The oxidation number of carbon in Freon, and hydroge cyanide is +x and +y. What is the value x+y.

Answer

To find the values of \( x \) and \( y \), which represent the oxidation numbers of carbon in Freon and hydrogen cyanide respectively, we will analyze each compound step by step.

### Step 1: Determine the oxidation number of carbon in Freon
1. **Identify the formula of Freon**: The common formula for Freon is \( \text{CCl}_2\text{F}_2 \).
2. **Assign oxidation numbers to other elements**:
- Chlorine (Cl) typically has an oxidation number of \(-1\).
- Fluorine (F) has an oxidation number of \(-1\).
3. **Set up the equation**:
- Let the oxidation number of carbon be \( x \).
- The overall charge of the molecule is neutral, so we can write the equation:
\[ x + 2(-1) + 2(-1) = 0 \]
- Simplifying this gives:
\[ x - 2 - 2 = 0 \implies x - 4 = 0 \implies x = +4 \]

### Step 2: Determine the oxidation number of carbon in hydrogen cyanide
1. **Identify the formula of hydrogen cyanide**: The formula is \( \text{HCN} \).
2. **Assign oxidation numbers to other elements**:
- Hydrogen (H) has an oxidation number of \( +1 \).
- Nitrogen (N) typically has an oxidation number of \(-3\).
3. **Set up the equation**:
- Let the oxidation number of carbon be \( y \).
- The overall charge of the molecule is neutral, so we can write the equation:
\[ y + 1 + (-3) = 0 \]
- Simplifying this gives:
\[ y + 1 - 3 = 0 \implies y - 2 = 0 \implies y = +2 \]

### Step 3: Calculate \( x + y \)
1. **Add the values of \( x \) and \( y \)**:
\[ x + y = 4 + 2 = 6 \]

### Final Answer
The value of \( x + y \) is \( 6 \).

---