C. C, Н "ии Н CI Н Н

Introductory Chemistry For Today
8th Edition
ISBN:9781285644561
Author:Seager
Publisher:Seager
Chapter6: The States Of Matter
Section: Chapter Questions
Problem 6.87E
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Determine the hybridization at each of the 3 labeled atoms.

C.
C,
Н
"ии Н
CI
Н
Н
Transcribed Image Text:C. C, Н "ии Н CI Н Н
Expert Solution
Step 1

atom labelled 1.

Since here C is the centre atom and it has 4 valence electron 

since 3 of the valence electrons are used in making 3 sigma bonds and 1 is used in making pi bond

Hence bonding electron pair = 3

and since no electron is remaining in C => non bonding electron pair = 0

Hence total number of hybridised orbitals = 3+0 = 3

hence hybridisation = sp2 

Step 2

atom labelled 2.

Since here C is the centre atom and it has 4 valence electron 

since all 4 of the valence electrons are used in making 4 sigma bonds

Hence bonding electron pair = 4

and since no electron is remaining in C => non bonding electron pair = 0

Hence total number of hybridised orbitals = 4+0 = 4

hence hybridisation = sp3 

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