Unit 1: Atomic Structure and Properties

The Periodic Table and Atomic Structure

The periodic table is a compact map of atomic structure and behavior. Each element’s “box” provides essential information you use constantly in AP Chemistry.

Reading an element’s information

Using carbon as an example, the periodic table typically shows:

• Symbol (C): shorthand name used in formulas and equations.
• Atomic number (6): the number of protons in the nucleus. In a neutral atom, the number of electrons also equals 6.
• Average atomic mass / molar mass (12.011):
  • the weighted-average mass of a single atom (in atomic mass units, amu)
  • numerically the mass of 1 mole of atoms (in grams per mole)

Because the periodic-table mass is a weighted average of isotopes, it usually is not a whole number. It can also hint at the most common isotope; for example, carbon’s average mass is about 12.01, and about 99% of naturally occurring carbon is carbon-12.

Periods, groups, and named families

• Periods are horizontal rows.
• Groups are vertical columns.

Groups may be labeled as 1–18 or with older Roman-numeral style labels. Common names:

• Group IA/1: Alkali metals
• Group IIA/2: Alkaline earth metals
• Groups B/3–12: Transition metals
• Group VIIA/17: Halogens
• Group VIIIA/18: Noble gases

The two rows shown below the main table are the lanthanides and actinides, often called the rare earth elements or inner transition metals.

Subatomic particles, isotopes, and mass

Atoms consist of:

• Protons (positive) in the nucleus
• Neutrons (neutral) in the nucleus
• Electrons (negative) in a region around the nucleus

Key identifiers:

• Atomic number = number of protons (defines the element)
• Mass number = protons + neutrons (varies by isotope)

Electrons have much less mass than protons and neutrons, so atomic mass is mostly nuclear mass.

Isotopes are atoms of the same element (same protons) with different numbers of neutrons. Example: carbon-12 has 6 protons and 6 neutrons, while carbon-14 has 6 protons and 8 neutrons.

Exam Focus

• Typical question patterns
  • Read an element’s box to determine protons/electrons (and connect the average atomic mass to isotopes).
  • Identify periods vs. groups and recognize family names (alkali metals, halogens, noble gases, etc.).
  • Use atomic number and mass number to determine subatomic particle counts.
• Common mistakes
  • Treating the atomic number as if it gives neutrons directly (neutrons require mass number).
  • Assuming periodic-table atomic masses must be whole numbers.
  • Mixing up periods (rows) and groups (columns).

The Mole and Molar Mass

Chemistry connects the microscopic world of atoms and molecules to the macroscopic world you can measure in a lab. The bridge is the mole, a counting unit—like “a dozen,” but for unbelievably tiny particles.

What a mole means (and why it matters)

A mole is defined as exactly:

6.02214076 \times 10^{23}

representative particles. This is Avogadro’s number. “Representative particles” depend on context and may be atoms, molecules, ions, or formula units.

The mole also connects directly to chemical equations: coefficients in a balanced equation represent mole ratios between substances.

Molar mass as a mass-to-particle conversion

The molar mass is the mass of 1 mole of a substance (g/mol). It comes from the periodic table:

• An element’s atomic mass (amu) corresponds numerically to its molar mass (g/mol).
• For compounds, add atomic masses for all atoms in the formula.

This sets up conversion chains:

• grams ↔ moles using molar mass
• moles ↔ particles using Avogadro’s number

How to do mole conversions (step by step)

A consistent approach is: “What do I have?” and “What do I want?”

1) Mass to moles:

\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}

2) Moles to particles:

\text{particles} = \text{moles} \times 6.02214076 \times 10^{23}

Equivalently, particles to moles:

\text{moles} = \frac{\text{particles}}{6.02214076 \times 10^{23}}

A common mistake is using the wrong type of particle. For example, 1 mole of sodium chloride contains 1 mole of NaCl formula units, which corresponds to 1 mole of Na ions and 1 mole of Cl ions once dissociated.

