This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the electron configurations for each species, presented in spdf notation, orbital diagram, and noble-gas-core abbreviation. a) Phosphorus (P) Phosphorus has an atomic number of 15, meaning it has 15 electrons. spdf notation: 1s^2 2s^2 2p^6 3s^2 3p^3 Orbital diagram: 1s: 2s: 2p: 3s: 3p: Noble-gas-core abbreviation: [Ne] 3s^2 3p^3 The noble gas preceding phosphorus is Neon (Ne), which has 10 electrons (1s^2 2s^2 2p^6). b) Chlorine (Cl) Chlorine has an atomic number of 17, meaning it has 17 electrons. spdf notation: 1s^2 2s^2 2p^6 3s^2 3p^5 Orbital diagram: 1s: 2s: 2p: 3s: 3p: Noble-gas-core abbreviation: [Ne] 3s^2 3p^5 The noble gas preceding chlorine is Neon (Ne), which has 10 electrons (1s^2 2s^2 2p^6). c) Chloride ion (Cl^-) A chloride ion has gained one electron compared to a neutral chlorine atom. So, it has 17 + 1 = 18 electrons. spdf notation: 1s^2 2s^2 2p^6 3s^2 3p^6 Orbital diagram: 1s: 2s: 2p: 3s: 3p: Noble-gas-core abbreviation: [Ne] 3s^2 3p^6 or simply [Ar] The noble gas preceding Cl^- (with 18 electrons) is Neon (Ne). The configuration is also identical to that of Argon (Ar). d) Aluminum ion (Al^2+) Aluminum has an atomic number of 13, meaning a neutral atom has 13 electrons. The Al^2+ ion has lost two electrons, so it has 13 - 2 = 11 electrons. Electrons are removed from the highest principal quantum number shell first, and then from the subshell with the highest l value within that shell. Neutral Al: 1s^2 2s^2 2p^6 3s^2 3p^1. First electron removed from 3p. Second electron removed from 3s. spdf notation: 1s^2 2s^2 2p^6 3s^1 Orbital diagram: 1s: 2s: 2p: 3s: Noble-gas-core abbreviation: [Ne] 3s^1 The noble gas preceding Al^2+ (with 11 electrons) is Neon (Ne), which has 10 electrons (1s^2 2s^2 2p^6).