What is the oxidation number and valency of carbon in methanal `(HCHO)`?

Explanation

To determine the oxidation number and valency of carbon in methanal (HCHO), we can follow these steps:

### Step 1: Understand the structure of methanal
Methanal has the molecular formula HCHO. It consists of one carbon atom (C), one oxygen atom (O), and two hydrogen atoms (H). The structure can be represented as:
```
H
|
H—C=O
```

### Step 2: Assign oxidation numbers
To find the oxidation number of carbon in methanal, we assign oxidation numbers to each atom based on common rules:
- The oxidation number of hydrogen (H) is +1.
- The oxidation number of oxygen (O) is -2.

Let the oxidation number of carbon be represented as \( x \).

### Step 3: Set up the equation
Since methanal is a neutral molecule, the sum of the oxidation numbers must equal zero. We can write the equation as follows:
\[ (+1) + x + (+1) + (-2) = 0 \]

### Step 4: Simplify the equation
Now, we simplify the equation:
\[ 1 + x + 1 - 2 = 0 \]
\[ x = 0 \]

### Step 5: Conclusion for oxidation number
Thus, the oxidation number of carbon in methanal is **0**.

### Step 6: Determine the valency of carbon
Valency is defined as the number of electrons in the outermost shell that an atom can lose, gain, or share to form chemical bonds.

For carbon:
- The atomic number of carbon is 6, which means it has 4 electrons in its outermost shell (2s² 2p²).
- Therefore, the valency of carbon is **4** because it can form four bonds by either sharing or losing its four outer electrons.

### Final Answer
- **Oxidation number of carbon in methanal (HCHO): 0**
- **Valency of carbon in methanal (HCHO): 4**