Atomic structure and electron configuration — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The nuclear model of the atom, quantum number rules, subshell and orbital definitions, Aufbau principle, Pauli exclusion principle, Hund’s rule, and writing full/condensed electron configurations for neutral atoms and monatomic ions.

You should already know: Basic structure of protons, neutrons, and electrons. The concepts of atomic number and mass number. Quantization of energy in atomic systems.

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 Atomic structure and electron configuration?

Atomic structure and electron configuration describes how electrons are arranged around the nucleus of an atom or ion, and it is the foundational topic for all of modern chemistry. Electron arrangement directly controls chemical reactivity, bonding behavior, and periodic properties, making this one of the most heavily tested core topics on the AP Chemistry exam. Per the AP Chemistry CED, this topic makes up roughly half of Unit 1 (Atomic Structure and Properties), which contributes 7-10% of the total AP exam score. Concepts from this topic appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ often asks to identify valid quantum number sets or correct electron configurations, while FRQ may require writing configurations to explain periodic trends or magnetic properties. The core idea is that electrons occupy discrete energy levels divided into subshells of individual orbitals, with standard notation that lists the energy level, subshell, and electron count per subshell as $1s^{2}2s^{2}2p^6$.

2. Quantum Numbers and Orbital Structure

Quantum numbers are a set of four values that describe the unique location and spin of any electron in an atom, consistent with the quantum mechanical model of the atom. Each quantum number narrows down the probability region (orbital) an electron occupies:

1. Principal quantum number ($n$): Defines the main energy level (shell) and average distance from the nucleus. $n$ is always a positive integer ($n=1,2,\mathrm{3...}$), and larger $n$ corresponds to higher energy and larger orbital size.
2. Azimuthal (angular momentum) quantum number ($l$): Defines the subshell and shape of the orbital. $l$ can take integer values from $0$ to $n-1$, so the maximum value of $l$ is always one less than $n$. The mapping of $l$ to subshell names is: $l=0=s$, $l=1=p$, $l=2=d$, $l=3=f$.
3. Magnetic quantum number ($m_l$): Defines the orientation of the orbital in space. $m_l$ can take integer values from $-l$ to $+l$, so the number of orbitals per subshell is $2l+1$.
4. Spin quantum number ($m_s$): Defines the spin state of the electron, which can only be $+1/2$ or $-1/2$.

The maximum number of electrons that can fit in a shell with principal quantum number $n$ is: $$\text{Total electrons per n}=2n^2$$

Worked Example

Which of the following sets of quantum numbers $(n,l,m_l,m_s)$ is not allowed for an electron in a ground-state atom? A) $(2,1,0,+1/2)$ B) $(3,2,-1,-1/2)$ C) $(4,3,+3,-1/2)$ D) $(3,3,-2,+1/2)$

1. Recall the first rule for valid quantum numbers: $l$ can only range from $0$ to $n-1$, so $l<n$ always.
2. Check each option against this rule first: Option A: $l=1<n=2$, allowed. Option B: $l=2<n=3$, allowed. Option C: $l=3<n=4$, allowed. Option D: $l=3$ is equal to $n=3$, which violates the rule.
3. Confirm: For D, $m_l=-2$ is within the range $-l$ to $+l$ (-3 to +3), but the violation of the $l<n$ rule makes the entire set invalid.
4. The invalid set is D.

Exam tip: On AP MCQ, always check if $l\ge n$ first. This is the most common violation tested, so you can eliminate wrong answers in seconds without checking other quantum numbers.

3. Rules for Filling Orbitals

To write the ground-state electron configuration of an atom, three core rules govern the order electrons fill orbitals:

1. Aufbau Principle: Electrons fill lower-energy orbitals before higher-energy orbitals. The standard energy order for first-row transition metals and lighter elements is $1s<2s<2p<3s<3p<4s<3d<4p$, which can be remembered with the diagonal rule.
2. Pauli Exclusion Principle: No two electrons in the same atom can have identical sets of four quantum numbers. This means each orbital can hold a maximum of two electrons, which must have opposite spins.
3. Hund's Rule: When filling degenerate (equal-energy) orbitals (e.g., the three 2p orbitals), electrons occupy each orbital singly with parallel spin before any orbital gets a paired electron. This minimizes electron-electron repulsion between electrons in the same subshell.

