Doc Practice Questions (MCQs & FRQs) Atomic Structure and Properties -

SECTION I: MULTIPLE CHOICE

Directions

Answer the following 20 multiple-choice questions. Each question has four answer choices (A, B, C, D). Select the best answer for each question. Some questions may instruct you to select TWO correct answers; these will be clearly marked. You may use a periodic table and a calculator where appropriate.

Questions 1-3 refer to the following photoelectron spectrum (PES) for an unknown element.

1. Based on the photoelectron spectrum above, which element is most likely represented?

1. Lithium (Li)
2. Boron (B)
3. Carbon (C)
4. Nitrogen (N)

2. Which of the following best explains why Peak 3 appears at the highest binding energy?

1. These electrons are in orbitals with the highest principal quantum number.
2. These electrons experience the greatest effective nuclear charge and are closest to the nucleus.
3. These electrons are in the outermost shell and are easiest to remove.
4. These electrons are in orbitals with the highest energy and therefore require the most energy to remove.

3. The relative height of Peak 1 compared to Peak 2 indicates that

1. Peak 1 represents electrons in a higher energy level than Peak 2
2. Peak 1 represents half as many electrons as Peak 2
3. Peak 1 electrons are more tightly bound than Peak 2 electrons
4. Peak 1 represents electrons in p orbitals while Peak 2 represents s orbital electrons

Table 1: First Ionization Energies

4. The data in Table 1 show that aluminum has a lower first ionization energy than magnesium, despite aluminum having a greater nuclear charge. Which of the following best explains this observation?

1. Aluminum has more electrons than magnesium, which increases electron-electron repulsion and makes electrons easier to remove.
2. The outermost electron in aluminum is in a 3p orbital, which is higher in energy and more shielded from the nucleus than the 3s orbital in magnesium.
3. Aluminum has a smaller atomic radius than magnesium, making its electrons harder to remove.
4. The additional proton in aluminum causes greater repulsion between the nucleus and the outermost electrons.

5. The first ionization energy of sulfur is slightly lower than that of phosphorus. Which of the following best accounts for this exception to the general trend?

1. Sulfur has a larger atomic radius than phosphorus.
2. One of the 3p electrons in sulfur is paired, and electron-electron repulsion in that orbital makes it easier to remove an electron.
3. Sulfur has more protons than phosphorus, which increases shielding.
4. The nuclear charge of sulfur is insufficient to hold the outermost electrons tightly.

6. Which of the following electron configurations represents an atom in an excited state?

1. 1s² 2s² 2p⁶
2. 1s² 2s² 2p⁶ 3s¹
3. 1s² 2s² 2p⁵ 3s¹
4. 1s² 2s² 2p⁶ 3s² 3p¹

7. Select TWO answers. Which of the following statements correctly describe the relationship between Coulomb's law and atomic structure?

1. According to Coulomb's law, the force of attraction between the nucleus and an electron increases as the distance between them decreases.
2. Coulomb's law explains why electrons in the same shell experience identical attractive forces from the nucleus.
3. The energy required to remove an electron from an atom is inversely proportional to the distance between the electron and the nucleus, as predicted by Coulomb's law.
4. Coulomb's law predicts that ions with greater charge will have stronger interactions with surrounding particles.

8. What is the approximate energy of the photon emitted during this transition?

1. 4.1 × 10^-19 J
2. 6.1 × 10^-19 J
3. 2.7 × 10^-19 J
4. 1.6 × 10^-19 J

9. Which region of the electromagnetic spectrum does the photon from Question 8 belong to?

1. Ultraviolet
2. Visible
3. Infrared
4. X-ray

10. Consider the isoelectronic species: O²⁻, F⁻, Na⁺, and Mg²⁺. Which of the following correctly ranks these ions in order of increasing ionic radius?

1. Mg²⁺ < Na⁺ < F⁻ < O²⁻
2. O²⁻ < F⁻ < Na⁺ < Mg²⁺
3. Na⁺ < Mg²⁺ < F⁻ < O²⁻
4. F⁻ < O²⁻ < Na⁺ < Mg²⁺

11. Which of the following best explains the trend in ionic radii for the isoelectronic series in Question 10?

1. As nuclear charge increases, the electrons are pulled closer to the nucleus, decreasing the ionic radius.
2. As the number of electrons increases, electron-electron repulsion increases, increasing the ionic radius.
3. As nuclear charge increases, shielding increases, which decreases the ionic radius.
4. Ions with more protons have more energy levels, which increases the ionic radius.

