The ratio of number of lone pairs on the central atom of CIF, and `XeF_4` is……

Answer

To determine the ratio of the number of lone pairs on the central atom of ClF (Chlorine Fluoride) and XeF4 (Xenon Tetrafluoride), we can follow these steps:

### Step 1: Identify the central atom in ClF
In ClF, the central atom is chlorine (Cl). Chlorine is a halogen and has 7 valence electrons.

**Hint:** Remember that the central atom in a molecule is usually the least electronegative atom.

### Step 2: Determine the number of bonds in ClF
In ClF, chlorine forms one bond with one fluorine atom. Since each fluorine atom contributes one electron to the bond, chlorine uses one of its 7 valence electrons for bonding.

**Hint:** Count the number of bonds formed by the central atom to find out how many electrons are used for bonding.

### Step 3: Calculate the lone pairs on Cl
After forming one bond with fluorine, chlorine has 6 electrons left (7 - 1 = 6). These 6 electrons will be arranged as 3 lone pairs.

**Hint:** Lone pairs are the remaining valence electrons that are not involved in bonding.

### Step 4: Identify the central atom in XeF4
In XeF4, the central atom is xenon (Xe). Xenon is a noble gas and has 8 valence electrons.

**Hint:** Noble gases typically have a complete octet, which means they have 8 valence electrons.

### Step 5: Determine the number of bonds in XeF4
In XeF4, xenon forms 4 bonds with 4 fluorine atoms. Each fluorine atom contributes one electron, so xenon uses 4 of its 8 valence electrons for bonding.

**Hint:** Each bond consumes two electrons (one from the central atom and one from the bonded atom).

### Step 6: Calculate the lone pairs on Xe
After forming 4 bonds with fluorine, xenon has 4 electrons left (8 - 4 = 4). These 4 electrons will be arranged as 2 lone pairs.

**Hint:** Again, lone pairs are the remaining valence electrons that are not involved in bonding.

### Step 7: Calculate the ratio of lone pairs
Now we can find the ratio of the number of lone pairs on the central atom of ClF and XeF4. ClF has 3 lone pairs and XeF4 has 2 lone pairs.

The ratio is:
\[ \text{Ratio} = \frac{\text{Lone pairs in ClF}}{\text{Lone pairs in XeF4}} = \frac{3}{2} \]

### Final Answer
The ratio of the number of lone pairs on the central atom of ClF and XeF4 is \( \frac{3}{2} \).

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