Two common exceptions to the Aufbau principle for neutral first-row transition metals (frequently tested on the AP exam) are chromium (Z=24) and copper (Z=29). Half-filled ($d^5$) and fully filled ($d^{10}$) d subshells have extra stability, so Cr is $[Ar]4s^{1}3d^5$ (not $4s^{2}3d^4$) and Cu is $[Ar]4s^{1}3d^{10}$ (not $4s^{2}3d^9$).

Worked Example

Draw the ground-state orbital diagram for neutral oxygen (Z=8) and state the number of unpaired electrons.

1. Neutral oxygen has 8 electrons. Fill orbitals in order: 1s holds 2 electrons, 2s holds 2 electrons, leaving 4 electrons for the 2p subshell.
2. Apply Pauli exclusion: 1s and 2s orbitals each have two paired electrons with opposite spin.
3. Apply Hund's rule to the 2p subshell (three degenerate orbitals): place one unpaired electron in each of the three orbitals first, then pair the fourth electron in one orbital.
4. The final orbital diagram is: $1s:[\uparrow \downarrow ]$, $2s:[\uparrow \downarrow ]$, $2p:[\uparrow \downarrow ][\uparrow ][\uparrow ]$. There are 2 unpaired electrons.

Exam tip: When drawing orbital diagrams for FRQ, always label each subshell and explicitly show the spin direction of every electron to earn full credit.

4. Electron Configurations for Atoms and Ions

Electron configurations can be written as full configurations (listing all subshells) or condensed (noble gas core) configurations, where the preceding noble gas is placed in brackets to represent inner-shell electrons, and only outer electrons are listed. The most common point of confusion for students is writing configurations for transition metal cations: the $ns$ electrons are always lost before the $(n-1)d$ electrons during ionization, even though $ns$ fills before $(n-1)d$ in neutral atoms. For anions, electrons are added to the lowest available energy subshell, following the same rules as neutral atoms. Valence electrons (the outermost electrons available for bonding) are counted as all electrons with the highest principal quantum number $n$ for main group elements; for transition metals, valence electrons include $ns$ and $(n-1)d$ electrons.

Worked Example

Write the condensed electron configuration for the $N{i}^{2+}$ cation (Ni, Z=28).

1. First write the configuration for neutral Ni: Z=28, so 28 electrons. The preceding noble gas is Ar (Z=18), so neutral Ni is $[Ar]4s^{2}3d^8$.
2. Recall the transition metal ionization rule: $4s$ electrons are lost before $3d$ electrons. $N{i}^{2+}$ has lost 2 electrons total.
3. Remove both electrons from the $4s$ subshell first, leaving the $3d$ subshell unchanged.
4. The final condensed configuration for $N{i}^{2+}$ is $[Ar]3d^8$.

Exam tip: After writing any electron configuration, count the total number of electrons to confirm they match the expected number ($Z$ for neutral, $Z-\text{charge}$ for cations, $Z+\text{charge}$ for anions). This catches 90% of common counting errors.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Marking the quantum number set $(2,2,0,+1/2)$ as allowed. Why: Students forget the upper limit of $l$ is $n-1$, so they confuse the maximum $l$ with $n$. Correct move: Always check $l<n$ first when validating quantum number sets.
• Wrong move: Writing the configuration of $T{i}^{2+}$ as $[Ar]4s^{2}3d^0$. Why: Students memorize that 4s fills before 3d, so they incorrectly assume 3d electrons are lost first. Correct move: Always remove all $ns$ electrons before removing any $(n-1)d$ electrons for transition metal cations.
• Wrong move: Drawing the 2p orbital diagram for nitrogen as $[\uparrow \downarrow ][\uparrow ][]$. Why: Students forget Hund's rule and pair electrons early to finish faster. Correct move: Always place one unpaired electron into each degenerate orbital before pairing any electrons.
• Wrong move: Writing the electron configuration of neutral copper (Z=29) as $[Ar]4s^{2}3d^9$. Why: Students ignore the stability of half-filled and fully filled d subshell exceptions. Correct move: Memorize the two AP-tested exceptions: Cr = $[Ar]4s^{1}3d^5$, Cu = $[Ar]4s^{1}3d^{10}$.
• Wrong move: Counting 18 valence electrons for calcium (Z=20, $[Ar]4s^2$). Why: Students count all electrons instead of only the highest n valence electrons. Correct move: For main group elements, only count electrons with the largest principal quantum number n (2 valence electrons for calcium, not 20).
• Wrong move: Assigning $l=1$ to an s subshell. Why: Students mix up the mapping of l values to subshell names. Correct move: Memorize $l=0=s,l=1=p,l=2=d,l=3=f$.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following is the correct condensed electron configuration for neutral arsenic (Z=33)? A) $[Ar]4s^{2}3d^{10}4p^3$ B) $[Ar]4s^{2}4p^3$ C) $[Ar]3d^{10}4p^3$ D) $[Ge]4p^3$

