General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 10, Problem 10.102SP
Interpretation Introduction

Interpretation: The given species have to be arranged in the increasing order of their bond order.

Concept Introduction:

Bond order:

Bond order determines the number of bonds in the pair of two atoms. So, it is the quantitative measure of a bond.

The bond order can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Trends in the bond order:

The species with the negative charge means that electrons are getting added in the anti-bonding molecular orbital. So the number of electrons in the anti-bonding molecular orbital increases accordingly. Ultimately, the bond order decreases. Whereas, the species with the positive charge means that electrons are being removed from the anti-bonding molecular orbital. So the number of electrons in the anti-bonding molecular orbital decreases accordingly. Ultimately, the bond order increases.

Expert Solution & Answer
Check Mark

Answer to Problem 10.102SP

The given species can be arranged in the increasing order of their bond order as follows:

NO2-<NO-<NO=NO2+<NO+

Explanation of Solution

Given: The order of molecular orbitals of NO is similar to that of O2.

The molecular orbitals of O2:

The molecular orbital configuration of O2 molecule is:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px1=π*2py1)*Antibondingmolecularorbital

Bond order of NO:

The molecular orbitals of NO:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px1)(π*2py0)*Antibondingmolecularorbital

The bond order of NO can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)52=52=2.5

Similarly, the bond orders can be calculated for each of the given species as follows:

Bond order of NO2-:

The molecular orbitals of NO2-:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px2)(π*2py1)*Antibondingmolecularorbital

The bond order of NO2- can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)72=32=1.5

Bond order of NO-:

The molecular orbitals of NO-:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px1)(π*2py1)*Antibondingmolecularorbital

The bond order of NO- can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)62=42=2

Bond order of NO+:

The molecular orbitals of NO+:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2=π2py2)(π*2px0)(π*2py0)*Antibondingmolecularorbital

The bond order of NO+ can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)42=62=3

Bond order of NO2+:

The molecular orbitals of NO2+:

(σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px2)(π2py1)(π*2px0)(π*2py0)*Antibondingmolecularorbital

The bond order of NO2+ can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(9)42=52=2.5

The calculated bond orders can be tabulated as follows:

Given moleculeBond order.
NO2-1.5
NO-2
NO2.5
NO+3
NO2+2.5

Based on this tabulation, the given species can be arranged in the increasing order of their bond order as follows:

NO2-<NO-<NO=NO2+<NO+

Conclusion

The given species have been arranged in the increasing order of their bond order.

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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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