Determine the molecular geometry about each interior atom and draw each molecule. (Skeletal structure is indicated in parentheses.) a. C 2 H 2 (skeletal structure HCCH) b. C 2 H 4 (skeletal structure H 2 CCH 2 ) c. C 2 H 6 (skeletal structure H 3 CCH 3 )

Hey everyone today, we've been given the chemical formula and skeletal structure of a compound and were asked to draw the Lewis structure and identified the geometry of each internal atom. So first things first to draw out the Lewis structure. The very first step is to identify how many valence electrons are present within the compound. Carbon is a group for molecule I'm sorry, Group for element which means it will have four valence electrons And we have four carbons hydrogen is the very first element and it will have one valence electron and we have 10 of those. Finally, oxygen is a group six element. It has £6 electrons and we have two of those adding it all up. We get 16 plus 10 plus Or 38 total valence electrons. Now that we have the valence electrons, I'd like to really quickly go back and take a look at this council structure because this will help us right out our drought or lewis structure. Here we see we have a central carbon that is bonded to three metal groups, three CH 3 groups. And to that same central carbon, it is also directly bonded to an O. H. Or sorry to an O. An oxidant which itself is then bonded to an O. H. Group. So with that in mind let's go ahead and start drawing, we have a central carbon. It is attached to three carbons itself and to an O. Which is also connected to another O. Another oxygen. And finally that is to a hydrogen. And each of these external or yeah external carbons are attached to hydrogen themselves. H. H. H. H. And H. Now let's count up how many valence electrons we have so far. We have eight here, hate here and eight here. So that brings us to 24. Then we also have 26, and 30. Remember each single Bond has two electrons in it. However, We've only reached 30 valence electrons. We still need the other eight. This can be really easily added as lone pairs to each of the oxygen's. So this fulfills their octet and We reach our total of eight or 38 valence electrons now to find the geometry of each internal atom. The internal atoms being the oxygen's and the carbons. Well, we can actually group them by element. Now each carbon has four bonds, Right? Each carbon is bonded towards four other molecules. It has no lone pairs. So let's write that out. Carbon has four bonds, four bonds and zero lone pairs. Zero piece. This means that each carbon will have a tetra hydro geometry. It'll look essentially like a plus. It'll be plainer. It'll be within the plane not pointing out in any odd directions. So this will have a tetrahedron geometry. Tetra he drill. And let me just change it to blue. Where is she? Uh How about green? Yeah, we'll change that to green for the oxygen's though. It gets a little trickier. She eats oxygen as two lone pairs. Right? So and each oxygen has two bonds. So let's write that out. Oxygen has two bonds for each and two lone pairs piers. So let's draw that out. Oxygen has two lone pairs and two bonds, while the lone pairs will cause it to bend. So, in other words, and we'll just put R and R. For any any molecule period. So because of this, because of the repulsion that is being induced by these lone pairs, oxygen or oxygen has a bent geometry, it'll be bent. So therefore the geometry for the carbons will all be tetrahedron. The geometry for both oxygen's will be bent. I hope this helps. And I look forward to seeing you on the next one.

Ch.10 - Chemical Bonding II: Molecular Shapes & Valence Bond Theory

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Ch.10 - Chemical Bonding II: Molecular Shapes & Valence Bond Theory

Problem 41

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Chapter 10, Problem 41

Determine the molecular geometry about each interior atom and draw each molecule. (Skeletal structure is indicated in parentheses.)
a. C₂H₂ (skeletal structure HCCH)
b. C₂H₄ (skeletal structure H₂CCH₂)
c. C₂H₆ (skeletal structure H₃CCH₃)

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Identify the type of hybridization for each interior carbon atom in the molecule.

For C2H2 (HCCH), recognize that each carbon is sp hybridized, leading to a linear geometry around each carbon atom.

For C2H4 (H2CCH2), recognize that each carbon is sp2 hybridized, leading to a trigonal planar geometry around each carbon atom.

For C2H6 (H3CCH3), recognize that each carbon is sp3 hybridized, leading to a tetrahedral geometry around each carbon atom.

Draw the molecular structure for each molecule, ensuring to represent the correct geometry around each carbon atom based on the hybridization identified.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) Theory is a model used to predict the geometry of individual molecules based on the repulsion between electron pairs in the valence shell of the central atom. According to VSEPR, electron pairs will arrange themselves as far apart as possible to minimize repulsion, leading to specific molecular shapes such as linear, trigonal planar, or tetrahedral.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding requirements of a molecule. For example, in ethyne (C2H2), the carbon atoms undergo sp hybridization, resulting in a linear geometry, while in ethylene (C2H4), sp2 hybridization leads to a trigonal planar arrangement around each carbon atom.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding pairs and lone pairs of electrons around the central atom. Understanding molecular geometry is crucial for predicting the physical and chemical properties of substances, including reactivity and polarity.