Answer
See the explanation
Work Step by Step
To determine the orbitals that overlap to form the various bonds in urea, we need to consider the hybridization of the carbon, nitrogen, and oxygen atoms.
Given the Lewis structure of urea:
$H:O:H\\ I\quad II\quad I\\ H-N-C-N-H$
1. Carbon (C) hybridization:
The carbon atom in urea is bonded to one nitrogen atom and one oxygen atom. To accommodate these bonding requirements, the carbon atom undergoes sp² hybridization.
- The sp² hybrid orbitals of carbon overlap with the s orbital of the hydrogen atoms and the p orbitals of the nitrogen and oxygen atoms to form the sigma (σ) bonds.
- The remaining p orbital of the carbon atom participates in the formation of the pi (π) bond with the oxygen atom.
2. Nitrogen (N) hybridization:
The nitrogen atoms in urea are each bonded to one carbon atom and two hydrogen atoms.
- The nitrogen atoms undergo sp³ hybridization.
- The sp³ hybrid orbitals of nitrogen overlap with the s orbitals of the hydrogen atoms and the p orbital of the carbon atom to form the sigma (σ) bonds.
3. Oxygen (O) hybridization:
The oxygen atom in urea is bonded to one carbon atom.
- The oxygen atom undergoes sp² hybridization.
- The sp² hybrid orbitals of oxygen overlap with the p orbital of the carbon atom to form the sigma (σ) bonds.
- The remaining p orbital of the oxygen atom participates in the formation of the pi (π) bond with the carbon atom.
In summary, the overlapping orbitals in the formation of the various bonds in urea are:
- Sigma (σ) bonds:
- sp² hybrid orbitals of carbon overlapping with s orbitals of hydrogen and p orbitals of nitrogen and oxygen
- sp³ hybrid orbitals of nitrogen overlapping with s orbitals of hydrogen and p orbital of carbon
- sp² hybrid orbitals of oxygen overlapping with p orbital of carbon
- Pi (π) bond:
- Remaining p orbital of carbon overlapping with p orbital of oxygen
