Match the LIST-I with LIST-II.
| LIST-I (Salt/ Complex) | LIST-II (Van’t Hoff factor) |
| (A) K3[Fe(CN)6] | (I) 5 |
| (B) Al2(SO4)3 | (II) 3 |
| (C) MgSO4 | (III) 4 |
| (D) K2SO4 | (IV) 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
The van’t Hoff factor (i) is the number of particles that one formula unit gives in solution. For a strong electrolyte that dissociates completely, i = total number of ions formed. A complex ion stays intact; only the ions outside the square bracket separate.
(A) \( \ce{K3[Fe(CN)6]} \) — This matches (III) 4. It gives \( \ce{3K^{+}} \) and one \( \ce{[Fe(CN)6]^{3-}} \): 3 + 1 = 4 ions, so i = 4.
\( \displaystyle \ce{K3[Fe(CN)6] -> 3K^+ + [Fe(CN)6]^3-} \)
(B) \( \ce{Al2(SO4)3} \) — This matches (I) 5. It gives \( \ce{2Al^{3+}} \) and \( \ce{3SO4^{2-}} \): 2 + 3 = 5 ions, so i = 5.
\( \displaystyle \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \)
(C) \( \ce{MgSO4} \) — This matches (IV) 2. It gives \( \ce{Mg^{2+}} \) and \( \ce{SO4^{2-}} \): 1 + 1 = 2 ions, so i = 2.
\( \displaystyle \ce{MgSO4 -> Mg^2+ + SO4^2-} \)
(D) \( \ce{K2SO4} \) — This matches (II) 3. It gives \( \ce{2K^{+}} \) and \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, so i = 3.
\( \displaystyle \ce{K2SO4 -> 2K^+ + SO4^2-} \)
Shortcut: (B)–(I) appears in Options (2) and (3), and (D)–(II) then leaves only Option (3).