The magnetic moment for two complexes of empirical formula `Ni(NH_(3))_(4)(NO_(3))_(2).2H_(2)O` is zero and 2.84 BM respectively. The second complex is not a netural complex.
The number of water molecules of crystallization are respectively

Options

A. zero, two
B. zero, zero
C. two, zero
D. two ,two

Answer

To solve the problem regarding the magnetic moments of the two complexes of empirical formula `Ni(NH₃)₄(NO₃)₂·2H₂O`, we will analyze the information given step by step.

### Step 1: Analyze the first complex
The first complex has a magnetic moment (μ) of 0 BM.

- **Magnetic moment formula**:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

- Since μ = 0, we can conclude that:
\[ n(n + 2) = 0 \implies n = 0 \]
This means that there are no unpaired electrons in the first complex.

### Step 2: Determine the oxidation state of Nickel in the first complex
In the first complex, we have:
- Ligands: 4 NH₃ (neutral) and 2 NO₃⁻ (each with a charge of -1).
- Let the oxidation state of Nickel (Ni) be \( x \).

The overall charge of the complex is neutral (0):
\[ x + 0 + (-2) = 0 \implies x = +2 \]
Thus, Nickel is in the +2 oxidation state.

### Step 3: Analyze the second complex
The second complex has a magnetic moment (μ) of 2.84 BM.

- Using the magnetic moment formula again:
\[ 2.84 = \sqrt{n(n + 2)} \]
Squaring both sides:
\[ 2.84^2 = n(n + 2) \implies 8.0656 = n(n + 2) \]
This can be approximated to:
\[ n(n + 2) \approx 8 \implies n^2 + 2n - 8 = 0 \]
Solving this quadratic equation using the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \]
This gives us:
\[ n = 2 \quad (\text{since } n \text{ must be non-negative}) \]
Thus, there are 2 unpaired electrons in the second complex.

### Step 4: Determine the oxidation state of Nickel in the second complex
The second complex is not neutral, which means it has a net charge.

Assuming the same ligands:
- 4 NH₃ (neutral) and 2 NO₃⁻ (each with a charge of -1).
- The oxidation state of Nickel is still \( x \).

Let’s denote the charge of the second complex as \( +1 \) (since it is not neutral):
\[ x + 0 + (-2) = +1 \implies x - 2 = 1 \implies x = +3 \]
Thus, Nickel is in the +3 oxidation state in the second complex.

### Step 5: Identify the number of water molecules of crystallization
- In the first complex, the 2 water molecules (2H₂O) are outside the coordination sphere and are considered as water of crystallization.
- In the second complex, the 2 water molecules are not mentioned to be outside the coordination sphere, which means they are likely part of the coordination sphere and thus not counted as water of crystallization.

### Conclusion
The number of water molecules of crystallization for the two complexes are:
- First complex: 2
- Second complex: 0

### Final Answer
The number of water molecules of crystallization are respectively **2 and 0**.

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