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Unit 1: Atomic Structure and Properties · Free-response practice

AP Chemistry Unit 1 FRQ Practice

One long and two short free-response questions on Unit 1, each with a point-by-point scoring guide so you can grade yourself.

These are original questions written in the style of the exam, not released College Board questions. Answer each one on paper with a calculator and periodic table before opening its scoring guide, then award yourself only the points your written answer actually earns. The long question is worth 10 points and each short question 4, the same values as on the exam. If you have not yet tried the Unit 1 practice test, its 15 multiple-choice questions are a quicker check of the same content.

Long Question: Chlorine From Isotopes to Bonding (10 Points)

Naturally occurring chlorine is a mixture of two isotopes. Chlorine-35 makes up 75.8% of chlorine atoms and chlorine-37 makes up 24.2%.

Part A1 point

Using whole-number isotope masses, calculate the average atomic mass of chlorine. Show your work.

Part B1 point

Identify the difference in atomic structure between a chlorine-35 atom and a chlorine-37 atom.

Part C2 points

(i) Write the ground-state electron configuration of a chlorine atom. (ii) Write the electron configuration of the chloride ion, Cl−.

Part D2 points

(i) Describe the photoelectron spectrum of chlorine: how many peaks it has and which peak is tallest. (ii) Explain why the peak for the 1s electrons has the highest binding energy.

Part E2 points

The first ionization energy of chlorine is greater than that of sulfur. Explain this difference in terms of atomic structure and Coulombic attraction.

Part F1 point

Calculate the number of chlorine atoms in a 5.00 g sample of chlorine gas, Cl2.

Part G1 point

Magnesium reacts with chlorine to form an ionic compound. Using the number of valence electrons in each atom, write its formula.

Scoring guide for the long question
  • Part A (1 point): 35(0.758) + 37(0.242) = 26.53 + 8.95 = 35.5. The answer must be closer to 35 because chlorine-35 is more abundant. Practice more in average atomic mass.
  • Part B (1 point): A chlorine-37 atom has 2 more neutrons (20 vs 18). Both have 17 protons, and as neutral atoms both have 17 electrons.
  • Part C(i) (1 point): 1s2 2s2 2p6 3s2 3p5, or [Ne] 3s2 3p5.
  • Part C(ii) (1 point): 1s2 2s2 2p6 3s2 3p6, the same as argon. Check configurations like these with the electron configuration calculator.
  • Part D(i) (1 point): Five peaks, one for each occupied subshell (1s, 2s, 2p, 3s, 3p). The 2p peak is tallest because it holds 6 electrons, more than any other subshell in chlorine.
  • Part D(ii) (1 point): 1s electrons are closest to the nucleus and experience the least shielding, so the Coulombic attraction on them is greatest and the most energy is needed to remove them. See photoelectron spectroscopy.
  • Part E (2 points): 1 point for structure: chlorine has one more proton than sulfur (17 vs 16) while its valence electrons are in the same shell (n = 3) with the same core shielding. 1 point for the link to attraction: the greater nuclear charge pulls the 3p electrons more strongly, so more energy is needed to remove one. Review ionization energy.
  • Part F (1 point): 5.00 g ÷ 70.90 g/mol = 0.0705 mol Cl2; × 2 = 0.141 mol Cl atoms; × 6.022 × 1023 = 8.49 × 1022 Cl atoms.
  • Part G (1 point): MgCl2. Magnesium has 2 valence electrons and forms Mg2+; chlorine has 7 and gains 1 to form Cl−, so two chloride ions balance one magnesium ion.

The most common lost point is Part E: “chlorine is smaller” restates the trend. The point requires the reason, more protons in the same shell, and its effect on the attraction.

Short Question 1: Hydrogen Peroxide Formula (4 Points)

A pure compound of hydrogen and oxygen is 5.93% hydrogen and 94.07% oxygen by mass. Its molar mass is 34.02 g/mol.

