Question

Determine the molecular geometry about the left carbon atom in ch3co2ch3.

Answer 1

The molecular geometry about the left carbon atom in ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}$ is $\boxed{{\text{Tetrahedral}}}$ (Refer to the image attached).

Further explanation

The VSEPR theory is also known as valence shell electron pair repulsion theory. This theory is used to determine the geometry of a central atom or ion. It is based on the repulsion between bond pairs and lone pairs in the valence shell of the central atom.

The ${\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}$ classification is used to expresses the geometry of the central atom in VSEPR theory. In ${\text{A}}{{\text{X}}_{\text{m}}}{{\text{E}}_{\text{n}}}$ classification, A is a central atom, X is covalent bond, E is the lone pair, m is the total number of covalent bond present on central atom, and n is the total number of lone pair present on the central atom.

VSEPR model is employed to predict the shape of compounds with a nonmetal as the central atom.

Bond Angle: Bond Angle is the angle between the two bonds that are attached to the same central atom.

Lone Pair: It refers to a pair of valence electrons that that is not shared with another atom.

Molecular Geometry: This is the 3-D arrangement of bonded atoms in a polyatomic ion or molecule.

Electron Pair Geometry: This is the 3-D arrangement of electron pairs around the central atom of a polyatomic ion or molecule.

The main difference between molecular geometry and electron pair geometry is that molecular geometry does not include unpaired electrons, whereas electron pair geometry includes both bonded atoms and unpaired electrons.

Steps of VSEPR to predict the shape of a covalent molecule are as follows:

1. Draw a Lewis structure for the ion or molecule in question.

2. Determine the number of electron pairs around the central atom.

3. Determine the basic geometry of the molecule.

The left carbon is ${\text{C}}{{\text{H}}_{\text{3}}}$. As there are four atoms around carbon and there is no lone pair present. Also, the left carbon forms four covalent bonds. One bond with adjacent carbon atom and one bond with each hydrogen atoms respectively. One 2s and three 2p orbitals of the carbon hybridize to form four ${\text{s}}{{\text{p}}^{\text{3}}}$ orbitals. Therefore, its electron geometry and its molecular geometry both will be tetrahedral.

Tetrahedral geometries result when there are four ligands bonded to a central atom without any lone electron pairs around that center atom.

Hence, the molecular geometry about the left carbon atom in ${\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}{\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}$ is tetrahedral.

Learn more:

1. Complete and balance the following equation: brainly.com/question/5526988

2. Number of covalent bonds does nitrogen formed with its unpaired: brainly.com/question/5974553

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Molecular structure and bonding

Keywords: vsepr, bond pair, lone pair, electron, tetrahedral, CH3, left carbon atom, molecular geometry, CH3CO2CH3.

Answer 2

Molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.

The geometry around left carbon that is CH₃ is tetrahedral.

As the hybridization around left carbon is sp³ that shows its geometry should be tetrahedral and as there are 4 ligands around carbon and there is no lone pair present so the geometry is tetrahedral. So, the molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.