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Helps us determine the most probable Lewis structure when there are several correct possibilities Determines the charge on a bonded atom if there was no.

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Presentation on theme: "Helps us determine the most probable Lewis structure when there are several correct possibilities Determines the charge on a bonded atom if there was no."— Presentation transcript:

1 Helps us determine the most probable Lewis structure when there are several correct possibilities Determines the charge on a bonded atom if there was no electronegativity To find the formal charge 1. Count the electrons around the atom Only count one electron from each bond H | H – C – H | H Carbon has 4 e- Each H has only 1 e-

2 2. Subtract the number of electrons from the normal number of valence electrons Carbon 4 – 4 = 0 Hydorgen 1-1 = 0 Since CH 4 is neutral, all formal charges must equal zero CO 3 -2 O | O – C – O C has 4 e- 4 – 4 = 0 4 – 4 = 0 Each O has 7 electrons 6 – 7 = -1 total = -2 6 – 7 = -1 total = -2 Since CO 3 totals to -2, this is the proper formal charges. Formal charge is not the actual charge. Actual charge is based more on electronegativity, but method helps us determine the proper Lewis structure. One O has 6 6-6 = 0 -2

3 Formal charges and Lewis structures If the molecule has more than one valid Lewis structure, determine the formal charges on all atoms The most likely Lewis structure is the one that 1. Atoms have formal charges closest to zero 2. Negative charges are on the most electronegative elements Examples CO 2 O – C  O O = C = O C is 0 O is -1 O is +1 C is 0 O is 0 Most likely structure!

4 Write out possible Lewis structures for H 2 O 2. Determine the most likely structure based on formal charges. H – O – O | H H – O – O – H Each H is1-10 Each H is1-10 This O is6-5+1 This O is 6-7 Each O is1-10 Lower charges Best structure Both follow all the rules, and formal charges add to zero

5 Write out possible Lewis structures for H 2 CO 2. Determine the most likely structure based on formal charges. O | H – C = O – H O || H – C – O – H 0 +1 0 0 0 0 0


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