What is the hybridisation of cation of liquid `BrF_(3)`, which undergoes self-dissociation.

Options

A. `sp^(2)`
B. `sp^(3)`
C. `sp`
D. `sp^(3)d`

Explanation

To determine the hybridization of the cation of liquid BrF3 that undergoes self-dissociation, we can follow these steps:

### Step 1: Understand the Self-Dissociation of BrF3
BrF3 (bromine trifluoride) can undergo self-dissociation to form cations and anions. The dissociation can be represented as:
\[ \text{BrF}_3 \rightleftharpoons \text{BrF}_2^+ + \text{BrF}_4^- \]

### Step 2: Identify the Cation
From the dissociation, we see that the cation formed is \(\text{BrF}_2^+\).

### Step 3: Determine the Valence Electrons of Bromine
Bromine (Br) has 7 valence electrons.

### Step 4: Count the Surrounding Atoms
In \(\text{BrF}_2^+\), there are 2 fluorine atoms surrounding the bromine atom.

### Step 5: Calculate the Effective Electrons
Using the formula to determine hybridization:
\[ \text{Hybridization} = \frac{(V + S - C)}{2} \]
Where:
- \(V\) = number of valence electrons of the central atom (Br = 7)
- \(S\) = number of surrounding atoms (2 F atoms)
- \(C\) = charge on the cation (since it's a cation, we subtract 1)

Substituting the values:
\[ \text{Hybridization} = \frac{(7 + 2 - 1)}{2} = \frac{8}{2} = 4 \]

### Step 6: Determine the Hybridization Type
A hybridization value of 4 corresponds to \(sp^3\) hybridization.

### Conclusion
Thus, the hybridization of the cation \(\text{BrF}_2^+\) is \(sp^3\).

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