Chapter 12, Problem 12D.7P

(a)

Interpretation Introduction

Interpretation:

The proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2s}$ orbital has to be stated.  The proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2p}$ orbital has to be stated.  The total probability that the electron is found on $\text{N}$ atoms has to be stated.

Concept introduction:

EPR stands for electronic paramagnetic resonance.  It is used to study species that contain unpaired electrons.  EPR can be used to study both solids and liquids.  However, the study of the gas phase samples is difficult due to the free rotation of the molecules.  The $g-$ value represents the ease with which the applied field can generate currents through the molecule.

Answer to Problem 12D.7P

The proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2s}$ orbital is $\underset{\_}{10.3\%}$.

The proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2p}$ orbital is $\underset{\_}{38.2\%}$.

The total probability that the electron is found on $\text{N}$ atoms is $\underset{\_}{48.5\%}$.

Explanation of Solution

The given hyperfine interaction of $\text{2s}$ orbital of $\text{N}$ atom with nucleus is $55.2\text{mT}$.

The given isotropic hyperfine interaction of $\text{2s}$ orbital of $\text{N}$ atom with nucleus is $5.7\text{mT}$.

So, the proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2s}$ orbital is calculated by the expression given below.

    $P\left({\text{N}}_{\text{2s}}\right)=\frac{\text{Isotropic hyperfine interaction}\text{of }\text{2s}\text{orbital}}{\text{Hyperfine interaction of }\text{2s}\text{orbital}}$

Substitute the values of hyperfine interaction of $\text{2s}$ orbital and isotropic hyperfine interaction of $\text{2s}$ orbital in the above expression.

    $\begin{array}{c}P\left({\text{N}}_{\text{2s}}\right)=\frac{5.7\text{mT}}{55.2\text{mT}}\\ =0.103\\ =0.103\times 100\%\\ =\underset{\_}{10.3\%}\end{array}$

Therefore, the proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2s}$ orbital is $\underset{\_}{10.3\%}$.

The given hyperfine interaction of $\text{2p}$ orbital of $\text{N}$ atom with nucleus is $3.4\text{mT}$.

The given anisotropic hyperfine interaction of $\text{2p}$ orbital of $\text{N}$ atom with nucleus is $1.3\text{mT}$.

So, the proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2p}$ orbital is calculated by the expression given below.

    $P\left({\text{N}}_{\text{2p}}\right)=\frac{\text{Anisotropic hyperfine interaction}\text{of }\text{2p}\text{orbital}}{\text{Hyperfine interaction of }\text{2p}\text{orbital}}$

Substitute the values of hyperfine interaction of $\text{2p}$ orbital and isotropic hyperfine interaction of $\text{2p}$ orbital in the above expression.

    $\begin{array}{c}P\left({\text{N}}_{\text{2p}}\right)=\frac{1.3\text{mT}}{3.4\text{mT}}\\ =0.382\\ =0.382\times 100\%\\ =\underset{\_}{38.2\%}\end{array}$

Therefore, the proportion of time corresponding to the unpaired electron of ${\text{NO}}_{\text{2}}$ which occupies $\text{2p}$ orbital is $\underset{\_}{38.2\%}$.

The total probability that the electron is found on $\text{N}$ atoms is calculated by adding $P\left({\text{N}}_{\text{2p}}\right)$ and $P\left({\text{N}}_{\text{2s}}\right)$ as follow.

    $P\left(\text{N}\right)=P\left({\text{N}}_{\text{2s}}\right)+P\left({\text{N}}_{\text{2p}}\right)$

Substitute the values of $P\left({\text{N}}_{\text{2p}}\right)$ and $P\left({\text{N}}_{\text{2s}}\right)$ in the above expression

    $\begin{array}{c}P\left(\text{N}\right)=0.103+0.382\\ =0.485\\ =0.485\times 100\%\\ =\underset{\_}{48.5\%}\end{array}$

Therefore, the total probability that the electron is found on $\text{N}$ atoms is $\underset{\_}{48.5\%}$.

(b)

Interpretation Introduction

Interpretation:

The total probability that the electron is found on $\text{O}$ atoms has to be stated. The hybridization ratio of $\text{N}$ atom has to be stated. Whether the hybridization predicts that ${\text{NO}}_{\text{2}}$ is angular or not has to be stated.

Concept introduction:

EPR stands for electronic paramagnetic resonance.  It is used to study species that contain unpaired electrons.  EPR can be used to study both solids and liquids.  However, the study of the gas phase samples is difficult due to the free rotation of the molecules.  The $g-$ value represents the ease with which the applied field can generate currents through the molecule.

Answer to Problem 12D.7P

The total probability that the electron is found on $\text{O}$ atoms is $\underset{\_}{51.5\%}$.

The hybridization ratio of $\text{N}$ atom is $\underset{\_}{3.7}$.

The hybridization predicts that ${\text{NO}}_{\text{2}}$ is angular in shape.

Explanation of Solution

The total probability that the electron is found on $\text{N}$ atoms is $0.485$.

The total probability that the electron is found on $\text{O}$ atoms is calculated by subtracting $P\left(\text{N}\right)$ from $1$  as follow.

    $P\left(\text{O}\right)=1-P\left(\text{N}\right)$

Substitute the value of $P\left(\text{N}\right)$ in the above expression

    $\begin{array}{c}P\left(\text{O}\right)=1-0.485\\ =0.515\\ =0.515\times 100\%\\ =\underset{\_}{51.5\%}\end{array}$

Therefore, the total probability that the electron is found on $\text{O}$ atoms is $\underset{\_}{51.5\%}$.

The expression to calculate the hybridization ratio of $\text{N}$ atom is given below.

    $\text{Hybridization}\text{ratio}\text{of}\text{N}\text{atom}=\frac{P\left({\text{N}}_{\text{2p}}\right)}{P\left({\text{N}}_{\text{2s}}\right)}$

Substitute the values of $P\left({\text{N}}_{\text{2p}}\right)$ and $P\left({\text{N}}_{\text{2s}}\right)$ in the above expression.

    $\begin{array}{c}\text{Hybridization}\text{ratio}\text{of}\text{N}\text{atom}=\frac{0.382}{0.103}\\ =\underset{\_}{3.7}\end{array}$

Therefore, the hybridization ratio of $\text{N}$ atom is $\underset{\_}{3.7}$.

As the shape of ${\text{NO}}_{\text{2}}$ radical is non-linear, thus, ${\text{NO}}_{\text{2}}$ is angular in shape.