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100+ Free La Rioja PAU Chemistry Practice Questions
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Sample La Rioja PAU Chemistry Practice Questions
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1Which set of quantum numbers (n, l, m_l, m_s) correctly describes the highest-energy valence electron in a ground-state sodium atom (Na, Z = 11)?
A.n = 3, l = 0, m_l = 0, m_s = +1/2
B.n = 3, l = 1, m_l = 0, m_s = +1/2
C.n = 2, l = 1, m_l = +1, m_s = -1/2
D.n = 3, l = 2, m_l = 0, m_s = +1/2
Explanation: The ground-state electronic configuration of sodium (Z = 11) is 1s² 2s² 2p⁶ 3s¹. The highest-energy valence electron occupies the 3s orbital, corresponding to principal quantum number n = 3, azimuthal quantum number l = 0 (s-orbital), magnetic quantum number m_l = 0, and spin quantum number m_s = +1/2.
2How does atomic radius change across Period 3 elements from sodium (Na) to chlorine (Cl), and what is the primary cause?
A.It increases because additional electron shells are filled.
B.It decreases because effective nuclear charge increases while core electron shielding remains nearly constant.
C.It remains constant because shielding by inner electrons completely cancels the increased nuclear charge.
D.It decreases because electron-electron repulsions force electrons closer to the nucleus.
Explanation: Across Period 3 (Na to Cl), protons are added to the nucleus while electrons are added to the same principal energy level (n = 3). Because core electron shielding remains approximately constant, the effective nuclear charge (Z_eff) increases, drawing the valence shell closer to the nucleus and decreasing atomic radius.
3What is the ground-state electronic configuration of copper (Cu, Z = 29)?
A.[Ar] 4s² 3d⁹
B.[Ar] 4s¹ 3d¹⁰
C.[Ar] 4s⁰ 3d¹¹
D.[Ar] 4s² 3d¹⁰
Explanation: Copper exhibits an anomalous electronic configuration due to the enhanced thermodynamic stability associated with a completely filled d-subshell. One electron is promoted from the 4s orbital to the 3d subshell, yielding [Ar] 3d¹⁰ 4s¹.
4The first ionization energy of nitrogen (N, Z = 7) is 1402 kJ/mol, whereas that of oxygen (O, Z = 8) is 1314 kJ/mol. Which statement accounts for this anomaly?
A.Nitrogen has a smaller atomic radius than oxygen.
B.Oxygen has a higher effective nuclear charge than nitrogen.
C.Nitrogen possesses a half-filled 2p subshell (2p³), providing extra exchange energy stability, while oxygen has paired electrons in a 2p orbital that experience inter-electron repulsion.
D.Nitrogen's 2s orbital is lower in energy than oxygen's 2s orbital.
Explanation: Nitrogen's valence configuration is 2s² 2p³, possessing a half-filled 2p subshell with parallel spins that maximizes exchange energy. Oxygen (2s² 2p⁴) has one doubly occupied 2p orbital. Electron-electron repulsion between the paired electrons in oxygen lowers the energy required to remove one electron, resulting in a lower first ionization energy for oxygen despite its higher Z.
5Which of the following orders correctly arranges the isoelectronic species S²⁻, Cl⁻, K⁺, and Ca²⁺ by decreasing ionic radius?
A.S²⁻ > Cl⁻ > K⁺ > Ca²⁺
B.Ca²⁺ > K⁺ > Cl⁻ > S²⁻
C.Cl⁻ > S²⁻ > Ca²⁺ > K⁺
D.K⁺ > Ca²⁺ > S²⁻ > Cl⁻
Explanation: All four species possess 18 electrons (isoelectronic with argon). For isoelectronic ions, ionic radius decreases as nuclear charge (Z) increases. S²⁻ (Z = 16) has the smallest Z and weakest pull on electrons (largest radius), followed by Cl⁻ (Z = 17), K⁺ (Z = 19), and Ca²⁺ (Z = 20, smallest radius).
6Calculate the energy of the photon emitted during an electronic transition from n = 3 to n = 1 in a hydrogen atom. (Rydberg constant R_H = 2.18 × 10⁻¹⁸ J)
A.1.94 × 10⁻¹⁸ J
B.2.42 × 10⁻¹⁹ J
C.1.45 × 10⁻¹⁸ J
D.4.84 × 10⁻¹⁹ J
Explanation: Using the Bohr/Rydberg formula ΔE = R_H (1/n_final² - 1/n_initial²): ΔE = 2.18 × 10⁻¹⁸ J × (1/1² - 1/3²) = 2.18 × 10⁻¹⁸ J × (1 - 1/9) = 2.18 × 10⁻¹⁸ J × (8/9) = 1.938 × 10⁻¹⁸ J ≈ 1.94 × 10⁻¹⁸ J.
7Using Slater's rules, what is the effective nuclear charge (Z_eff) experienced by a 3p valence electron in a chlorine atom (Z = 17)?
A.6.10
B.7.00
C.10.90
D.5.25
Explanation: Chlorine (Z = 17) configuration is (1s²)(2s² 2p⁸)(3s² 3p⁵). The target electron is in the (3s, 3p) group. Shielding constant S = (6 electrons in same group × 0.35) + (8 electrons in n-1 shell × 0.85) + (2 electrons in n-2 shell × 1.00) = (6 × 0.35) + (8 × 0.85) + (2 × 1.00) = 2.10 + 6.80 + 2.00 = 10.90. Thus Z_eff = Z - S = 17 - 10.90 = 6.10.
8What is the de Broglie wavelength of an electron (mass m = 9.11 × 10⁻³¹ kg) traveling at a velocity of 2.00 × 10⁶ m/s? (Planck's constant h = 6.626 × 10⁻³⁴ J·s)
A.0.364 nm
B.3.64 nm
C.0.182 nm
D.1.21 nm
Explanation: Using the de Broglie relation λ = h / (m · v): λ = (6.626 × 10⁻³⁴ J·s) / (9.11 × 10⁻³¹ kg × 2.00 × 10⁶ m/s) = 6.626 × 10⁻³⁴ / 1.822 × 10⁻²⁴ = 3.637 × 10⁻¹⁰ m = 0.364 nm.
9Which set of quantum numbers represents a valid state for an electron in an atom?
A.n = 3, l = 2, m_l = -1, m_s = +1/2
B.n = 2, l = 2, m_l = 0, m_s = -1/2
C.n = 4, l = 1, m_l = +2, m_s = +1/2
D.n = 1, l = 0, m_l = 0, m_s = +1
Explanation: For n = 3, l can take values 0, 1, 2. When l = 2 (d-orbital), m_l can range from -l to +l (-2, -1, 0, +1, +2). The spin m_s can be +1/2 or -1/2. Thus n = 3, l = 2, m_l = -1, m_s = +1/2 is fully valid.
10The first four successive ionization energies (IE_1 to IE_4 in kJ/mol) of a Period 3 element are 738, 1450, 7730, and 10540. To which Group of the periodic table does this element belong?
A.Group 2 (Alkaline earth metals)
B.Group 1 (Alkali metals)
C.Group 13 (Boron group)
D.Group 14 (Carbon group)
Explanation: The ionization energies increase moderately from IE_1 (738) to IE_2 (1450), but there is a massive jump of 6280 kJ/mol between IE_2 and IE_3. This huge energy gap signifies that the third electron is being removed from a stable, filled inner noble-gas core. Therefore, the element has 2 valence electrons and belongs to Group 2 (magnesium).
About the La Rioja PAU Chemistry Practice Questions
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