Simple Procedure for writing Lewis Structures – Lewis Structures for nitric acid (HNO3)

 Lewis Structures and the Octet Rule: A Simple Method to write Lewis Structures >  Lewis Structures for NO2+ and HCN - Example #3  >  Lewis Structures for nitric acid (HNO3)  - Example #4

A simple procedure for writing Lewis structures is given in a previous article entitled “Lewis Structures and the Octet Rule”. Relevant worked examples were given in the following articles: Examples #1, #2, #3.

A few more examples for writing Lewis structures following the above procedure are given bellow:

Let us consider the case of nitric acid HNO₃ . Nitric acid is a strong oxidizing agent and it dissolves practically all metals except gold and platinum and some other precious metals. As such, is an important raw material for the chemical and pharmaceutical industry. It is mainly used for etching and for the production of pure nitrates. Even though nitric acid was known since the 9th century - alchemists used it to separate gold and silver - its mass production started in 1902 when a German chemist Wilhelm Ostwald developed an industrial process.  The German corporation BASF start producing it in an industrial scale by 1915. Initially it was used for the production of explosives but today its main use is for the production of fertilizers such as ammonium nitrate. Other main applications is for the production of explosives, nylon precursors and substituted organic compounds.
In elemental analysis by atomic absorption spectroscopy, ICP, graphite furnace atomic spectroscopy dilute nitric acid is used as a "solvent" for the determination of metal traces in solution.

Step 1: The central atom will be the N atom since it is the less electronegative (H is a terminal atom – it cannot be a central atom).  Connect the atoms with single bonds

Step 2:  Calculate the # of electrons in π bonds (multiple bonds) using formula (1) in the article entitled “Lewis Structures and the Octet Rule”. 

Where n in this case is 4 since  HNO₃ consists of five atoms but one of them is a H atom.

Where V = (1 + 6 + 5 + 6 + 6) = 24  

Therefore, P = 6n + 2 – V = 6 * 4 + 2 – 24 = 2      \  there are 2 π electrons in HNO₃  \

1 double bond must be added to the structure of Step 1.

Step 3 & 4: One double bond must therefore be placed through the 3 N-O bonds. Therefore, the Lewis structures for HNO₃ are as follows:

Figure 1: Lewis structures for HNO₃