JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Surface Chemistry

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :

(A) $AlCl_{3} > K_{3} [Fe (CN)_{6}] > K_{2} CrO_{4} > KBr = KNO_{3}$
(B) $K_{3} [Fe (CN)_{6}] < K_{2} CrO_{4} < KBr = KNO_{3} = AlCl_{3}$ ✓ Correct
(C) $K_{3} [Fe (CN)_{6}] < K_{2} CrO_{4} < AlCl_{3} < KBr < KNO_{3}$
(D) $K_{3} [Fe (CN)_{6} > AlCl_{3} > K_{2} CrO_{4} > KBr > KNO_{3}$

Answer: (B)

Solution

Ferric hydroxide sol is positively charged, so its coagulation is driven by anions, and by the Hardy-Schulze rule the coagulating power rises steeply with anion charge while the flocculation value runs inversely to that power. The anions here are

[Fe (CN)_{6}]^{3 -}

(trivalent),

Cr O_{4}^{2 -}

(divalent) and the monovalent

B r^{-}

N O_{3}^{-}

and

C l^{-}

, the last supplied by

AlC l_{3}

. The most highly charged anion coagulates most effectively and so has the smallest flocculation value, giving the order

K_{3} [Fe (CN)_{6}] < K_{2} Cr O_{4} < KBr = KN O_{3} = AlC l_{3}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Surface Chemistry →

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Surface Chemistry

Solution

More JEE Main questions on Surface Chemistry