How to Draw the Lewis Structure of Nitric Acid (HNO₃) and Calculate Required Electrons

To draw the Lewis structure of nitric acid (HNO₃), start by determining the total number of valence electrons available. Nitric acid consists of one nitrogen (N) atom, three oxygen (O) atoms, and one hydrogen (H) atom. The valences are as follows:

  • N: 5 valence electrons
  • O: 6 valence electrons x 3 = 18 valence electrons
  • H: 1 valence electron

This gives us a total of 5 + 18 + 1 = 24 valence electrons to work with.

Next, we need to arrange the atoms. Nitrogen is the central atom, bonded to three oxygen atoms and one hydrogen atom. The most common arrangement is to have one double bond between the nitrogen and one of the oxygen atoms, a single bond to another oxygen, and then the single bond to hydrogen. Thus, the nitrogen atom is bonded as follows:

  • One N=O double bond
  • One N-O single bond with an -OH group (hydroxyl)
  • One N-O single bond to a negatively charged oxygen atom (which will hold an extra lone pair)

Now, let’s distribute the remaining electrons. Start by fulfilling the octet rule for oxygen, keeping in mind that the double-bonded oxygen will have two lone pairs, the singly bonded (hydroxyl) oxygen will have two lone pairs, and the negatively charged oxygen will have three lone pairs. This arrangement minimizes the formal charges:

  • For the nitrogen, the formal charge is 0.
  • The double-bonded oxygen has a formal charge of 0.
  • The hydroxyl oxygen has a formal charge of 0.
  • The negatively charged oxygen has a formal charge of -1.

This structure effectively minimizes the overall formal charges, yielding the following Lewis structure for nitric acid:

N
  ||
 O - H
  |
 O-

To calculate the electrons required for this structure, remember that we initially calculated a total of 24 valence electrons. Since all atoms satisfy their octet requirements, we have effectively used all the valence electrons in our Lewis structure of HNO₃.

This means the calculation of electrons required for the structure is simply the total number of available valence electrons, which is 24. Hence, the electrons required (ER) for the proper bonding in nitric acid is also 24.

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