Draw the Lewis structure of \(\mathrm{NO}_{2}^{-}\).
Count the total number of valence electrons.
Nitrogen (N) has 5 valence electrons, each oxygen (O) has 6 valence electrons, and there is an extra electron due to the negative charge. Therefore, the total number of valence electrons is \(5 + 2 \times 6 + 1 = 18\) electrons.
Determine the central atom and arrange the atoms.
Nitrogen (N) is the central atom, with the two oxygen (O) atoms bonded to it.
Distribute the electrons to form bonds and lone pairs.
Place a single bond between nitrogen and each oxygen. This uses 4 electrons (2 bonds \(\times\) 2 electrons per bond). Distribute the remaining 14 electrons as lone pairs to satisfy the octet rule for each atom.
Check for formal charges and adjust if necessary.
Initially, each oxygen has 3 lone pairs and one single bond, and nitrogen has one lone pair. This results in formal charges: nitrogen has a formal charge of +1, and each oxygen has a formal charge of -1. To minimize formal charges, form a double bond between nitrogen and one oxygen, resulting in one oxygen with a double bond (0 formal charge) and one oxygen with a single bond and 3 lone pairs (-1 formal charge). Nitrogen will have a formal charge of 0.
\(\boxed{\text{Lewis structure of } \mathrm{NO}_{2}^{-}}\)
\(\boxed{\text{Lewis structure of } \mathrm{NO}_{2}^{-}}\)
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