A transition metal M can exist in two oxidation states +2 and +3 It forms an oxide whose experiments formula is given by `M_xO` where `xlt1` Then the ratio of metal ions in +3 state to those in +2 states in oxide given by :

Options

A. `(1-x)/(1+x)`
B. 1+2x
C. `1+x/2`
D. `(2(1-x))/(3x-2)`

Explanation

To solve the problem, we need to find the ratio of metal ions in the +3 oxidation state to those in the +2 oxidation state in the oxide \( M_xO \), where \( x < 1 \).

### Step-by-Step Solution:

1. **Understanding the Oxide Formula**:
The oxide is represented as \( M_xO \). Here, \( x \) is less than 1, which indicates that there are fewer metal ions than oxygen ions.

2. **Defining Variables**:
Let:
- \( y \) = number of moles of \( M^{2+} \) ions
- \( x - y \) = number of moles of \( M^{3+} \) ions (since the total moles of metal ions is \( x \))

3. **Setting Up the Charge Balance**:
The total positive charge contributed by the metal ions must balance the negative charge from the oxide ions. Since each oxide ion \( O^{2-} \) contributes a charge of -2, we can set up the equation:
\[ 2y + 3(x - y) = 2 \]
This equation represents the total charge balance.

4. **Simplifying the Charge Balance Equation**:
Expanding the equation:
\[ 2y + 3x - 3y = 2 \]
Simplifying this gives:
\[ -y + 3x = 2 \]
Rearranging gives:
\[ y = 3x - 2 \]

5. **Finding the Ratio**:
Now, we need to find the ratio of \( M^{3+} \) to \( M^{2+} \):
\[ \text{Ratio} = \frac{M^{3+}}{M^{2+}} = \frac{x - y}{y} \]
Substituting \( y = 3x - 2 \):
\[ M^{3+} = x - (3x - 2) = -2x + 2 \]
Thus, the ratio becomes:
\[ \text{Ratio} = \frac{-2x + 2}{3x - 2} \]

6. **Final Expression**:
The final expression for the ratio of metal ions in +3 state to those in +2 state is:
\[ \text{Ratio} = \frac{2(1 - x)}{3x - 2} \]

### Conclusion:
The ratio of metal ions in the +3 oxidation state to those in the +2 oxidation state in the oxide \( M_xO \) is given by:
\[ \frac{2(1 - x)}{3x - 2} \]