Atomic Structure and Electron Configuration原子结构与电子排布

One force explains a whole unit

An atom is a positively charged nucleus (protons and neutrons) surrounded by negatively charged electrons. Nearly everything in U1.5 through U1.8 can be explained by one force: the Coulombic attraction between the nucleus and the electrons, together with the repulsion between electrons. That covers how electrons are arranged, where PES peaks appear, how radius and ionization energy change, and what charge an ion takes.

Lesson 2.3.2 used Coulomb’s law qualitatively to explain periodic trends. AP asks for more: proportional reasoning with the law, and explanations that name both factors, charge and distance.

Proportional reasoning with Coulomb’s law

The force between two charged particles is

For a nucleus and one electron, $q_1$ is the nuclear charge the electron feels (more precisely, the effective nuclear charge, since core electrons shield part of it), $q_2$ is the electron’s charge, and $r$ is the distance between them. Proportional reasoning only needs the factors:

• Double a charge and the force doubles.
• Double the distance and the force drops to 1/4.
• When both change, multiply the factors: charge × 2 and distance × 2 give 2 × 1/4 = 1/2 of the force.

So whenever you compare the attraction on two electrons, ask two questions: What nuclear charge does it feel? How far from the nucleus is it (which shell)? Drag the charge and distance on sides A and B, then try the two presets that use real data:

Electron configurations, the AP way

Lesson 2.2.5 covered the aufbau principle, the Pauli exclusion principle, and Hund’s rule. AP questions come at them from these angles:

• Core and valence electrons. Core electrons match a noble-gas configuration and sit close to the nucleus; valence electrons are in the outermost shell and control an element’s chemistry.
• Ions. Anions add electrons in aufbau order: $\ce{O^2-}$ is 1s²2s²2p⁶. Cations lose electrons from the highest n first, so transition metals lose 4s before 3d: Mn is [Ar]4s²3d⁵, $\ce{Mn^2+}$ is [Ar]3d⁵, and $\ce{Mn^3+}$ is [Ar]3d⁴.
• Unpaired electrons. Draw the orbital diagram (by Hund’s rule, electrons occupy a sublevel’s orbitals singly before pairing) and count the single arrows: O has 2, N has 3, and $\ce{Fe^3+}$ has 5.
• Highest-energy electron. For main-group elements, look at the last sublevel filled: for Ga it’s 4p. For transition metals, separate “filled last” (3d) from “lost first” (4s).

US textbooks and AP usually write configurations in filling order ([Ar]4s²3d⁶), while Chinese textbooks write them by shell ([Ar]3d⁶4s²); AP accepts both. AP doesn’t ask you to write the exceptions such as Cr and Cu.

Pick an element and a charge to see the configuration, orbital diagram, and unpaired electrons. Be sure to try Mn, $\ce{Mn^3+}$, and $\ce{Zn^2+}$:

Ground state, excited state, or impossible?

• Ground state: follows the aufbau principle, the Pauli principle, and Hund’s rule; it is the lowest-energy arrangement.
• Excited state: the right number of electrons, but at least one sits in a higher sublevel while a lower one still has room, such as Be as 1s²2s¹2p¹ or Mg as [Ne]3s¹3p¹. Absorbing energy produces an excited state, and returning to the ground state emits light (lesson 2.2.3).
• Impossible: breaks a sublevel’s capacity (2s³, 2p⁷) or uses a sublevel that doesn’t exist (1p, 2d, 3f).

A common trap: writing $\ce{Fe^2+}$ as [Ar]4s²3d⁴. The electron count is right (24), but it isn’t the ground state, because a cation loses its 4s electrons first. The ground state is [Ar]3d⁶.

Sizes of isoelectronic species

Species with the same number of electrons are isoelectronic. Each of these six has 10 electrons, in n = 1 and n = 2:

(Shannon effective ionic radii; four-coordinate for $\ce{N^3-}$, six-coordinate for the rest.)

The Coulombic explanation: the electrons and shells are the same, so electron–electron repulsion is about the same; more protons pull harder on that same set of electrons and draw them in closer, so the radius is smaller. That’s the reasoning AP accepts. These two claims earn no credit: “a neutral atom has zero net charge, so its electrons feel no attraction” (every electron is attracted by the nucleus), and “a cation’s positive charge increases the repulsion between its electrons” (repulsion depends on the number of electrons, which is the same).

Removing electrons from different sublevels

Within one atom, the energy needed to remove an electron depends strongly on its sublevel. Core electrons are close to the nucleus and hardly shielded, so they are attracted far more strongly; valence electrons are far away and shielded by the core, so they are the easiest to remove. By Coulomb’s law, a 1s electron in sodium is the hardest to remove, then 2s and 2p, and the 3s electron is the easiest. Photoelectron spectroscopy, in U1.6, measures these energies directly.

Example 1: The configuration of Mn³⁺

Write the ground-state configuration of $\ce{Mn^3+}$ and give its number of unpaired electrons.

Example 2: Ranking an isoelectronic series

Rank $\ce{S^2-}$, $\ce{Cl-}$, $\ce{K+}$, and $\ce{Ca^2+}$ from largest to smallest radius, and explain.

Common mistakes

Quick check

AP-style free response

Summary

• Coulomb’s law: $F \propto q_1 q_2 / r^2$. The attraction an electron feels depends on effective nuclear charge and distance; in proportional reasoning, multiply the two factors.
• Anions add electrons in aufbau order; cations lose electrons from the highest n first, so transition metals lose 4s first: $\ce{Mn^3+}$ is [Ar]3d⁴.
• Draw orbital diagrams by Hund’s rule and count single arrows for unpaired electrons. An electron in a higher sublevel while a lower one has room means an excited state; overfilled or nonexistent sublevels mean an impossible configuration.
• Isoelectronic species: same electrons and shells, so more protons means stronger attraction and a smaller radius.