Question

Draw the structure
of HNO3(chemical bonding)​

Answer 1

ANSWER IS IN THE ATTACHMENT...

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Answer 2

Answer:

Thus, there are 1+5+(3×6=)24 valence electrons to account for in HNO3. 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. ... Thus, the number of electrons left to to complete all octets is (24-8=)16.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 (2×4=)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