CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL
CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL
12th Edition
ISBN: 9781259292422
Author: Chang
Publisher: MCG
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Chapter 10, Problem 10.108QP

The molecules cis-dichloroethylene and trans-dichloroethylene shown in Section 10.2 can be interconverted by heating or irradiation. (a) Starting with cis-dichloroethylene, show that rotating the C═C bond by 180° will break only the pi bond but will leave the sigma bond intact. Explain the formation of trans-dichloroethylene from this process. (Treat the rotation as two stepwise 90° rotations.) (b) Account for the difference in the bond enthalpies for the pi bond (about 270 kJ/mol) and the sigma bond (about 350 kJ/mol). (c) Calculate the longest wavelength of light needed to bring about this conversion.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

In 1,2-dichloroethylene,while rotating the C=C bond by 180o, only pi-bond breaks whereas sigma bond will be intact. The formation of trans-dichloroethylene from this process has to be explained.

Concept Introduction:

σ-bonds are covalent bonds formed by the end-to-end overlapping of two atomic orbitals. In the σ-bonding, the electronic concentration will be only within the internuclear axis and because of that the σ-bond is being a localized bond.

π-bonds are also covalent bonds formed by the sideways overlapping of two atomic orbitals. In the π-bonding, the electronic concentration will be above and below the molecular plane of the σ-bond and because of that the π-bond is being a delocalized bond.

Rotation of 1,2-dichloroethylene:

1,2-dichloroethylene has two distinct isomers such as cis- and trans- isomers. The double bond between the two carbon atoms will have one sigma bond and one pi- bond. During 180o rotation, the Pi-bond restricts the rotation about the sigma bond. To convert the isomer from one form to other, both sigma and pi bonds have to rotate together. A significant energy is required to carry out this process.

Explanation of Solution

It is known that sigma bond is formed by end-to-end overlap. So, rotation about 180o cannot break sigma bond easily. Since a pi- bond is formed by sideways overlapping, it can be broken easily by the rotation about 180o. If the rotation takes place in two steps like 90o rotations, then in the 1st rotation the pi-bond gets broken and in the 2nd rotation it gets reformed. Consequently, the cis-form gets transformed into the trans-form. So only the breaking of pi-bond brings the conversion. This is how the trans-dichloroethylene is being formed from the rotation about 180o.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The difference in the bond enthalpies for the pi and the sigma bond has to be accounted.

Concept Introduction:

Bond enthalpy is the amount of energy required to break one mole of a particular type of bond. Hence, bond enthalpy decides the bond strength.

Trends of bond enthalpy:

The larger bond enthalpy value for a type of bond means that the bond requires more energy for breaking it which implies that the particular bond is being a strong bond. Whereas the smaller bond enthalpy value for a type of bond means that the bond requires less energy for breaking it which implies that the particular bond is being a weak bond.

Explanation of Solution

The bond enthalpy for the sigma bond is given as 350kJ/mol

The bond enthalpy for the pi bond is given as 270kJ/mol

Clearly, the bond enthalpy value of sigma bond is being higher than that of the pi bond. This difference in the bond enthalpy implies the difference in the bond strength. From this information it can be concluded that sigma bond is stronger bond whereas pi-bond is weaker bond. It is known that sigma bond is formed by end-to-end overlap whereas pi-bond is formed by sideways overlap. The extent of the sideways overlap is less than the end-to-end overlap. Hence, pi-bond is weaker than the sigma bond.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

In the conversion 1,2-dichloroethylene from cis- to trans-form, the longest wavelength of light needed to bring about the conversion, has to be calculated.

Concept Introduction:

In 1,2-dichloroethylene , conversion from cis- to trans- can be achieved by the rotation of the double bond for about 180o.

Rotation of 1,2-dichloroethylene:

1,2-dichloroethylene has two distinct isomers such as cis- and trans- isomers. The double bond between the two carbon atoms will have one sigma bond and one pi- bond. During 180o rotation, the Pi-bond restricts the rotation about the sigma bond. To convert the isomer from one form to other, both sigma and pi bonds have to rotate together. A significant energy is required to carry out this process.