Worked Solution: First, neutral arsenic has 33 total electrons. Argon (the preceding noble gas) has 18 electrons, so we need 15 remaining electrons after the Ar core. Option B only has 5 electrons after the core, option C only has 13, so both are wrong. Option D uses germanium (a non-noble gas) as a core, which is never allowed for condensed configurations. Option A has $2+10+3=15$ electrons after the Ar core, $18+15=33$ total electrons matching Z=33. The correct answer is A.

Question 2 (Free Response)

Answer the following questions about gallium (Ga, Z=31): (a) Write the full ground-state electron configuration for neutral gallium. (b) How many unpaired electrons are in neutral gallium? Justify your answer. (c) What is the maximum number of electrons in gallium that have $n=3$ and $l=1$? Explain.

Worked Solution: (a) Gallium has 31 electrons. Filling in order: $$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^1$$ Counting total electrons: $2+2+6+2+6+2+10+1=31$, which matches Z=31.

(b) All subshells except 4p are fully filled (all electrons paired). The 4p subshell has 1 electron, which is unpaired per Hund's rule. Total unpaired electrons = 1.

(c) $n=3$ and $l=1$ corresponds to the 3p subshell. For $l=1$, $m_l$ ranges from $-1$ to $+1$, so there are 3 orbitals in the 3p subshell. Each orbital holds 2 electrons per the Pauli exclusion principle, so maximum $3\times 2=6$ electrons. In gallium, the 3p subshell is fully filled with 6 electrons, so the answer is 6.

Question 3 (Application / Real-World Style)

Paramagnetic materials are attracted to external magnetic fields and have at least one unpaired electron; diamagnetic materials have all electrons paired and are weakly repelled. A sample of titanium (Ti, Z=22) oxide is found to be diamagnetic. What is the charge of the titanium cation in this compound? Justify your answer.

Worked Solution:

1. Neutral titanium has electron configuration $[Ar]4s^{2}3d^2$.
2. For any titanium cation, 4s electrons are lost first. We need all electrons paired for diamagnetism, so zero unpaired electrons.
3. If Ti loses 4 electrons: $T{i}^{4+}$ has configuration $[Ar]$, which has all electrons paired (no unpaired electrons), matching the diamagnetic observation.
4. Check other common oxidation states: $T{i}^{2+}$ is $[Ar]3d^2$ (two unpaired electrons, paramagnetic), $T{i}^{3+}$ is $[Ar]3d^1$ (one unpaired electron, paramagnetic). In context, the titanium cation in the diamagnetic oxide has a +4 charge.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the absolute foundation for all subsequent topics in AP Chemistry, because electron arrangement directly determines every chemical property of an element, from its reactivity to its bonding behavior. Next, you will apply the rules of electron configuration to understand periodic trends, including atomic radius, ionization energy, and electron affinity, which are heavily tested on the AP exam. Without correctly writing electron configurations and counting valence electrons, you cannot explain why these trends exist or predict the properties of unfamiliar elements. This topic also feeds directly into the study of chemical bonding, molecular geometry, and intermolecular forces later in the course, as the arrangement of valence electrons controls how atoms bond to one another.

• Photoelectron spectroscopy
• Periodic trends
• Ionic and covalent bonding
• Valence bond theory