Following order is observed in oxidising power of certain ions :
`VO_2^(+) lt Cr_2O_7^(2-) lt MnO_4^(-)`
The reason for this increasing order of oxidising power is

Options

A. increasing stability of the lower species to which they are reduced
B. increasing stability of the higher species to which they are oxidised
C. increasing stability of the higher species to which they are reduced
D. increasing stability of the lower species to which they are oxidised.

Answer

To understand the increasing order of oxidizing power of the ions \( VO_2^+ \), \( Cr_2O_7^{2-} \), and \( MnO_4^{-} \), we will analyze the oxidation states, electronic configurations, and the ability of these ions to gain or lose electrons.

### Step-by-Step Solution:

1. **Identify the Ions and Their Central Atoms**:
- The ions in question are:
- Vanadate ion: \( VO_2^+ \) (Vanadium)
- Chromate ion: \( Cr_2O_7^{2-} \) (Chromium)
- Manganate ion: \( MnO_4^{-} \) (Manganese)

2. **Determine the Oxidation States**:
- For \( VO_2^+ \):
- Let the oxidation state of Vanadium be \( x \).
- The equation is \( x + 2(-2) = +1 \) (since each oxygen is -2).
- Solving gives \( x = +5 \).
- For \( Cr_2O_7^{2-} \):
- Let the oxidation state of Chromium be \( x \).
- The equation is \( 2x + 7(-2) = -2 \).
- Solving gives \( x = +6 \).
- For \( MnO_4^{-} \):
- Let the oxidation state of Manganese be \( x \).
- The equation is \( x + 4(-2) = -1 \).
- Solving gives \( x = +7 \).

3. **Write the Electronic Configurations**:
- Vanadium (\( V \)): \( [Ar] 3d^3 4s^2 \)
- Chromium (\( Cr \)): \( [Ar] 3d^5 4s^1 \) (note the half-filled stability)
- Manganese (\( Mn \)): \( [Ar] 3d^5 4s^2 \)

4. **Analyze the Ability to Gain Electrons**:
- The oxidizing power of an ion is inversely related to its ability to gain electrons:
- **Vanadate ion (\( VO_2^+ \))**: Has \( 3d^3 4s^2 \) configuration; it can easily gain electrons, thus has the least oxidizing power.
- **Chromate ion (\( Cr_2O_7^{2-} \))**: Has \( 3d^5 4s^1 \); it can gain one more electron, thus has moderate oxidizing power.
- **Manganate ion (\( MnO_4^{-} \))**: Has \( 3d^5 4s^2 \); it is less likely to gain electrons (due to half-filled stability), thus has the highest oxidizing power.

5. **Conclusion**:
- The increasing order of oxidizing power is due to the increasing ability of these ions to gain electrons:
- \( VO_2^+ < Cr_2O_7^{2-} < MnO_4^{-} \)
- Therefore, the correct reasoning for this order is that the more stable the lower oxidation state is, the less oxidizing power the ion has.

### Final Answer:
The increasing order of oxidizing power is due to the increasing stability of the lower oxidation states to which these ions can be reduced. Thus, \( VO_2^+ \) has the least oxidizing power, followed by \( Cr_2O_7^{2-} \), and \( MnO_4^{-} \) has the highest oxidizing power.