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Question

Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2.6H2O(X) and 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 the complex Y. Among the following options, which statement(s) is/are correct?

A
The hybridization of the central metal ion in Y is d2sp3
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B
Z is a tetrahedral complex
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C
Addition of silver nitrate to Y gives only two equivalents of silver chloride
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D
When X and Z are in equilibrium at 0C, the colour of the solution is pink
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Solution

The correct options are
A The hybridization of the central metal ion in Y is d2sp3
B Z is a tetrahedral complex
D When X and Z are in equilibrium at 0C, the colour of the solution is pink
[Co(H2O)6(X)]Octahedral (pink)(μ=3.87 B.M.)Cl2Aq.NH3+NH4Cl−−−−−−−−−air[Co(NH3)6]Cl3(Y)(μ=0)

1:3 elctrolyte means 3 Cland one[Co(NH3)6]3+

Y is inner-orbital complex having d2sp3 hybridization and thus, all the electrons are paired in the complex.

[Co(H2O)6(X)]2++HCl(excess)0C[CoCl4]2tetrahedral(blue)(μ=3.87 B.M.),ΔH=+ve

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