How to Draw the Lewis Structure for CH3NH2? How Many Sigma and Pi Bonds Does It Have, and How Many Bond Pairs and Lone Pairs Are There?

The Lewis structure for CH₃NH₂ can be drawn by following these steps:

1. Count the total number of valence electrons: Carbon (C) has 4, each Hydrogen (H) has 1 (3 H total = 3), and Nitrogen (N) has 5. So, the total is 4 + 3 + 5 = 12 valence electrons.
2. Start by placing the Carbon atom in the center, surrounded by three Hydrogens and one Nitrogen atom. The structure would look like H₃C-NH₂.
3. Connect the Carbon to three Hydrogens with single bonds (these are sigma bonds). Connect the Carbon to the Nitrogen with a single bond. This accounts for 4 sigma bonds so far.
4. Now, connect the Nitrogen to the two remaining Hydrogens with single bonds as well, adding 2 more sigma bonds. In total, we have 6 sigma bonds.

The final structure will look like this:

   H
   |
H–C–N–H
   |
   H

Next, let’s break down the bonds:

• Sigma bonds: There are 6 sigma bonds in total (3 C-H, 1 C-N, and 2 N-H).
• Pi bonds: There are no pi bonds present in CH₃NH₂ since all bonds are single.

Regarding bond pairs and lone pairs:

• Bond pairs: There are 6 bond pairs (all the bonds counted above).
• Lone pairs: Nitrogen has 1 lone pair since it uses 3 of its valence electrons to bond (2 with H and 1 with C), leaving 2 valence electrons as a lone pair.

In summary, CH₃NH₂ has:

• 6 sigma bonds
• 0 pi bonds
• 6 bond pairs
• 1 lone pair