AP Atomic Structure and Periodicity 2 — Questions and Answers

Question 1: An electron in a hydrogen atom transitions from n = 4 to n = 2. What is the wavelength of the emitted photon? (RH = 2.179 × 10⁻¹⁸ J, h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s)

486 nm (Correct answer)
656 nm
410 nm
434 nm

Correct answer: 486 nm

ΔE = RH(1/n₁² − 1/n₂²) = 2.179 × 10⁻¹⁸(1/4 − 1/16) = 2.179 × 10⁻¹⁸ × 0.1875 = 4.086 × 10⁻¹⁹ J. λ = hc/E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(4.086 × 10⁻¹⁹) = 486 nm.

The Balmer series consists of electron transitions to n = 2 in hydrogen, emitting visible light. For the n = 4 → n = 2 transition: ΔE = RH(1/n_lower² − 1/n_upper²) = 2.179 × 10⁻¹⁸ J × (1/4 − 1/16) = 2.179 × 10⁻¹⁸ × (4/16 − 1/16) = 2.179 × 10⁻¹⁸ × (3/16) = 4.086 × 10⁻¹⁹ J. λ = hc/ΔE = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s)/(4.086 × 10⁻¹⁹ J) = 4.86 × 10⁻⁷ m = 486 nm. This is the blue-green line of the Balmer series, visible in hydrogen's emission spectrum. The Balmer series (n → 2) is in the visible range; the Lyman series (n → 1) is in the UV; and the Paschen series (n → 3) is in the infrared. These discrete spectral lines were key evidence for the quantized nature of atomic energy levels.

Question 2: Which set of quantum numbers is NOT allowed for an electron in an atom?

n=3, l=2, ml=+2, ms=+½
n=2, l=1, ml=−1, ms=−½
n=4, l=0, ml=0, ms=+½
n=3, l=3, ml=+2, ms=+½ (Correct answer)

Correct answer: n=3, l=3, ml=+2, ms=+½

For n = 3, l can be 0, 1, or 2 only (l must be 0 to n−1). l = 3 is not permitted when n = 3, making this set of quantum numbers invalid.

The four quantum numbers (n, l, ml, ms) follow strict rules: 1. Principal quantum number n: positive integers (1, 2, 3, ...) 2. Angular momentum quantum number l: integers from 0 to n−1 3. Magnetic quantum number ml: integers from −l to +l 4. Spin quantum number ms: +½ or −½ Evaluating the options: - n=3, l=2, ml=+2: l=2 is valid (0≤l≤2), ml=+2 is valid (−2≤ml≤+2). ALLOWED. - n=2, l=1, ml=−1: l=1 valid (0≤l≤1), ml=−1 valid (−1≤ml≤+1). ALLOWED. - n=4, l=0, ml=0: s orbital, fully valid. ALLOWED. - n=3, l=3, ml=+2: l=3 requires n≥4. For n=3, maximum l=2. NOT ALLOWED. This type of analysis is fundamental to understanding electron configuration and orbital occupancy rules (Aufbau principle, Pauli exclusion, Hund's rule).

Question 3: The first ionization energies (in kJ/mol) for elements in Period 3 include an unusually high value for Mg (738) compared to Al (577). What best explains why Al has a lower first ionization energy than Mg?

Al has a larger atomic radius than Mg
Al's outer electron is in a 3p orbital, which is higher in energy and better shielded than Mg's 3s electron (Correct answer)
Al has more electrons, increasing electron-electron repulsion
Mg has a higher nuclear charge that attracts electrons more strongly

Correct answer: Al's outer electron is in a 3p orbital, which is higher in energy and better shielded than Mg's 3s electron

Al's highest-energy electron occupies a 3p orbital, which has higher energy and experiences more shielding from the 3s electrons than the 3s² configuration of Mg, making it easier to remove.

Ionization energy generally increases across Period 3, but there are two notable exceptions: between Mg and Al, and between P and S. Mg has the configuration [Ne]3s²; its outermost electrons are in the 3s subshell. Al has the configuration [Ne]3s²3p¹; its outermost electron is in the 3p subshell. The 3p orbital is: (1) higher in energy than 3s, (2) more diffuse and farther from the nucleus on average, and (3) better shielded by the 3s² electrons below it. Together, these factors mean the 3p electron experiences a lower effective nuclear charge and is more easily removed. The second exception (P vs. S) occurs because S has a paired electron in a 3p orbital, introducing extra electron-electron repulsion that lowers its ionization energy below that of P. These anomalies are important for AP Chemistry and demonstrate that periodic trends have underlying quantum mechanical explanations.

Question 4: Which of the following correctly lists the ions in order of increasing radius: Na⁺, Mg²⁺, Al³⁺, O²⁻, F⁻?

Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ (Correct answer)
O²⁻ < F⁻ < Na⁺ < Mg²⁺ < Al³⁺
Na⁺ < Mg²⁺ < Al³⁺ < F⁻ < O²⁻
F⁻ < O²⁻ < Na⁺ < Mg²⁺ < Al³⁺

Correct answer: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻

All five are isoelectronic (10 electrons). Larger nuclear charge contracts the electron cloud more, so larger Z → smaller radius. Z: Al(13) > Mg(12) > Na(11) > F(9) > O(8), giving Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻.

Isoelectronic species all have the same number of electrons (here, 10 electrons — neon configuration) but differ in nuclear charge. The effective nuclear charge Zeff felt by each electron depends primarily on the actual nuclear charge Z, since all species have the same shielding configuration. Atomic numbers: O = 8, F = 9, Na = 11, Mg = 12, Al = 13. Higher nuclear charge pulls the 10 electrons inward more strongly, reducing ionic radius: - O²⁻: Z=8, largest (least nuclear pull) - F⁻: Z=9 - Na⁺: Z=11 - Mg²⁺: Z=12 - Al³⁺: Z=13, smallest (most nuclear pull) So in order of increasing radius: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻. This isoelectronic series analysis is a key tool in comparing ionic sizes across the periodic table.

Question 5: What is the de Broglie wavelength of an electron (mass = 9.11 × 10⁻³¹ kg) moving at 2.00 × 10⁶ m/s? (h = 6.626 × 10⁻³⁴ J·s)

3.64 × 10⁻¹⁰ m (Correct answer)
7.28 × 10⁻¹⁰ m
1.82 × 10⁻¹⁰ m
3.64 × 10⁻⁷ m

Correct answer: 3.64 × 10⁻¹⁰ m

λ = h/mv = (6.626 × 10⁻³⁴ J·s)/((9.11 × 10⁻³¹ kg)(2.00 × 10⁶ m/s)) = 6.626 × 10⁻³⁴/(1.822 × 10⁻²⁴) = 3.64 × 10⁻¹⁰ m.

Louis de Broglie proposed in 1924 that particles have wave properties, with wavelength λ = h/p = h/(mv). For an electron: λ = (6.626 × 10⁻³⁴ J·s) / (9.11 × 10⁻³¹ kg × 2.00 × 10⁶ m/s). Denominator = 9.11 × 10⁻³¹ × 2.00 × 10⁶ = 1.822 × 10⁻²⁴ kg·m/s. λ = 6.626 × 10⁻³⁴ / 1.822 × 10⁻²⁴ = 3.64 × 10⁻¹⁰ m = 0.364 nm. This wavelength (0.364 nm) is on the order of atomic bond lengths and X-ray wavelengths, explaining why electron diffraction can probe crystal structures. The de Broglie relationship is fundamental to the wave-mechanical model of the atom and underlies modern electron microscopy, diffraction, and quantum computing concepts.

Question 6: An element has the electron configuration [Kr]4d¹⁰5s²5p⁴. What is the element, and how many unpaired electrons does it have?

Tellurium (Te), 2 unpaired electrons (Correct answer)
Selenium (Se), 2 unpaired electrons
Tellurium (Te), 4 unpaired electrons
Tin (Sn), 0 unpaired electrons

Correct answer: Tellurium (Te), 2 unpaired electrons

Kr ends at Z=36. Adding 4d¹⁰(10) + 5s²(2) + 5p⁴(4) = 16 more gives Z=52, which is tellurium. In the 5p⁴ configuration, two p orbitals are singly occupied and one is doubly occupied, giving 2 unpaired electrons.

Starting from the krypton core at Z = 36: [Kr] 4d¹⁰ 5s² 5p⁴ adds 10 + 2 + 4 = 16 electrons. Z = 36 + 16 = 52. Element 52 is tellurium (Te), a chalcogen in Period 5. For the 5p⁴ configuration, distribute 4 electrons among 3 p orbitals (px, py, pz) using Hund's rule (maximize unpaired spins): px↑↓, py↑, pz↑. This gives 2 unpaired electrons. Tellurium is analogous to sulfur and oxygen in its electron configuration and chemistry. The 2 unpaired electrons make Te paramagnetic and explain its oxidation states of −2, +2, +4, and +6. Understanding electron configurations and their relationship to orbital diagrams is essential for predicting magnetic properties, reactivity, and the position of elements in the periodic table.

AP Atomic Structure and Periodicity 2

An electron in a hydrogen atom transitions from n = 4 to n = 2.
What is the wavelength of the emitted photon? (RH = 2.179 × 10⁻¹⁸ J, h = 6.626 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s)