The longest wavelength of light needed to bring about the conversion, can be calculated using the formula shown below:

E=hcλλ=hcEE=Energyrequiredfortheconversionofcistotrans- form.λ=Wavelengthofthelightthatcancausetheconversion.

h=planck'sconstantc=velocity oflight

Answer to Problem 10.108QP

The longest wavelength of light that is needed for the isomeric conversion of 1,2-dichloroethylene is λ=443.43nm.

Explanation of Solution

For the conversion from cis- to trans-form in 1,2-dichloroethylene, it is known that only the breaking of pi-bond brings the conversion. The bond enthalpy value of pi-bond is the amount of energy required to break the pi-bond. The bond enthalpy value of pi-bond is

270kJ/mol.

Converting the bond enthalpy value from kJ/mol to J/mol.

1 kJ=1000J270kJ/mol=1000J1 kJ×270kJ/mol= 270000J/mol

Converting the bond enthalpy value from J/mol into J/molecule

1 mole=6.023×1023molecules270000J/mol=270000J1 mol 270000J6.023×1023molecules=4.4828×1019J/molecule

This is the energy required to for the conversion of cis-to trans-form in one molecule. The wavelength corresponding to this energy can be calculated using the formula as follows:

E=hcλλ=hcEE=Energyrequiredfortheconversionofcistotrans- form.=4.4828×1019J/molecule

λ=Wavelengthofthelightthatcancausetheconversion.

h=planck'sconstant= 6.626×1034J/sc=velocity oflight=3×108m/s

Substituting all the known values in the formula and evaluating it:

λ=hcE=6.626×1034J/s×3×108m/s4.4828×1019J=4.4343×107m

Converting the wavelength from meter into nanometre:

1m=10-9nm4.4343×107m=109nm1m×4.4343×107m=443.43nm

Therefore, λ=443.43nm is the longest wavelength of light that is needed for the isomeric conversion of 1,2-dichloroethylene through rotation.

Conclusion

The longest wavelength of light that is needed for the isomeric conversion of 1,2-dichloroethylene, has been calculated.

