Step 1: The valence shell electron configurations of H, N, and O are `1s^1, 2s^2 2p^3`, and `2s^2 2p^4`, respectively. Thus, there are `1+5+(3xx6=)24` valence electrons to account for in `HNO_3`.
Step 2: Since `HNO_3` is a ternary oxoacid, it is neither a peroxide nor a cyclic compound. Thus, the most likely skeleton structure is
i.e., the three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms. Note that the above skeleton avoids `O-O` bonds, ring structures, and includes the `O-H` bond expected for an oxyacid.
Step 3: The number of valence electrons needed to form the four single bonds in teh skeleton is `(2xx4=)8`. Thus, the number of electrons left to to complete all octets is `(24-8=)16`.
Step 4: The number of valence electrons required to complete all octet is 0 for the H atom, 4 for the O atom bonded to H, 2 for the N atom (it has formed 3 single bonds thereby getting `6e^(-)s`), and 6 for each of the two O atoms with one bond, i.e., 18 electrons.
Step 5: Since the number of remaining valence electrons (16) is two less than the number of electrons (18) required to complete all octets, one additional covalent bond must be introduced in such a way that the normal valency of every element (if possible) gets satisfied. Placing the extra bond in a way that avoids two bonds to H atom or a ring structure or three bonds to an O atom gives
First drawing (I) places the extra bond between the N atom and the upper O atom. The extra bond could just as well be placed between the N atom and the lower O atom, since these two O atoms are identical. Completing the octets by placing the dots around the appropriate symbols gives the Lewis structure:
Step 6: The number of valence electrons in the Lewis structure (I or II), 24 (7 lone pairs +5 bonding pairs), equals the number of valence electrons counted in step 1.