Moles and gases

Gas problems often use the ideal gas law:

PV = nRT

where:

• P is pressure (atm)
• V is volume (L)
• n is moles
• T is temperature (K)
• R = 0.0821 (L·atm)/(mol·K)

Many introductory gas questions use STP (standard temperature and pressure), where P = 1 atm and T = 273 K. At STP, 1 mole of an ideal gas occupies 22.4 L, so:

\text{moles} = \frac{\text{liters}}{22.4 \text{ L/mol}}

Worked example: mass to particles

Problem: How many molecules are in 18.0 g of water, H2O?

Step 1: Find molar mass.

Molar mass:

\approx 2(1.008) + 16.00 = 18.016

Step 2: Convert grams to moles.

\text{moles H2O} = \frac{18.0}{18.016} \approx 0.999

Step 3: Convert moles to molecules.

\text{molecules} = 0.999 \times 6.02214076 \times 10^{23} \approx 6.02 \times 10^{23}

The result is sensible because 18 g of water is very close to 1 mole.

Worked example: particles to mass

Problem: What mass of aluminum contains:

3.00 \times 10^{23}

Al atoms?

Step 1: Convert atoms to moles.

\text{moles Al} = \frac{3.00 \times 10^{23}}{6.02214076 \times 10^{23}} \approx 0.498

Step 2: Convert moles to grams.

\text{mass} = 0.498 \times 26.98 \approx 13.4

Exam Focus

• Typical question patterns
  • Convert between grams, moles, and number of particles (atoms, molecules, formula units).
  • Compute molar mass from a formula, including parentheses and subscripts.
  • Multi-step conversion chains (grams → moles → particles).
  • Use PV = nRT to solve for moles; use 22.4 L/mol at STP when appropriate.
• Common mistakes
  • Treating ionic compounds as “molecules” instead of formula units.
  • Forgetting to multiply atomic masses by subscripts (especially with parentheses).
  • Using Avogadro’s number backward (dividing when you should multiply, or vice versa).
  • Mixing STP shortcuts into non-STP conditions.

Mass Spectrometry and Isotopes

Atoms of the same element do not always have the same mass because elements can exist as isotopes—same protons, different neutrons. Mass spectrometry is a key tool that reveals isotopic masses and abundances.

Isotopes: what they are and why they matter

An element is defined by its number of protons (atomic number). Changing neutrons changes mass but not identity. Isotopes matter because:

• The periodic-table atomic mass is a weighted average based on isotope abundances.
• Isotopes have nearly the same chemistry but different masses, affecting physical properties and enabling applications in medicine and geochemistry.

How mass spectrometry works (conceptually)

In a typical mass spectrometer, atoms are ionized (often to +1), accelerated, and separated by mass-to-charge ratio. The detector produces a spectrum with peaks:

• Peak position corresponds to isotopic mass (measured precisely).
• Peak height/area corresponds to relative abundance.

Example interpretation idea: selenium has a most abundant isotope with mass about 80, and four other naturally occurring isotopes. A weighted average of all naturally occurring isotopes gives selenium’s average atomic mass.

Average atomic mass as a weighted average

If isotopes have masses m_1, m_2, \dots and fractional abundances f_1, f_2, \dots (fractions sum to 1), then:

\text{average atomic mass} = f_1 m_1 + f_2 m_2 + \dots

Percent abundances must be converted to decimals first.

Worked example: calculating average atomic mass

Problem: An element has two isotopes: 10.012 amu (19.9%) and 11.009 amu (80.1%). Find the average atomic mass.

Convert to fractions: 0.199 and 0.801.

\text{avg} = (0.199)(10.012) + (0.801)(11.009)

\text{avg} \approx 1.992 + 8.817 = 10.809

Worked example: finding abundance from average mass

Problem: Chlorine’s average atomic mass is about 35.45 amu. Suppose it has two isotopes: 35.00 amu and 37.00 amu. Let x be the fraction that is 35.00 amu. Find x.

