The number of isomers for the compound with molecular formula `C_(2)BrClFI` is:

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To determine the number of isomers for the compound with the molecular formula C₂BrClFI, we can follow these steps:

### Step 1: Identify the Structure
The compound C₂BrClFI consists of two carbon atoms (C) and four different halogen atoms (Br, Cl, F, I). The presence of different groups attached to the carbon atoms suggests that we can have geometric (cis-trans) isomerism as well as structural isomerism.

### Step 2: Draw Possible Structures
We can start by drawing the possible structural arrangements of the compound. The two carbon atoms can be connected by a double bond (C=C) or a single bond (C-C).

1. **C=C Structure**:
- C=C with different halogens on each carbon.
- Example: C=C with Br and Cl on one carbon and F and I on the other.

2. **C-C Structure**:
- C-C with different halogens attached to each carbon.
- Example: One carbon has Br and Cl, and the other has F and I.

### Step 3: Consider Geometric Isomerism
For the C=C structure, we can have geometric isomers (E/Z configurations) based on the priority of the substituents:
- Assign priority based on atomic number (higher atomic number gets higher priority).
- For example, in the case of C=C, if we have Br and Cl on one carbon and F and I on the other, we can determine the E/Z configurations.

### Step 4: Count Isomers
1. **For C=C**:
- Z configuration (both higher priority groups on the same side).
- E configuration (higher priority groups on opposite sides).
- Each arrangement can yield two configurations (E and Z).

2. **For C-C**:
- Similar analysis can be done for the C-C structure, leading to additional isomers.

### Step 5: Total Isomers
After considering both the C=C and C-C structures, we can count the total number of unique isomers:
- From the C=C structure, we can derive 4 isomers (2 E/Z configurations).
- From the C-C structure, we can derive additional isomers.

### Conclusion
After analyzing all possible configurations and isomers, we find that the total number of isomers for the compound C₂BrClFI is **6**.

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