Question

use the references to access important values if needed for this question.
predict the larger ion in each of the following pairs.
(a) ba²+ or i-
(b) cu²+ or cu+
(c) se²- or se-
(d) ca²+ or ba²+

Explanation:

• (a) \(I^-\) has more electron - electron repulsion and less nuclear charge attraction compared to \(Ba^{2 +}\) (since \(I^-\) is an anion and \(Ba^{2+}\) is a cation). Anions are generally larger than cations.
• (b) \(Cu^{2+}\) has a higher positive charge than \(Cu^{+}\). Higher positive charge leads to more contraction of the electron cloud (due to greater nuclear pull on the electrons). So \(Cu^{+}\) is larger.
• (c) \(Se^{2 -}\) has a greater negative charge than \(Se^{-}\). More negative charge means more electron - electron repulsion, which causes the electron cloud to expand. So \(Se^{2-}\) is larger.
• (d) \(Ca^{2+}\) and \(Ba^{2+}\) are in the same group. As we move down a group, the atomic (and ionic) radius increases. \(Ba\) is below \(Ca\) in Group 2, so \(Ba^{2+}\) is larger.

Answer:

(a) \(I^-\), (b) \(Cu^{+}\), (c) \(Se^{2-}\), (d) \(Ba^{2+}\)