Question

Rank the ions in each set in order of decreasing size, and explain your ranking:

a) Se², S², O²
b) Te²-, Cs⁺, I⁻
c) Sr²+, Ba²+, Cs⁺

Answer 1

Sizes of ions can be ranked by considering periodic trends and isoelectronicity, with Se²⁻ being larger than S²⁻ and O²⁻, Te²⁻ larger than I⁻ and Cs⁻, and Cs⁻ larger than Ba²⁻ and Sr²⁻.

Step-by-step explanation:

When ranking the sizes of ions, it's important to consider both the periodic trends and the principle of isoelectronicity (ions sharing the same electron configuration). In each set, ions are ranked from largest to smallest:

1. Se²⁻, S²⁻, O²⁻: Each ion is isoelectronic, having the same number of electrons as an argon atom (Ar). However, their nuclear charges differ. Selenium has the least nuclear charge (Z=34) compared to sulfur (Z=16) and oxygen (Z=8), making Se²⁻ the largest and O²⁻ the smallest.
2. Te²⁻, I⁻, Cs⁻: Tellurium and iodine are in the same period with Te²⁻ having more electrons but the same nuclear charge as I⁻, making Te²⁻ larger. Cs⁻, however, is a cation and is smaller than anions in the same period.
3. Cs⁻, Ba²⁻, Sr²⁻: Cesium is an alkali metal with a larger size due to its lower effective nuclear charge and additional electron shell compared to barium and strontium, which are both alkaline earth metals with higher charges leading to smaller sizes.