When a carbon atom has sp3 hybridization, it has
A. four π bonds.
B. three π bonds and one σ bond.
C. two π bonds and two σ bonds.
D. one π bond and three σ bonds.
E. four σ bonds.
The correct answer and explanation is :
The correct answer is:
E. four σ bonds.
Explanation:
Hybridization is the process by which atomic orbitals mix to form new hybrid orbitals that influence molecular bonding. The hybridization of a carbon atom depends on the number of regions of electron density (bonds or lone pairs) surrounding it.
- Understanding sp³ Hybridization:
- Carbon has an atomic number of 6 and an electron configuration of 1s² 2s² 2p².
- In its ground state, carbon has two unpaired electrons in the 2p orbitals, which would only allow it to form two bonds.
- To form four bonds (as in methane, CH₄), carbon undergoes sp³ hybridization, where one 2s orbital and three 2p orbitals mix to form four sp³ hybrid orbitals.
- Each hybrid orbital forms a σ (sigma) bond, leading to a tetrahedral geometry with bond angles of approximately 109.5°.
- Nature of Bonds in sp³ Hybridization:
- Since all four hybrid orbitals are used for bonding, only sigma (σ) bonds are present.
- π (pi) bonds arise from unhybridized p orbitals overlapping sideways, but in sp³ hybridization, there are no unhybridized p orbitals left.
- Therefore, sp³ hybridized carbon forms only sigma bonds (σ bonds) and has no π bonds.
- Examples of sp³ Hybridized Carbon Compounds:
- Methane (CH₄) – Each C-H bond is a sigma bond.
- Ethane (C₂H₆) – Each C-H and C-C bond is a sigma bond.
- Tetrahedral geometry is a characteristic of sp³ hybridization.
Image Representation:
I will generate an image showing the sp³ hybridization of carbon, tetrahedral shape, and sigma bonds.
Here is an image illustrating the sp³ hybridization of a carbon atom, showing the tetrahedral geometry and sigma (σ) bonds.