Atomic structure and electron configuration

Structure of the Atom

Every atom contains a positively charged nucleus surrounded by negatively charged electrons that determine the atom’s chemical behavior.

The nucleus consists of protons and neutrons. Protons carry a +1 charge and determine the element’s identity (atomic number, Z). Neutrons carry no charge and contribute to the atom’s mass. Together, protons and neutrons are called nucleons.

Electrons carry a −1 charge and have negligible mass compared to nucleons. In a neutral atom, the number of electrons equals the number of protons, so the overall charge is zero. When electrons are removed, a cation [positively charged ion] forms; when electrons are added, an anion [negatively charged ion] forms.

The electrostatic attraction between the positive nucleus and the negative electrons holds the atom together.

Coulomb’s Law and Electrostatic Force

Coulomb’s law describes the force of attraction or repulsion between two charged particles.

$${F}_{\text{coulombic}}∝\frac{{q}_{1}{q}_{2}}{{r}^{2}}$$

Where: ${q}_{1}$ and ${q}_{2}$ = charges on the two particles, and $r$ = distance between their centers.

Key relationships from this proportionality:

• Larger charges produce a stronger force. An electron near a nucleus with more protons experiences a greater attractive force because ${q}_{1}{q}_{2}$ increases in magnitude.
• Shorter distance produces a stronger force. An electron closer to the nucleus is held more tightly because $r$ is smaller, and force increases as ${r}^{2}$ decreases.
• The sign of the force indicates direction: opposite charges (positive nucleus, negative electron) produce an attractive force, while like charges (electron–electron) produce a repulsive force.

MisconceptionStudents sometimes treat Coulomb’s law as requiring exact numerical calculations on the AP exam. In this course, you apply it qualitatively — comparing magnitudes of charge and distance to predict which force is stronger.

Electron Shells, Subshells, and Configuration

Electrons in atoms occupy shells (energy levels) and subshells (sublevels), and the arrangement of electrons across these is called the electron configuration.

Shells are designated by the principal energy level number $n$ = 1, 2, 3, … . Higher values of $n$ correspond to electrons that are, on average, farther from the nucleus and at higher energy.

Each shell contains one or more subshells, labeled s, p, d, and f:

Shell $n$ = 1 contains only the 1s subshell. Shell $n$ = 2 contains 2s and 2p. Shell $n$ = 3 contains 3s, 3p, and 3d. Shell $n$ = 4 contains 4s, 4p, 4d, and 4f.

The Aufbau principle states that electrons fill subshells in order of increasing energy. The standard filling order is:

1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p

Notice that 4s fills before 3d, and 5s fills before 4d — this crossover occurs because subshell energies depend on both $n$ and the subshell type.

To write a ground-state electron configuration, fill subshells in Aufbau order, placing the number of electrons as a superscript. For example:

• Nitrogen (Z = 7): 1s² 2s² 2p³
• Iron (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Core electrons are those in inner, fully occupied shells; valence electrons are those in the outermost shell (highest $n$). For main-group elements, valence electrons are in the outermost s and p subshells. Valence electrons determine an element’s chemical reactivity and bonding behavior.

For ions, adjust the electron count:

• Cations: remove electrons starting from the highest $n$ (and within the same $n$, from the subshell with the highest $l$ — in practice, this means removing from the outermost s subshell before the d subshell for transition metals).
• Anions: add electrons to the next available subshell in the Aufbau order.

Example — Fe²⁺ (Z = 26, 24 electrons):

Neutral Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

Fe²⁺: Remove 2 electrons from 4s first → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

Out of ScopeThe assignment of quantum numbers (n, l, mₗ, mₛ) to electrons in subshells will not be assessed on the AP Exam. Do not memorize this for the exam.

MisconceptionStudents often remove d electrons first when writing transition metal cation configurations. Always remove electrons from the highest principal energy level ($n$) first — for transition metals, this means removing 4s electrons before 3d electrons.

Examiner InsightAP FRQs frequently ask students to write electron configurations for transition metal cations such as Fe²⁺, Fe³⁺, Cu²⁺, or Cr³⁺. Getting the ion configuration correct is a recurring point-earning opportunity.

Ionization Energy and Effective Nuclear Charge

The ionization energy of an atom is the minimum energy required to remove an electron from the atom in the gas phase, and it can be estimated qualitatively using Coulomb’s law.

According to $F∝{q}_{1}{q}_{2}/{r}^{2}$, two factors determine how tightly an electron is held:

1. Effective nuclear charge ($Z_{\mathrm{eff}}$) — the net positive charge experienced by a given electron after accounting for shielding [the partial cancellation of nuclear attraction by repulsion from core electrons]. More core electrons produce more shielding, which decreases $Z_{\mathrm{eff}}$. A higher $Z_{\mathrm{eff}}$ means a larger effective ${q}_{1}$, so the force on the electron increases and more energy is required to remove it.
2. Distance from the nucleus — electrons in higher shells ($n$) are farther from the nucleus. A larger $r$ decreases the force according to $1/{r}^{2}$, so the electron is easier to remove and ionization energy is lower.

Combining these two factors explains key comparisons:

The large jump in successive ionization energies when moving from valence electrons to core electrons is a direct consequence of both decreased distance and dramatically increased effective nuclear charge — the core electron is in a lower shell and has far fewer electrons shielding it from the nucleus.

Electrons in different subshells of the same shell also have different ionization energies. An electron in a 2s subshell is, on average, closer to the nucleus than an electron in a 2p subshell, so 2s electrons experience a slightly higher effective nuclear charge and require more energy to remove.

Examiner InsightAP exam questions frequently present successive ionization energy data and ask students to identify the group of an element based on where the largest jump occurs. The jump happens when removal transitions from valence electrons to core electrons.