Calculate the number of `p_(pi)-d_(pi)` bond(s) present in `SO_(4)^(2-)` :

Explanation

To calculate the number of `p_(pi)-d_(pi)` bonds present in `SO4^(2-)`, we can follow these steps:

### Step 1: Determine the Lewis Structure of `SO4^(2-)`
- Start by calculating the total number of valence electrons in `SO4^(2-)`.
- Sulfur (S) has 6 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are 4 oxygen atoms, the total number of valence electrons is:
\[ 6 + (4 \times 6) + 2 = 32 \text{ electrons} \]
- Draw the Lewis structure by placing sulfur in the center and surrounding it with the four oxygen atoms.

### Step 2: Assign Bonds in the Lewis Structure
- In the Lewis structure, sulfur forms 4 bonds: 2 double bonds with 2 oxygen atoms and 2 single bonds with the other 2 oxygen atoms.
- Each double bond consists of one sigma bond and one pi bond, while each single bond consists of one sigma bond.

### Step 3: Identify the Types of Bonds
- Count the total number of sigma and pi bonds:
- There are 4 sigma bonds (2 from double bonds and 2 from single bonds).
- There are 2 pi bonds (one from each double bond).

### Step 4: Determine the Nature of Pi Bonds
- The pi bonds in `SO4^(2-)` are formed by the overlap of the p orbitals of oxygen with the d orbitals of sulfur.
- Thus, each pi bond is classified as a `p_(pi)-d_(pi)` bond.

### Conclusion
- Since there are 2 pi bonds in the structure, the number of `p_(pi)-d_(pi)` bonds present in `SO4^(2-)` is **2**.

### Final Answer
The number of `p_(pi)-d_(pi)` bonds present in `SO4^(2-)` is **2**.

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