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 (N**.

_{2}O_{4})__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 N

_{2}O_{2}consists of six atoms.
Where V = (6 + 5
+ 5 + 6 + 6 + 6 ) = 34, where V the number of the

**valence electrons**of N_{2}O_{2}
Therefore, P = 6n
+ 2 – V = 6 * 6 + 2 – 34 = 4

**There are 4****π electrons in**N_{2}O_{4}**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 N

_{2}O

_{4}is as follows:

Figure 1: Lewis electron dot structures for N2O4 |

This is a very beautiful and interesting research

ReplyDeleteThe most educating one i have read today!

High School Diploma

Thank you. This is the best explanation on the web :)

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