35.45 = x(35.00) + (1 - x)(37.00)

35.45 = 37.00 - 2.00x

2.00x = 1.55

x = 0.775

So about 77.5% is chlorine-35 and 22.5% is chlorine-37.

What goes wrong: common isotope misconceptions

A frequent error is assuming the periodic-table atomic mass must match one isotope’s mass number. It usually doesn’t because it’s an average. Another common error is forgetting abundances must total 100% (or 1.00 as decimals).

Exam Focus

• Typical question patterns
  • Interpret mass spectra: identify isotopes from peak positions and determine relative abundances.
  • Calculate average atomic mass from isotopic masses and abundances.
  • Solve for an unknown abundance given the average mass and isotopic masses.
• Common mistakes
  • Using percent numbers directly in the weighted average without converting to decimals.
  • Mixing up “mass number” (an integer) with “isotopic mass” (measured, not always an integer).
  • Forgetting the second isotope abundance is 1 - x (or 100\% - x).

Elemental Composition of Pure Substances (Percent Composition)

Chemical formulas don’t just name substances; they quantify composition. A formula gives atom ratios, which lets you compute mass percent of each element.

Chemical formulas as particle ratios

A chemical formula gives a fixed composition for a pure substance:

• CO2 means 1 carbon atom for every 2 oxygen atoms.
• CaCl2 means 1 calcium ion for every 2 chloride ions.

These are counting ratios, not mass ratios. Because atoms have different masses, an element can contribute most of the mass even if it contributes fewer atoms.

Mass percent composition: what it is

Mass percent of an element is the fraction of a compound’s mass due to that element, multiplied by 100%.

How to calculate mass percent from a formula

1) Compute the compound’s molar mass.
2) Compute the total mass contributed by the element of interest.
3) Divide and multiply by 100%.

If an element appears multiple times (subscripts or parentheses), count all of them.

Worked example: percent composition (NH4NO3)

Problem: Find the mass percent of nitrogen in ammonium nitrate, NH4NO3.

NH4NO3 contains 2 N, 4 H, 3 O.

M = 2(14.01) + 4(1.008) + 3(16.00)

M = 80.052

m_N = 2(14.01) = 28.02

\%N = \frac{28.02}{80.052} \times 100\% \approx 35.0\%

Worked example: percent composition (Ca(NO3)2)

Problem: Calculate the percent composition of each element in calcium nitrate, Ca(NO3)2.

First count atoms: Ca1, N2, O6.

Compute molar mass using typical atomic masses Ca 40.08, N 14.01, O 16.00:

M = 1(40.08) + 2(14.01) + 6(16.00)

M = 40.08 + 28.02 + 96.00 = 164.10

Now compute each element’s mass contribution and percent:

\%\text{Ca} = \frac{40.08}{164.10} \times 100\% \approx 24.4\%

\%\text{N} = \frac{28.02}{164.10} \times 100\% \approx 17.1\%

\%\text{O} = \frac{96.00}{164.10} \times 100\% \approx 58.5\%

A helpful check is that the percentages should add to about 100% (small rounding differences are normal).

Real-world connection

Fertilizer labels often emphasize percent nitrogen (and other nutrients). The chemistry is the same idea: composition determines how much nutrient mass you’re getting.

Exam Focus

• Typical question patterns
  • Compute percent composition from a formula.
  • Compare compounds by percent composition (which has higher percent oxygen, etc.).
  • Correctly interpret parentheses and subscripts when counting atoms.
  • Use “percent sum to 100%” as a reasonableness check.
• Common mistakes
  • Forgetting to multiply by subscripts (especially outside parentheses).
  • Using atomic masses inconsistently (mixing rounded and unrounded values mid-problem).
  • Treating a formula as if it gives mass ratios directly rather than atom ratios.

Composition of Mixtures and Solutions (Mass Percent, ppm/ppb, Mole Fraction, Molarity)

Many real substances are mixtures—physical combinations of substances. Unlike pure substances, mixture composition can vary, so chemistry uses several ways to describe mixture composition quantitatively.

