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:

• For part d: Electronegativity increases across a period (from left to right) and decreases down a group. Among \(P\), \(S\), \(Cl\), \(F\), \(F\) is the most electronegative as it is in the upper - right position in the periodic table (except for noble gases).
• For part e: Metallic character decreases across a period (from left to right) and increases down a group. \(F\) has the least metallic character as it is a non - metal and is located in the upper - right of the periodic table (far from the metallic regions).
• For part f: When comparing ions, consider the electron - shell structure. For ions with the same electron configuration (isoelectronic ions), the ion with the smallest nuclear charge (i.e., the one with the least number of protons) has the largest radius. \(P^{3 -}\) has the largest ion. \(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\) are isoelectronic (\(18\) electrons). The number of protons in \(P^{3 -}\) is \(15\), in \(S^{2 -}\) is \(16\), and in \(Cl^{-}\) is \(17\). Since \(P^{3 -}\) has the least number of protons to attract the \(18\) electrons, it has the largest ionic radius. \(F^{-}\) has \(10\) electrons. Comparing with the other ions (\(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\)) which have more electron shells (\(n = 3\) for \(P^{3 -}\), \(S^{2 -}\), \(Cl^{-}\) and \(n=2\) for \(F^{-}\)), \(P^{3 -}\) is larger.

Answer:

d. \(F\)
e. \(F\)
f. \(P\) (specifically \(P^{3 -}\))