Question

Number of molecules/ions from the following in which the central atom is involved in sp3 hybridization is ________.\( \text{NO}_3^-, \, \text{BCl}_3, \, \text{ClO}_2^-, \, \text{ClO}_3 \)

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

The Correct Option is A

Determine the Hybridization of Each Species:

For \( \text{NO}_3^- \):
The structure of \( \text{NO}_3^- \) (nitrate ion) shows that nitrogen is involved in three sigma bonds and has no lone pairs on it.
The hybridization of nitrogen in \( \text{NO}_3^- \) is \( sp^2 \).

For \( \text{BCl}_3 \):
Boron in \( \text{BCl}_3 \) forms three sigma bonds with no lone pairs, resulting in a planar triangular structure.
The hybridization of boron in \( \text{BCl}_3 \) is \( sp^2 \).

For \( \text{ClO}_2^- \):
In \( \text{ClO}_2^- \) (chlorite ion), chlorine has two sigma bonds with oxygen atoms and two lone pairs of electrons.
The hybridization of chlorine in \( \text{ClO}_2^- \) is \( sp^3 \), resulting in a bent or V-shaped geometry.

For \( \text{ClO}_3^- \):
In \( \text{ClO}_3^- \) (chlorate ion), chlorine forms three sigma bonds with oxygen atoms and has one lone pair of electrons.
The hybridization of chlorine in \( \text{ClO}_3^- \) is \( sp^3 \), resulting in a pyramidal structure.

Count the Species with \( sp^3 \) Hybridization:

From the analysis above, \( \text{ClO}_2^- \) and \( \text{ClO}_3^- \) have \( sp^3 \) hybridization for the central atom.

Conclusion:

The number of molecules/ions in which the central atom is involved in \( sp^3 \) hybridization is 2, corresponding to Option (1).