100+ Free Balearic Islands PAU Chemistry / Química 2º Bachillerato UIB Practice Questions

Explanation: For a 3p electron, the principal quantum number n = 3, and the azimuthal (orbital angular momentum) quantum number l = 1 (since l = 0 for s, 1 for p, 2 for d). The magnetic quantum number ml can take integer values from -l to +l, so for l = 1, ml ∈ {-1, 0, +1}. The spin quantum number ms can be +1/2 or -1/2. Therefore, (3, 1, -1, +1/2) is valid.

Explanation: Chromium (Z = 24) is a well-known exception to the standard Aufbau ordering. To achieve extra stability associated with a half-filled 3d subshell (5 electrons), an electron is promoted from the 4s orbital to the 3d subshell. Thus, its ground-state configuration is [Ar] 4s¹ 3d⁵ rather than [Ar] 4s² 3d⁴.

Explanation: First ionization energy increases across a period from left to right and decreases down a group. Neon (Ne, Z = 10) is a noble gas at the extreme right of Period 2 with a stable closed-shell octet (2s² 2p⁶), giving it the highest first ionization energy among the options (2080 kJ/mol).

Explanation: Mg²⁺, Na⁺, F⁻, and O²⁻ are isoelectronic species, all possessing 10 electrons ([Ne] configuration). In an isoelectronic series, radius decreases as nuclear charge (Z) increases. Nuclear charges are: Mg (Z=12), Na (Z=11), F (Z=9), O (Z=8). Thus, radius increases as Z decreases: Mg²⁺ (smallest) < Na⁺ < F⁻ < O²⁻ (largest).

Explanation: For n = 4, the principal energy level contains subshells l = 0 (4s), l = 1 (4p), l = 2 (4d), and l = 3 (4f). Total number of orbitals in shell n = 4 is n² = 4² = 16 orbitals. Each orbital can hold at most 2 electrons: one with ms = +1/2 and one with ms = -1/2. Therefore, exactly 16 electrons in the n = 4 level have ms = -1/2.

Explanation: Carbon has electron configuration 1s² 2s² 2p², so adding an electron yields a half-filled 2p³ subshell ([1s² 2s² 2p³]), which is energetically favorable. Nitrogen already has a stable half-filled 2p³ configuration; adding an electron forces it into an already occupied 2p orbital (2p⁴), incurring electron-electron pairing repulsion. Hence, N has an electron affinity close to zero (~0 kJ/mol), less exothermic than C (-122 kJ/mol).

Explanation: According to de Broglie's equation: λ = h / (m · v). Substituting values: λ = (6.626 × 10⁻³⁴ J·s) / [(9.11 × 10⁻³¹ kg) × (2.0 × 10⁶ m/s)] = (6.626 × 10⁻³⁴) / (1.822 × 10⁻²⁴) = 3.6366 × 10⁻¹⁰ m ≈ 3.64 × 10⁻¹⁰ m (0.364 nm).

Explanation: Neutral iron atom (Z = 26) has the configuration [Ar] 4s² 3d⁶. When transition metal cations are formed, electrons are removed first from the outermost valence shell, which is the 4s subshell. Removing 3 electrons means removing 2 electrons from 4s and 1 electron from 3d, leaving [Ar] 4s⁰ 3d⁵, simplified as [Ar] 3d⁵.

Explanation: Electronegativity measures an atom's ability to attract shared bonding electrons. Across Period 3 (Na to Cl), nuclear charge increases from Z = 11 to Z = 17 while core shielding remains nearly constant (10 core electrons). Consequently, effective nuclear charge (Z_eff) increases and atomic radius contracts, drawing bonding electrons more strongly toward the nucleus. Thus, electronegativity increases from Na (0.93) to Cl (3.16).

Explanation: For any atomic orbital: Angular nodes = l. For a p orbital, l = 1, so it has 1 angular (planar) node. Radial nodes = n - l - 1. For a 4p orbital (n = 4, l = 1), radial nodes = 4 - 1 - 1 = 2. Total nodes = n - 1 = 3 (2 radial + 1 angular).