The valancy of carbons is generally `4`, but its oxidation state may be `-4, -2, 0, +2, -1`, etc. In the compounds containing `C, H`, and `O`, the oxidation number of `C` is calculated as
Oxidation number of `C= (2n_(O)-n_(H))/(n_(C ))`
Where `n_(O), n_(H)` and `n_(C )` are the numbers of oxygen, hydrogen, and carbons, atoms, respectively.
In which of the following compounds is the oxidation state of carbon is zero?

Options

A. `CH_(4)`
B. `CH_(3)OH`
C. `HCOOH`
D. `C_(6)H_(12)O_(6)`

Explanation

To determine the oxidation state of carbon in the given compounds, we will use the formula provided:

\[ \text{Oxidation number of C} = \frac{(2n_O - n_H)}{n_C} \]

where:
- \( n_O \) = number of oxygen atoms
- \( n_H \) = number of hydrogen atoms
- \( n_C \) = number of carbon atoms

We will evaluate the oxidation state of carbon for each compound step by step.

### Step 1: Analyze each compound

**A. CH₄ (Methane)**
- \( n_O = 0 \) (no oxygen atoms)
- \( n_H = 4 \) (four hydrogen atoms)
- \( n_C = 1 \) (one carbon atom)

Using the formula:
\[ \text{Oxidation number of C} = \frac{(2 \times 0 - 4)}{1} = \frac{-4}{1} = -4 \]

**B. CH₃OH (Methanol)**
- \( n_O = 1 \) (one oxygen atom)
- \( n_H = 4 \) (four hydrogen atoms)
- \( n_C = 1 \) (one carbon atom)

Using the formula:
\[ \text{Oxidation number of C} = \frac{(2 \times 1 - 4)}{1} = \frac{(2 - 4)}{1} = \frac{-2}{1} = -2 \]

**C. HCOOH (Formic Acid)**
- \( n_O = 2 \) (two oxygen atoms)
- \( n_H = 2 \) (two hydrogen atoms)
- \( n_C = 1 \) (one carbon atom)

Using the formula:
\[ \text{Oxidation number of C} = \frac{(2 \times 2 - 2)}{1} = \frac{(4 - 2)}{1} = \frac{2}{1} = +2 \]

**D. C₆H₁₂O₆ (Glucose)**
- \( n_O = 6 \) (six oxygen atoms)
- \( n_H = 12 \) (twelve hydrogen atoms)
- \( n_C = 6 \) (six carbon atoms)

Using the formula:
\[ \text{Oxidation number of C} = \frac{(2 \times 6 - 12)}{6} = \frac{(12 - 12)}{6} = \frac{0}{6} = 0 \]

### Conclusion
The oxidation state of carbon is zero in **C₆H₁₂O₆ (Glucose)**.