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

Answer: (A)

2

Answer 2

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).