For each of the following molecules, draw the Lewis structure and fill in the table as indicated. It is advisable to work in pencil. A dash in the table means that item is not relevant for that molecule. Molecular Geometry Bond Lewis Structure Sketch of Model # Val e- Formula Angle Fa N- Н,о PF CO2 CH HCN C H 79 WORKSHE Question 2 For the molecule CH.O, draw two structurally different Lewis structures that obey the octet rule. Calcu- late the formal charges on each atom (except H) and indicate the value above the atom. Based on the formal charges, circle the preferred structure and then construct and sketch a model of it. Also, indi cate the bond angles and hybridization of the atoms indicated. Lewis Structure 1 Lewis Structure 2 Sketch the model of the preferred structure below and fill in the required information in the spaces provided based on the preferred structure: H-C-H bond angle H-O-C bond angle H-C-O bond angle Hybridization of C Hybridization of 0 Molecular geometry around C: Molecular geometry around 0: 80 WORKSHEE Question 3 For the molecules/ions below, draw Lewis structures that obey the octet rule. Indicate any reasonable resonance structures, geometry, bond angles, and hybridization as indicated. Lewis Structure Formula 0H Detem Gas Law Cons CN Bond angle NH. Hybrid. of N Geometry: Bond angle $02 Hybrid. of S Geometry Bond angles: HNO3 (H is bonded to an O atom) H-0-N 0-N-0 Hybrid. of N Bond angle Co2 Hybrid. of C Geometry: Bond angles: HCO, H-C-O 0-C-0 Hybrid. of C 81
The Correct Answer and Explanation is :
Let’s break down the tasks and concepts related to drawing Lewis structures and calculating formal charges, hybridization, and bond angles for various molecules. I’ll first guide you through the steps for completing the exercise for CH₃OH (methanol), then I’ll briefly explain the other molecules.
Question 2: CH₃OH (Methanol)
To draw the two valid Lewis structures for methanol, we need to obey the octet rule for the atoms involved, except for hydrogen (which can only form one bond). The formal charges must also be calculated to determine the preferred structure.
Lewis Structure 1:
In the first structure, the carbon atom forms single bonds with three hydrogens and one oxygen atom, while the oxygen atom forms a single bond with hydrogen. There are no lone pairs on carbon, and the oxygen has two lone pairs.
Lewis Structure 2:
For the second structure, the bonding pattern remains the same, but we might adjust where the lone pairs are placed, ensuring that the formal charges are minimized.
Formal Charge Calculation:
The formal charge (FC) on an atom is calculated as:
[ FC = V – (L + \frac{B}{2}) ]
Where:
- V is the number of valence electrons of the atom.
- L is the number of lone electrons on the atom.
- B is the number of bonding electrons.
For example, in Lewis Structure 1:
- Carbon (C) has 4 valence electrons, and it’s bonded to four atoms. Its formal charge is 0.
- Oxygen (O) has 6 valence electrons, and after bonding, it has two lone pairs and two bonds. Its formal charge is 0.
- Each hydrogen (H) atom has one bond, and since it has no lone pairs, its formal charge is 0.
Preferred Structure:
- The preferred structure is the one with the fewest formal charges (ideally all atoms with zero formal charge).
Bond Angles and Hybridization:
- Bond Angles:
- The H-C-H bond angle: approximately 109.5° (due to the sp³ hybridization of carbon).
- The H-O-C bond angle: approximately 109.5°.
- The H-C-O bond angle: approximately 109.5°.
- Hybridization:
- The carbon (C) atom is sp³ hybridized because it forms four single bonds.
- The oxygen (O) atom is sp³ hybridized because it forms two single bonds and has two lone pairs.
- Molecular Geometry:
- Around C: Tetrahedral.
- Around O: Bent (due to the lone pairs on oxygen).
Sketch of Preferred Model:
The sketch would show a central carbon atom bonded to three hydrogens and one oxygen atom. The oxygen atom would be bonded to a hydrogen atom, with lone pairs on oxygen.
Question 3: Other Molecules and Ions
For the other molecules/ions in Question 3, the process of drawing Lewis structures and determining the hybridization and bond angles follows similar steps:
- CN⁻ (Cyanide Ion):
- Carbon is sp hybridized (since it forms a triple bond with nitrogen).
- The bond angle is approximately 180° (linear geometry).
- NH₃ (Ammonia):
- Nitrogen is sp³ hybridized (forming three bonds and one lone pair).
- The bond angle is 107°, which is slightly less than the ideal tetrahedral angle due to the lone pair.
- SO₂ (Sulfur Dioxide):
- Sulfur is sp² hybridized.
- The bond angle is 120°, as sulfur forms two double bonds with oxygen in a bent geometry.
- HNO₃ (Nitric Acid):
- Nitrogen is sp² hybridized.
- Bond angles around nitrogen are about 120°, with a trigonal planar geometry.
- CO₂ (Carbon Dioxide):
- Carbon is sp hybridized.
- The bond angle is 180°, with a linear geometry.
- HCO₃⁻ (Bicarbonate Ion):
- The carbon is sp² hybridized.
- The bond angle is 120°, with trigonal planar geometry.
Final Notes:
- Resonance Structures: Some molecules, like SO₂ and HCO₃⁻, may have resonance structures, where electrons are delocalized between atoms.
- Molecular Geometry: The geometry can be determined using VSEPR (Valence Shell Electron Pair Repulsion) theory, considering bonding and lone pairs.
- Bond Angles: Ideal angles correspond to hybridization (sp³ ≈ 109.5°, sp² ≈ 120°, sp ≈ 180°).
This exercise involves applying concepts like the octet rule, formal charge calculations, hybridization theory, and VSEPR to accurately describe molecular structures.