Pure substances vs. mixtures

A pure substance has fixed composition (every sample of H2O is H2O). A mixture can vary (saltwater can be dilute or concentrated). Mixtures can be:

• Homogeneous (uniform throughout, like air)
• Heterogeneous (nonuniform, like sand in water)

Mass percent in mixtures

Mass percent is:

\% \text{by mass} = \frac{\text{mass of component}}{\text{total mass of mixture}} \times 100\%

Parts per million (ppm) and parts per billion (ppb)

For very dilute mixtures:

\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6

\text{ppb} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^9

A useful interpretation: 1 ppm is often treated as 1 mg solute per 1 kg solution (because 10^6 mg = 1 kg), but units must match the problem’s definition.

A key conversion is:

1\% = 10{,}000 \text{ ppm}

Mole fraction

Mole fraction focuses on particle counts:

\chi_A = \frac{n_A}{n_{\text{total}}}

Mole fractions sum to 1.

Molarity

Molarity (M) expresses concentration using solution volume and is used heavily later (equilibrium, acids-bases, electrochemistry, etc.). Brackets indicate molarity, such as [Na+] meaning “molar concentration of sodium ions.”

M = \frac{\text{moles solute}}{\text{liters solution}}

Worked example: mass percent

Problem: A mixture is made from 15.0 g of NaCl and 135.0 g of water. What is the mass percent NaCl?

m_{\text{total}} = 15.0 + 135.0 = 150.0

\%\text{NaCl} = \frac{15.0}{150.0} \times 100\% = 10.0\%

Worked example: ppm

Problem: Drinking water contains 2.5 mg of fluoride ion in 1.0 kg of water. Find fluoride concentration in ppm.

For a dilute solution, treat 1.0 kg water as approximately 1.0 kg solution:

\text{ppm} = \frac{2.5 \text{ mg}}{1.0 \text{ kg}} = 2.5 \text{ ppm}

What goes wrong: common mixture errors

A frequent mistake is using mass of solvent instead of mass of solution when the question specifically defines concentration using the solution mass. In very dilute cases the difference is small, but on exams you should follow the stated definition.

Exam Focus

• Typical question patterns
  • Calculate mass percent, ppm, or ppb from given masses.
  • Convert between mass percent and ppm using 1\% = 10{,}000 ppm.
  • Compute mole fraction from given moles or from masses after converting to moles.
  • Interpret bracket notation like [Na+] as molarity and use M = \frac{n}{V}.
• Common mistakes
  • Forgetting mole fraction is based on moles, not grams.
  • Using the wrong total mass (solution vs. solvent) when the problem specifies one.
  • Treating ppm as “per mL” automatically instead of as a mass ratio with consistent units.

Empirical and Molecular Formulas

When you are given masses or percent composition, you should be able to determine the empirical formula and, with molar mass, the molecular formula.

Empirical vs. molecular formula

• The empirical formula is the simplest whole-number ratio of elements (example: CH2O).
• The molecular formula is the actual number of atoms of each element in the molecule (example: C6H12O6).

Worked example: empirical formula from percent composition

Problem: A compound is found to contain 56.5% carbon, 7.11% hydrogen, and 36.4% phosphorus. Determine the empirical formula.

Assume 100 g of compound:

• C: 56.5 g
• H: 7.11 g
• P: 36.4 g

Convert to moles:

n_C = \frac{56.5}{12.01} \approx 4.70

n_H = \frac{7.11}{1.008} \approx 7.05

n_P = \frac{36.4}{30.97} \approx 1.175

Divide by the smallest (1.175):

• C: about 4.00
• H: about 6.00
• P: 1.00

Empirical formula: C4H6P

Worked example: molecular formula from molar mass

Problem: If the molar mass is 170.14 g/mol, what is the molecular formula?

