What is the van’t Hoff factor (i) for a dilute aqueous solution of Na[Ag(CN)2]?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
The van’t Hoff factor i counts how many particles one formula unit of the solute gives in solution. In a dilute solution a salt dissociates completely.
Ions formed: \( \ce{Na[Ag(CN)2]} \) is a coordination compound. The complex ion stays intact; only the sodium ion separates:
\( \displaystyle \ce{Na[Ag(CN)2] -> Na^+ + [Ag(CN)2]^-} \)
Particles: one formula unit gives 1 \( \ce{Na^+} \) + 1 \( \ce{[Ag(CN)2]^-} \) = 2 ions, so i = 2.
NCERT (2026-27 edition):
5.1 Werner’s Theory of Coordination Compounds – “… However, complex ions such as [Fe(CN)6]4– of K4[Fe(CN)6] do not dissociate into Fe2+ and CN– ions.”