To determine the number of valence electrons in the Lewis structure of HNO2 (nitrous acid), we start by identifying the individual contributions from each atom:
- Hydrogen (H) has 1 valence electron.
- Nitrogen (N) has 5 valence electrons.
- Each Oxygen (O) has 6 valence electrons, and there are two of them.
Now, we can sum these up:
- 1 (from H) + 5 (from N) + 2 x 6 (from O) = 1 + 5 + 12 = 18 valence electrons.
Therefore, the Lewis structure of HNO2 contains a total of 18 valence electrons. In constructing the structure, you would distribute these electrons to satisfy the octet rule for the oxygen atoms while maintaining proper bonding with nitrogen and hydrogen. This results in one nitrogen-oxygen double bond, one nitrogen-oxygen single bond, and one nitrogen-hydrogen bond, effectively using all 18 valence electrons.