Table 2: Selected Atomic Properties

12. Based on the data in Table 2, which of the following statements best describes the trend in atomic properties down Group 1?

1. Atomic radius decreases because effective nuclear charge increases.
2. First ionization energy decreases because the outermost electron is farther from the nucleus and more shielded.
3. Electronegativity increases because the atoms become more capable of attracting electrons.
4. First ionization energy increases because nuclear charge increases.

13. An atom of element X has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵. Which of the following is most likely true about element X?

1. Element X is a noble gas with very low reactivity.
2. Element X is a halogen that readily forms a -1 anion.
3. Element X is a transition metal with variable oxidation states.
4. Element X is an alkali metal that readily forms a +1 cation.

14. Which of the following species has the greatest number of unpaired electrons?

1. N (atomic number 7)
2. O (atomic number 8)
3. Fe²⁺ (atomic number 26)
4. Cr (atomic number 24)

15. Select TWO answers. The emission spectrum of hydrogen contains discrete lines rather than a continuous spectrum. Which of the following statements correctly explain this observation?

1. Electrons in hydrogen atoms occupy specific, quantized energy levels.
2. When an electron transitions between energy levels, it emits a photon with energy equal to the difference between the two levels.
3. All electrons in hydrogen atoms have the same energy regardless of their distance from the nucleus.
4. The energy of emitted photons increases continuously as electrons move closer to the nucleus.

16. Based on the data above, which transition releases the most energy?

1. n = 3 to n = 2
2. n = 4 to n = 2
3. n = 5 to n = 2
4. All transitions release the same amount of energy

17. The student predicts that a transition from n = 6 to n = 2 will emit light with a wavelength shorter than 434 nm. Which of the following best justifies this prediction?

1. Transitions from higher energy levels release less energy and therefore produce photons with shorter wavelengths.
2. Transitions from higher energy levels release more energy, and higher energy photons have shorter wavelengths.
3. The energy difference between levels decreases as n increases, resulting in shorter wavelength photons.
4. Wavelength is independent of the energy levels involved in the transition.

18. Consider the following atoms: ¹²C, ¹³C, and ¹⁴C. Which of the following statements is correct?

1. These atoms have different numbers of protons.
2. These atoms have identical chemical properties but different mass numbers.
3. These atoms have different numbers of electrons.
4. These atoms will appear at different positions in the periodic table.

19. The mass spectrum of naturally occurring chlorine shows two peaks at mass-to-charge ratios of 35 and 37, with relative abundances of approximately 75% and 25%, respectively. What is the approximate average atomic mass of chlorine?

1. 35.0 amu
2. 35.5 amu
3. 36.0 amu
4. 36.5 amu

Data Table for Question 20

A research team measures the effective nuclear charge (Z_eff) experienced by the outermost electron in several elements:

20. Which of the following best explains why Z_eff increases down a group, even though shielding also increases?

1. The increase in nuclear charge outweighs the increase in shielding, resulting in a net increase in Z_eff.
2. Electrons in higher energy levels experience less shielding from core electrons.
3. The atomic radius decreases down a group, bringing the outermost electrons closer to the nucleus.
4. Z_eff does not actually increase down a group; the data table contains an error.

SECTION II: FREE RESPONSE

Directions

Answer both of the following questions. The first question is a long free-response question that should take approximately 23 minutes to complete. The second question is a short free-response question that should take approximately 9 minutes to complete. Show all your work for each part of each question. Answers without supporting work may not receive credit. You may use a periodic table and a calculator.

Question 1 (Long Free-Response Question)

Sodium (Na) and potassium (K) are both Group 1 elements. A student investigates the atomic properties of these elements using photoelectron spectroscopy (PES) and analyzes ionization energy data.

Photoelectron Spectroscopy Data:

Sodium (Na):

• Peak 1: 0.50 MJ/mol (relative height: 1)
• Peak 2: 3.67 MJ/mol (relative height: 2)
• Peak 3: 6.84 MJ/mol (relative height: 6)
• Peak 4: 104 MJ/mol (relative height: 2)

Potassium (K):

• Peak 1: 0.42 MJ/mol (relative height: 1)
• Peak 2: 2.38 MJ/mol (relative height: 2)
• Peak 3: 3.93 MJ/mol (relative height: 6)
• Peak 4: 18.2 MJ/mol (relative height: 2)
• Peak 5: 31.1 MJ/mol (relative height: 8)

Additional Data:

1. Write the complete electron configuration for a neutral sodium atom. (1 point)
2. Explain why the PES peak with the highest binding energy for sodium (104 MJ/mol) corresponds to only 2 electrons. Your explanation should reference both the electron configuration and the quantum mechanical model of the atom. (2 points)
3. Calculate the energy required to remove one mole of electrons from the 3s orbital of sodium atoms. Express your answer in kilojoules per mole. Show your work. (2 points)
4. The second ionization energy of sodium is nearly 10 times greater than its first ionization energy. Justify this observation using principles of atomic structure and Coulomb's law. (2 points)
5. Compare the first ionization energies of sodium and potassium. Identify which element has the lower first ionization energy and explain this difference in terms of atomic structure, electron shielding, and distance from the nucleus. (3 points)
6. A student proposes that the effective nuclear charge (Z_eff) experienced by the outermost electron in potassium is greater than that experienced by the outermost electron in sodium because potassium has more protons. Evaluate this claim and explain whether the data support or refute it. (2 points)

Question 2 (Short Free-Response Question)

1. Calculate the initial energy level (n_i) from which the electron transitioned. Show all work and round your answer to the nearest whole number. (3 points)
2. Draw a diagram showing the electron transition described above. Your diagram should include labeled energy levels for n = 1, n = 2, and the level you calculated in part (A), with an arrow indicating the direction of the transition. (2 points)
3. Predict whether a transition from n = 4 to n = 1 would emit light of greater or lesser energy than the transition in part (A), and justify your prediction using the relationship between energy level differences and photon energy. (2 points)

ANSWER KEY

Part A: Multiple-Choice Answer Table

Part B: Free-Response Detailed Answers

FRQ 1 - Answer Key

Part A: Complete electron configuration for sodium

Answer: 1s² 2s² 2p⁶ 3s¹

Scoring note: The electron configuration must be complete and correct. Alternative noble gas notation [Ne] 3s¹ is also acceptable.

Part B: Explanation of highest binding energy peak

Model Answer:

The peak at 104 MJ/mol corresponds to the two electrons in the 1s orbital (the innermost shell, n = 1). These electrons experience the greatest effective nuclear charge because they are closest to the nucleus and have no inner electrons to shield them from the full +11 nuclear charge. According to Coulomb's law, the force of attraction between the nucleus and these electrons is extremely strong because the distance between them is minimal. Therefore, these electrons require the most energy to remove, resulting in the highest binding energy peak. The peak height of 2 matches the two electrons that occupy the 1s orbital according to sodium's electron configuration.

Scoring: 1 point for identifying that these are 1s electrons closest to the nucleus; 1 point for explaining using Coulomb's law or effective nuclear charge and relating distance/shielding to binding energy.

Part C: Calculate energy to remove electrons from 3s orbital

Calculation:

The binding energy of the 3s electron is given as 0.50 MJ/mol in Peak 1.
Converting to kJ/mol:
\( 0.50 \text{ MJ/mol} \times \frac{1000 \text{ kJ}}{1 \text{ MJ}} = 500 \text{ kJ/mol} \)

Answer: 500 kJ/mol

Scoring: 1 point for correct conversion; 1 point for correct answer with units.

Part D: Justify large increase in second ionization energy

Model Answer:

The first ionization of sodium removes the single 3s electron, leaving the electron configuration 1s² 2s² 2p⁶, which is isoelectronic with neon. The second ionization energy removes an electron from the 2p orbital, which is in a much lower energy level (n = 2) and is significantly closer to the nucleus. This electron experiences much greater effective nuclear charge because there is now less shielding (the 3s electron is gone), and the remaining 10 electrons are more tightly held by the 11 protons. According to Coulomb's law, the force of attraction is inversely proportional to the square of the distance, so removing an electron from the second shell requires dramatically more energy than removing the loosely held outermost electron. Additionally, removing a second electron from an already positively charged ion (Na⁺) requires overcoming greater electrostatic attraction.

Scoring: 1 point for recognizing that the second electron comes from a lower energy level (n = 2) and is closer to nucleus; 1 point for explaining using Coulomb's law, effective nuclear charge, or shielding and connecting to the large energy difference.

