Which of the following elements forms an ion that is smaller in ionic radius than a calcium ion (Ca^{2+})?

• A. Strontium (Sr^{2+})
• B. Potassium (K^+)
• C. Magnesium (Mg^{2+})
• D. Cesium (Cs^+)

Answer: (C)

Verified step by step guidance

1. Identify the charge and electron configuration of the calcium ion (Ca^{2+}). Calcium loses two electrons to form Ca^{2+}, resulting in the electron configuration of the noble gas argon (Ar), which has 18 electrons.
2. Compare the ions given (Sr^{2+}, K^{+}, Mg^{2+}, Cs^{+}) by considering their charges and positions in the periodic table. Ions with the same charge and fewer electron shells generally have smaller ionic radii.
3. Recognize that all the ions listed have a +2 or +1 charge, but their number of protons (nuclear charge) differs. A higher nuclear charge with the same electron configuration pulls electrons closer, reducing ionic radius.
4. Since Mg^{2+} has fewer electron shells than Ca^{2+} and a higher effective nuclear charge per electron, it will have a smaller ionic radius than Ca^{2+}.
5. Use periodic trends: moving up a group decreases ionic radius for ions with the same charge, and increasing positive charge also decreases ionic radius. This explains why Mg^{2+} is smaller than Ca^{2+}.