Addition of excess aqueous ammonia to a pink coloured aqueous solution of (X) and gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. Among the following options, which statement(s) is(are) correct?
- (A)Addition of silver nitrate to Y gives only two equivalents of silver chloride
- (B)The hybridization of the central metal ion in Y is ✓ Correct
- (C)Z is a tetrahedral complex✓ Correct
- (D)When X and Z are in equilibrium at , the colour of the solution is pink✓ Correct
Answer: (B), (C), (D)
Solution
X is (pink, octahedral, B.M. for Co²⁺ with 3 unpaired electrons).
With excess and air, Co²⁺ is oxidized to Co³⁺, forming Y = (1:3 electrolyte, for low-spin Co³⁺, hybridization).
With excess HCl, Z = (blue, tetrahedral, B.M.).
Option A: Y gives 3 equivalents of AgCl (all three Cl⁻ are outside), not 2.
Option D: The equilibrium is exothermic (), so at (low temperature), the equilibrium shifts left to the pink octahedral complex.
Correct answers: B, C, D
With excess and air, Co²⁺ is oxidized to Co³⁺, forming Y = (1:3 electrolyte, for low-spin Co³⁺, hybridization).
With excess HCl, Z = (blue, tetrahedral, B.M.).
Option A: Y gives 3 equivalents of AgCl (all three Cl⁻ are outside), not 2.
Option D: The equilibrium is exothermic (), so at (low temperature), the equilibrium shifts left to the pink octahedral complex.
Correct answers: B, C, D
JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options