonding. For main group elements (s- and p-block, groups 1A-8A), only electrons in the outermost s and p sublevels count as valence; d and f electrons in lower energy levels are never counted, even if they appear after the outermost s orbital in condensed configurations. A quick shortcut: the number of valence electrons equals the element's group number for main group elements.

**Valence Electron** — Outermost, highest energy electron(s) that participate in chemical bonding

*Example:* Bromine has 7 valence electrons in its $n=4$ shell

Lewis dot notation is a standard convention to represent valence electrons: draw the element's chemical symbol, then add one dot per valence electron, placing one dot on each of the four sides before pairing any dots. This makes it easy to see how many electrons an atom will lose or gain to form an ion.

**Worked example:** Write the Lewis dot structure for a neutral bromine (Br) atom, and state how many valence electrons it has.

1. Locate bromine on the periodic table: it is a main group element in group 7A, period 4. Its condensed electron configuration is
2. $$[\text{Ar}] 4s^2 3d^{10} 4p^5$$
3. Count valence electrons by highest $n$: only electrons in $n=4$ count, so $4s^2 4p^5 = 2 + 5 = 7$ total valence electrons. The $3d^{10}$ electrons are in $n=3$, so they are excluded.
4. Arrange dots: place one dot on each of the four sides, then add the three remaining dots to get three paired sides and one unpaired side, for 7 total dots.
5. Final result: 7 valence electrons, with the Lewis dot structure:
6. $$\cdot \underset{\cdot \cdot}{\overset{\cdot \cdot}{\text{Br}}} \cdot$$

> **Exam tip:** If you are asked to count valence electrons for a main group ion, add one electron for each negative charge and subtract one for each positive charge, always starting from the neutral atom count.

## Predicting Ionic Charges and the Octet Rule

**Octet Rule** — Main group atoms lose or gain valence electrons to achieve a full valence shell of 8 electrons, matching the stable electron configuration of the nearest noble gas. Atoms with valence in the $n=1$ shell follow a duet rule of 2 electrons.

Metals (left of the metalloid staircase on the periodic table) have low ionization energy, so they lose all their valence electrons to form positively charged cations. For main group metals, the charge of the most stable cation equals the group number: 1A = +1, 2A = +2, 3A = +3. Nonmetals (right of the staircase) gain electrons to fill their valence shell, forming negatively charged anions. The charge of the most stable main group anion is $-(8 - \text{group number})$: 7A = -1, 6A = -2, 5A = -3. Transition metals are an exception: they form multiple stable cations with different charges, so their charge cannot be predicted from group number alone.

**Worked example:** Predict the charge of the most stable ion formed by (a) barium (Ba), (b) iodine (I), (c) gallium (Ga).

1. Barium is a main group metal in group 2A. It loses its 2 valence electrons to match the electron configuration of xenon, so it forms
2. $$\text{Ba}^{2+}$$
3. with a +2 charge.
4. Iodine is a nonmetal in group 7A. It gains 1 electron to fill its valence shell to 8 electrons, matching xenon, so it forms
5. $$\text{I}^-$$
6. with a -1 charge.
7. Gallium is a main group metal in group 3A. It loses its 3 valence electrons to match the electron configuration of argon, so it forms
8. $$\text{Ga}^{3+}$$
9. with a +3 charge.
10. All ions follow the octet rule, so these are the most stable charges.

> **Exam tip:** When asked for the most stable ion, always default to the octet rule prediction for main group elements; do not leave the ion neutral or give a non-standard charge unless explicitly prompted.

## Writing Formulas for Neutral Ionic Compounds

All stable ionic compounds are electrically neutral, meaning the total positive charge from cations equals the total negative charge from anions, for a net charge of zero. The criss-cross method is a simple technique to get the correct formula:

1. Write the cation first, then the anion, with their correct charges.
2. The absolute value of the cation charge becomes the subscript for the anion, and the absolute value of the anion charge becomes the subscript for the cation.
3. Reduce the subscripts to the lowest whole number ratio by dividing by their greatest common factor.
4. Enclose polyatomic ions in parentheses if their subscript is greater than 1, to indicate the subscript applies to the entire ion.

**Worked example:** Write the correct empirical formula for the ionic compound formed between aluminum ions and sulfate ions.