First find the empirical-formula molar mass:

M_{\text{emp}} = 4(12.01) + 6(1.008) + 1(30.97) \approx 85.07

Now find the multiple:

\text{multiple} = \frac{170.14}{85.07} \approx 2

Multiply subscripts by 2:

Molecular formula: C8H12P2

Exam Focus

• Typical question patterns
  • Convert percent composition to empirical formula by assuming 100 g and converting to moles.
  • Determine molecular formula using the empirical formula mass and given molar mass.
• Common mistakes
  • Forgetting to divide all mole amounts by the smallest.
  • Rounding too early and missing a simple whole-number ratio.
  • Forgetting the molecular formula is a whole-number multiple of the empirical formula.

Electron Configuration and Atomic Orbitals

Chemistry happens because of electrons: where they are, how much energy they have, and how they are arranged. Electron configuration systematically describes how electrons occupy orbitals.

Quantized energy, attraction, and potential energy

The positively charged nucleus attracts negatively charged electrons. As an electron is farther from the nucleus, its potential energy is higher. Electron energies are quantized, meaning electrons can only exist at specific energy levels separated by specific energy differences.

The Bohr model (conceptual usefulness)

The Bohr model helps you think about energy levels even though it is not structurally accurate. Key ideas:

• Each principal energy level is associated with a period (row) of the periodic table.
• Electrons can move between energy levels by absorbing or emitting electromagnetic radiation.
  • Absorb radiation → jump to a higher energy level.
  • Drop to a lower energy level → emit radiation.

Orbitals and subshells

An orbital is a region of space with high probability of finding an electron at a particular energy. Sublevels (subshells) are labeled s, p, d, f and have characteristic capacities:

• s holds 2
• p holds 6
• d holds 10
• f holds 14

The periodic table’s blocks reflect which subshell is being filled:

Configuration rules

Electron configurations follow three key principles:

1. Aufbau principle: electrons fill lower-energy orbitals before higher-energy ones.
2. Pauli exclusion principle: each orbital holds at most 2 electrons with opposite spins.
3. Hund’s rule: in equal-energy orbitals, electrons spread out singly before pairing.

Writing electron configurations

Two common formats are used:

• Full configuration
• Noble-gas shorthand

Example (Na, 11):

• Full: 1s2 2s2 2p6 3s1
• Shorthand: [Ne] 3s1

A practical filling order (as you “walk” the periodic table) is:

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

AP Chemistry expects you to recognize that 4s fills before 3d in this scheme.

Worked example: electron configuration

Problem: Write the shorthand electron configuration for bromine (Br, 35).

Starting from Ar: fill 4s2 (20), then 3d10 (30), then 4p5 (35).

Answer: [Ar] 4s2 3d10 4p5

Cations and electron configurations (especially transition metals)

When transition metals form cations, electrons are removed from the highest principal energy level first (often the outer s electrons), even if a d sublevel was filled later.

Example: Fe is often written [Ar] 4s2 3d6. For \text{Fe}^{2+} remove the 4s electrons first to get [Ar] 3d6.

Exam Focus

• Typical question patterns
  • Write full or shorthand electron configurations for neutral atoms.
  • Use periodic table blocks to determine the outer electron configuration.
  • Apply Aufbau, Pauli, and Hund’s rule in orbital diagrams.
  • Write electron configurations for cations, especially transition-metal ions.
• Common mistakes
  • Misplacing electrons between 4s and 3d.
  • Violating Hund’s rule by pairing electrons too early in p orbitals.
  • Removing electrons from the “last written” subshell instead of from the highest principal energy level.

Photoelectron Spectroscopy (PES)

Electron configuration is a model; photoelectron spectroscopy provides experimental evidence by measuring how tightly electrons are held.

What PES measures

In PES, high-energy light ejects electrons. Electrons in different subshells are held with different strengths.

• Peak position corresponds to binding energy (how strongly electrons are held).
• Peak area/height corresponds to the number of electrons in that subshell.

Electrons closer to the nucleus (like 1s) typically have higher binding energies than outer electrons (like 3s or 3p).

