The homologous series and IUPAC name of `C_2 H_4 ` is

Options

A. alkyne and ethane
B. alkyne and ethyne
C. alkene and ethene
D. alkene and ethyne

Explanation

To determine the homologous series and IUPAC name of the compound \( C_2H_4 \), we can follow these steps:

### Step 1: Identify the number of carbon and hydrogen atoms
The molecular formula \( C_2H_4 \) indicates that there are 2 carbon atoms and 4 hydrogen atoms.

### Step 2: Determine the type of hydrocarbon
To classify the compound, we can use the general formulas for different types of hydrocarbons:
- Alkanes: \( C_nH_{2n+2} \)
- Alkenes: \( C_nH_{2n} \)
- Alkynes: \( C_nH_{2n-2} \)

For \( C_2H_4 \):
- If it were an alkane, it would have \( C_2H_{2(2)+2} = C_2H_6 \).
- If it were an alkene, it would have \( C_2H_{2(2)} = C_2H_4 \).
- If it were an alkyne, it would have \( C_2H_{2(2)-2} = C_2H_2 \).

Since \( C_2H_4 \) matches the formula for alkenes, we conclude that it is an **alkene**.

### Step 3: Draw the structure of the compound
We can draw the structure of \( C_2H_4 \):
```
H H
\ /
C = C
/ \
H H
```
This structure shows that there is a double bond between the two carbon atoms.

### Step 4: Assign the IUPAC name
To name the compound:
- The longest carbon chain has 2 carbon atoms, which corresponds to the prefix "eth".
- Since there is a double bond between the carbon atoms, we use the suffix "ene".
- The double bond starts at carbon 1, but since it is the first position, we can simply name it "ethene".

### Conclusion
The homologous series of \( C_2H_4 \) is **alkenes**, and its IUPAC name is **ethene**.

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