1. Identify ion charges: Aluminum is group 3A, so
2. $$\text{Al}^{3+}$$
3. ; sulfate is a polyatomic ion with formula
4. $$\text{SO}_4^{2-}$$
5. and charge -2.
6. Apply the criss-cross method: The absolute value of aluminum’s charge (3) becomes the subscript for sulfate, and the absolute value of sulfate’s charge (2) becomes the subscript for aluminum.
7. Add parentheses for sulfate, since we have more than one:
8. $$\text{Al}_2(\text{SO}_4)_3$$
9. Check for common factors: 2 and 3 share no common whole number factor, so this is the final formula. Check neutrality: $(2 \times +3) + (3 \times -2) = +6 -6 = 0$, which is correct.

> **Exam tip:** Always check for common factors after criss-cross; for example, $\text{Ca}_2\text{O}_2$ must be reduced to $\text{CaO}$, which is the correct formula for calcium oxide.

## Lattice Energy Trends from Coulomb's Law

**Lattice Energy** — The energy released when one mole of gaseous ions combines to form a solid ionic compound, or equivalently the energy required to separate one mole of solid ionic compound into gaseous ions. Higher magnitude lattice energy means stronger ionic attraction.

Per Coulomb’s law, the magnitude of the electrostatic force between two charged particles is proportional to:

$$E \propto \frac{q_1 q_2}{r}$$

where $q_1$ and $q_2$ are the charges of the two ions, and $r$ is the interionic distance (sum of the two ionic radii). This gives two key trends: (1) lattice energy increases as the product of the ion charges increases, and (2) for the same charge product, lattice energy increases as interionic distance decreases. Charge product has a much larger effect on lattice energy than radius, so always compare charge first.

**Worked example:** Which compound has the higher magnitude lattice energy: $\text{NaBr}$ or $\text{MgS}$? Justify your answer.

1. Identify ion charges: $\text{NaBr}$ is $\text{Na}^+$ (+1) and $\text{Br}^-$ (-1), so the product $q_1 q_2 = (1)(1) = 1$. $\text{MgS}$ is $\text{Mg}^{2+}$ (+2) and $\text{S}^{2-}$ (-2), so product $q_1 q_2 = (2)(2) = 4$.
2. Compare interionic distance: $\text{Na}^+$ radius = 102 pm, $\text{Br}^-$ = 196 pm, sum $r = 298$ pm. $\text{Mg}^{2+}$ = 72 pm, $\text{S}^{2-}$ = 184 pm, sum $r = 256$ pm, which is smaller than $\text{NaBr}$’s distance.
3. Both the higher charge product and smaller interionic distance mean electrostatic attraction between ions is much stronger in $\text{MgS}$.
4. Conclusion: $\text{MgS}$ has a higher magnitude lattice energy than $\text{NaBr}$.

> **Exam tip:** On FRQ, you must explicitly reference Coulomb’s law and compare both charge (first) and radius (second) to earn full justification points; vague statements about "stronger bonding" are not enough.

## AP-Style Concept Check

**Check your understanding**

Test your understanding of core concepts with these AP-style multiple choice questions:

1. Which of the following gives the correct number of valence electrons, most stable ion charge, and correct ionic compound formula for indium (In, group 3A) and sulfur (S, group 6A)?

   - A) 3 valence electrons, $\text{In}^{3+}$, $\text{In}_2\text{S}_3$
   - B) 13 valence electrons, $\text{In}^{3+}$, $\text{In}_2\text{S}_3$
   - C) 3 valence electrons, $\text{In}^{3-}$, $\text{InS}_2$
   - D) 3 valence electrons, $\text{In}^{3+}$, $\text{InS}_2$

   *Answer:* A) 3 valence electrons, $\text{In}^{3+}$, $\text{In}_2\text{S}_3$

   *Why:* Correct! Indium is a main group group 3A element, so only highest $n$ electrons count as valence, giving 3 total. It forms a +3 cation, and neutral charge requires 2 indium ions for every 3 sulfide ions, giving the formula $\text{In}_2\text{S}_3$.

2. Energy companies use molten ionic salts for high-temperature thermal energy storage, where higher melting point (directly correlated with higher lattice energy) is required. Which candidate ($\text{NaF}$ vs $\text{CaO}$) is better, and why?

   - A) $\text{NaF}$, because its ions have smaller combined radius than $\text{CaO}$
   - B) $\text{CaO}$, because the product of its ion charges is 4 times larger than $\text{NaF}$
   - C) $\text{NaF}$, because it has lower molar mass than $\text{CaO}$
   - D) $\text{CaO}$, because it is composed of more abundant elements than $\text{NaF}$

   *Answer:* B) $\text{CaO}$, because the product of its ion charges is 4 times larger than $\text{NaF}$

   *Why:* Correct! The product of charges for $\text{CaO}$ is $(+2)(-2) = 4$, compared to 1 for $\text{NaF}$. This large difference in charge product dominates over the small difference in ionic radius, leading to much higher lattice energy and a higher melting point, which meets the requirement for high-temperature operation.

