Question

1. based on the concept of periodic trends, answer the following questions for these atoms: p, s, cl, f. be prepared to defend your answers. d which element has the highest electronegativity? __ e which element has the least metallic character? f which element has the largest ion? __

Explanation:

d. Which element has the highest electronegativity?

Electronegativity generally increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table. The elements given (assuming the full set is P, S, Cl, F as per the partial text) are P (Phosphorus), S (Sulfur), Cl (Chlorine), and F (Fluorine). All are in the same period (period 3 for P, S, Cl; period 2 for F) or adjacent periods. Moving from left to right across a period, electronegativity increases. Among these, F is in group 17 (halogens) and is the most electronegative element in the periodic table due to its small atomic radius and high effective nuclear charge, which strongly attracts electrons. Comparing P, S, Cl, and F: F > Cl > S > P in terms of electronegativity.

Metallic character decreases across a period (left to right) and increases down a group (top to bottom). Metallic character is associated with the ability to lose electrons; non - metals have less metallic character. Among P, S, Cl, F: F is a non - metal, Cl is a non - metal, S is a non - metal, and P is a non - metal/metalloid. As we move from P to S to Cl to F across the period (or considering their positions), the ability to lose electrons decreases. F has the least tendency to lose electrons, so it has the least metallic character.

When forming ions, for these non - metals (P, S, Cl, F), they generally gain electrons to form anions. The size of an anion increases as we go down a group and also depends on the number of electrons gained and the effective nuclear charge. The ions formed are \(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\), and \(F^{-}\). For ions with the same electron configuration (or similar electron - electron repulsion and nuclear charge considerations), the ion with the least nuclear charge (number of protons) will be the largest. P has 15 protons, S has 16, Cl has 17, and F has 9. But when comparing \(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\): they are isoelectronic (have the same number of electrons, \(18\)). For isoelectronic species, the ion with the smallest nuclear charge (least number of protons) will have the largest radius because the protons pull the electrons less strongly. So \(P^{3 -}\) has 15 protons, \(S^{2 -}\) has 16, \(Cl^{-}\) has 17. So \(P^{3 -}\) is larger than \(S^{2 -}\) and \(Cl^{-}\). And \(F^{-}\) has 10 electrons, while \(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\) have 18 electrons. The \(P^{3 -}\) ion has more electron shells (3 shells for the ion, as P is in period 3) compared to \(F^{-}\) (2 shells). So \(P^{3 -}\) is larger.

Answer:

F (Fluorine)

e. Which element has the least metallic character?