Give an example of compound which has `ppi-dpi` bonding?

Explanation

To find an example of a compound that exhibits p-pi d-pi bonding, we can analyze the sulfate ion (SO₄²⁻). Here’s a step-by-step breakdown of the reasoning:

### Step 1: Identify the Compound
The compound we will consider is the sulfate ion, SO₄²⁻.

### Step 2: Determine the Valence Electrons
Sulfur (S) has 6 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are four oxygen atoms, we have:
- Sulfur: 6 electrons
- Oxygen: 4 × 6 = 24 electrons
- Total: 6 + 24 = 30 electrons
- Considering the 2- charge of the sulfate ion, we add 2 more electrons: 30 + 2 = 32 electrons.

### Step 3: Calculate the Steric Number
The steric number can be calculated using the formula:
\[ \text{Steric Number} = \text{Number of atoms bonded to the central atom} + \text{Number of lone pairs on the central atom} \]

For SO₄²⁻:
- Sulfur is bonded to 4 oxygen atoms.
- There are no lone pairs on sulfur in this ion.

Thus, the steric number is:
\[ \text{Steric Number} = 4 + 0 = 4 \]

### Step 4: Determine the Hybridization
A steric number of 4 corresponds to sp³ hybridization. In the case of SO₄²⁻, sulfur undergoes sp³ hybridization to form four equivalent hybrid orbitals.

### Step 5: Analyze Bonding
In the sulfate ion, sulfur forms sigma bonds with the oxygen atoms using the sp³ hybridized orbitals. Additionally, sulfur can utilize its unhybridized d-orbitals (3d) to form p-pi d-pi bonds with the oxygen atoms.

### Step 6: Identify the p-pi d-pi Bonding
In the sulfate ion:
- The oxygen atoms have unpaired p orbitals (2p) that can overlap with the d orbitals of sulfur.
- This overlap results in p-pi d-pi bonding, which is a characteristic of the resonance structures of the sulfate ion.

### Conclusion
Thus, the sulfate ion (SO₄²⁻) is an example of a compound that exhibits p-pi d-pi bonding.

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