Energy accounting in PES

To eject an electron, the incoming radiation must overcome the electron’s binding energy (often discussed alongside ionization energy ideas). If the incoming radiation exceeds the binding energy, the leftover energy becomes kinetic energy of the ejected electron:

E_{\text{photon}} = BE + KE

More kinetic energy means a faster ejected electron. Outer electrons generally require less energy to remove.

Energy units:

1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}

For single atoms or small samples, energies are commonly expressed in eV; for moles of atoms, energies are often expressed in kJ/mol or MJ/mol.

Interpreting PES spectra

A PES spectrum plots relative number of electrons (vertical) versus binding energy (horizontal). Binding energy axes are not always oriented the same way; a common convention is that binding energy decreases left to right, but you must always read the axis.

Spectral features connect to structure:

• Each major “cluster” of peaks corresponds to a principal energy level.
• Multiple peaks within a level reflect different subshells (s vs p vs d, etc.).
• Peak heights indicate electron counts (p peaks can be about three times s peaks because p holds 6 while s holds 2).
• A shorter-than-expected peak can indicate a partially filled subshell.

Trends in binding energy: nuclear charge and shielding

Binding energy depends strongly on:

• Nuclear charge (more protons generally increases binding energy)
• Shielding (inner electrons reduce the attraction felt by outer electrons)

Across a period (left to right), binding energies for valence electrons generally increase because nuclear charge increases while shielding changes only modestly.

What goes wrong: common PES interpretation errors

Students may confuse peak height with binding energy. Height is abundance; horizontal position is energy. Also, the leftmost peak is not automatically the highest binding energy unless the axis shows that.

Exam Focus

• Typical question patterns
  • Match PES peaks to subshells (1s, 2s, 2p, etc.) and infer electron configuration.
  • Use relative peak heights/areas to determine electron counts in subshells.
  • Compare two spectra to infer which element has higher nuclear charge or is further right in a period.
  • Use E_{\text{photon}} = BE + KE conceptually to connect binding energy and electron kinetic energy.
• Common mistakes
  • Ignoring axis direction (binding energy may increase to the left or right depending on the graph).
  • Treating peak height as energy rather than electron count.
  • Assuming shielding changes drastically across a period for valence electrons (it changes only modestly across a period).

Coulomb’s Law and Effective Nuclear Charge

Many periodic patterns become logical when you view the atom as charged particles interacting through electrostatic forces.

Coulomb’s law: the core relationship

Electrostatic force between charges is:

F = k\frac{q_1 q_2}{r^2}

Qualitatively:

• More charge (more protons in the nucleus) increases attraction.
• Greater distance decreases attraction.

In atomic reasoning, electrons closer to the nucleus tend to have stronger attraction and lower potential energy. To remove an electron, you must supply enough energy to overcome this electrostatic attraction (a binding-energy idea).

Shielding and effective nuclear charge

Inner electrons repel outer electrons and reduce the nuclear attraction felt by valence electrons. The simplified AP Chemistry idea is effective nuclear charge:

• Z_{\text{eff}} is the net positive charge felt by an electron after shielding.
• Across a period, Z_{\text{eff}} increases because proton count increases while core shielding stays roughly constant.
• Down a group, shielding increases significantly because additional inner shells are added.

Worked reasoning example: comparing attraction

Question: Which valence electron experiences stronger attraction: the valence electron in Na or in Cl (same period)?

Both have valence electrons in the third energy level, so distance is comparable. Chlorine has more protons, and both have similar core shielding ([Ne]). Therefore Z_{\text{eff}} is higher for Cl, so its valence electrons are more strongly attracted.

Exam Focus

• Typical question patterns
  • Use Coulomb’s law qualitatively to explain periodic trends.
  • Explain why Z_{\text{eff}} increases across a period and how that affects properties.
  • Compare attraction in two atoms or ions using charge and distance.
• Common mistakes
  • Claiming shielding increases significantly across a period for valence electrons (core shielding is similar).
  • Treating Z_{\text{eff}} as identical to nuclear charge.
  • Forgetting the distance factor when comparing electrons in different principal energy levels.

