Question:medium

How many of the following compounds are sp3 hybridized

Updated On: Feb 25, 2026
  • CIO-3
  • CIO-2
  • NH3
  • NO2
Show Solution

The Correct Option is C

Solution and Explanation

To determine which of the given compounds are \(\text{sp}^3\) hybridized, we need to consider the molecular geometry and the hybridization of the central atom in each compound:

  • The central atom is chlorine (Cl). 
  • The ion has three oxygen atoms bonded to the chlorine atom and one lone pair of electrons.
  • The hybridization of chlorine in

CIO-3

  • is \(\text{sp}^3\) as it exhibits a tetrahedral geometry due to the presence of one lone pair and three bond pairs, which requires four hybrid orbitals.

CIO-3:

  • The central atom is chlorine (Cl).
  • The ion has two oxygen atoms and two lone pairs of electrons.
  • The hybridization of chlorine in

CIO-2

  • is \(\text{sp}^3\) as it has a bent shape, associated with two lone pairs and two bond pairs, thus requiring four hybrid orbitals.

CIO-2:

  • The central atom is nitrogen (N).
  • The compound has three hydrogen atoms bonded to nitrogen and one lone pair of electrons.
  • The hybridization of nitrogen in

NH3

  • is also \(\text{sp}^3\), resulting in a trigonal pyramidal geometry with one lone pair and three bond pairs requiring four hybrid orbitals.

NH3:

  • The central atom is nitrogen (N).
  • The molecule has two oxygen atoms bonded to the nitrogen and one lone pair of electrons.
  • The hybridization of nitrogen in

NO2

  • is \(\text{sp}^2\) as it has a trigonal planar shape due to one lone pair and two bond pairs, requiring only three hybrid orbitals.

NO2:

From the analysis above, the compounds that are \(\text{sp}^3\) hybridized are

CIO-3

,

CIO-2

, and

NH3

, making it 3 compounds. However, as per the provided answer in question,

NH3

is highlighted. Nevertheless, correcting according to chemical knowledge,

CIO-3

and

CIO-2

are also \(\text{sp}^3\).

 

Therefore, the correct answer indicating that

NH3

is \(\text{sp}^3\) hybridized is correct, but it should include other \(\text{sp}^3\) hybridized compounds specified before from group.

 

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