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 silver 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 problem step by step, we need to analyze the information given in the question and the video transcript.

### Step 1: Identify the Complex X
The pink colored aqueous solution is given as \( MCl_2 \cdot 6H_2O \) (denoted as X). The color pink indicates that the metal ion is likely cobalt (Co), which is known for forming pink complexes in certain oxidation states.

### Step 2: Determine the Oxidation State of M in X
Since the complex is \( MCl_2 \cdot 6H_2O \), we can assume:
- Let \( M \) be cobalt (Co).
- The oxidation state of Co in this complex can be determined as follows:
\[ x + 2(-1) + 6(0) = 0 \implies x - 2 = 0 \implies x = +2 \]
Thus, \( M \) is in the +2 oxidation state.

### Step 3: Formation of Complex Y
When excess aqueous ammonia is added to X in the presence of air, an octahedral complex Y is formed. The ammonia is a strong field ligand, which leads to pairing of electrons. The reaction can be represented as:
\[ \text{Co}^{2+} + 6 \text{NH}_3 \rightarrow \text{[Co(NH}_3\text{)}_6]^{2+} \]
This complex Y has zero unpaired electrons, leading to a zero magnetic moment.

### Step 4: Behavior of Complex Y
The complex Y behaves as a 1:3 electrolyte in aqueous solution, meaning it dissociates into one cation and three anions:
\[ \text{[Co(NH}_3\text{)}_6]^{3+} + 3 \text{Cl}^- \rightarrow \text{[Co(NH}_3\text{)}_6]^{3+} + 3 \text{Cl}^- \]

### Step 5: Formation of Complex Z
The reaction of X with excess HCl at room temperature results in the formation of a blue colored complex Z. This can be represented as:
\[ \text{Co}^{2+} + 4 \text{Cl}^- \rightarrow \text{[CoCl}_4]^{2-} \]
This complex Z is tetrahedral and has unpaired electrons, leading to a spin-only magnetic moment of 3.87 B.M.

### Step 6: Analyze the Magnetic Moments
- For complex X and Z, the spin-only magnetic moment is calculated as:
\[ \mu = \sqrt{n(n+2)} \text{ B.M.} \]
Given that the magnetic moment is 3.87 B.M., we can solve for \( n \):
\[ 3.87 = \sqrt{n(n+2)} \implies n = 3 \text{ (indicating 3 unpaired electrons)} \]
- For complex Y, since it has zero unpaired electrons, the magnetic moment is 0 B.M.

### Step 7: Evaluate the Options
Now we need to evaluate the statements provided in the options to find the incorrect one:
1. Hybridization of the central metal ion in Y is \( d^2sp^3 \) - This is correct.
2. Addition of silver nitrate to Y gives only two equivalents of silver chloride - This is incorrect; it should give three equivalents.
3. When X and Z are in equilibrium at 0°C, the color of the solution is pink - This is correct.
4. Z is a tetrahedral complex - This is correct.

### Conclusion
The incorrect statement is option 2.