Mastering s-Block Elements: Formulas, Traps & Exam Strategy | ExamiFly

Thermal stability depends on the polarizing power ( ionic potential $\phi = \frac{z}{r}$ ) of the cation. $\text{Be}^{2+}$ has a very small ionic radius, resulting in high $\phi$. It heavily polarizes the electron cloud of the large $\text{CO}_3^{2-}$ anion, weakening the $\text{C-O}$ bond and causing $\text{BeCO}_3$ to decompose at lower temperatures ($< 100^\circ\text{C}$). As cation size increases down to $\text{Ba}^{2+}$, polarizing power decreases, stabilizing the crystal lattice of $\text{BaCO}_3$ up to high temperatures ($> 1300^\circ\text{C}$).

Analyze solubility using lattice energy ($\Delta_{\text{lattice}}H$) and hydration energy ($\Delta_{\text{hyd}}H$). For hydroxides, the anion $\text{OH}^-$ is small, so lattice energy decreases much faster than hydration energy down the group, increasing solubility (e.g., $\text{Mg(OH)}_2 < \text{Ba(OH)}_2$). For sulfates, the anion $\text{SO}_4^{2-}$ is very large, so lattice energy stays nearly constant down the group while cation hydration energy drops significantly, making solubility decrease down the group (e.g., $\text{BeSO}_4 > \text{BaSO}_4$).