Simple method for writing Lewis Structures for N2O4 - Ex. #22

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 , #4, #5, #6,  #7,  #8, #9, #10, #11, #12, #13, #14,  #15, #16, #17, #18, #19, #20 and #21.

Another example  for writing Lewis structures following the above procedure is given bellow:

Let us consider the case of dinitrogen tetroxide (N₂O₄).
 

Step 1: Connect the atoms with single bonds. The central atoms are the nitrogen atoms.

Step 2: Calculate the # of electrons in π bonds (multiple bonds) using formula (1): 

Where n in this case is 6 since N₂O₂ consists of six atoms.

Where V = (6 + 5 + 5 + 6 + 6 + 6 ) = 34, where V the number of the valence electrons of  N₂O₂

Therefore, P = 6n + 2 – V = 6 * 6 + 2 – 34 = 4     There are 4π electrons in N₂O₄ and that means 2 double bonds or a triple bond must be added to the structure in Step 1.

However, all the atoms present are second row elements and cannot accommodate more than 8 valence electrons. Therefore, a triple bond cannot be used as it would place ten electrons around the atoms bonded. The only possible case is that of two double bonds.

 

Step 3 & 4: The Lewis structure for N₂O₄  is as follows:

Figure 1: Lewis electron dot structures for N2O4