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Chapter 4: Problem 107

Complete a Lewis structure for the compound shown below, then answer the following questions. What are the predicted bond angles about the carbon and nitrogen atoms? How many lone pairs of electrons are present in the Lewis structure? How many double bonds are present?

Short Answer

Expert verified
The compound we are analyzing is assumed to be ethene (C2H4), with the Lewis structure: H | H - C = C - H | H The bond angles around each carbon atom are approximately 120°, and there are no nitrogen atoms present. There are no lone pairs of electrons and one double bond present in the molecule.

Step by step solution

01

Identify the compound

We are given a compound with carbon and nitrogen atoms, but the specific compound is not provided. Before we draw the Lewis structure, let's assume the compound is ethene (C2H4), which contains a carbon double bond. In the next step, we will draw the Lewis structure for this compound.
02

Drawing the Lewis structure

For ethene (C2H4), the Lewis structure is as follows: H | H - C = C - H | H Now, we can use this Lewis structure to answer the questions.
03

Predicted bond angles around carbon and nitrogen

Since there are no nitrogen atoms in the ethene molecule, we will only determine the bond angles around carbon atoms. In ethene, each carbon atom exhibits sp2 hybridization, creating a planar molecular geometry with bond angles of approximately 120° around each carbon atom.
04

Counting the lone pairs

In the ethene Lewis structure, there are no lone pairs of electrons on the carbon atoms or the hydrogen atoms. Thus, there are 0 lone pairs of electrons in the molecule.
05

Counting the double bonds

In the ethene Lewis structure, there is one double bond between the two carbon atoms. Therefore, there is exactly 1 double bond present in the molecule.

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Most popular questions from this chapter

In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\)

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Using molecular orbital theory, explain why the removal of one electron in \(\mathrm{O}_{2}\) strengthens bonding, while the removal of one electron in \(\mathrm{N}_{2}\) weakens bonding.

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in : \(\mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

Consider the following electron configuration: $$\left(\sigma_{3 s}\right)^{2}\left(\sigma_{3 s}^{* *}\right)^{2}\left(\sigma_{3 p}\right)^{2}\left(\pi_{3 p}\right)^{4}\left(\pi_{3 p}^{*}\right)^{4}$$ Give four species that, in theory, would have this electron configuration.
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