Arrange the ions Rb⁺, Br^-, and Sr²⁺ from the smallest to the largest. (LO 6.4) (a) Br^- < Rb⁺ < Sr²⁺ (b) Sr²⁺ < Br^- < Rb⁺ (c) Rb⁺ < Sr²⁺ < Br^- (d) Sr²⁺ < Rb⁺ < Br^-

Verified step by step guidance

1. Step 1: Understand that ionic size is influenced by the number of electrons and the effective nuclear charge. Cations (positive ions) are generally smaller than their parent atoms, while anions (negative ions) are larger.
2. Step 2: Consider the electron configuration of each ion. Rb⁺ has lost one electron compared to its neutral atom, Br⁻ has gained one electron, and Sr²⁺ has lost two electrons.
3. Step 3: Recognize that losing electrons (as in Rb⁺ and Sr²⁺) results in a smaller ionic radius due to increased effective nuclear charge, while gaining electrons (as in Br⁻) results in a larger ionic radius.
4. Step 4: Compare the effective nuclear charge and electron configuration: Sr²⁺, having lost two electrons, will be smaller than Rb⁺, which has lost one electron. Br⁻, having gained an electron, will be larger than both.
5. Step 5: Arrange the ions based on the analysis: Sr²⁺ < Rb⁺ < Br⁻. This corresponds to option (d) in the given choices.