Knowing that radius of Na + Mg 2+ and radius of S 2- Cl -, predict which compound will be least soluble in polar solvent?

Question

Knowing that radius of Na +> Mg 2+ and radius of S 2-> Cl -, predict which compound will be least soluble in polar solvent? (A) MgS (B) Na 2 S (C) MgCl 2 (D) NaCl

Tardigrade Admin · Accepted Answer

Smaller size of cation, larger size of anion and high charge favours polarisation thus more covalenf character. As covalent character increases, solubility in polar solvent decreases because covalent compounds are non-polar.

Q. Knowing that radius of $N a^{+} > M g^{2 +}$ and radius of $S^{2 -} > C l^{-},$ predict which compound will be least soluble in polar solvent?

$M g S$

$N a_{2} S$

$M g C l_{2}$

$N a Cl$

Smaller size of cation, larger size of anion and high charge favours polarisation thus more covalenf character. As covalent character increases, solubility in polar solvent decreases because covalent compounds are non-polar.

Q. Knowing that radius of Na+>Mg2+ and radius of S2−>Cl−, predict which compound will be least soluble in polar solvent?

MgS

Na2​S

MgCl2​

NaCl

Smaller size of cation, larger size of anion and high charge favours polarisation thus more covalenf character. As covalent character increases, solubility in polar solvent decreases because covalent compounds are non-polar.

Q. Knowing that radius of $N a^{+} > M g^{2 +}$ and radius of $S^{2 -} > C l^{-},$ predict which compound will be least soluble in polar solvent?

$M g S$

$N a_{2} S$

$M g C l_{2}$

$N a Cl$