JEE Advanced 2017 · Paper 1 · Chemistry Q5

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ChemistryMultiple correct+4 / -2Question 23 of 54

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $\ce{MCl2.6H2O}$ (X) and $\ce{NH4Cl}$ 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?

Addition of silver nitrate to Y gives only two equivalents of silver chloride

The hybridization of the central metal ion in Y is

d^2sp^3

Z is a tetrahedral complex

When X and Z are in equilibrium at 0°C, the colour of the solution is pink

Answer and worked solution

Official answer: B, C, D

Identify the species.

X is pink $\ce{[Co(H2O)6]Cl2}$ : $\ce{Co^{2+}}$ , $d^7$ high spin, 3 unpaired, $\mu=3.87$ BM.
Y: with $\ce{NH3/NH4Cl}$ and air, Co(II) is oxidised to $\ce{[Co(NH3)6]Cl3}$ . It is a 1:3 electrolyte, $d^6$ low spin, so diamagnetic.
Z: excess HCl gives blue $\ce{[CoCl4]^{2-}}$ , tetrahedral with 3 unpaired electrons.

Check the options.

(A) All three $\ce{Cl-}$ in Y are ionisable, giving three equivalents of AgCl. False.
(B) Low-spin octahedral Co(III) is $d^2sp^3$ . True.
(C) True.
(D) $\ce{[Co(H2O)6]^{2+} + 4Cl- <=> [CoCl4]^{2-} + 6H2O}$ is endothermic. At 0 °C it lies to the left, so the solution is pink. True.

(B), (C), (D)

Topics:Valence bond theory Magnetic properties and colour Crystal field theory

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JEE Advanced 2017 · Paper 1, Chemistry question 5