What is the oxidation number of the underlined atoms in the following ?
`Kul(Mn)O_4,Na_2ul(Cr)_2O_7, ul(Fe)_3O_4,Kul(Cl)O_3,ulPO_4^(3-)ul(C)Cl_4`

Answer

To determine the oxidation numbers of the underlined atoms in the given compounds, we will follow a systematic approach for each compound.

### Step-by-Step Solution:

1. **For `KMnO4`:**
- Let the oxidation number of Mn be \( x \).
- Potassium (K) is in Group 1, so its oxidation number is +1.
- Oxygen (O) has an oxidation number of -2.
- The compound is neutral, so the sum of oxidation numbers must equal 0.
- The equation is:
\[ +1 + x + 4(-2) = 0 \]
\[ +1 + x - 8 = 0 \]
\[ x - 7 = 0 \implies x = +7 \]
- **Oxidation number of Mn = +7**

2. **For `Na2Cr2O7`:**
- Let the oxidation number of Cr be \( x \).
- Sodium (Na) is also in Group 1, so its oxidation number is +1.
- The equation is:
\[ 2(+1) + 2x + 7(-2) = 0 \]
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \implies 2x = 12 \implies x = +6 \]
- **Oxidation number of Cr = +6**

3. **For `Fe3O4`:**
- Let the oxidation number of Fe be \( x \).
- The equation is:
\[ 3x + 4(-2) = 0 \]
\[ 3x - 8 = 0 \]
\[ 3x = 8 \implies x = \frac{8}{3} \approx +2.67 \]
- **Average oxidation number of Fe = +2.67**

4. **For `KClO3`:**
- Let the oxidation number of Cl be \( x \).
- The equation is:
\[ +1 + x + 3(-2) = 0 \]
\[ +1 + x - 6 = 0 \]
\[ x - 5 = 0 \implies x = +5 \]
- **Oxidation number of Cl = +5**

5. **For `PO4^(3-)`:**
- Let the oxidation number of P be \( x \).
- The equation is:
\[ x + 4(-2) = -3 \]
\[ x - 8 = -3 \]
\[ x = +5 \]
- **Oxidation number of P = +5**

6. **For `CCl4`:**
- Let the oxidation number of C be \( x \).
- The equation is:
\[ x + 4(-1) = 0 \]
\[ x - 4 = 0 \]
\[ x = +4 \]
- **Oxidation number of C = +4**

### Summary of Results:
- Oxidation number of Mn in `KMnO4` = +7
- Oxidation number of Cr in `Na2Cr2O7` = +6
- Average oxidation number of Fe in `Fe3O4` = +2.67
- Oxidation number of Cl in `KClO3` = +5
- Oxidation number of P in `PO4^(3-)` = +5
- Oxidation number of C in `CCl4` = +4