CHEMISTRY 1111 LAB MANUAL >C<
CHEMISTRY 1111 LAB MANUAL >C<
1st Edition
ISBN: 9781307092097
Author: Chang
Publisher: MCG/CREATE
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Chapter 10, Problem 10.113QP

Consider a N2 molecule in its first excited electronic state, that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular orbital. (a) Identify the molecular orbitals involved and sketch a diagram to show the transition. (b) Compare the bond order and bond length of N2* with N2, where the asterisk denotes the excited molecule. (c) Is N2* diamagnetic or paramagnetic? (d) When N2* loses its excess energy and converts to the ground state N2, it emits a photon of wavelength 470 nm, which makes up part of the auroras lights. Calculate the energy difference between these levels.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  1 and CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  2 should be found and the bond length should be compared.  The magnetic properties of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  3should 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.
  • CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  4
  • 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

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  5

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  6

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

Answer to Problem 10.113QP

The transition sketch is,

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  7

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 CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  8 is given below

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  9

Figure 1

In the ground state of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  10 the electrons are in CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  11 orbital when the CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  12 gets excited by getting energy the electron move to CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  13 orbitals.

The diagram that showing transition is given below,

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  14

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  15 and CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  16 should be found and the bond length should be compared.  The magnetic properties of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  17should 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.
  • CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  18
  • 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

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  19

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  20

To identify: Bond of order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  21andCHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  22. Also to compare its bond length

Answer to Problem 10.113QP

Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  23 and CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  24 is 3 and 2 respectively. Also the bond length of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  25 is longer than CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  26.

Explanation of Solution

Electronic configuration of excited nitrogen molecule CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  27 is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  28

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.

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  29

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  30

Electronic configuration of excited nitrogen molecule CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  31 is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  32

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.

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  33

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  34

Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  35 is 3 whereas CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  36 is 2.

Therefore, the bond length of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  37 is longer than CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  38.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  39 and CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  40 should be found and the bond length should be compared.  The magnetic properties of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  41should 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.
  • CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  42
  • 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

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  43

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  44

To identify: The magnetic properties of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  45

Answer to Problem 10.113QP

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  46is 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 CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  47 is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  48

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)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  49 and CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  50 should be found and the bond length should be compared.  The magnetic properties of CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  51should 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.
  • CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  52
  • 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

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  53

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  54

To determine: The energy difference of the given transition.

Answer to Problem 10.113QP

The energy difference of the given transition is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  55

Explanation of Solution

The energy of light is calculated below.

Given,

The wavelength of light is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  56.

Planck’s constant is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  57

Speed of the light is CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  58

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

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  59

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  60

Substituting the given values in the equation,

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  61

The energy difference of the given transition is

CHEMISTRY 1111 LAB MANUAL >C<, Chapter 10, Problem 10.113QP , additional homework tip  62

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Chapter 10 Solutions

CHEMISTRY 1111 LAB MANUAL >C<

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? How does it differ...Ch. 10 - Use valence bond theory to explain the bonding in...Ch. 10 - Prob. 10.27QPCh. 10 - Prob. 10.28QPCh. 10 - What is the angle between the following two hybrid...Ch. 10 - How would you distinguish between a sigma bond and...Ch. 10 - Describe the bonding scheme of the AsH3 molecule...Ch. 10 - What is the hybridization state of Si in SiH4 and...Ch. 10 - Describe the change in hybridization (if any) of...Ch. 10 - Consider the reaction BF3+NH3F3BNH3 Describe the...Ch. 10 - What hybrid orbitals are used by nitrogen atoms in...Ch. 10 - What are the hybrid orbitals of the carbon atoms...Ch. 10 - Specify which hybrid orbitals are used by carbon...Ch. 10 - Prob. 10.38QPCh. 10 - The allene molecule H2CCCH2 is linear (the three C...Ch. 10 - Prob. 10.40QPCh. 10 - How many sigma bonds and pi bonds are there in...Ch. 10 - How many pi bonds and sigma bonds are there in the...Ch. 10 - Give the formula of a cation comprised of iodine...Ch. 10 - Give the formula of an anion comprised of iodine...Ch. 10 - What is molecular orbital theory? How does it...Ch. 10 - Sketch the shapes of the following molecular...Ch. 10 - 10.47 Compare the Lewis theory, valence bond...Ch. 10 - Explain the significance of bond order. Can bond...Ch. 10 - Explain in molecular orbital terms the changes in...Ch. 10 - The formation of H2 from two H atoms is an...Ch. 10 - Prob. 10.51QPCh. 10 - Arrange the following species in order of...Ch. 10 - Prob. 10.53QPCh. 10 - Which of these species has a longer bond, B2 or...Ch. 10 - Acetylene (C2H2) has a tendency to lose two...Ch. 10 - Compare the Lewis and molecular orbital treatments...Ch. 10 - Explain why the bond order of N2 is greater than...Ch. 10 - Compare the relative stability of the following...Ch. 10 - Use molecular orbital theory to compare the...Ch. 10 - A single bond is almost always a sigma bond, and a...Ch. 10 - In 2009 the ion N23 was isolated. 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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