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AP Chemistry Student Binder Companion · Unit 1: Topic 1.7

Print this dedicated study guide and place it in your 3-ring binder under Unit 1: Atomic Structure & Properties. Refer to the worked Coulombic models before answering free-response trend questions.

Topic 1.7 Dedicated Guide

Periodic Trends & Coulombic Mechanics

Coulomb’s Law, Atomic & Ionic Radii, Isoelectronic Series, Ionization Energy Anomalies & Successive IE Tables
Course: AP Chemistry Unit: Unit 1 (Atomic Structure & Properties) Instruction: Jeremy Krug Framework Platform: SABIS-CHEMISTRY.COM

1. The Governing Framework: Coulomb's Law

Every single periodic trend on the AP Chemistry Exam must be explained through Coulomb's Law:

F \propto \frac{q_1 \cdot q_2}{r^2} \quad \text{or} \quad F \propto \frac{Z_{\text{eff}} \cdot (-1)}{r^2}

Where:

  • $q_1 = Z_{\text{eff}}$: Effective nuclear charge experienced by valence electrons ($Z_{\text{eff}} \approx Z - S$).
  • $q_2 = -1$: Charge of an electron.
  • $r$: Internuclear distance between valence electrons and the nucleus (governed by principal quantum level $n$).
The Cardinal Rule for AP Free Response Questions
NEVER justify a trend by citing table position! Statements such as "fluorine has higher ionization energy because it is higher and further right" earn ZERO points on the AP Exam. ALWAYS explain the physical mechanism:
  1. Down a Group: Distance ($r$) / Number of Shells dominates! Adding energy levels places valence electrons at larger $r$, weakening Coulombic attraction.
  2. Across a Period: Effective Nuclear Charge ($Z_{\text{eff}}$) dominates! Same shell $n$, constant shielding, but $+1$ proton per step pulls electrons tighter.

2. Atomic Radius & Ionic Radii Trends

Periodic Trends Summary Map
Figure 1.7-1 (From AP 1.7 Slide 5): Periodic Trends Summary Map — Atomic radius increases down and to the left (increasing $n$ shells); First Ionization Energy, Electronegativity, and Electron Affinity increase up and to the right (increasing effective nuclear charge $Z_{\text{eff}}$).
First Ionization Energy of the Elements Graph
Figure 1.7-2 (Wikimedia Commons): First Ionization Energy vs. Atomic Number ($Z$). Note the sharp drops at Beryllium $\rightarrow$ Boron ($2s \rightarrow 2p$) and Nitrogen $\rightarrow$ Oxygen ($2p^3 \rightarrow 2p^4$ pairing repulsion).
Empirical Atomic Radius Trends
Figure 1.7-3 (Wikimedia Commons): Empirical Atomic Radii across the Periodic Table. Contraction occurs across rows due to increasing effective nuclear charge ($Z_{\text{eff}}$), while expansion occurs down columns due to additional principal energy shells.

Cations vs. Anions: The Coulombic Explanation

  • Cations ($+$) are ALWAYS SMALLER than their parent neutral atoms:
    When $\text{Na} \rightarrow \text{Na}^+ + e^-$, the entire outer $n = 3$ energy level is lost. With fewer electrons (10 vs 11), electron-electron repulsions decrease, and the 11 protons pull the remaining electrons closer to the nucleus.
  • Anions ($-$) are ALWAYS LARGER than their parent neutral atoms:
    When $\text{Cl} + e^- \rightarrow \text{Cl}^-$, an electron is added to the valence shell. The nuclear charge remains unchanged ($Z = 17$), but localized electron-electron repulsions among the 18 electrons increase, causing the electron cloud to expand outward.

3. The Isoelectronic Series: Radius Controlled by Protons

An isoelectronic series consists of species having the identical number of electrons and ground-state electron configurations.

