Determine the hybridization (sp3, sp2, or sp) of the central atom in: a. CH4 b. NH3 c. CO2 d. H2CO
The Correct Answer and Explanation is:
Here are the correct hybridizations of the central atom in each compound:
a. CH₄ (Methane)
Hybridization: sp³
b. NH₃ (Ammonia)
Hybridization: sp³
c. CO₂ (Carbon Dioxide)
Hybridization: sp
d. H₂CO (Formaldehyde)
Hybridization: sp²
Explanation
Hybridization is the process of mixing atomic orbitals in an atom to form new hybrid orbitals that can explain the geometry and bonding in molecules. The type of hybridization depends on the number of electron groups (bond pairs and lone pairs) around the central atom.
a. CH₄ (Methane)
Carbon is the central atom. It forms four sigma bonds with hydrogen atoms. These four bonding pairs result in four electron groups. Since carbon needs four hybrid orbitals, it uses one s orbital and three p orbitals to form four sp³ hybrid orbitals. The geometry is tetrahedral with bond angles of 109.5°.
b. NH₃ (Ammonia)
Nitrogen is the central atom. It has three bonding pairs (with hydrogen) and one lone pair. This gives a total of four electron groups. Just like CH₄, four orbitals are needed, so nitrogen hybridizes its orbitals into sp³. The molecular geometry is trigonal pyramidal, with bond angles slightly less than 109.5° due to lone pair repulsion.
c. CO₂ (Carbon Dioxide)
Carbon is the central atom, double bonded to two oxygen atoms on opposite sides. There are two regions of electron density (each double bond counts as one region). Two electron groups require two orbitals, so carbon uses one s orbital and one p orbital, resulting in sp hybridization. The molecule is linear with a bond angle of 180°.
d. H₂CO (Formaldehyde)
Carbon is the central atom. It is bonded to two hydrogen atoms with single bonds and one oxygen atom with a double bond. This results in three electron groups. For three regions, carbon hybridizes one s orbital and two p orbitals, forming sp² hybrid orbitals. The geometry is trigonal planar with bond angles of about 120°.