## Common pitfalls

- **Wrong:** Counting d-electrons as valence for main group elements, e.g., counting 13 valence electrons for gallium ([Ar]4s²3d¹⁰4p¹) instead of 3
  - Why it fails: Students confuse total electrons outside the noble gas core with the highest $n$ definition of valence electrons used by AP Chemistry
  - Correct: Only count electrons in the highest principal energy level $n$, regardless of sublevel, when counting valence for main group elements
- **Wrong:** Writing the anion first in an ionic compound formula, e.g., $\text{ClK}$ for potassium chloride
  - Why it fails: Students mix up the order from electron transfer diagrams that show nonmetals gaining electrons first
  - Correct: Always write the cation first, then the anion, per IUPAC convention for all ionic compounds
- **Wrong:** Forgetting to enclose polyatomic ions in parentheses when the subscript is greater than 1, e.g., writing $\text{BaOH}_2$ instead of $\text{Ba(OH)}_2$ for barium hydroxide
  - Why it fails: Students do not recognize that the subscript applies to the entire polyatomic ion, not just the last atom
  - Correct: Always wrap polyatomic ions in parentheses if you have more than one of them in the formula unit
- **Wrong:** Failing to reduce subscripts to the lowest whole number ratio, e.g., writing $\text{Ca}_2\text{O}_2$ instead of $\text{CaO}$
  - Why it fails: Students stop after applying the criss-cross method and do not check for common factors
  - Correct: After criss-cross, divide both subscripts by their greatest common factor to get the correct empirical formula
- **Wrong:** Comparing lattice energy based only on ionic radius before checking charge product, e.g., claiming LiF has higher lattice energy than MgO because Li⁺ and F⁻ are smaller
  - Why it fails: Students prioritize size over charge, but charge has a much larger effect on electrostatic force
  - Correct: Always compare the product of ion charges first; only compare interionic distance if the charge products are equal
- **Wrong:** Predicting transition metal ion charges from group number, e.g., claiming iron (group 8) forms an Fe⁸⁺ ion
  - Why it fails: Students extend the main group charge rule to all elements
  - Correct: Remember transition metals form multiple stable cations, so their charge must be deduced from the corresponding anion in the compound, not predicted by group number

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Main group valence count | Number of valence electrons = group number (1A-8A) | Only count electrons in highest $n$; ignore d/f electrons |
| Octet rule | Atoms lose/gain electrons to reach 8 valence electrons (2 for n=1) | Applies to most stable main group ions |
| Main group cation charge | Charge = +(group number) | Metals lose electrons to form positive cations |
| Main group anion charge | Charge = -(8 - group number) | Nonmetals gain electrons to form negative anions |
| Ionic neutrality rule | Total positive charge = Total negative charge, net charge = 0 | Required for all stable ionic compounds |
| Criss-cross method | Swap absolute value of ion charges to get subscripts | Cation first; reduce subscripts to lowest whole number ratio |
| Polyatomic ion notation | Enclose in parentheses if subscript > 1 | Subscript applies to entire ion, not last atom |
| Lattice energy trend | $E \propto \frac{q_1 q_2}{r}$ | Compare charge product first, then interionic distance |

## What's next

Mastery of valence electrons and ionic compounds is the foundation for all subsequent bonding, nomenclature, and stoichiometry topics in AP Chemistry. Next, you will apply these concepts to covalent bonding, molecular geometry, and naming inorganic compounds, all of which rely on correct valence electron counting and charge prediction. You will also reuse Coulomb's law and lattice energy concepts when studying enthalpy of solution and Born-Haber cycles in thermodynamics. Errors in ionic formula writing lead to incorrect stoichiometric calculations, molar mass determinations, and limiting reactant problems across the entire exam, so reinforcing this topic early pays off with higher scores across all units.

- [AP Chemistry Unit 1 Overview](https://www.owlsprep.com/study/ap-chemistry-u1-overview/)
- [Molecular and Ionic Compound Structure and Properties Overview](https://www.owlsprep.com/study/ap-chemistry-u2-overview/)
- [Types of chemical bonds](https://www.owlsprep.com/study/ap-chemistry-u2-types-of-chemical-bonds/)

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