Consider following compounds A to E :
(A) `XeF_(n) " " (B) XeF_((n+1))^(+) " " (C ) XeF_((n+1))^(-) " " (D) XeF_((n+2))`
(E ) `XeF_((n+4))^(2-)`,
If value of n is 4, then calculate value of `p div q` here, 'p' is total number of bond pair and 'q' is total number of lone pair on central atoms of compounds (A) to (E ).

Explanation

To solve the problem, we will analyze each compound (A to E) based on the value of n, which is given as 4. We will calculate the total number of bond pairs (p) and lone pairs (q) for each compound and then find the ratio \( \frac{p}{q} \).

### Step 1: Analyze Compound A - \( \text{XeF}_4 \)

1. **Determine the steric number**:
- Valence electrons of Xe = 8
- Number of fluorine atoms = 4
- Charge = 0 (neutral compound)
- Steric number = \( \frac{(8 + 4 + 0)}{2} = \frac{12}{2} = 6 \)

2. **Calculate bond pairs and lone pairs**:
- Bond pairs = 4 (from 4 fluorine atoms)
- Lone pairs = 6 - 4 = 2

### Step 2: Analyze Compound B - \( \text{XeF}_5^+ \)

1. **Determine the steric number**:
- Valence electrons of Xe = 8
- Number of fluorine atoms = 5
- Charge = +1
- Steric number = \( \frac{(8 + 5 - 1)}{2} = \frac{12}{2} = 6 \)

2. **Calculate bond pairs and lone pairs**:
- Bond pairs = 5 (from 5 fluorine atoms)
- Lone pairs = 6 - 5 = 1

### Step 3: Analyze Compound C - \( \text{XeF}_5^- \)

1. **Determine the steric number**:
- Valence electrons of Xe = 8
- Number of fluorine atoms = 5
- Charge = -1
- Steric number = \( \frac{(8 + 5 + 1)}{2} = \frac{14}{2} = 7 \)

2. **Calculate bond pairs and lone pairs**:
- Bond pairs = 5 (from 5 fluorine atoms)
- Lone pairs = 7 - 5 = 2

### Step 4: Analyze Compound D - \( \text{XeF}_6 \)

1. **Determine the steric number**:
- Valence electrons of Xe = 8
- Number of fluorine atoms = 6
- Charge = 0
- Steric number = \( \frac{(8 + 6 + 0)}{2} = \frac{14}{2} = 7 \)

2. **Calculate bond pairs and lone pairs**:
- Bond pairs = 6 (from 6 fluorine atoms)
- Lone pairs = 7 - 6 = 1

### Step 5: Analyze Compound E - \( \text{XeF}_8^{2-} \)

1. **Determine the steric number**:
- Valence electrons of Xe = 8
- Number of fluorine atoms = 8
- Charge = -2
- Steric number = \( \frac{(8 + 8 + 2)}{2} = \frac{18}{2} = 9 \)

2. **Calculate bond pairs and lone pairs**:
- Bond pairs = 8 (from 8 fluorine atoms)
- Lone pairs = 9 - 8 = 1

### Step 6: Summarize the Results

Now we will summarize the bond pairs (p) and lone pairs (q) for each compound:

- **Compound A**: Bond pairs = 4, Lone pairs = 2
- **Compound B**: Bond pairs = 5, Lone pairs = 1
- **Compound C**: Bond pairs = 5, Lone pairs = 2
- **Compound D**: Bond pairs = 6, Lone pairs = 1
- **Compound E**: Bond pairs = 8, Lone pairs = 1

### Step 7: Calculate Total p and q

- Total bond pairs (p) = \( 4 + 5 + 5 + 6 + 8 = 28 \)
- Total lone pairs (q) = \( 2 + 1 + 2 + 1 + 1 = 7 \)

### Step 8: Calculate \( \frac{p}{q} \)

\[ \frac{p}{q} = \frac{28}{7} = 4 \]

### Final Answer

The value of \( \frac{p}{q} \) is **4**.

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