General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 10, Problem 10.92SP

(a)

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.

(a)

Expert Solution
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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)

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.

(b)

Expert Solution
Check Mark

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)

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

(c)

Expert Solution
Check Mark

Answer to Problem 10.92SP

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

General Chemistry

Ch. 10.6 - Prob. 1RCCh. 10.6 - Prob. 2RCCh. 10.6 - Prob. 1PECh. 10 - Prob. 10.1QPCh. 10 - Prob. 10.2QPCh. 10 - 10.3 How many atoms arc directly bonded to the...Ch. 10 - 10.4 Discuss the basic features of the VSEPR...Ch. 10 - 10.5 In the trigonal bipyramidal arrangement, why...Ch. 10 - 10.6 The geometry of CH4 could be square planar,...Ch. 10 - Prob. 10.7QPCh. 10 - Prob. 10.8QPCh. 10 - Prob. 10.9QPCh. 10 - Prob. 10.10QPCh. 10 - 10.11 Describe the geometry around each of the...Ch. 10 - 10.12 Which of these species are tetrahedral?...Ch. 10 - 10.13 Define dipole moment. What are the units and...Ch. 10 - 10.14 What is the relationship between the dipole...Ch. 10 - 10.15 Explain why an atom cannot have a permanent...Ch. 10 - 10.16 The bonds in beryllium hydride (BeH2)...Ch. 10 - 10.17 Referring to Table 10.3. arrange the...Ch. 10 - 10.18 The dipole moments of the hydrogen halides...Ch. 10 - 10.19 List these molecules in order of increasing...Ch. 10 - 10.20 Docs the molecule OCS have a higher or lower...Ch. 10 - 10.21 Which of these molecules has a higher dipole...Ch. 10 - 10.22 Arrange these compounds in order of...Ch. 10 - 10.23 What is valence bond theory? How does it...Ch. 10 - 10.24 Use valence bond theory to explain the...Ch. 10 - 10.25Draw a potential energy curve for the bond...Ch. 10 - 10.26 What is the hybridization of atomic...Ch. 10 - 10.27 How does a hybrid orbital differ from a pure...Ch. 10 - 10.28 What is the angle between these two hybrid...Ch. 10 - 10.29 How would you distinguish between a sigma...Ch. 10 - 10.30 Which of these pairs of atomic orbitals of...Ch. 10 - 10.31 The following potential energy curve...Ch. 10 - 10.32 What is the hybridization state of Si in...Ch. 10 - 10.33 Describe the change in hybridization (if...Ch. 10 - 10.34 Consider the reaction Describe the changes...Ch. 10 - 10.35 What hybrid orbitals are used by nitrogen...Ch. 10 - Prob. 10.36QPCh. 10 - 10.37 Specify which hybrid orbitals are used by...Ch. 10 - 10.38 What is the hybridization state of the...Ch. 10 - 10.39 The allene molecule H2C=C=CH2 is linear (the...Ch. 10 - 10.40 Describe the hybridization of phosphorus in...Ch. 10 - 10.41 How many sigma bonds and pi bonds are there...Ch. 10 - 10.42 How many pi bonds and sigma bonds are there...Ch. 10 - 10.43 Give the formula of a cation comprised of...Ch. 10 - 10.44 Give the formula of an anion comprised of...Ch. 10 - 10.45 What is molecular orbital theory? How does...Ch. 10 - 10.46 Define these terms: bonding molecular...Ch. 10 - 10.47 Sketch the shapes of these molecular...Ch. 10 - 10.48 Explain the significance of bond order. Can...Ch. 10 - 10.49 Explain in molecular orbital terms the...Ch. 10 - Prob. 10.50QPCh. 10 - Prob. 10.51QPCh. 10 - Prob. 10.52QPCh. 10 - Prob. 10.53QPCh. 10 - Prob. 10.54QPCh. 10 - Prob. 10.55QPCh. 10 - 10.56 Compare the Lewis and molecular orbital...Ch. 10 - Prob. 10.57QPCh. 10 - 10.58 Compare the relative stability of these...Ch. 10 - Prob. 10.59QPCh. 10 - Prob. 10.60QPCh. 10 - Prob. 10.61QPCh. 10 - Prob. 10.62QPCh. 10 - Prob. 10.63QPCh. 10 - Prob. 10.64QPCh. 10 - Prob. 10.65QPCh. 10 - Prob. 10.66QPCh. 10 - Prob. 10.67QPCh. 10 - Prob. 10.68QPCh. 10 - 10.69 Draw Lewis structures and give the other...Ch. 10 - Prob. 10.70QPCh. 10 - Prob. 10.71QPCh. 10 - Prob. 10.72QPCh. 10 - Prob. 10.73QPCh. 10 - Prob. 10.74QPCh. 10 - Prob. 10.75QPCh. 10 - Prob. 10.76QPCh. 10 - Prob. 10.77QPCh. 10 - Prob. 10.78QPCh. 10 - Prob. 10.79QPCh. 10 - Prob. 10.80QPCh. 10 - Prob. 10.81QPCh. 10 - Prob. 10.82QPCh. 10 - Prob. 10.83QPCh. 10 - 10.84 The ionic character of the bond in a...Ch. 10 - Prob. 10.85QPCh. 10 - 10.86 Aluminum trichloride (AlCl3) is an...Ch. 10 - Prob. 10.87QPCh. 10 - Prob. 10.88QPCh. 10 - 10.90 Progesterone is a hormone responsible for...Ch. 10 - Prob. 10.91SPCh. 10 - Prob. 10.92SPCh. 10 - Prob. 10.93SPCh. 10 - 10.94 The molecule benzyne (C6H4) is a very...Ch. 10 - Prob. 10.95SPCh. 10 - 10.96 As mentioned in the chapter, the Lewis...Ch. 10 - Prob. 10.97SPCh. 10 - Prob. 10.98SPCh. 10 - Prob. 10.99SPCh. 10 - Prob. 10.100SPCh. 10 - Prob. 10.101SPCh. 10 - Prob. 10.102SP
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