Chapter 11, Problem 11.3PR

(a)

Interpretation Introduction

Interpretation:

Reason for using Raman spectroscopy for studying binding of $O_2$ to haemerythrin not infrared spectroscopy has to be described.

Concept Introduction:

Gross selection rule for a normal mode to be infrared activity:

A normal mode is infrared active, if the motion causes the electric dipole moment of the molecule to change.

Gross selection rule for vibrational Raman spectra:

A normal mode is vibrational Raman active, if the motion causes the change in polarizability of the molecule.

Explanation of Solution

The $O-O$ stretching mode does not change the dipole moment in $O_2$. Hence, in free $O_2$ this mode is not infrared active. It may give a feeble signal. But this mode is active in the vibrational Raman spectroscopy. However, if $O_2$ is bound with a complex, then the $O-O$ stretching mode might change its dipole moment. It is not certain that it may change dipole moment to give a good signal.

(b)

Interpretation Introduction

Interpretation:

Fundamental vibrational wavenumber of the ${}^{18}O{-}^{18}O$ stretching mode in the sample of haemerythrin has to be predicted.

Concept Introduction:

Fundamental vibrational wavenumbers of a molecule can be determined using the given formula.

  $Vibrationalfrequency,\stackrel{\sim }{\nu }\text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\boxed{\begin{array}{c}\text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}}$

Explanation of Solution

Vibrational wavenumber depends on the frequency and effective mass.

  $\stackrel{\sim }{\nu }\alpha {\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}$

Therefore,

  $\begin{array}{c}\frac{\stackrel{\sim }{\nu }\left({}^{18}{O}_2\right)}{\stackrel{\sim }{\nu }\left({}^{16}{O}_2\right)}\text{ = }\frac{m\left({}^{16}{O}_2\right)}{m\left({}^{18}{O}_2\right)}\\ \\ ={\left(\frac{16}{18}\right)}^{\text{1/2}}\\ \\ =\text{ 0}\text{.943}\end{array}$

Vibrational wavenumber of ${}^{16}{O}_2$ bound with haemerythrin is $844{\text{ cm}}^{-1}$. Hence, vibrational wavenumber of ${}^{16}{O}_2$ bound with haemerythrin is $\left(844{\text{ cm}}^{-1}\times 0.943\right){\text{ = 796 cm}}^{-1}$. Thus it is clear that effective masses and isotopic masses are directly proportional. This is valid for free $O_2$ molecule where its effective mass is twice of mass of $O$ atom. It is valid also when $O_2$ molecule strongly bound with a complex at one end whereas the other atom is free.

(c)

Interpretation Introduction

Interpretation:

Reason for the given trend in terms of the electronic structures of ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{-}{\text{ and O}}_2{}^{2-}$ has to be explained. Bond orders of ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{-}{\text{ and O}}_2{}^{2-}$ has to be determined.

Concept Introduction:

Fundamental vibrational wavenumbers of a molecule can be determined using the given formula.

  $Vibrationalfrequency,\stackrel{\sim }{\nu }\text{ = }\frac{1}{2\pi c}{\left(\frac{k_f}{\mu }\right)}^{\text{1/2}}\boxed{\begin{array}{c}\text{μ = Effective mass}\\ {\text{k}}_{\text{f}}\text{= Force constant}\end{array}}$

Bond order: It is the measure of number of electron pairs shared between two atoms.

$\text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs}-\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)$

Explanation of Solution

1. (i) Vibration wavelength depends directly to the square root of force constant. Force constant itself is a measurement used for determining the bond strength that depends on the bond order. Vibration wavelength increases as the bond order increases.
2. (ii) Bond orders of ${\text{O}}_{\text{2}}{\text{, O}}_{\text{2}}{}^{-}{\text{ and O}}_2{}^{2-}$ can be given as,

The electronic configuration of oxygen molecule $O_2$ can be represented as follows,

    ${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^2$

The * represent the antibonding orbital

Therefore,

Bond order of $O_2$ is,

  $\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs}-\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \\ =\frac{\text{1}}{\text{2}}\left(\text{10 }-\text{ 6}\right)\\ \\ =\text{ 2}\end{array}$

The electronic configuration of oxygen molecule $O_2{}^{-}$ can be represented as follows,

    ${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^3$

The * represent the antibonding orbital

Therefore,

Bond order of $O_2{}^{-}$ is,

  $\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs}-\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \\ =\frac{\text{1}}{\text{2}}\left(\text{10 }-\text{ 7}\right)\\ \\ =\text{ 1}\frac{1}{2}\end{array}$

The electronic configuration of oxygen molecule $O_2{}^{2-}$ can be represented as follows,

    ${\left({\text{σ}}_{\text{1s}}\right)}^2{\left({\text{σ*}}_{\text{1s}}\right)}^2{\left({\text{σ}}_{\text{2s}}\right)}^2{\left({\text{σ*}}_{\text{2s}}\right)}^2{\left({\text{σ}}_{\text{2p}}\right)}^2{\left( {\pi }_{\text{2p}}\right)}^4{\left( \pi {*}_{\text{2p}}\right)}^4$

The * represent the antibonding orbital

Therefore,

Bond order of $O_2{}^{2-}$ is,

  $\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\text{1}}{\text{2}}\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\text{bondoing}\text{MOs}-\\ \text{Number}\text{of}\text{electrons}\text{in}\text{antibonding}\text{MOs}\end{array}\right)\\ \\ =\frac{\text{1}}{\text{2}}\left(\text{10 }-\text{ 8}\right)\\ \\ =\text{ 1}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Best described species in haemerythrin has to be identified from ${\text{Fe}}_2^{\text{2+}}{\text{O}}_{\text{2}},{\text{ Fe}}_2^{\text{2+}}{\text{Fe}}_2^{\text{3+}}{\text{O}}_{\text{2}}{\text{ and Fe}}_2^{\text{3+}}{\text{O}}_2^{2-}$.

Explanation of Solution

The wavelength of $O-O$ stretch and that of peroxide anion is similar. Therefore, the best described species in haemerythrin is ${\text{Fe}}^{3+}{O}_2^{2-}$.

(e)

Interpretation Introduction

Interpretation:

The way in which one or more proposed structures are excluded have to be given where two bands are obtained in Raman spectrum for $O-O$ stretching when haemerythrin mixed with ${}^{16}{O}^{18}O$.

Concept Introduction:

Gross selection rule for vibrational Raman spectra:

A normal mode is vibrational Raman active, if the motion causes the change in polarizability of the molecule.

Explanation of Solution

Detecting two bands because of ${}^{16}{O}^{18}O$ indicates that the both atoms are occupied at the non-equivalent positions in the complex. Therefore, structures that are consistent with the given observation are 6 and 7 whereas structure 4 and 5 are inconsistent.