Ion / Atom Protons ($Z$) Electrons Configuration Ionic Radius Coulombic Justification
$\mathbf{\text{O}^{2-}}$ 8 10 $1s^2 2s^2 2p^6$ 140 pm (Largest) Fewest protons pulling on 10 electrons $\rightarrow$ weakest Coulombic attraction $\rightarrow$ largest cloud.
$\mathbf{\text{F}^-}$ 9 10 $1s^2 2s^2 2p^6$ 133 pm 9 protons pull 10 electrons slightly tighter than in $\text{O}^{2-}$.
$\mathbf{\text{Na}^+}$ 11 10 $1s^2 2s^2 2p^6$ 102 pm 11 protons exert substantially stronger pull on 10 electrons.
$\mathbf{\text{Mg}^{2+}}$ 12 10 $1s^2 2s^2 2p^6$ 72 pm 12 protons pull 10 electrons even tighter.
$\mathbf{\text{Al}^{3+}}$ 13 10 $1s^2 2s^2 2p^6$ 54 pm (Smallest) Most protons (13) with identical core shielding $\rightarrow$ strongest Coulombic attraction $\rightarrow$ smallest cloud.

4. The Two Essential Ionization Energy Exceptions

Across Period 2 ($\text{Li} \rightarrow \text{Ne}$), First Ionization Energy generally increases. However, there are two distinct drops tested repeatedly on the AP Exam:

Anomaly 1: Beryllium ($Z = 4$) vs. Boron ($Z = 5$)

Observation: Boron ($801\text{ kJ/mol}$) has a lower 1st IE than Beryllium ($899\text{ kJ/mol}$), despite having one more proton.

AP Justification: Beryllium's electron is removed from a full, stable $2s$ subshell ($1s^2 2s^2$). Boron's electron is removed from the $2p$ subshell ($1s^2 2s^2 2p^1$). The $2p$ subshell is at a higher average energy level and is partially shielded by the $2s^2$ electrons, requiring less energy to ionize.

Anomaly 2: Nitrogen ($Z = 7$) vs. Oxygen ($Z = 8$)

Observation: Oxygen ($1314\text{ kJ/mol}$) has a lower 1st IE than Nitrogen ($1402\text{ kJ/mol}$), despite higher $Z$.

AP Justification: Nitrogen ($1s^2 2s^2 2p^3$) has each $2p$ orbital singly occupied with parallel spins ($\uparrow \ \uparrow \ \uparrow$). Oxygen ($1s^2 2s^2 2p^4$) has one pair of electrons sharing the same $2p$ orbital ($\uparrow\downarrow \ \uparrow \ \uparrow$). The two electrons sharing an orbital experience localized electron-electron repulsion, which destabilizes the electron and lowers the energy required to remove it.

5. Successive Ionization Energies & Identifying Unknown Elements

When successive electrons are removed from the same atom, a giant leap (typically $4\times$ to $10\times$) occurs as soon as all valence electrons are exhausted and the first core electron is ionized.

Element $\text{IE}_1$ $\text{IE}_2$ $\text{IE}_3$ $\text{IE}_4$ $\text{IE}_5$ Location of Giant Jump Valence e⁻ Group
Element X 578 1817 2745 11577 14842 Between $\text{IE}_3$ & $\text{IE}_4$ 3 Group 13 ($\text{Al}$)
Element Y 738 1451 7733 10540 13630 Between $\text{IE}_2$ & $\text{IE}_3$ 2 Group 2 ($\text{Mg}$)

6. Worked AP Walkthrough Model

Model: Deducing Formula from Successive IE Data AP Exam FRQ

Problem: A Period 3 element exhibits successive ionization energies in $\text{kJ/mol}$ of: $\text{IE}_1 = 786$, $\text{IE}_2 = 1577$, $\text{IE}_3 = 3232$, $\text{IE}_4 = 4356$, $\mathbf{\text{IE}_5 = 16091}$. Deduce the identity of the element and write the formula of its compound with Fluorine.

Solution:
  1. Identify Jump: The massive increase occurs between $\text{IE}_4$ and $\text{IE}_5$ ($4356 \rightarrow 16091\text{ kJ/mol}$, nearly a $4\times$ leap).
  2. Valence Count: Removing the 5th electron breaks into the stable inner core shell ($n = 2$). Thus, the atom has exactly 4 valence electrons.
  3. Element Identity: The Period 3 element with 4 valence electrons (Group 14) is Silicon ($\text{Si}$).
  4. Compound Formula: Silicon shares 4 valence electrons with 4 Fluorine atoms to form $\mathbf{\text{SiF}_4}$.
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