Periodic Trends and Element Categories

Periodic trends are patterns in properties across periods and down groups. On the AP exam you are often asked to rank, predict, or explain these trends using nuclear charge, shielding, and distance.

Big ideas that drive trends

Three core reasoning statements show up repeatedly:

• Electrons are attracted to protons in the nucleus.
• Attraction increases with more protons and smaller electron-nucleus distance.
• Electron shielding: inner electrons repel outer electrons, decreasing nuclear attraction felt by valence electrons.

A related stability idea is that completed shells are relatively stable, so many atoms tend to gain/lose electrons (when possible) to reach a noble-gas-like valence configuration.

Metals, nonmetals, and metalloids

Elements are often categorized as:

• Metals (left side): tend to give up electrons in bonding.
• Nonmetals (right side): tend to gain electrons in bonding.
• Metalloids: in between, with mixed metallic and nonmetallic character; they form the “stair-step” boundary region.

Metallic character generally decreases moving to the right. For example, lithium behaves more metallic than zinc.

Atomic radius

Atomic radius is a measure of atom size.

• Across a period (left to right): generally decreases because Z_{\text{eff}} increases and pulls electrons closer in the same energy level.
• Down a group: generally increases because electrons occupy higher energy levels (farther from the nucleus) and shielding increases.

Ionic radius

Ion size changes when atoms form ions:

• Cations are smaller than their neutral atoms (electrons removed, less repulsion, sometimes loss of an entire shell).
• Anions are larger than their neutral atoms (more electron-electron repulsion in the same shell).

Isoelectronic series (same number of electrons): more protons means smaller radius.

Example: O2−, F−, Ne, Na+, Mg2+ all have 10 electrons; Mg2+ is smallest and O2− is largest.

Ionization energy

First ionization energy is the energy required to remove the outermost electron from a gaseous atom:

\text{M(g)} \to \text{M}^+\text{(g)} + e^-

Trends:

• Across a period: generally increases as Z_{\text{eff}} increases.
• Down a group: generally decreases due to larger distance and more shielding.

Ionization energies are sequential: second ionization energy is the energy to remove a second electron, and so on.

• The second ionization energy is greater than the first because after one electron is removed, the remaining electrons experience a higher effective attraction (a stronger proton-to-electron ratio).
• Once an electron shell has been emptied, ionization energy increases dramatically because the next electron removed is much closer to the nucleus.

Exception reasoning (subshell structure): small “dips” occur due to s vs p subshell energies and electron pairing.

Electronegativity

Electronegativity is how strongly an atom attracts shared bonding electrons. It depends strongly on:

• atomic size (smaller tends to be more electronegative)
• closeness to a complete valence shell

Trends:

• Across a period: generally increases.
• Down a group: generally decreases.

Fluorine is the most electronegative element. Noble gases are often not assigned electronegativities on many scales because they do not commonly form bonds.

Electron affinity

Electron affinity is the energy change when an electron is added to a gaseous atom:

\text{X(g)} + e^- \to \text{X}^-\text{(g)}

Across a period, electron affinity generally becomes more favorable (more energy released) as atoms approach a stable valence configuration, but there are notable exceptions (especially for filled/half-filled subshells). AP emphasizes qualitative comparisons more than memorizing every exception.

Worked comparison: ordering atomic radius

Problem: Order Na, Mg, Al from largest to smallest atomic radius.

All are in Period 3. Radius decreases left to right as Z_{\text{eff}} increases.

Largest to smallest: Na > Mg > Al.

Worked comparison: ordering ionization energy (a classic exception)

Problem: Which has higher first ionization energy: Mg or Al?

Although the general trend across a period increases, Al’s first electron removed is a 3p electron, which is higher in energy (easier to remove) than Mg’s 3s electron. Therefore Mg has the higher first ionization energy.

