Question

solubility of NaF in water is less than that of RbF

Answer 1

here,Na+ is a much smaller cation,as compared to Rb+,so,Na+ more effectively follows Fajan's rule,n thereby,make NaF partially covalent,by distorting the electron cloud of F,towards itself...Rb being a bulkier cation n also due to shielding effect,n low effective nuclear charge,does not have such effect...n maintains its electrovalent character,thereby easily dissociates on hydration,becoming more soluble than NaF