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Chapter 9: Problem 71

Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule. a. \(\mathrm{BH}_{3}\) b. \(\mathrm{N}_{2} \mathrm{~F}_{2}\) c. \(\mathrm{C}_{4} \mathrm{H}_{6}\)

Short Answer

Expert verified
In summary: a. BH3: Lewis structure → trigonal planar molecular structure → nonpolar → 120° bond angles → boron with sp2 hybrid orbitals. b. N2F2: Lewis structure → linear molecular structure → polar → 180° bond angles → nitrogen with sp hybrid orbitals. c. C4H6 (1,3-butadiene): Lewis structure → chain molecular structure → nonpolar → 120° and 109.5° bond angles → carbon with sp2 and sp3 hybrid orbitals.

Step by step solution

01

Find the total number of valence electrons.

Boron (B) has 3 valence electrons and hydrogen (H) has 1 valence electron. Since there are 3 hydrogen atoms, the total number of valence electrons in BH3 is \(3 + 3 \times 1 = 6\).
02

Complete the Lewis structure.

Place Boron in the center and surround it with three hydrogen atoms. Distribute the 6 valence electrons as bonds (shared electron pairs) between boron and the hydrogen atoms. Now, the Lewis structure for BH3 is: H | B - H
03

Molecular structure.

The molecular structure of BH3 is trigonal planar, with bond angles of 120°.
04

Polarity.

BH3 is a nonpolar molecule because the electronegativity difference between boron and hydrogen is small (about 0.4) and the molecule is symmetric.
05

Hybrid orbitals.

Boron uses sp2 hybrid orbitals to form sigma bonds with three hydrogen atoms. #Part b: N2F2#
06

Find the total number of valence electrons.

Nitrogen (N) has 5 valence electrons and fluorine (F) has 7 valence electrons. Since there are 2 nitrogen and 2 fluorine atoms, the total number of valence electrons in N2F2 is \(2 \times 5 + 2 \times 7 = 24\).
07

Complete the Lewis structure.

Start by connecting the two nitrogen atoms with a triple bond, and then add a single bond to each fluorine atom. Distribute the remaining valence electrons as lone pairs on the nitrogen and fluorine atoms. The Lewis structure for N2F2 is: F - N ≡ N - F
08

Molecular structure.

The molecular structure of N2F2 is linear, with bond angles of 180°.
09

Polarity.

N2F2 is a polar molecule because there is a significant electronegativity difference between nitrogen (about 3.0) and fluorine (about 4.0). The symmetry of the molecule does not cancel out the polarity.
10

Hybrid orbitals.

Nitrogen (marked by an asterisk) uses sp hybrid orbitals to form sigma bonds with another nitrogen atom and a fluorine atom. #Part c: C4H6#
11

Find the total number of valence electrons.

Carbon (C) has 4 valence electrons and hydrogen (H) has 1 valence electron. Since there are 4 carbon atoms and 6 hydrogen atoms, the total number of valence electrons in C4H6 is \(4 \times 4 + 6 \times 1 = 22\).
12

Complete the Lewis structure.

There are multiple possible structures for C4H6. We will consider the simplest structure, 1,3-butadiene, for this exercise. Arrange the carbon atoms in a chain and add a double bond between the first and second carbons and between the third and fourth carbons. Attach the hydrogen atoms to the carbon atoms. The Lewis structure for 1,3-butadiene is: H2C = CH - CH = CH2
13

Molecular structure.

The molecular structure of 1,3-butadiene is a chain, with bond angles of 120° around the sp2 hybridized carbons and 109.5° around the sp3 hybridized carbons.
14

Polarity.

1,3-butadiene is a nonpolar molecule because the electronegativity difference between carbon and hydrogen is small (about 0.4) and the molecule is symmetric.
15

Hybrid orbitals.

Carbon atoms marked by asterisks use sp2 hybrid orbitals to form sigma bonds with other carbon atoms and hydrogen atoms.

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Most popular questions from this chapter

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? a. \(\mathrm{Li}_{2}\) b. \(\bar{C}_{2}\) c. \(S_{2}\)

Consider the following molecular orbitals formed from the combination of two hydrogen \(1 s\) orbitals: a. Which is the bonding molecular orbital and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \\ \text { Cyanamide } \end{aligned} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: Dicyandiamide not shown) a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

One of the first drugs to be approved for use in treatment of acquired immune deficiency syndrome (AIDS) was azidothymidine (AZT). Complete the Lewis structure for AZT. a. How many carbon atoms are \(s p^{3}\) hybridized? b. How many carbon atoms are \(s p^{2}\) hybridized? c. Which atom is \(s p\) hybridized? d. How many \(\sigma\) bonds are in the molecule? e. How many \(\pi\) bonds are in the molecule? f. What is the \(\mathrm{N}=\mathrm{N}=\mathrm{N}\) bond angle in the azide \(\left(-\mathrm{N}_{3}\right)\) group? g. What is the \(\mathrm{H}-\mathrm{O}-\mathrm{C}\) bond angle in the side group attached to the five-membered ring? h. What is the hybridization of the oxygen atom in the \(-\mathrm{CH}_{2} \mathrm{OH}\) group?
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