Question

Which of the following has p\(_\pi\) – d\(_\pi\) bonding

• A. NO₃^-
• B. SO₃^-2
• C. BO₃^-3
• D. CO₃^-2

Answer 1

The Correct Option is B

 To determine which of the given compounds has \(\pi\) \(- d_\pi\) bonding, we need to explore the electron configuration and bonding characteristics of each compoun\)

1. Explanation of \(p_\pi - d_\pi\) bonding: This type of bonding occurs when the p orbitals of a lighter atom overlap with the d orbitals of a heavier atom, forming a double or triple bond. This is typically observed in compounds where a non-metal is bonded to an element capable of expanding its octet by using empty d orbitals.
2. Compound Analysis:
   • NO₃^-: Nitrogen has no available d orbitals to participate in \(p_\pi - d_\pi\) bonding.
   • SO₃^-2: Sulfur is capable of expanding its octet as it has empty 3d orbitals. In sulfite ion, \(p_\pi - d_\pi\) bonding can occur between oxygen's p orbitals and sulfur's d orbitals, leading to resonance stabilization.
   • BO₃^-3: Boron is from the second period and lacks d orbitals. Therefore, it cannot engage in \(p_\pi - d_\pi\) bonding.
   • CO₃^-2: Carbon also does not have d orbitals to form \(p_\pi - d_\pi\) bonding.
3. Conclusion: Only SO₃^-2 has the possibility of \(p_\pi - d_\pi\) bonding due to sulfur's ability to use its d orbitals for bonding with oxygen atoms.

Therefore, the correct answer is SO₃^-2.