Using the Lewis electron dot structure for #NO_3^-#, what is the total number of non-shared electrons?

1 Answer
May 7, 2016

Given the ONE resonance isomer, there are 8 (unshared) lone pairs on the oxygen atoms.


#(O=)N^(+)(-O)_2^(-)# is the typical resonance isomer. Such a structure demands that the doubly bound oxygen has 2 lone pairs, and each of the singly bound oxygen have 3 lone pairs.

Because there are 7 valence electrons around each singly bound oxygen atom, these atoms have a formal negative charge. To balance this #-2# charge, the quaternized nitrogen, which has 4 valence electrons, has a formal #+1# charge. This is entirely consistent with the ionic charge of the nitrate ion. Why?