What VSEPR “shapes” and bond angles do you find in Toluene?
The C-H bonds are sp and linear.
The C-C bonds are trigonal planar and 120 degrees according to this site:(Broken Link Removed)
To determine the VSEPR shapes and bond angles in toluene (C7H8), we first need to determine the Lewis structure of the molecule.
1. Draw the skeletal structure of toluene:
CH3-CH2-CH2-CH2-CH2-CH3
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C6H5
2. Count the total number of valence electrons:
Carbon (C) has 4 valence electrons, Hydrogen (H) has 1 valence electron. So,
7 (from Carbon) + 8 (from Hydrogen) = 15 valence electrons
3. Distribute the valence electrons around the atoms:
Place a single bond between each hydrogen atom (H) and the carbon atoms (C). Each carbon atom should have 4 bonds.
The remaining valence electrons should be placed around the carbon atoms to complete their octets.
Put the remaining valence electrons on the central carbon atom (C) that is bonded to the phenyl group (C6H5).
The final Lewis structure of toluene is:
H H
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H-C-C-C-C-C-C-C-H
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C6H5
Now, let's analyze the VSEPR shapes and bond angles based on the Lewis structure:
- The carbon atoms (C) are all surrounded by four bonded pairs of electrons, resulting in a tetrahedral electron domain geometry.
- The central carbon atom (C) is also attached to a phenyl group, which is a trigonal planar structure. This introduces a planar distortion to the shape.
- The bond angles are approximately 109.5 degrees for the tetrahedral arrangement and 120 degrees for the trigonal planar arrangement.
Therefore, in toluene, the VSEPR shapes are primarily tetrahedral for the carbon atoms and trigonal planar for the central carbon atom attached to the phenyl group. The bond angles are around 109.5 degrees and 120 degrees, respectively.
To determine the VSEPR shapes and bond angles in toluene (C7H8), we need to look at the Lewis structure of the molecule first.
Step 1: Draw the Lewis structure of toluene (C7H8)
Toluene consists of a benzene ring (a hexagon with alternating double bonds) with a methyl group (-CH3) attached to it.
H H H
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H–C–C–C–C–C–C–C
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CH3
Methyl group
Step 2: Identify the electron pairs around each atom
In the benzene ring, each carbon atom is connected to a single hydrogen atom. Therefore, each carbon atom has a total of 4 electron pairs: 1 for each of the three sigma bonds and 1 for the lone pair of electrons. The methyl group has 4 bonding pairs (C-H bonds) and no lone pairs.
Step 3: Determine the electron domain geometry
The electron domain geometry is the three-dimensional arrangement of electron domains (bonding pairs and lone pairs) around the central atom or each atom in the molecule.
In toluene, each carbon atom in the benzene ring has a trigonal planar electron domain geometry (3 bond pairs and 1 lone pair), while the carbon atom in the methyl group has a tetrahedral electron domain geometry (4 bond pairs).
Step 4: Determine the molecular geometry
The molecular geometry describes the arrangement of atoms in space, considering only the positions of the atoms and not the bonding electrons. To determine the molecular geometry, we need to consider the number of bonding and lone pairs around the central atom.
In benzene:
Each carbon atom has a trigonal planar electron domain geometry. However, since there is no lone pair, the molecular geometry is also trigonal planar. Therefore, the benzene ring has a trigonal planar shape, and all bond angles are around 120 degrees.
In the methyl group:
The carbon in the methyl group has a tetrahedral electron domain geometry. Since there are no lone pairs, the molecular geometry remains tetrahedral. Therefore, the methyl group has a tetrahedral shape, with bond angles of approximately 109.5 degrees (typical of tetrahedral arrangement).
In conclusion, the VSEPR shapes and bond angles in toluene are as follows:
- The benzene ring has a trigonal planar shape with bond angles around 120 degrees.
- The methyl group has a tetrahedral shape with bond angles approximately 109.5 degrees.