General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 10, Problem 10.101SP
Interpretation Introduction

Interpretation: Using a molecular orbital diagram, the properties such as the bond order and magnetism of N23- and O2- have to be compared.

Concept Introduction:

Molecular orbitals:

Linear combinations of atomic orbitals are called as molecular orbitals. These molecular orbitals are generally divided into two types:

  1. (i) Bonding orbitals
  2. (ii) Antibonding orbitals

Molecular orbital theory:

During the formation of bond, the atomic orbitals of individual atom will combine together to form equal number of molecular orbital. These orbitals are arranged in the order of increasing energy level, while in case of valence bond theory, the individual atomic orbital will directly involve in bond formation.

Bond order:

Bond order of the molecule indicates the number of bonds present between the pair of atoms.

  • Bond order is the measurement of the bond strength of a molecule.
  • An increase in the bond order shows the increase in the bond strength and therefore, a stable compound is formed.
  • Bond order can be calculated by using molecular orbital theory and molecular energy diagram.
  • The formula for calculating bond order is:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Magnetic properties:

Paramagnetic and diamagnetic:

The compounds in which all the electrons are paired in their orbitals are considered as diamagnetic compounds these compounds are repelled by the magnetic field. The paramagnetic compound contains one or more unpaired electrons in their orbital, and these compounds are attracted by the magnetic field.

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Explanation of Solution

General Chemistry, Chapter 10, Problem 10.101SP , additional homework tip  1

General Chemistry, Chapter 10, Problem 10.101SP , additional homework tip  2

Comparison of N23- and O2- based on the molecular orbital diagram:

  • Isoelectric nature: From the molecular orbital diagrams of N23- and O2-, it is clear N23- ion is being isoelectric with O2- ion which means both the ions have exactly the same electronic arrangement in their molecular orbitals.
  • Magnetic properties: From the molecular orbital diagrams of N23- and O2-, it can be clearly observed that both N23- and O2- have an unpaired electron in the anti-bonding pi-2p-orbitals. So, these two ions are paramagnetic in nature.
  • Bond order of N23- ion:

The molecular orbitals of N23-:

The molecular orbital configuration of N23- molecule is:

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

The bond order of N23- can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)72=32=1.5

  • Bond order of O2- ion:

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*2px2)(π*2py1)*Antibondingmolecularorbital

The bond order of O2- can be calculated as follows:

Bondorder=(Numberofelectronsinbondingmolecularorbital)(Numberofelectronsinantibondingmolecularorbital)2

Bondorder=(10)72=32=1.5

Therefore, the bond orders of N23- and O2- is being the same since both the ions are isoelectronic to each other.

The bond orders of N23- and O2- is 1.5. So both N23- and O2- ions can form only half of a single bond.

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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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