General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 10, Problem 10.95SP

(a)

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of N2 and N2* should be found and the bond length should be compared.  The magnetic properties of N2* should be found out.  The energy difference of the given transition should be determined

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals
  • The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.
  • Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2
  • Paramagnetic species contains at least one unpaired electrons and can be attracted towards magnetic fields. Diamagnetic species does have any unpaired electrons. That is spins of all the electrons are paired. It slightly repelled towards the magnetic fields

  E=hcλ

  where,E=Energyh=Planck'sconstantc=speedoflightλ=wavelength of the emittedlight 

To identify: molecular orbital involved in transition and to sketch the transition.

(a)

Expert Solution
Check Mark

Answer to Problem 10.95SP

The transition sketch is,

General Chemistry, Chapter 10, Problem 10.95SP , additional homework tip  1

Explanation of Solution

In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals and only contain a maximum of two electrons. The number of newly formed molecular orbital is equal to the number of atomic orbitals involved in the bonding.

There are two types of molecular orbitals,

  1. a) Bonding molecular orbitals: sharing of electron density is between the nuclei and has comparatively lower energy and fills first.
  2. b) Antibonding molecular orbitals: Two nuclei is pulled by the electrons density in opposite direction and has higher energy comparing to bonding molecular orbital.

Molecular orbital diagram of N2 is given below

General Chemistry, Chapter 10, Problem 10.95SP , additional homework tip  2

Figure 1

In the ground state of N2 the electrons are in σ2pz orbital when the N2 gets excited by getting energy the electron move to π*2pxorπ*2py orbitals.

The diagram that showing transition is given below,

General Chemistry, Chapter 10, Problem 10.95SP , additional homework tip  3

(b)

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of N2 and N2* should be found and the bond length should be compared.  The magnetic properties of N2* should be found out.  The energy difference of the given transition should be determined

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals
  • The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.
  • Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2
  • Paramagnetic species contains at least one unpaired electrons and can be attracted towards magnetic fields. Diamagnetic species does have any unpaired electrons. That is spins of all the electrons are paired. It slightly repelled towards the magnetic fields

  E=hcλ

  where,E=Energyh=Planck'sconstantc=speedoflightλ=wavelength of the emittedlight 

To identify: Bond of order of N2 and N2*. Also to compare its bond length

(b)

Expert Solution
Check Mark

Answer to Problem 10.95SP

Bond order of N2 and N2* is 3 and 2 respectively. Also the bond length of N2* is longer than N2.

Explanation of Solution

Electronic configuration of excited nitrogen molecule N2 is [He](σ2s)2(σ2s*)2(π2py)2(π2px)2(σ2pz)2

The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.

Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2

bondorder=12(60)=3

Electronic configuration of excited nitrogen molecule N2* is [He](σ2s)2(σ2s*)2(π2py)2(π2px)2(σ2pz)1(π2py*)1

The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.

Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2

bondorder=12(5-1)=2

Bond order of N2 is 3 whereas N2* is 2.

Therefore, the bond length of N2* is longer than N2.

(c)

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of N2 and N2* should be found and the bond length should be compared.  The magnetic properties of N2* should be found out.  The energy difference of the given transition should be determined

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals
  • The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.
  • Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2
  • Paramagnetic species contains at least one unpaired electrons and can be attracted towards magnetic fields. Diamagnetic species does have any unpaired electrons. That is spins of all the electrons are paired. It slightly repelled towards the magnetic fields

  E=hcλ

  where,E=Energyh=Planck'sconstantc=speedoflightλ=wavelength of the emittedlight 

To identify: The magnetic properties of N2*

(c)

Expert Solution
Check Mark

Answer to Problem 10.95SP

N2* is diamagnetic

Explanation of Solution

Paramagnetic species contains at least one unpaired electrons and can be attracted towards magnetic fields. Diamagnetic species does have any unpaired electrons. That is spins of all the electrons are paired. It slightly repelled towards the magnetic fields.

Electronic configuration of excited nitrogen molecule N2* is [He](σ2s)2(σ2s*)2(π2py)2(π2px)2(σ2pz)1(π2py*)1

Even though there are unpaired electrons, the spin of the electrons was not change in the time of transition. All the electrons are paired so it is diamagnetic.

(d)

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of N2 and N2* should be found and the bond length should be compared.  The magnetic properties of N2* should be found out.  The energy difference of the given transition should be determined

Concept Introduction:

  • In molecular orbital theory, when the bonding takes place the atomic orbitals that take part combine to get a new orbital that has the properties of the whole molecule. The newly formed orbitals are known as molecular orbitals
  • The bond order gives an idea about the stability of a molecule. It can be calculated using the molecular orbital theory. The stability of a molecule increase as the bond order increases.
  • Bondorder=(numberofelectronsinbondingmolecularorbitals)-(numberofelectronsinantibondingmolecularorbitals)2
  • Paramagnetic species contains at least one unpaired electrons and can be attracted towards magnetic fields. Diamagnetic species does have any unpaired electrons. That is spins of all the electrons are paired. It slightly repelled towards the magnetic fields

  E=hcλ

  where,E=Energyh=Planck'sconstantc=speedoflightλ=wavelength of the emittedlight 

To determine: The energy difference of the given transition.

(d)

Expert Solution
Check Mark

Answer to Problem 10.95SP

The energy difference of the given transition is 4.23×109J

Explanation of Solution

The energy of light is calculated below.

Given,

The wavelength of light is 470 nm=470 ×109m.

Planck’s constant is 6.63×1034Js

Speed of the light is 3×108ms1

The energy of light is calculated is calculated by the equation,

  E=hcλ

where,E=Energyh=Planck'sconstantc=speedoflightλ=wavelength of the emittedlight 

Substituting the given values in the equation,

  E=hcλ=6.63×1034Js×3×108ms1470 ×109m=4.23×109J

The energy difference of the given transition is

4.23×109J

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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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