Addition of excess aqueous ammonia to a pink coloured aqueous solution of `MCl_(2). 6H_(2)O(X) and NH_(4)Cl` 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 is incorrect ?

Options

A. The hybridization of the central metal ion in Y is `d^(2)sp^(3)`
B. Addition of solver nitrate to Y gives only two equivalents of silver chloride
C. When X and Z are in equilibrium at `0^(@)C`, the colour of the solution is pink
D. Z is a tetrahedral complex

Answer

To solve the given question, we need to analyze the information provided step by step and identify the incorrect statement among the options.

### Step 1: Identify the nature of compound X
- The compound X is given as MCl₂·6H₂O, which is pink in color.
- The metal ion (M) is in the +2 oxidation state, as indicated by the presence of two chloride ions (Cl⁻) and six water molecules (H₂O) which are neutral.

### Step 2: Reaction of X with excess aqueous ammonia
- When excess aqueous ammonia (NH₃) is added to the pink solution of X, it forms an octahedral complex Y in the presence of air.
- The complex formed is [M(NH₃)₆]Cl₂, which behaves as a 1:3 electrolyte in aqueous solution. This means it dissociates into one cation and three anions.

### Step 3: Magnetic properties of the complexes
- The spin-only magnetic moment of X and Z is given as 3.87 B.M. This indicates that both X and Z have unpaired electrons, suggesting they are likely to be high-spin complexes.
- The complex Y has a spin-only magnetic moment of zero, indicating that it has no unpaired electrons, which is typical for low-spin complexes.

### Step 4: Reaction of X with excess HCl
- The reaction of X with excess HCl results in the formation of a blue-colored complex Z, which is likely [MCl₄]²⁻.
- This complex is tetrahedral in geometry, and since Cl⁻ is a weak field ligand, it does not cause pairing of electrons.

### Step 5: Analyze the statements
- **Statement 1**: The hybridization of the central metal ion in Y is d²sp³. This is correct for an octahedral complex.
- **Statement 2**: Addition of silver nitrate to Y gives only two equivalents of silver chloride. This is incorrect because it should give three equivalents of silver chloride (AgCl) upon reaction.
- **Statement 3**: When X and Z are in equilibrium at 0°C, the color of the solution is pink. This is correct as X is pink and Z is blue, and equilibrium favors the pink color.
- **Statement 4**: Z is a tetrahedral complex. This is correct as Z is formed from MCl₄²⁻.

### Conclusion
The incorrect statement among the options is **Statement 2**.