The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. A) 18 B) 20 C) 26 D) 32 E) 36 symbol I33 In the Lewis paired and sulfur atom, there C C; the € bond CC and = unpaired electrons- F) $
The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. A) 18 B) 20 C) 26 D) 32 E) 36 symbol I33 In the Lewis paired and sulfur atom, there C C; the € bond CC and = unpaired electrons- F) $
The Correct Answer and Explanation is:
To determine the correct Lewis structure and number of valence electrons in CH₃OCH₂CH₃ (ethyl methyl ether), we follow a step-by-step approach. This molecule consists of carbon (C), hydrogen (H), and oxygen (O) atoms.
🔢 Step 1: Count Valence Electrons
We add the valence electrons from each atom:
- Carbon (C) has 4 valence electrons × 3 atoms = 12 e⁻
- Hydrogen (H) has 1 valence electron × 8 atoms = 8 e⁻
- Oxygen (O) has 6 valence electrons × 1 atom = 6 e⁻
Total valence electrons = 12 + 8 + 6 = 26 electrons
✅ Correct answer: C) 26
🧪 Step 2: Build the Structure
CH₃OCH₂CH₃ is an ether:
- One methyl group (CH₃) on the left
- One ethyl group (CH₂CH₃) on the right
- An oxygen atom connects both sides
Structure:
CH₃–O–CH₂CH₃
Connectivity:
- C (left) forms 4 bonds: 3 with H, 1 with O
- O forms 2 bonds: one with each C, and has 2 lone pairs
- Middle C (CH₂) forms 2 single bonds: one with O, one with the terminal C, and 2 Hs
- Terminal C (CH₃) forms 3 bonds with H and one with CH₂
🧲 Step 3: Distribute Electrons
- Each bond represents 2 shared electrons
- We ensure all atoms (except H) follow the octet rule
- Oxygen must have 2 lone pairs (4 electrons) to complete its octet
- All 8 hydrogen atoms get 1 bond each (2 electrons total), satisfying their duet rule
✍️ Final Notes
- The Lewis structure does not involve any double or triple bonds in this molecule.
- Oxygen contributes a major portion of the non-bonding (lone pair) electrons
- All atoms are in a stable, neutral state in this structure
✅ Answer: C) 26 valence electrons
