100+ Free Canary Islands PAU Chemistry Practice Questions

Explanation: For a 3p orbital, the principal quantum number is n = 3 and the azimuthal (angular momentum) quantum number is l = 1. The magnetic quantum number ml can take integer values from -l to +l (i.e., -1, 0, +1), and the spin quantum number ms can be +1/2 or -1/2. Thus, (3, 1, 0, +1/2) represents a completely valid set.

Explanation: Chromium exhibits an anomalous electron configuration due to the extra stability associated with a half-filled 3d subshell. Transferring one electron from the 4s orbital to the 3d subshell produces [Ar] 4s¹ 3d⁵, which minimizes electron-electron repulsion and maximizes exchange energy.

Explanation: First ionization energy generally increases across a period from left to right as nuclear charge (Z) increases while valence electrons are added to the same principal energy level (n = 3), resulting in a higher effective nuclear charge (Zeff). Chlorine, located at the far right of Period 3 among the options given, holds its 3p electrons most tightly.

Explanation: All four species are isoelectronic, possessing 10 electrons with the configuration 1s² 2s² 2p⁶. The radius of isoelectronic species decreases as nuclear charge (Z) increases. O²⁻ has the smallest atomic number (Z = 8), meaning its 8 protons exert the weakest electrostatic attraction on the 10 electrons, giving it the largest ionic radius.

Explanation: Using Planck's equation E = h·ν = h·c / λ, convert λ = 400 nm = 400 × 10⁻⁹ m. Substituting the constants gives E = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / (400 × 10⁻⁹ m) = (1.9878 × 10⁻²⁵) / (4.00 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J.

Explanation: According to de Broglie's equation λ = h / (m·v), substitute the electron's mass and velocity: 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.

Explanation: A massive jump occurs between I₂ (1451 kJ/mol) and I₃ (7733 kJ/mol), exceeding a 5-fold increase. This huge energy jump signifies that removing the third electron requires breaking into a highly stable core noble-gas electron shell. Therefore, element X possesses exactly 2 valence electrons and belongs to Group 2 (alkaline earth metals, e.g., Mg).

Explanation: Fluorine's 2p orbital is extremely compact in volume. When an incoming electron is added to form F⁻, it experiences unusually strong inter-electronic repulsion from the seven existing valence electrons crowded in the small n=2 shell. This repulsive force partially offsets the strong nuclear attraction, making F's electron affinity slightly less exothermic than Cl's (3p orbital is larger and less crowded).

Explanation: Neutral iron (Z = 26) has the configuration [Ar] 4s² 3d⁶. Ionization removes electrons first from the outermost principal quantum shell (4s), so forming Fe³⁺ involves losing two 4s electrons and one 3d electron, giving [Ar] 3d⁵. According to Hund's rule of maximum multiplicity, all five 3d electrons occupy separate d orbitals with parallel spins, resulting in 5 unpaired electrons.

Explanation: Wavelength λ and frequency ν are related by λ = c / ν. Substituting the speed of light and frequency gives λ = (3.00 × 10⁸ m/s) / (6.17 × 10¹⁴ s⁻¹) = 4.862 × 10⁻⁷ m = 486 nm. This corresponds to the blue-green H-beta Balmer spectral line of hydrogen (n=4 → n=2).