Part A1 point

Determine the empirical formula of the compound. Show your work.

Part B1 point

Determine the molecular formula of the compound.

Part C1 point

Calculate the number of moles of the compound in a 1.70 g sample.

Part D1 point

A second sample of the same compound, prepared by a different method, has a mass of 10.0 g. Predict the mass of hydrogen it contains and justify your answer.

Scoring guide for short question 1
  • Part A (1 point): In 100 g: 5.93 ÷ 1.008 = 5.88 mol H and 94.07 ÷ 16.00 = 5.88 mol O. The ratio is 1 : 1, so the empirical formula is HO. The empirical formula calculator shows the same steps.
  • Part B (1 point): The empirical formula mass is 17.01 g/mol, and 34.02 ÷ 17.01 = 2, so the molecular formula is H2O2.
  • Part C (1 point): 1.70 g ÷ 34.02 g/mol = 0.0500 mol H2O2.
  • Part D (1 point): 0.593 g of hydrogen (5.93% of 10.0 g). A pure compound has a fixed composition by mass whatever its source, the law of definite proportions, so the percent hydrogen is the same.

Short Question 2: Sodium and Magnesium Trends (4 Points)

Sodium and magnesium are next to each other in period 3. The first three ionization energies of one of them, element X, are 738, 1451, and 7733 kJ/mol.

Part A1 point

Which atom, Na or Mg, has the larger atomic radius? Justify your answer in terms of atomic structure.

Part B2 points

(i) Identify element X. (ii) Justify your answer using the ionization energy data.

Part C1 point

Is the radius of the Na+ ion greater than, less than, or equal to the radius of a Na atom? Justify your answer.

Scoring guide for short question 2
  • Part A (1 point): Na. Both atoms have valence electrons in the third shell, but Na has fewer protons (11 vs 12), so its valence electron is attracted less strongly and lies farther from the nucleus. See periodic trends.
  • Part B(i) (1 point): Magnesium.
  • Part B(ii) (1 point): The large jump comes between the second and third ionization energies (1451 to 7733 kJ/mol), so the third electron is removed from a filled inner shell. Element X has 2 valence electrons, which matches magnesium, not sodium (1 valence electron).
  • Part C (1 point): Less than. Na+ has lost its only n = 3 electron, so its outermost electrons are in the second shell, closer to the nucleus; the same 11 protons also attract 10 electrons instead of 11.

Answering “Mg, because it is in group 2” for Part B(ii) earns nothing: the justification must point to where the large jump occurs in the data.

How to Grade Your Unit 1 FRQ Answers

Grade strictly: a correct number without the work asked for, or a trend restated without a reason, earns nothing on the real exam. Add up your points out of 18, then look at where they went missing rather than at the total. Missed points on the configuration and PES parts point back to electron configurations; missed points on Parts E and short question 2 point to Coulomb’s law reasoning in the Unit 1 review. To see how free-response points combine with multiple choice, enter a full-length estimate in the FRQ Calculator or the AP Chem Score Calculator.

Frequently Asked Questions

How are Unit 1 topics tested in AP Chemistry free response?

Usually as parts of a larger question rather than a whole question: an average atomic mass from a mass spectrum, an electron configuration, a photoelectron spectrum, or a periodic trend explained with Coulomb's law. These practice questions group those parts the way the exam does.

How do you earn the point for explaining a periodic trend?

Name the structural difference (more protons in the same shell, or valence electrons in a shell farther from the nucleus) and connect it to the strength of the Coulombic attraction on the electron in question. Saying only that an atom is smaller, or that the trend goes up across a period, does not earn the point.

How long should these practice FRQs take?

About 23 minutes for the 10-point long question and about 9 minutes for each 4-point short question, the same pace as the real exam, which gives 105 minutes for seven questions.

These are original practice questions written for this site, not released College Board questions. AP® is a trademark registered by the College Board, which is not affiliated with and does not endorse this site.