Trend summary table (for reasoning, not memorization)

Exam Focus

• Typical question patterns
  • Rank elements by atomic radius, ionic radius, ionization energy, or electronegativity.
  • Explain trends using Z_{\text{eff}}, shielding, and distance.
  • Identify and justify trend exceptions (like Mg vs Al) using subshell reasoning.
  • Compare ion sizes in an isoelectronic series.
• Common mistakes
  • Using “more protons” as the only explanation without considering shielding and distance.
  • Ranking ionic sizes without checking charge type (cation vs anion) or isoelectronic relationships.
  • Assuming second ionization energy is similar to first (it is always larger, often much larger after a stable shell is reached).
  • Confusing electron affinity with electronegativity.

Valence Electrons, Ions, and Ionic Compounds

Unit 1 often culminates in using electron configuration and trends to explain ion formation and ionic compound formulas.

Valence electrons: the chemical “interface”

Valence electrons are the electrons in the outermost occupied energy level. They control bonding, reactivity, and common ion formation. For main-group elements (s- and p-block), group number strongly predicts valence electron count:

• Group 1: 1 valence electron
• Group 2: 2 valence electrons
• Group 17: 7 valence electrons
• Group 18: 8 valence electrons (He has 2)

For many main-group elements, a stable configuration corresponds to having 8 valence electrons (2 in s and 6 in p).

Ions: atoms with a charge

An ion forms when an atom gains or loses electrons:

• Losing electrons → cation (positive)
• Gaining electrons → anion (negative)

Ion formation does not change the nucleus, so it does not change the element’s identity.

Predicting common ion charges

Main-group patterns:

• Group 1 → +1
• Group 2 → +2
• Group 13 → +3
• Group 15 → −3
• Group 16 → −2
• Group 17 → −1

Halogens commonly form anions like Cl− (sometimes written as a −1 charge), and oxygen-group elements commonly form −2 ions (like O2−).

Transition metals are cations but often have multiple possible charges depending on the compound. Two common fixed-charge exceptions:

• zinc typically forms +2
• silver typically forms +1

Determining numbers of subatomic particles (including ions)

• Protons = atomic number
• Neutrons = mass number − atomic number
• Electrons (neutral) = atomic number
• Electrons (ion) = atomic number − charge (with sign)

Examples:

For \text{Fe}^{3+}:

• protons = 26
• electrons = 26 − 3 = 23

For \text{S}^{2-}:

• protons = 16
• electrons = 16 + 2 = 18

Building ionic compound formulas (charge balance)

Ionic compounds are electrically neutral: total positive charge equals total negative charge.

A reliable method:

1) Write ion charges.
2) Choose subscripts so charge sums to zero.

Example: aluminum and oxygen

• Al3+
• O2−

Least common multiple of 3 and 2 is 6, so:

• 2 Al3+ gives +6
• 3 O2− gives −6

Formula: Al2O3.

Worked example: ionic formula

Problem: Write the formula for magnesium chloride.

• Mg2+
• Cl−

Need two chloride ions to balance +2:

Formula: MgCl2.

Worked example: determining transition-metal charge from a formula

Problem: Determine copper’s charge in CuBr2 and name the compound.

Each bromide is −1; two bromides total −2, so copper must be +2 to make a neutral compound. Name: copper(II) bromide.

If the formula were CuBr, copper would be +1, and the name would be copper(I) bromide.

Connecting to periodic trends: why ion size matters

Forming a cation shrinks the atom; forming an anion expands it. In isoelectronic ions, more protons means a smaller radius. These size ideas later connect to lattice structure and melting points.

Exam Focus

• Typical question patterns
  • Determine valence electrons from group number or electron configuration.
  • Predict likely ion charges for main-group elements.
  • Find protons/neutrons/electrons from isotopic notation and ionic charge.
  • Write correct ionic formulas and determine unknown transition-metal charges from formulas (then name with Roman numerals).
• Common mistakes
  • Confusing mass number with atomic number.
  • Forgetting charge depends on electrons only (protons stay constant for an element).
  • Writing charges as subscripts without balancing total neutrality.
  • Assuming all transition metals have a single fixed charge.