The correct order of increasing oxidising power in the series is

Options

A. `VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)`
B. `Cr_(2)O_(7)^(2-)lt VO_(2)^(+) lt MnO_(4)^(-)`
C. `Cr_(2)O_(7)^(2-) ltMnO_(4)^(-) lt VO_(2)^(+)`
D. `MnO_(4)^(-) lt Cr_(2)O_(7)^(2-) ltVO_(2)^(+)`

Answer

To determine the correct order of increasing oxidizing power among the species \( VO_2^+ \), \( Cr_2O_7^{2-} \), and \( MnO_4^- \), we need to analyze the oxidation states of the central metal atoms in each species.

### Step-by-Step Solution:

1. **Identify the oxidation state of Vanadium in \( VO_2^+ \)**:
- The oxidation state of oxygen is \(-2\).
- Let the oxidation state of Vanadium be \( x \).
- The equation can be set up as:
\[ x + 2(-2) = +1 \]
- Simplifying gives:
\[ x - 4 = 1 \implies x = +5 \]

2. **Identify the oxidation state of Chromium in \( Cr_2O_7^{2-} \)**:
- The oxidation state of oxygen is \(-2\).
- Let the oxidation state of Chromium be \( y \).
- The equation can be set up as:
\[ 2y + 7(-2) = -2 \]
- Simplifying gives:
\[ 2y - 14 = -2 \implies 2y = 12 \implies y = +6 \]

3. **Identify the oxidation state of Manganese in \( MnO_4^- \)**:
- The oxidation state of oxygen is \(-2\).
- Let the oxidation state of Manganese be \( z \).
- The equation can be set up as:
\[ z + 4(-2) = -1 \]
- Simplifying gives:
\[ z - 8 = -1 \implies z = +7 \]

4. **Compare the oxidation states**:
- The oxidation states obtained are:
- \( VO_2^+ \): +5
- \( Cr_2O_7^{2-} \): +6
- \( MnO_4^- \): +7
- The higher the oxidation state, the stronger the oxidizing agent. Therefore, the order of increasing oxidizing power is:
\[ VO_2^+ < Cr_2O_7^{2-} < MnO_4^- \]

5. **Conclusion**:
- The correct order of increasing oxidizing power is:
\[ VO_2^+ < Cr_2O_7^{2-} < MnO_4^- \]

### Final Answer:
The correct order of increasing oxidizing power is:
\[ VO_2^+ < Cr_2O_7^{2-} < MnO_4^- \]