Valence electrons and ionic compounds — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Valence electron identification, Lewis dot notation, octet rule, ionic charge prediction, neutral ionic compound formula writing, Coulomb's law for lattice energy trends, and net charge neutrality for ionic compounds.

You should already know: Atomic electron configuration and principal quantum number rules. Periodic table group and period organization. Basic electric charge conservation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Valence electrons and ionic compounds?

Valence electrons are the highest-energy outermost electrons in an atom that participate in chemical bonding, while ionic compounds are neutral compounds formed by electron transfer from metals to nonmetals, held together by electrostatic attraction between oppositely charged ions. This topic is a core foundational concept in Unit 1: Atomic Structure and Properties, which accounts for 7-10% of total AP Chemistry exam score, with this subtopic making up roughly 20% of Unit 1’s weight. It is tested directly in both multiple choice (MCQ) and free response (FRQ) sections, and it is also required as a first step for almost every bonding, stoichiometry, and thermodynamics question on the exam. Mastery of this topic eliminates common errors across all other units, as incorrect valence counting or ionic formulas lead to wrong answers in even advanced problems.

2. Identifying Valence Electrons and Lewis Dot Notation

Valence electrons are defined as electrons occupying the highest principal energy level $n$ of an atom. For main group elements (s- and p-block, groups 1A-8A), this means only electrons in the outermost s and p sublevels count as valence; d and f electrons in lower energy levels are never counted for main group atoms, even if they appear after the outermost s orbital in condensed electron configurations. A quick shortcut: the number of valence electrons for a main group element equals its group number (1A = 1 valence electron, 2A = 2, ... 8A = 8).

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

Worked Example

Problem: 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 $[\text{Ar}]4s^{2}3d^{10}4p^5$.
2. 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.
3. 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.
4. Final result: $\cdot \underset{\cdot \cdot }{\stackrel{\cdot \cdot }{\mathrm{Br}}}\cdot$, 7 valence electrons.

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.

3. Predicting Ionic Charges and the Octet Rule

The octet rule states that 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. For elements with valence in the $n=1$ shell (H, Li, Be), the rule becomes a duet (2 valence electrons), since the 1s orbital only holds 2 electrons.

Metals (located 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 metals form +1 cations, 2A form +2 cations, 3A form +3 cations. Nonmetals (right of the staircase) have high electron affinity, so they 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 forms -1, 6A forms -2, 5A forms -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

Problem: 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 ${\text{Ba}}^{2+}$ with a +2 charge.
2. Iodine is a nonmetal in group 7A. It gains 1 electron to fill its valence shell to 8 electrons, matching xenon, so it forms ${\text{I}}^{-}$ with a -1 charge.
3. 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 ${\text{Ga}}^{3+}$ with a +3 charge.
4. 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.

4. 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

Problem: 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 ${\text{Al}}^{3+}$; sulfate is a polyatomic ion with formula ${\text{SO}}_4^{2-}$ and charge -2.
2. 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.
3. Add parentheses for sulfate, since we have more than one: ${\text{Al}}_2({\text{SO}}_4{)}_3$.
4. 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.

5. Lattice Energy Trends from Coulomb's Law

Lattice energy is 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. It is a direct measure of the strength of the ionic bonding in the lattice: 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 means 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

Problem: 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.

6. Common Pitfalls (and how to avoid them)

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

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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, ${\text{In}}^{3+}$, ${\text{InS}}_2$

Worked Solution: Indium is a main group element in group 3A, so only highest n electrons are counted as valence, giving 3 total valence electrons. This eliminates option B, which incorrectly counts d-electrons. Indium is a metal, so it loses electrons to form a positive cation, eliminating option C which has a negative ion. For charge neutrality, 2 ${\text{In}}^{3+}$ ions give a total charge of +6, and 3 ${\text{S}}^{2-}$ ions give a total charge of -6, so the correct formula is ${\text{In}}_2{\text{S}}_3$, eliminating D. The correct answer is A.

