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 valency of `C` two?

Options

A. Ketenes
B. Alkenes
C. Allenes
D. Carbenes

Explanation

To determine in which of the given compounds the valency of carbon is 2, we will use the formula for calculating the oxidation number of carbon in compounds containing carbon (C), hydrogen (H), and oxygen (O):

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

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

We will analyze each compound provided in the question.

### Step 1: Analyze Ketenes (C2H2O)

1. **Identify the number of atoms**:
- \(n_O = 1\) (1 oxygen)
- \(n_H = 2\) (2 hydrogens)
- \(n_C = 2\) (2 carbons)

2. **Calculate the oxidation number of carbon**:
\[ \text{Oxidation number of } C = \frac{(2 \times 1 - 2)}{2} = \frac{(2 - 2)}{2} = \frac{0}{2} = 0 \]
- The oxidation number is 0, so the valency of carbon is not 2.

### Step 2: Analyze Alkenes (C2H4)

1. **Identify the number of atoms**:
- \(n_O = 0\) (0 oxygen)
- \(n_H = 4\) (4 hydrogens)
- \(n_C = 2\) (2 carbons)

2. **Calculate the oxidation number of carbon**:
\[ \text{Oxidation number of } C = \frac{(2 \times 0 - 4)}{2} = \frac{(0 - 4)}{2} = \frac{-4}{2} = -2 \]
- The oxidation number is -2, so the valency of carbon is not 2.

### Step 3: Analyze Allenes (C3H4)

1. **Identify the number of atoms**:
- \(n_O = 0\) (0 oxygen)
- \(n_H = 4\) (4 hydrogens)
- \(n_C = 3\) (3 carbons)

2. **Calculate the oxidation number of carbon**:
\[ \text{Oxidation number of } C = \frac{(2 \times 0 - 4)}{3} = \frac{(0 - 4)}{3} = \frac{-4}{3} \]
- The oxidation number is \(-\frac{4}{3}\), so the valency of carbon is not 2.

### Step 4: Analyze Carbenes (CH2)

1. **Identify the number of atoms**:
- \(n_O = 0\) (0 oxygen)
- \(n_H = 2\) (2 hydrogens)
- \(n_C = 1\) (1 carbon)

2. **Calculate the oxidation number of carbon**:
\[ \text{Oxidation number of } C = \frac{(2 \times 0 - 2)}{1} = \frac{(0 - 2)}{1} = -2 \]
- The oxidation number is -2, indicating that the carbon here has a valency of 2.

### Conclusion

From the analysis, the compound in which the valency of carbon is 2 is **Carbene (CH2)**.

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