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 solve the problem, we need to determine the oxidation states of metals M1 and M2 in the given metal oxide, where oxide ions crystallize in a cubic close-packed (CCP) lattice.

### Step-by-Step Solution:

1. **Identify the number of oxide ions (O²⁻)**:
- In a CCP lattice (also known as face-centered cubic or FCC), the number of atoms (or ions) per unit cell is given by:
\[ Z = 4 \]
- Therefore, there are 4 oxide ions in the unit cell.

2. **Determine the number of octahedral voids**:
- The number of octahedral voids in a CCP lattice is equal to the number of atoms in the lattice:
\[ \text{Number of octahedral voids} = Z = 4 \]

3. **Calculate the number of M1 atoms**:
- It is given that metal M1 occupies 50% of the octahedral voids:
\[ \text{Number of M1 atoms} = 50\% \text{ of } 4 = 2 \]

4. **Determine the number of tetrahedral voids**:
- The number of tetrahedral voids in a CCP lattice is twice the number of atoms in the lattice:
\[ \text{Number of tetrahedral voids} = 2Z = 2 \times 4 = 8 \]

5. **Calculate the number of M2 atoms**:
- It is given that metal M2 occupies 12.5% of the tetrahedral voids:
\[ \text{Number of M2 atoms} = 12.5\% \text{ of } 8 = 1 \]

6. **Write the formula of the metal oxide**:
- From the calculations, we have:
- M1: 2 atoms
- M2: 1 atom
- O: 4 atoms
- The formula of the metal oxide can be written as:
\[ \text{M}_1^2\text{M}_2^1\text{O}_4 \]

7. **Set up the oxidation state equation**:
- Let the oxidation state of M1 be \( +x \) and that of M2 be \( +y \).
- The sum of the oxidation states must equal the total negative charge from the oxide ions:
\[ 2x + y + 4(-2) = 0 \]
- Simplifying this gives:
\[ 2x + y - 8 = 0 \quad \Rightarrow \quad 2x + y = 8 \]

8. **Test the options for oxidation states**:
- **Option 1**: \( x = 2, y = 4 \)
\[ 2(2) + 4 = 4 + 4 = 8 \quad \text{(Valid)} \]
- **Option 2**: \( x = 1, y = 3 \)
\[ 2(1) + 3 = 2 + 3 = 5 \quad \text{(Invalid)} \]
- **Option 3**: \( x = 3, y = 2 \)
\[ 2(3) + 2 = 6 + 2 = 8 \quad \text{(Valid)} \]
- **Option 4**: \( x = 3, y = 1 \)
\[ 2(3) + 1 = 6 + 1 = 7 \quad \text{(Invalid)} \]

### Conclusion:
The valid oxidation states are:
- For M1: \( +2 \) and for M2: \( +4 \) (Option 1)
- For M1: \( +3 \) and for M2: \( +2 \) (Option 3)

Thus, the oxidation states of metals M1 and M2 are \( +2 \) and \( +4 \) or \( +3 \) and \( +2 \) respectively.