Total acyclic optically active isomers of `C_(3)H_(2)D_(2)` are `:`

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To determine the total acyclic optically active isomers of the compound \( C_3H_2D_2 \), we will follow these steps:

### Step 1: Identify the possible structures
The molecular formula \( C_3H_2D_2 \) indicates that we have three carbon atoms, two hydrogen atoms, and two deuterium atoms. Since we are looking for acyclic (open-chain) structures, we can consider the following arrangements:

1. **Structure 1**: A straight-chain with a double bond:
- \( H_2C=CH-CH_2D \)
- This structure can also be represented as \( H_2C=CH-CHD_2 \).

2. **Structure 2**: Another arrangement with a double bond:
- \( H_2C=CHD-CH_2D \)
- This structure can also be represented as \( H_2C=CHD_2-CHD \).

### Step 2: Analyze the structures for optical activity
Next, we need to analyze these structures to determine if they are optically active or inactive.

- **Structure 1**: \( H_2C=CH-CH_2D \)
- This structure has a plane of symmetry, making it optically inactive.

- **Structure 2**: \( H_2C=CHD-CH_2D \)
- This structure does not have a plane of symmetry, making it optically active.

### Step 3: Count the optically active isomers
From our analysis:
- We found **one** optically active structure (Structure 2).
- The other structure (Structure 1) is optically inactive.

### Conclusion
Therefore, the total number of acyclic optically active isomers of \( C_3H_2D_2 \) is **1**.

### Final Answer
The total acyclic optically active isomers of \( C_3H_2D_2 \) are: **1**.