(a) Write a single Lewis structure for N 2 O, and determine the hybridization of the central N atom.

Welcome back everyone. The lewis structure of acetic acid. A byproduct of fermentation is illustrated in the following figure, determine the hybridization of each carbon atom in the molecule. So let's begin by numbering our carbons. We have the left most carbon as carbon one and the right most carbon as carbon number two. Recall that our number of electron domains on an atom will correspond to the hybridization And so beginning with carbon # one. So at carbon number one, we can count three single carbon hydrogen bonds and one single carbon carbon bond, meaning we would have at carbon number one, a total of four domains Electron domains. And this is attributed from our four sigma bonds on carbon number one. So therefore this corresponds to sp three hybridization when we have four domains. So now moving to carbon number two. So at carbon number two, we can see that we have a carbon oxygen double bond, a carbon carbon single bond and a carbon oxygen single bond. So at carbon # two We have a total Of three domains, 3 electron domains. We have to sigma bonds and one pi bond being our carbon oxygen double bond. And so therefore because we have three domains, that means that we have Sp two hybridization for that carbon. And so for our final answers, we can highlight for carbon number one, sorry, that's carbon number one. Here we highlight its sp three hybridization and carbon number two. It's sp two hybridization and this will correspond to choice be in the multiple choice as the correct answer. So I hope that this made sense. And let us know if you have any questions.

Ch.9 - Molecular Geometry and Bonding Theories

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Brown 14th Edition

Ch.9 - Molecular Geometry and Bonding Theories

Problem 64a

Chapter 9, Problem 64a

(a) Write a single Lewis structure for N₂O, and determine the hybridization of the central N atom.

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Identify the total number of valence electrons in N2O. Nitrogen (N) has 5 valence electrons and Oxygen (O) has 6 valence electrons. Since there are two nitrogen atoms and one oxygen atom, calculate the total valence electrons: 2(5) + 6 = 16 valence electrons.

Determine the skeletal structure of N2O. Typically, the least electronegative atom is central, so arrange the atoms as N-N-O.

Distribute the valence electrons to satisfy the octet rule. Start by forming a single bond between the central nitrogen and each of the terminal atoms. Then, distribute the remaining electrons to complete the octets, starting with the outer atoms.

Consider multiple bonds if necessary to satisfy the octet rule for each atom. Adjust the structure by forming double or triple bonds between the nitrogen atoms or between nitrogen and oxygen, ensuring that the total number of electrons remains 16.

Determine the hybridization of the central nitrogen atom. Count the number of regions of electron density (bonds and lone pairs) around the central nitrogen atom. The hybridization corresponds to the number of these regions: 2 regions = sp, 3 regions = sp², 4 regions = sp³, etc.

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Key Concepts

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

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They use dots to represent valence electrons and lines to represent bonds between atoms. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting the behavior of molecules.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding and lone pairs of electrons in a molecule. The type of hybridization (e.g., sp, sp2, sp3) is determined by the number of electron domains around a central atom, influencing the molecule's geometry and bond angles.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is influenced by the number of bonding pairs and lone pairs of electrons around the central atom, which can be predicted using the VSEPR (Valence Shell Electron Pair Repulsion) theory. Understanding molecular geometry is crucial for predicting the physical and chemical properties of substances.