Draw a valid Lewis structure for the molecule CH3NO in which there are no nonzero formal charges on any of the atoms.

Draw a valid Lewis structure for the molecule CH3NO in which there are no nonzero formal charges on any of the atoms. This structure should not include any bonds between N and O.

The Correct Answer and Explanation is:

To draw the Lewis structure for CH₃NO (formaldehyde oxime) where there are no nonzero formal charges on any of the atoms and without a bond between nitrogen and oxygen, follow these steps:

1. Count the total valence electrons:

  • Carbon (C): 4 valence electrons (since C is in group 14).
  • Hydrogen (H): 1 valence electron each, for 3 H atoms → 3 electrons.
  • Nitrogen (N): 5 valence electrons (since N is in group 15).
  • Oxygen (O): 6 valence electrons (since O is in group 16).

Total valence electrons = 4 (C) + 3 (H) + 5 (N) + 6 (O) = 18 valence electrons.

2. Determine the skeleton structure:

  • Carbon typically forms four bonds, so it will be the central atom.
  • Hydrogen atoms will be attached to carbon, as carbon can only form single bonds.
  • Nitrogen and oxygen will be attached to carbon, but we will not directly connect nitrogen and oxygen.

3. Place atoms and connect them with single bonds:

  • Attach each hydrogen atom to the carbon atom.
  • Attach nitrogen and oxygen atoms to the carbon atom via single bonds.

The skeleton structure looks like this:

H—C—N—O

4. Distribute the electrons:

  • Each bond represents 2 electrons. For 4 C—H bonds (3 H atoms), we use 8 electrons.
  • The C—N and C—O single bonds use 2 more electrons, making 10 electrons used so far.

5. Complete the octets:

  • Oxygen needs 2 more electrons to complete its octet. Place these electrons as lone pairs on oxygen.
  • Nitrogen needs 2 more electrons to complete its octet. Place these electrons as lone pairs on nitrogen.

Now, there are 18 total electrons used (8 for C—H bonds, 2 for C—N, 2 for C—O, and 6 for lone pairs).

6. Check for formal charges:

  • Carbon: The formal charge is calculated by subtracting the number of bonds (4) and lone pairs (0) from the valence electrons (4), resulting in a formal charge of 0.
  • Hydrogen: Each hydrogen has 1 bond, which is correct for its 1 valence electron, resulting in no formal charge on hydrogen.
  • Nitrogen: Nitrogen has 1 bond to carbon and 2 lone pairs. This gives a formal charge of 0 (5 valence electrons minus 1 bond and 2 lone pairs).
  • Oxygen: Oxygen has 1 bond to carbon and 3 lone pairs, giving a formal charge of 0 (6 valence electrons minus 1 bond and 3 lone pairs).

7. Final structure:

mathematicaCopyEditH—C—N—O
  |     |
  H     H
  |
  H

This structure has no formal charges on any atom and follows the given conditions.

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