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Report (any type)/Brief report
22/09/2021
Step I: Find valence e- for all atoms.
H:1 C:4 N:5
Add the total number of e-, 1+4+5 = 10
Step II: Find octet e- for each atom.
H:2 C:8 O:8
Add the e- together, 2+8+8 = 18
Step III: Calculate e-. Octet e- minus the total number of e-
(subtract step1 from step 2) 18-10=8e-
Step IV: Find the number of bonds in the HCN molecule.
Divide the bonding e- obtained in step 3 = 8e- by 2 e-
(remember, each bond is made up of 2e-). This gives us 4 bond pairs.
Step V: Find the number of nonbonding (lone pairs) e-.
Step I gives you total number of valence e- (e- to be used in bonding). They are 10
Step III gives you total number of bonding e-, they are 8
Therefore, subtract Step III from Step II, which is 10 – 8 = 2e- = 1 lone pair
10-8= 2e-=1 lone pair
Step VI: THE LEWIS STRUCTURE OF HCN. Use information from THE STEPS IV
AND V.
Carbon goes in the centre.Make sure carbon and Nitrogen get 8 electrons to fulfil .
H – CN a triple bond between C and N, a lone pair at N
For HCN, above is the Lewis Structure. But there are molecules where the Lewis structure can be written in different ways, still satisfying the Octet rule. So, how do you decide which is the correct Lewis Structure? So, I introduce the term formal charge which is different than the actual charge. In HCN, the valence e- are:
H: 1 C:4 N:5
The lone pair is assigned to N, there are 6 e-in a triple bond, assigned to C and 2 e- between H and C. So the formal charges are
N 5-5=0 C 4-4 = 0 H 1-1 = 0
Formal charge is different than actual charge, and above is a covalent compound so there is no actual charge on the atoms, they are not existing as cation or anion. If there is a compound that can have more than one Lewis structure while satisfying the Octet rule, the correct Lewis Structure would be the one with least amount of formal charge.
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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