Question

in which pair does the first atom or ion have a larger radius than the second atom or ion?
a) s, mg
b) br, br-
c) k, k+
d) na, rb

Explanation:

Step1: Recall atomic radius trends

Atomic radius increases down a group and decreases across a period. For ions, cations are smaller than their parent atoms (due to loss of electrons, increased effective nuclear charge), and anions are larger than their parent atoms (due to gain of electrons, decreased effective nuclear charge).

Step2: Analyze each option

• Option A: \(S\) and \(Mg\) are in the same period (\(3^{rd}\) period). As we move from left (\(Mg\)) to right (\(S\)) in a period, atomic radius decreases. So \(Mg > S\) (atomic radius of \(Mg\) is larger than \(S\)).
• Option B: \(Br\) (atom) and \(Br^-\) (ion). When an atom gains an electron to form an anion (\(Br^-\)), the electron - electron repulsion increases and the effective nuclear charge per electron decreases. So \(Br^->Br\) (radius of \(Br^-\) is larger than \(Br\)).
• Option C: \(K\) (atom) and \(K^+\) (ion). When an atom loses an electron to form a cation (\(K^+\)), the effective nuclear charge per electron increases. So \(K > K^+\) (radius of \(K\) is larger than \(K^+\)).
• Option D: \(Na\) and \(Rb\) are in the same group (\(1^{st}\) group). As we move down a group (from \(Na\) to \(Rb\)), the number of electron shells increases. So \(Rb>Na\) (atomic radius of \(Rb\) is larger than \(Na\)).

Answer:

B. \(Br, Br^-\)