The ratio of number of `d pi - p pi` bonds in `XeO_(4) and SO_(3)` is.

Explanation

To find the ratio of the number of `dπ - pπ` bonds in `XeO4` and `SO3`, we will analyze the bonding in both compounds step by step.

### Step 1: Determine the structure of `XeO4`
1. **Valence Electrons**: Xenon (Xe) has 8 valence electrons, and each oxygen (O) has 6 valence electrons. In `XeO4`, there are 4 oxygen atoms, contributing a total of 24 valence electrons. Therefore, the total number of valence electrons is:
\[ 8 + 4 \times 6 = 32 \text{ valence electrons} \]
2. **Lewis Structure**: In `XeO4`, xenon forms double bonds with each of the four oxygen atoms. This can be represented as:
\[ \text{Xe} = \text{O} \text{(double bond)} \text{O} \text{(double bond)} \text{O} \text{(double bond)} \text{O} \]
3. **Hybridization**: The hybridization of xenon in `XeO4` is `sp^3`, which means it uses all four of its p orbitals to form bonds.

### Step 2: Count the `dπ - pπ` bonds in `XeO4`
- In `XeO4`, all four bonds are `dπ - pπ` bonds because the xenon atom utilizes its d orbitals for bonding with oxygen. Thus, the number of `dπ - pπ` bonds in `XeO4` is:
\[ \text{Number of } dπ - pπ \text{ bonds in } XeO4 = 4 \]

### Step 3: Determine the structure of `SO3`
1. **Valence Electrons**: Sulfur (S) has 6 valence electrons, and each oxygen (O) has 6 valence electrons. In `SO3`, there are 3 oxygen atoms, contributing a total of 18 valence electrons. Therefore, the total number of valence electrons is:
\[ 6 + 3 \times 6 = 24 \text{ valence electrons} \]
2. **Lewis Structure**: In `SO3`, sulfur forms double bonds with each of the three oxygen atoms. This can be represented as:
\[ \text{S} = \text{O} \text{(double bond)} \text{O} \text{(double bond)} \text{O} \]
3. **Hybridization**: The hybridization of sulfur in `SO3` is `sp^2`, which means it uses its p orbitals for bonding.

### Step 4: Count the `dπ - pπ` bonds in `SO3`
- In `SO3`, the bonds formed are `pπ - pπ` bonds, as sulfur does not utilize d orbitals for bonding. Thus, the number of `dπ - pπ` bonds in `SO3` is:
\[ \text{Number of } dπ - pπ \text{ bonds in } SO3 = 0 \]

### Step 5: Calculate the ratio of `dπ - pπ` bonds in `XeO4` to `SO3`
- Now, we can find the ratio of the number of `dπ - pπ` bonds in `XeO4` to `SO3`:
\[ \text{Ratio} = \frac{\text{Number of } dπ - pπ \text{ bonds in } XeO4}{\text{Number of } dπ - pπ \text{ bonds in } SO3} = \frac{4}{0} \]
- Since we cannot divide by zero, we conclude that the ratio is undefined in this context. However, if we consider only the presence of `dπ - pπ` bonds, we can say that `XeO4` has `4` and `SO3` has `0`.

### Final Answer
The ratio of the number of `dπ - pπ` bonds in `XeO4` to `SO3` is:
\[ \text{Ratio} = 4:0 \text{ (or simply, } 4 \text{ since } SO3 \text{ has no } dπ - pπ \text{ bonds)} \]