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Chapter 10 Solutions

CHEMISTRY-ALEK 360 ACCES 1 SEMESTER ONL

Ch. 10.6 - One way to account for the fact that an O2...Ch. 10.7 - Estimate the bond enthalpy (kJ/mol) of the H2+...Ch. 10.7 - Which of the following species has a longer bond...Ch. 10.8 - Describe the bonding in the nitrate ion (NO3) in...Ch. 10 - How is the geometry of a molecule defined and why...Ch. 10 - Sketch the shape of a linear triatomic molecule, a...Ch. 10 - How many atoms are directly bonded to the central...Ch. 10 - Discuss the basic features of the VSEPR model....Ch. 10 - Prob. 10.5QPCh. 10 - Prob. 10.6QPCh. 10 - Predict the geometries of the following species...Ch. 10 - Predict the geometries of the following species:...Ch. 10 - Predict the geometry of the following molecules...Ch. 10 - Predict the geometry of the following molecules...Ch. 10 - Predict the geometry of the following molecules...Ch. 10 - Predict the geometries of the following ions: (a)...Ch. 10 - Describe the geometry around each of the three...Ch. 10 - Which of the following species are tetrahedral?...Ch. 10 - Prob. 10.15QPCh. 10 - Prob. 10.16QPCh. 10 - Prob. 10.17QPCh. 10 - The bonds in beryllium hydride (BeH2) molecules...Ch. 10 - Referring to Table 10.3, arrange the following...Ch. 10 - The dipole moments of the hydrogen halides...Ch. 10 - List the following molecules in order of...Ch. 10 - Does the molecule OCS have a higher or lower...Ch. 10 - Which of the molecules (a) or (b) has a higher...Ch. 10 - Prob. 10.24QPCh. 10 - What is valence bond theory? 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Use a molecular...Ch. 10 - The following potential energy curve represents...Ch. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Explain why the symbol on the left is a better...Ch. 10 - Determine which of these molecules has a more...Ch. 10 - Nitryl fluoride (FNO2) is very reactive...Ch. 10 - Describe the bonding in the nitrate ion NO3 in...Ch. 10 - Prob. 10.70QPCh. 10 - Which of the following species is not likely to...Ch. 10 - Draw the Lewis structure of mercury(II) bromide....Ch. 10 - Sketch the bond moments and resultant dipole...Ch. 10 - Although both carbon and silicon are in Group 4A,...Ch. 10 - Acetaminophen is the active ingredient in Tylenol....Ch. 10 - Caffeine is a stimulant drug present in coffee....Ch. 10 - Predict the geometry of sulfur dichloride (SCl2)...Ch. 10 - Antimony pentafluoride, SbF5, reacts with XeF4 and...Ch. 10 - Draw Lewis structures and give the other...Ch. 10 - Predict the bond angles for the following...Ch. 10 - Briefly compare the VSEPR and hybridization...Ch. 10 - Describe the hybridization state of arsenic in...Ch. 10 - Draw Lewis structures and give the other...Ch. 10 - Which of the following molecules and ions are...Ch. 10 - Prob. 10.85QPCh. 10 - The N2F2 molecule can exist in either of the...Ch. 10 - Cyclopropane (C3H6) has the shape of a triangle in...Ch. 10 - The compound 1,2-dichloroethane (C2H4Cl2) is...Ch. 10 - Does the following molecule have a dipole moment?...Ch. 10 - So-called greenhouse gases, which contribute to...Ch. 10 - The bond angle of SO2 is very close to 120, even...Ch. 10 - 3-azido-3-deoxythymidine, shown here, commonly...Ch. 10 - The following molecules (AX4Y2) all have...Ch. 10 - The compounds carbon tetrachloride (CCl4) and...Ch. 10 - Prob. 10.95QPCh. 10 - What are the hybridization states of the C and N...Ch. 10 - Use molecular orbital theory to explain the...Ch. 10 - Referring to the Chemistry in Action essay...Ch. 10 - Which of the molecules (a)(c) are polar?Ch. 10 - Prob. 10.100QPCh. 10 - The stable allotropic form of phosphorus is P4, in...Ch. 10 - Referring to Table 9.4, explain why the bond...Ch. 10 - Use molecular orbital theory to explain the...Ch. 10 - The ionic character of the bond in a diatomic...Ch. 10 - Prob. 10.105QPCh. 10 - Prob. 10.106QPCh. 10 - Aluminum trichloride (AlCl3) is an...Ch. 10 - The molecules cis-dichloroethylene and...Ch. 10 - Prob. 10.109QPCh. 10 - Prob. 10.110QPCh. 10 - The molecule benzyne (C6H4) is a very reactive...Ch. 10 - Assume that the third-period element phosphorus...Ch. 10 - Consider a N2 molecule in its first excited...Ch. 10 - Prob. 10.114QPCh. 10 - Prob. 10.116QPCh. 10 - Draw the Lewis structure of ketene (C2H2O) and...Ch. 10 - TCDD, or 2,3,7,8-tetrachlorodibenzo-p-dioxin, is a...Ch. 10 - Write the electron configuration of the cyanide...Ch. 10 - Prob. 10.120QPCh. 10 - The geometries discussed in this chapter all lend...Ch. 10 - Prob. 10.122QPCh. 10 - Which of the following ions possess a dipole...Ch. 10 - Given that the order of molecular orbitals for NO...Ch. 10 - Shown here are molecular models of SX4 for X = F,...Ch. 10 - Based on what you have learned from this chapter...Ch. 10 - How many carbon atoms are contained in one square...
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