`K_(4)[Fe(CN)_(6)]` is converted into `CO_(3)^(-2)`, `Fe^(+3)` ions and `NO_(3)^(-)` ions. Here_____________

Options

A. N is reduced
B. C is oxidised
C. Iorn is oxidised
D. Eq.wt `("Formula wt")/61`

Answer

To solve the problem of the conversion of \( K_4[Fe(CN)_6] \) into \( CO_3^{2-} \), \( Fe^{3+} \), and \( NO_3^{-} \), we will follow these steps:

### Step 1: Determine the oxidation states in \( K_4[Fe(CN)_6] \)

- Potassium (K) has an oxidation state of +1.
- The oxidation state of iron (Fe) in this complex is +2.
- For the cyanide ion \( CN^{-} \), let the oxidation state of carbon be \( x \) and nitrogen be -3. The equation can be set up as:
\[ x + 6(-3) = -1 \implies x - 18 = -1 \implies x = +17 \]
However, since \( CN^{-} \) is a complex ion, we need to consider it as a whole. The oxidation state of carbon in \( CN^{-} \) is +2 (as it is bonded to a nitrogen with a -3 charge).

### Step 2: Identify the products and their oxidation states

- In the products, \( CO_3^{2-} \):
- The oxidation state of carbon is +4.
- For \( Fe^{3+} \):
- The oxidation state of iron is +3.
- For \( NO_3^{-} \):
- The oxidation state of nitrogen is +5.

### Step 3: Determine the changes in oxidation states

- Carbon changes from +2 in \( [Fe(CN)_6]^{4-} \) to +4 in \( CO_3^{2-} \). This is an oxidation.
- Iron changes from +2 to +3. This is also an oxidation.
- Nitrogen changes from +3 in \( CN^{-} \) to +5 in \( NO_3^{-} \). This is an oxidation.

### Step 4: Calculate the n-factor for each element

- For carbon:
- Change from +2 to +4: \( n = 4 - 2 = 2 \).
- For iron:
- Change from +2 to +3: \( n = 3 - 2 = 1 \).
- For nitrogen:
- Change from +3 to +5: \( n = 5 - 3 = 2 \).

### Step 5: Total n-factor calculation

- If there are 6 cyanide ions, the total n-factor for carbon is:
\[ 6 \times 2 = 12 \]
- The total n-factor for nitrogen is:
\[ 6 \times 2 = 12 \]
- The total n-factor for iron is:
\[ 1 \text{ (for one Fe)} \]
- Therefore, the overall n-factor is:
\[ 12 + 12 + 1 = 25 \]

### Step 6: Calculate the equivalent weight

- The molecular weight of \( K_4[Fe(CN)_6] \) is calculated based on its constituent elements. However, we can use the n-factor to find the equivalent weight:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} \]

### Conclusion

Thus, the conversion of \( K_4[Fe(CN)_6] \) into \( CO_3^{2-} \), \( Fe^{3+} \), and \( NO_3^{-} \) involves oxidation of carbon, iron, and nitrogen.