In a metal oxide, oxide ions crystallises in CCP lattice in which metal M occupies 50% of octahedral voids and metal `M_2` occupies 12.5% of tetrahedral voids. then the oxidation state of metal `M_1 and M_2` respectively are:

Options

A. 2,4
B. 1,3
C. 3,2
D. 3,1

Answer

To determine the oxidation states of metals \( M_1 \) and \( M_2 \) in the given metal oxide with oxide ions crystallizing in a CCP lattice, we can follow these steps:

### Step 1: Determine the number of oxide ions in the unit cell
In a CCP (cubic close-packed) lattice, oxide ions are located at the corners and face centers of the unit cell.

- **Contribution from corner atoms:** There are 8 corners in a cube, and each corner contributes \( \frac{1}{8} \) of an ion. Therefore, the total contribution from corners is:
\[ 8 \times \frac{1}{8} = 1 \text{ oxide ion} \]

- **Contribution from face-centered atoms:** There are 6 faces in a cube, and each face center contributes \( \frac{1}{2} \) of an ion. Therefore, the total contribution from face centers is:
\[ 6 \times \frac{1}{2} = 3 \text{ oxide ions} \]

Thus, the total number of oxide ions in the unit cell is:
\[ 1 + 3 = 4 \text{ oxide ions} \]

### Step 2: Determine the number of voids in the unit cell
In a CCP structure:
- The number of octahedral voids is equal to the number of atoms, which is 4.
- The number of tetrahedral voids is double the number of atoms, which is 8.

### Step 3: Calculate the number of \( M_1 \) and \( M_2 \) ions
- **Metal \( M_1 \)** occupies 50% of the octahedral voids:
\[ \text{Number of } M_1 = 50\% \text{ of } 4 = 2 \]

- **Metal \( M_2 \)** occupies 12.5% of the tetrahedral voids:
\[ \text{Number of } M_2 = 12.5\% \text{ of } 8 = 1 \]

### Step 4: Write the empirical formula
From the above calculations, we have:
- \( M_1 = 2 \)
- \( M_2 = 1 \)
- \( O = 4 \)

Thus, the empirical formula of the compound can be written as:
\[ M_1^2 M_2^1 O_4 \]

### Step 5: Calculate the total oxidation state
The total charge contributed by the oxide ions is:
\[ 4 \text{ oxide ions} \times (-2) = -8 \]

Let the oxidation states of \( M_1 \) and \( M_2 \) be \( x \) and \( y \) respectively. The charge balance for the neutral compound can be expressed as:
\[ 2x + 1y + (-8) = 0 \]
This simplifies to:
\[ 2x + y = 8 \quad \text{(Equation 1)} \]

### Step 6: Analyze the coordination numbers
- \( M_1 \) is in an octahedral void with a coordination number of 6, suggesting a higher oxidation state.
- \( M_2 \) is in a tetrahedral void with a coordination number of 4, suggesting a lower oxidation state.

### Step 7: Solve for oxidation states
From the options given, we can test the possible values for \( x \) and \( y \).

1. If \( x = 3 \) and \( y = 2 \):
\[ 2(3) + 2 = 6 + 2 = 8 \quad \text{(satisfies Equation 1)} \]

2. If \( x = 2 \) and \( y = 4 \):
\[ 2(2) + 4 = 4 + 4 = 8 \quad \text{(satisfies Equation 1)} \]

3. If \( x = 1 \) and \( y = 3 \):
\[ 2(1) + 3 = 2 + 3 = 5 \quad \text{(does not satisfy)} \]

4. If \( x = 4 \) and \( y = 1 \):
\[ 2(4) + 1 = 8 + 1 = 9 \quad \text{(does not satisfy)} \]

Thus, the valid oxidation states are:
- \( M_1 = +3 \)
- \( M_2 = +2 \)

### Final Answer
The oxidation states of metals \( M_1 \) and \( M_2 \) are \( +3 \) and \( +2 \) respectively.

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