Chapter 7, Problem 7.138QP

Consider an 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 N*₂ with N₂, where the asterisk denotes the excited molecule, (c) Is N*₂ diamagnetic or paramagnetic? (d) When N*₂ 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.

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$ should be found and the bond length should be compared.  The magnetic properties of ${\text{N}}_{\text{2}}{}^{*}$ 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.
• $\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$
• 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

$\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

$\begin{array}{c}\text{where,}\text{E=}\text{Energy}\\ \text{h=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{λ}\text{=}\text{wavelength of the emitted}\text{light }\end{array}$

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

Answer to Problem 7.138QP

The transition sketch is,

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 ${\text{N}}_{\text{2}}$ is given below

Figure 1

In the ground state of ${\text{N}}_{\text{2}}$ the electrons are in ${\text{σ}}_{{\text{2p}}_{\text{z}}}$ orbital when the ${\text{N}}_{\text{2}}$ gets excited by getting energy the electron move to ${\text{π}}^{\text{*}}{}_{{\text{2p}}_{\text{x}}}\text{or}{\text{π}}^{\text{*}}{}_{{\text{2p}}_{\text{y}}}$ orbitals.

The diagram that showing transition is given below,

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$ should be found and the bond length should be compared.  The magnetic properties of ${\text{N}}_{\text{2}}{}^{*}$ 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.
• $\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$
• 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

$\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

$\begin{array}{c}\text{where,}\text{E=}\text{Energy}\\ \text{h=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{λ}\text{=}\text{wavelength of the emitted}\text{light }\end{array}$

To identify: Bond of order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$. Also to compare its bond length

Answer to Problem 7.138QP

Bond order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$ is 3 and 2 respectively. Also the bond length of ${\text{N}}_{\text{2}}{}^{*}$ is longer than ${\text{N}}_{\text{2}}$.

Explanation of Solution

Electronic configuration of excited nitrogen molecule ${\text{N}}_{\text{2}}$ is $\left[\text{He}\right]{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}{}^{\text{*}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_x}\right)}^{\text{2}}{\left({\sigma }_{{\text{2p}}_{\text{z}}}\right)}^2$

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.

$\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$

$\begin{array}{c}\text{bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\text{6}-\text{0}\right)\\ \text{=}3\end{array}$

Electronic configuration of excited nitrogen molecule ${\text{N}}_{\text{2}}{}^{*}$ is $\left[\text{He}\right]{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}{}^{\text{*}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_x}\right)}^{\text{2}}{\left({\sigma }_{{\text{2p}}_{\text{z}}}\right)}^1{\left({\text{π}}_{{\text{2p}}_{\text{y}}}{}^{*}\right)}^1$

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.

$\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$

$\begin{array}{c}\text{bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\text{5-1}\right)\\ \text{=}2\end{array}$

Bond order of ${\text{N}}_{\text{2}}$ is 3 whereas ${\text{N}}_{\text{2}}{}^{*}$ is 2.

Therefore, the bond length of ${\text{N}}_{\text{2}}{}^{*}$ is longer than ${\text{N}}_{\text{2}}$.

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$ should be found and the bond length should be compared.  The magnetic properties of ${\text{N}}_{\text{2}}{}^{*}$ 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.
• $\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$
• 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

$\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

$\begin{array}{c}\text{where,}\text{E=}\text{Energy}\\ \text{h=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{λ}\text{=}\text{wavelength of the emitted}\text{light }\end{array}$

To identify: The magnetic properties of ${\text{N}}_{\text{2}}{}^{*}$

Answer to Problem 7.138QP

${\text{N}}_{\text{2}}{}^{*}$ 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 ${\text{N}}_{\text{2}}{}^{*}$ is $\left[\text{He}\right]{\left({\text{σ}}_{\text{2s}}\right)}^{\text{2}}{\left({\text{σ}}_{\text{2s}}{}^{\text{*}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_{\text{y}}}\right)}^{\text{2}}{\left({\text{π}}_{{\text{2p}}_x}\right)}^{\text{2}}{\left({\sigma }_{{\text{2p}}_{\text{z}}}\right)}^1{\left({\text{π}}_{{\text{2p}}_{\text{y}}}{}^{*}\right)}^1$

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.

Interpretation Introduction

Interpretation:

The molecular orbital involved in transition should be identified and to sketch the transition.  Bond order of ${\text{N}}_{\text{2}}$ and ${\text{N}}_{\text{2}}{}^{*}$ should be found and the bond length should be compared.  The magnetic properties of ${\text{N}}_{\text{2}}{}^{*}$ 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.
• $\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\text{molecular}\text{orbitals}\end{array}\right)\text{-}\left(\begin{array}{l}\text{number}\text{of}\text{electrons}\text{in}\\ \text{antibonding}\text{molecular}\text{orbitals}\end{array}\right)}{\text{2}}$
• 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

$\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

$\begin{array}{c}\text{where,}\text{E=}\text{Energy}\\ \text{h=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{λ}\text{=}\text{wavelength of the emitted}\text{light }\end{array}$

To determine: The energy difference of the given transition.

Answer to Problem 7.138QP

The energy difference of the given transition is $4.23\times {\text{10}}^{-9}\text{J}$

Explanation of Solution

The energy of light is calculated below.

Given,

The wavelength of light is $470\text{ nm}=470\times {\text{10}}^{-9}\text{m}$.

Planck’s constant is $\text{6}\text{.63}\text{×}{\text{10}}^{-\text{34}}\text{Js}$

Speed of the light is $3\times {10}^8{\text{ms}}^{-1}$

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

$\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$

$\begin{array}{c}\text{where,}\text{E=}\text{Energy}\\ \text{h=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{λ}\text{=}\text{wavelength of the emitted}\text{light }\end{array}$

Substituting the given values in the equation,

$\begin{array}{c}\text{E}\text{=}\frac{\text{hc}}{\text{λ}}\\ =\frac{\text{6}\text{.63}\text{×}{\text{10}}^{-\text{34}}\text{Js}\times 3\times {10}^8{\text{ms}}^{-1}}{470\times {\text{10}}^{-9}\text{m}}\\ =4.23\times {\text{10}}^{-9}\text{J}\end{array}$

The energy difference of the given transition is

$4.23\times {\text{10}}^{-9}\text{J}$