Hybridisation in `CIF_(3)` is

Options

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

Answer

To determine the hybridization of chlorine in CIF₃ (Chlorine Trifluoride), we can follow these steps:

### Step 1: Identify the central atom
The central atom in CIF₃ is chlorine (Cl).

### Step 2: Determine the number of valence electrons
Chlorine has 7 valence electrons (as it is in group 17 of the periodic table). Each fluorine atom (F) has 7 valence electrons as well.

### Step 3: Count the number of atoms bonded to the central atom
In CIF₃, chlorine is bonded to three fluorine atoms. Each bond contributes one electron to the chlorine.

### Step 4: Calculate the total number of bonding electrons
Since there are three Cl-F bonds, we have:
- 3 bonds × 1 electron (from Cl) = 3 bonding electrons.

### Step 5: Calculate the number of lone pairs on the central atom
Chlorine starts with 7 valence electrons. After using 3 for bonding with fluorine, the remaining electrons are:
- 7 (total valence electrons) - 3 (bonding electrons) = 4 electrons.

These remaining 4 electrons will form 2 lone pairs.

### Step 6: Determine the hybridization index
The hybridization index is calculated as follows:
- Number of sigma bonds + Number of lone pairs = 3 (from Cl-F bonds) + 2 (lone pairs) = 5.

### Step 7: Identify the hybridization type
A hybridization index of 5 corresponds to sp³d hybridization.

### Conclusion
Thus, the hybridization of chlorine in CIF₃ is **sp³d**.