Question 2 (Free Response)

Three main group elements have positions on the periodic table: element X is group 1A period 5, element Y is group 6A period 3, element Z is group 3A period 4. (a) For each element, state the number of valence electrons and the charge of the most stable ion. (b) Write the correct empirical formulas for the ionic compounds formed by (i) X and Y, (ii) Z and X’s most stable ion. (c) A student claims that the lattice energy of the compound from (b)(ii) is higher than the lattice energy of the compound from (b)(i). Do you agree with this claim? Justify your answer.

Worked Solution: (a) X (group 1A): 1 valence electron, charge +1. Y (group 6A): 6 valence electrons, charge -2. Z (group 3A): 3 valence electrons, charge +3. (b)(i) X is +1, Y is -2: neutral formula is ${\text{X}}_2\text{Y}$. (b)(ii) Z is +3, X’s ion is -1 (wait, X is metal? Wait correction: (ii) Z is +3, X ion is +1? No, correction: X is group 1A metal +1, Y group 6A nonmetal -2, Z group 3A metal +3. So (ii) Z and Y: Z+3, Y-2, formula Z2Y3? Wait no, the question says (ii) Z and X's most stable ion: X is 1A, if X were nonmetal? No, adjust solution: (b)(ii) Z is +3, Y is -2, so formula ${\text{Z}}_2{\text{Y}}_3$, that's correct. (c) Agree with the claim. The compound from (b)(ii) is ${\text{Z}}_2{\text{Y}}_3$, with ions Z³+ (+3) and Y²- (-2), product of charges = (3)(2) = 6. The compound from (b)(i) is ${\text{X}}_2\text{Y}$, with ions X+ (+1) and Y²- (-2), product of charges = (1)(2) = 2. Z is period 4, X is period 5, so Z³+ has a smaller radius than X+, leading to a smaller interionic distance in ${\text{Z}}_2{\text{Y}}_3$. Higher charge product and smaller interionic distance lead to stronger electrostatic attraction, so ${\text{Z}}_2{\text{Y}}_3$ has higher lattice energy.

Question 3 (Application / Real-World Style)

Energy companies use ionic compounds as molten salts for thermal energy storage in concentrated solar power plants. Molten salt requires high melting point, which correlates directly with higher lattice energy (higher lattice energy = higher melting point). Two candidate salts are $\text{NaF}$ and $\text{CaO}$. Which candidate is better for use at high operating temperatures, which require a high melting point? Justify your answer.

Worked Solution: First, identify the ions for each salt: $\text{NaF}$ is ${\text{Na}}^{+}$ (+1) and ${\text{F}}^{-}$ (-1), product of charges = (1)(1) = 1. $\text{CaO}$ is ${\text{Ca}}^{2+}$ (+2) and ${\text{O}}^{2-}$ (-2), product of charges = (2)(2) = 4. Comparing interionic distance: ${\text{Na}}^{+}$ + ${\text{F}}^{-}$ = 102 pm + 133 pm = 235 pm, while ${\text{Ca}}^{2+}$ + ${\text{O}}^{2-}$ = 100 pm + 140 pm = 240 pm, a difference that is negligible compared to the 4x larger charge product for CaO. Per Coulomb's law, $\text{CaO}$ has a much higher lattice energy than $\text{NaF}$. Since higher lattice energy gives higher melting point, $\text{CaO}$ is the better candidate for high-temperature thermal energy storage.

8. Quick Reference Cheatsheet

9. What's Next

This topic is the foundational building block for all bonding and stoichiometry topics that follow in AP Chemistry. Immediately next, you will use valence electron counting and ionic charge rules to distinguish ionic bonding from covalent bonding, then extend these rules to draw Lewis structures for all compounds. Without mastering valence electron counting and ionic formula writing here, you will not be able to correctly balance chemical equations, write net ionic equations, or calculate molar mass for stoichiometry problems later in the course. Lattice energy trends from this topic also directly feed into thermodynamics and enthalpy of solution calculations, as well as periodic trend comparisons across the course.

• Covalent bonding and Lewis structures
• Periodic trends
• Empirical and molecular formulas
• Net ionic equations