Draw a Lewis structure for, and indicate the number of sigma and pi bonds in H2O.
The Correct Answer and Explanation is:
To draw the Lewis structure of water (H₂O) and determine the number of sigma and pi bonds, follow these steps:
Step 1: Count Valence Electrons
- Hydrogen (H) has 1 valence electron, and there are 2 hydrogen atoms, so the total valence electrons for hydrogen are 2.
- Oxygen (O) has 6 valence electrons, so the total valence electrons for oxygen are 6.
- Total valence electrons = 2 (from H) + 6 (from O) = 8 valence electrons.
Step 2: Arrange the Atoms
- Oxygen (O) is the central atom because it is more electronegative than hydrogen.
- Attach each hydrogen atom to the oxygen atom using a single bond.
Step 3: Distribute Electrons
- Each single bond (O-H) consists of 2 electrons, one from oxygen and one from hydrogen.
- Place the remaining electrons around oxygen to satisfy the octet rule for oxygen. After bonding with hydrogen, oxygen has 4 valence electrons left, which form 2 lone pairs.
Lewis Structure of H₂O:
cppCopyEdit H — O — H
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(lone pair)
- Oxygen has 2 lone pairs, and each hydrogen is bonded to oxygen with a single bond.
Step 4: Count the Bonds
- Sigma (σ) Bonds: A sigma bond is formed by the head-on overlap of atomic orbitals. In H₂O, there are two O-H single bonds, each consisting of one sigma bond.
- Pi (π) Bonds: Pi bonds are formed by the sideways overlap of p orbitals, and they are present only in double or triple bonds. Since water has no double or triple bonds, it has zero pi bonds.
Conclusion:
- Number of sigma bonds = 2 (one between each H and O).
- Number of pi bonds = 0 (since there are no double or triple bonds in H₂O).
This structure follows the octet rule for oxygen (8 electrons around oxygen) and satisfies the hydrogen rule (2 electrons around each hydrogen).
