JEE Advanced · Chemistry · 23. Coordination Compounds

Addition of excess aqueous ammonia to a pink coloured aqueous solution of $M C l_{2} . 6 H_{2} O (X)$ and $N H_{4} C l$ 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 When X and Z are in equilibrium at 0°C, the colour of the solution is pink
B The hybridization of the central metal ion in Y is $d^{2} s p^{3}$
C Z is atetrahedral complex
D Addition of silver nitrate to Y gives only two equivalents of silver chloride

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Cobalt salts are pink in anhydrous & blue in hydrated form

$\underset{( X )}{[ Co ( H_{2} O)_{5}]^{2+}}+\underset{\text { (excess) }}{4 Cl ^{-}} \longrightarrow \underset{\text { (Z) blue colour }}{\left[ CoCl _4\right]^{2-}}$ (i) Hybridisation of (Y) is

d^{2} s p^{3}

N H_{3}

is a strong field ligand

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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