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Question: Knowing that radius of $Na^+ > Mg^{2+}$ and radius of $S^{2-} > Cl^-$ predict which compound will be...

Knowing that radius of Na+>Mg2+Na^+ > Mg^{2+} and radius of S2−>Cl−S^{2-} > Cl^- predict which compound will be least soluble in polar solvent?

A

MgS

B

Na2SNa_2S

C

MgCl2MgCl_2

D

NaCl

Answer

MgS

Explanation

Solution

To determine which compound will be least soluble in a polar solvent, we need to consider the degree of covalent character in each compound. Compounds with higher covalent character tend to be less soluble in polar solvents. Fajan's rules help predict the extent of covalent character in an ionic compound:

  1. Smaller cation: A smaller cation has a higher charge density and thus greater polarizing power, leading to more covalent character.
  2. Larger anion: A larger anion is more easily distorted (polarized) by the cation, leading to more covalent character.
  3. Higher charge on cation: A higher positive charge on the cation increases its polarizing power, leading to more covalent character.
  4. Higher charge on anion: A higher negative charge on the anion increases its polarizability, leading to more covalent character.

The question provides the following information:

  • Radius of Na+>Mg2+Na^+ > Mg^{2+} (meaning Mg2+Mg^{2+} is smaller than Na+Na^+).
  • Radius of S2−>Cl−S^{2-} > Cl^- (meaning S2−S^{2-} is larger than Cl−Cl^-).

Based on Fajan's Rules, Mg2+Mg^{2+} (smaller size and higher charge) will have greater polarizing power than Na+Na^+, and S2−S^{2-} (larger size and higher charge) will be more polarizable than Cl−Cl^-.

Therefore, the compound formed by Mg2+Mg^{2+} and S2−S^{2-} (MgS) will have the highest covalent character among the given options and will be the least soluble in a polar solvent.