Match Column I with Column II for the oxidation states of the central atoms .
`{:("Column I","Column II"),((i) Cr_2O_7^(2-),(a)+3),((ii) MnO_4^(-),(b)+4),((iii) VO_3^(-),(c)+5),((iv) FeF_6^(3-),(e)+7):}`

Answer

To solve the problem of matching the oxidation states of the central atoms in Column I with their corresponding values in Column II, we will calculate the oxidation states for each compound step by step.

### Step 1: Calculate the oxidation state of Chromium in \( Cr_2O_7^{2-} \)

1. Let the oxidation state of Chromium (Cr) be \( x \).
2. There are 2 Chromium atoms, so the contribution from Chromium is \( 2x \).
3. Oxygen has an oxidation state of -2, and there are 7 Oxygen atoms, contributing \( 7 \times (-2) = -14 \).
4. The overall charge of the dichromate ion \( Cr_2O_7^{2-} \) is -2.

Now, we can set up the equation:
\[ 2x - 14 = -2 \]
5. Solving for \( x \):
\[ 2x = 12 \implies x = 6 \]
Thus, the oxidation state of Chromium in \( Cr_2O_7^{2-} \) is +6.

### Step 2: Calculate the oxidation state of Manganese in \( MnO_4^{-} \)

1. Let the oxidation state of Manganese (Mn) be \( x \).
2. The contribution from Mn is \( x \).
3. Again, Oxygen has an oxidation state of -2, and there are 4 Oxygen atoms, contributing \( 4 \times (-2) = -8 \).
4. The overall charge of the permanganate ion \( MnO_4^{-} \) is -1.

Setting up the equation:
\[ x - 8 = -1 \]
5. Solving for \( x \):
\[ x = 7 \]
Thus, the oxidation state of Manganese in \( MnO_4^{-} \) is +7.

### Step 3: Calculate the oxidation state of Vanadium in \( VO_3^{-} \)

1. Let the oxidation state of Vanadium (V) be \( x \).
2. The contribution from V is \( x \).
3. Oxygen has an oxidation state of -2, and there are 3 Oxygen atoms, contributing \( 3 \times (-2) = -6 \).
4. The overall charge of the vanadate ion \( VO_3^{-} \) is -1.

Setting up the equation:
\[ x - 6 = -1 \]
5. Solving for \( x \):
\[ x = 5 \]
Thus, the oxidation state of Vanadium in \( VO_3^{-} \) is +5.

### Step 4: Calculate the oxidation state of Iron in \( FeF_6^{3-} \)

1. Let the oxidation state of Iron (Fe) be \( x \).
2. The contribution from Fe is \( x \).
3. Fluorine has an oxidation state of -1, and there are 6 Fluorine atoms, contributing \( 6 \times (-1) = -6 \).
4. The overall charge of the hexafluoroferrate ion \( FeF_6^{3-} \) is -3.

Setting up the equation:
\[ x - 6 = -3 \]
5. Solving for \( x \):
\[ x = 3 \]
Thus, the oxidation state of Iron in \( FeF_6^{3-} \) is +3.

### Summary of Results

- \( Cr_2O_7^{2-} \) → +6
- \( MnO_4^{-} \) → +7
- \( VO_3^{-} \) → +5
- \( FeF_6^{3-} \) → +3

### Final Matching

Now we can match the results with Column II:

- (i) \( Cr_2O_7^{2-} \) → (a) +6
- (ii) \( MnO_4^{-} \) → (e) +7
- (iii) \( VO_3^{-} \) → (c) +5
- (iv) \( FeF_6^{3-} \) → (b) +3

### Final Answer

- (i) → (a)
- (ii) → (e)
- (iii) → (c)
- (iv) → (b)