The problems are on the photoelectron spectroscopy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electrons fill subshells in order of energy, giving each atom a configuration. These worked answers correct the 20 problems of the AP Chemistry worksheet on photoelectron spectroscopy, from configuration of sulfur to paragraph argument: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the result in bold with a short reason. For explanation problems, check that you named the charge, the distance and the shielding, because those three ideas earn the marks.
Level: AP Chemistry · Curriculum: AP Chem Unit 1
Problem 1 Configuration of sulfur
1. 1s² 2s² 2p⁶ 3s² 3p⁴.2. 6 valence electrons (3s² 3p⁴).
3. [Ne] 3s² 3p⁴.
Problem 2 Configuration of a cation
1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² = [Ar] 4s².2. The two 4s electrons leave: 1s² 2s² 2p⁶ 3s² 3p⁶.
3. Argon.
Problem 3 Chromium and copper
1. [Ar] 4s¹ 3d¹⁰.2. A completely full 3d subshell lowers the energy, so one 4s electron moves to 3d to complete it.
Problem 4 Unpaired electrons in nitrogen
1. 2p: ↑ ↑ ↑ (three orbitals, one electron each, all with the same spin).2. 3 unpaired electrons.
3. Hund’s rule: orbitals of equal energy are filled singly before pairing.
Problem 5 Identify the element from PES
1. From the left: 1s, 2s, 2p, 3s.2. Heights 2 : 2 : 6 : 1 give 1s² 2s² 2p⁶ 3s¹, so the element is sodium.
3. The 3s peak at 5.1 eV, the lowest binding energy.
Problem 6 Coulomb’s law quick checks
1. F ∝ 1/r², so the force is one quarter.2. F ∝ q, so the force is doubled.
3. Factor 2 × 4 = 8 times bigger.
Problem 7 Which atom is bigger?
1. Sodium has the larger radius.2. Both have three shells, but Zeff is about 1 for Na and about 7 for Cl. The larger Zeff pulls the electrons of Cl closer.
Problem 8 Effective nuclear charge
1. Mg: 12 − 10 = 2. Si: 14 − 10 = 4. Cl: 17 − 10 = 7.2. Chlorine has the smallest radius (largest Zeff).
3. Chlorine has the greatest ionisation energy of these.
Problem 9 Successive ionisation energies
1. The jump comes after the third electron (2745 → 11577).2. 3 valence electrons.
3. Aluminium.
Problem 10 Comparing two spectra
1. A is aluminium (13 electrons); B is phosphorus (15 electrons).2. B has 15 protons against 13, so the core 1s electrons feel a stronger Coulomb attraction.
3. B’s 3p peak is three electrons high against one, so B has three valence p electrons.
Problem 11 Isoelectronic ions
1. Each has 10 electrons.2. F⁻, then Na⁺, then Mg²⁺ (largest first).
3. They have identical electron structure but nuclear charge 9, 11 and 12. The larger nuclear charge pulls the same electron cloud in more.
Problem 12 Ionisation energy exceptions
1. The Al valence electron is in 3p, which is higher in energy and shielded by the 3s electrons, so it is easier to remove.2. In O the fourth 2p electron pairs with another in the same orbital. The repulsion between the paired electrons makes it easier to remove.
Problem 13 Transition-metal ions
1. Fe³⁺: [Ar] 3d⁵, 5 unpaired.2. Zn²⁺: [Ar] 3d¹⁰, 0 unpaired.
3. Cu⁺: [Ar] 3d¹⁰, 0 unpaired.
Problem 14 Kinetic energy of photoelectrons
1. KE = photon energy − binding energy.2. KE(2s) = 1500 − 118 = 1382 eV.
3. KE(2p) = 1500 − 73 = 1427 eV.
Problem 15 Ionisation energy across period 3
1. IE₁ generally increases from Na to Ar as Zeff grows and the radius shrinks.2. The electron in Al is in 3p, which sits higher than 3s and is partly shielded by the 3s electrons, so IE₁ is below Mg.
3. In S two electrons share a 3p orbital, and their repulsion makes removal easier, so IE₁ is below P.
Problem 16 Force between ions
1. F = k e²/r² = 8.99 × 10⁹ × (1.602 × 10⁻¹⁹)² / (2.76 × 10⁻¹⁰)² = 3.03 × 10⁻⁹ N.2. F = 8.99 × 10⁹ × 4 × (1.602 × 10⁻¹⁹)² / (2.10 × 10⁻¹⁰)² = 2.09 × 10⁻⁸ N.
3. Ratio = 2.09 × 10⁻⁸ / 3.03 × 10⁻⁹ = 6.9. Stronger ionic attraction gives higher melting points for MgO than for NaCl.
Problem 17 Ranking ionisation energies
1. Be, Mg, Li, Na, K (highest first).2. Down a group the valence electron is in a higher shell, farther from the nucleus and shielded, so IE falls.
3. Beryllium and magnesium have a larger nuclear charge than their group-1 neighbours in the same period, with the electron in the same shell, so the pull is stronger.
Problem 18 Interpreting a sulfur spectrum
1. The heights 2 : 2 : 6 : 2 : 4 give 1s² 2s² 2p⁶ 3s² 3p⁴, which is sulfur.2. The 1s peak for chlorine would appear at a higher binding energy, about 2820 eV.
3. Chlorine has one more proton, so the 1s electrons are held more strongly by the nucleus.
Problem 19 Why two electrons then a jump?
1. The first electron comes from the 3s valence shell. The second for sodium comes from the 2p shell, closer to the nucleus with less shielding, so it needs far more energy.2. Magnesium loses two 3s electrons, both in the same shell, so IE₂ is only moderately larger than IE₁ as the ion is positive.
3. IE₃ takes an electron from the 2p shell, so it is larger than 4562 kJ/mol (the real value is about 7730 kJ/mol).
Problem 20 Paragraph argument
Across a period (Na to Cl) electrons are added to the same shell and shield each other only weakly, while the nuclear charge increases from +11 to +17. By Coulomb’s law the attraction between the nucleus and the valence electrons rises, so the radius decreases. Down a group (Li to Cs) each new element adds a shell, so the valence electron is farther away and shielded by more core electrons. The attraction, which falls with 1/r², is weaker and the radius increases. Radius is set by Zeff across a period and by shell number down a group.
