Answer :
To determine the total number of valence electrons in the Lewis structure of \(CH_3CHCl_2\), we need to consider the contributions from each atom in the molecule:
1. Carbon (C):
- Each carbon atom has 4 valence electrons.
- In \(CH_3CHCl_2\), there are 2 carbon atoms.
- Therefore, the total number of valence electrons contributed by carbon atoms is \(2 \times 4 = 8\).
2. Hydrogen (H):
- Each hydrogen atom has 1 valence electron.
- In \(CH_3CHCl_2\), there are 5 hydrogen atoms.
- Therefore, the total number of valence electrons contributed by hydrogen atoms is \(5 \times 1 = 5\).
3. Chlorine (Cl):
- Each chlorine atom has 7 valence electrons.
- In \(CH_3CHCl_2\), there are 2 chlorine atoms.
- Therefore, the total number of valence electrons contributed by chlorine atoms is \(2 \times 7 = 14\).
Now, we sum up all the valence electrons from each type of atom:
[tex]\[8 \text{ (from C)} + 5 \text{ (from H)} + 14 \text{ (from Cl)} = 27 \text{ valence electrons}\][/tex]
Therefore, the total number of valence electrons in the Lewis structure of [tex]\(CH_3CHCl_2\)[/tex] is [tex]\(\boxed{27}\)[/tex].
1. Carbon (C):
- Each carbon atom has 4 valence electrons.
- In \(CH_3CHCl_2\), there are 2 carbon atoms.
- Therefore, the total number of valence electrons contributed by carbon atoms is \(2 \times 4 = 8\).
2. Hydrogen (H):
- Each hydrogen atom has 1 valence electron.
- In \(CH_3CHCl_2\), there are 5 hydrogen atoms.
- Therefore, the total number of valence electrons contributed by hydrogen atoms is \(5 \times 1 = 5\).
3. Chlorine (Cl):
- Each chlorine atom has 7 valence electrons.
- In \(CH_3CHCl_2\), there are 2 chlorine atoms.
- Therefore, the total number of valence electrons contributed by chlorine atoms is \(2 \times 7 = 14\).
Now, we sum up all the valence electrons from each type of atom:
[tex]\[8 \text{ (from C)} + 5 \text{ (from H)} + 14 \text{ (from Cl)} = 27 \text{ valence electrons}\][/tex]
Therefore, the total number of valence electrons in the Lewis structure of [tex]\(CH_3CHCl_2\)[/tex] is [tex]\(\boxed{27}\)[/tex].