Part E: Compare first ionization energies of Na and K

Model Answer:

Potassium has the lower first ionization energy (419 kJ/mol compared to 496 kJ/mol for sodium). This occurs because the outermost electron in potassium is in the 4s orbital (n = 4), while sodium's outermost electron is in the 3s orbital (n = 3). The 4s electron in potassium is farther from the nucleus-the atomic radius data show potassium (227 pm) is significantly larger than sodium (186 pm). Although potassium has more protons (19 vs. 11), the outermost electron experiences greater shielding from the additional inner electron shells (three complete inner shells in K vs. two in Na). The increased distance and increased shielding both reduce the effective nuclear charge experienced by the outermost electron in potassium, making it easier to remove despite the larger nuclear charge.

Scoring: 1 point for correctly identifying K has lower first IE; 1 point for explaining using distance from nucleus and/or atomic radius data; 1 point for discussing shielding and effective nuclear charge.

Part F: Evaluate student claim about Z_eff

Model Answer:

The student's claim is partially correct but incomplete. While it is true that potassium has more protons (19) than sodium (11), the effective nuclear charge (Z_eff) experienced by the outermost electron does not simply equal the number of protons. The data actually refute the claim that K has a greater Z_eff at its outermost electron than Na. The first ionization energy data show that it takes less energy to remove the outermost electron from K than from Na (419 kJ/mol vs. 496 kJ/mol), which indicates that the outermost electron in potassium actually experiences less effective nuclear charge than sodium's outermost electron. This is because potassium's outermost electron is shielded by many more core electrons (18 inner electrons vs. 10 in Na), and the increased distance reduces the attractive force according to Coulomb's law. Therefore, although nuclear charge increases, the Z_eff experienced by the valence electron does not increase proportionally.

Scoring: 1 point for stating the claim is incorrect or incomplete using ionization energy data as evidence; 1 point for explaining that increased shielding in K reduces Z_eff despite increased nuclear charge.

FRQ 2 - Answer Key

Part A: Calculate initial energy level

Calculation:

Given: λ = 410 nm = 410 × 10^-9 m, n_f = 2, R_H = 1.097 × 10⁷ m^-1

Using the Rydberg equation:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

Substituting values:
\[ \frac{1}{410 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{n_i^2} \right) \]

\[ 2.439 \times 10^6 = 1.097 \times 10^7 \left( 0.25 - \frac{1}{n_i^2} \right) \]

\[ \frac{2.439 \times 10^6}{1.097 \times 10^7} = 0.25 - \frac{1}{n_i^2} \]

\[ 0.2224 = 0.25 - \frac{1}{n_i^2} \]

\[ \frac{1}{n_i^2} = 0.25 - 0.2224 = 0.0276 \]

\[ n_i^2 = \frac{1}{0.0276} = 36.2 \]

\[ n_i = \sqrt{36.2} \approx 6 \]

Answer: n_i = 6

Scoring: 1 point for correct setup of Rydberg equation with proper substitution; 1 point for correct algebraic manipulation; 1 point for correct final answer rounded to nearest whole number.

Part B: Draw electron transition diagram

Model Diagram Description:

The diagram should show:

• A vertical axis representing increasing energy (upward direction)
• Horizontal lines representing energy levels labeled n = 1 (lowest), n = 2 (middle), and n = 6 (highest)
• A downward arrow from n = 6 to n = 2, clearly indicating the direction of electron transition
• Optional but helpful: label "photon emitted" or "λ = 410 nm" near the arrow

Scoring: 1 point for correct energy level diagram with proper labels; 1 point for arrow showing correct transition direction (n = 6 → n = 2).

Part C: Predict and justify energy comparison

Model Answer:

Prediction: A transition from n = 4 to n = 1 would emit light of greater energy than the transition from n = 6 to n = 2.

Justification: The energy of an emitted photon is equal to the difference in energy between the initial and final levels. The energy difference between n = 4 and n = 1 is much larger than the energy difference between n = 6 and n = 2 because lower energy levels are spaced farther apart than higher energy levels. The n = 1 level is the ground state and has the lowest (most negative) energy, so any transition ending at n = 1 releases more energy than transitions to higher levels. Additionally, the energy difference can be seen from the Rydberg equation: \( \frac{1}{1^2} - \frac{1}{4^2} = 1 - 0.0625 = 0.9375 \), while \( \frac{1}{2^2} - \frac{1}{6^2} = 0.25 - 0.0278 = 0.222 \). Since 0.9375 > 0.222, the n = 4 → n = 1 transition releases more energy and therefore produces a higher energy (shorter wavelength) photon.

Scoring: 1 point for correct prediction (greater energy); 1 point for justification using energy level differences and/or mathematical comparison showing larger ΔE for the 4 → 1 transition.