Exams › NEET › Chemistry

Which of the following has pπ–dπ bonding?

1. NO₃⁻
2. SO₃²⁻
3. BO₃³⁻
4. CO₃²⁻

Correct answer: SO₃²⁻

Solution

In SO₃²⁻, the sulfur atom (from Period 3) can utilize its vacant d-orbitals to form pπ–dπ bonds with oxygen atoms. This type of bonding is not possible in the other options as they lack suitable d-orbitals.

Related NEET Chemistry questions

• Which of the following have identical bond order? (I) \( \boldsymbol{C} \boldsymbol{N}^{-} \) (II) \( \boldsymbol{O}_{\mathbf{2}}^{-} \) \( (\|\|) N O^{+} \) \( (\mid V) C N^{+} \)
• What type of bond does \( M g \) form with \( C l ? \)
• Shape of \( C O_{2} \) is:
• The shape of \( s p^{3} d^{2} \) hybridized molecule with 2 lone pairs will be:
• The hybrid orbitals with \( 33.33 \% \) Scharacter are involved in the bonding of one of the crystalline allotropes of carbon. The allotrope is :
• In certain polar solvents \( \boldsymbol{P} \boldsymbol{C l}_{5} \) undergoes an ionisation reaction in which \( C l^{-} \) ion leaves one \( P C l_{5} \) molecule and attaches itself to another \( 2 P C l_{5} \rightleftharpoons P C l_{4}^{+}+P C l_{6}^{-} \) What are the changes in the geometrical shapes that occur in this ionisation?

⚔️ Practice NEET Chemistry free + battle 1v1 →