Chapter 16, Problem 1DE

Mercury in the environment can exist in oxidation states 0, +1, and +2. One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the coreelectrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4f orbitals at 105 eV, from an X-ray source that provided 1253.6 eV of energy (1 ev = 1.602 x 10 ¹⁹J). The oxygen on the mineral surface gave emitted electron energies at 531 eV, corresponding to the 1s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for Hg and -2 for 0. (a) Calculate the wavelength of the X rays used in this experiment. (b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground-state electron configurations for He²+ and 0²- ; which electrons are the valence electrons in each case?

Interpretation Introduction

To determine: The wavelength of the X rays used in the experiment.

Answer to Problem 1DE

Solution: The wavelength of the X rays used in the experiment is $\text{9}\text{.89}\stackrel{\circ }{\text{A}}$ .

Explanation of Solution

Given

The energy of the X rays is $\text{1253}\text{.6}\text{eV}$ .

The conversion of $\text{eV}$ into joules $\left(\text{J}\right)$ is done as,

$\text{1}\text{eV}=\text{1}{\text{.602×10}}^{\text{19}}\text{J}$

Hence, the conversion of $\text{1253}\text{.6}\text{eV}$ into joules $\left(\text{J}\right)$ is,

$\begin{array}{c}\text{1253}\text{.6}\text{eV}=\frac{\text{1}{\text{.602×10}}^{-19}\text{J}}{\text{1}\text{eV}}\times \text{1253}\text{.6}\text{eV}\\ =\text{2}{\text{.01×10}}^{-16}\text{J}\end{array}$

The wavelength of the X rays is calculated by the formula,

$\lambda =\frac{hc}{E}$

Where,

• $E$ is the energy of theX rays.
• $h$ is the planck’s constant $\left(\text{6}\text{.626}\times {\text{10}}^{-34}\text{J}\cdot \text{s}\right)$ .
• $c$ is the speed of light $\left(\text{3}\text{.0}\times {\text{10}}^{\text{8}}\text{m/s}\right)$ .
• $\lambda$ is the wavelength of the photon.

Substitute the value of $h$ , $c$ , and $E$ in the above equation,

$\begin{array}{c}\lambda =\frac{\text{6}\text{.626}\times {\text{10}}^{-34}\text{J}\cdot \text{s}\times \text{3}\text{.0}\times {\text{10}}^{\text{8}}\text{m/s}}{\text{2}{\text{.01×10}}^{-16}\text{J}}\\ =\text{9}{\text{.89×10}}^{-10}\text{m}\left(\frac{1\stackrel{\circ }{\text{A}}}{{\text{10}}^{-10}\text{m}}\right)\\ =\text{9}\text{.89}\stackrel{\circ }{\text{A}}\end{array}$

Conclusion

The wavelength of the X rays used in the experiment is $\text{9}\text{.89}\stackrel{\circ }{\text{A}}$ .

Interpretation Introduction

To determine: The difference found in the first ionization energies of mercury and oxygen to that of XPS energy for the removal of $\text{4f}$ electron of mercury and $\text{1s}$ electron of oxygen.

Answer to Problem 1DE

Solution: For mercury, $10$ times more energy is required to remove electron from $\text{4f}$ orbital as compared to the $\text{6s}$ orbital.

For oxygen, $40$ times more energy is required to remove electron from $\text{1s}$ orbital as compared to the $\text{2p}$ orbital of oxygen.

Explanation of Solution

The XPS energy for the removal of $\text{4f}$ electron of mercury is $\text{105}\text{eV}$ .

The conversion of $\text{eV}$ into joules $\left(\text{J}\right)$ is done as,

$\text{1}\text{eV}=\text{1}{\text{.602×10}}^{\text{-19}}\text{J}$

Hence, the conversion of $\text{105}\text{eV}$ into joules $\left(\text{J}\right)$ is,

$\begin{array}{c}\text{105}\text{eV}=\frac{\text{1}{\text{.602×10}}^{-19}\text{J}}{\text{1}\text{eV}}\times \text{105}\text{eV}\\ =\text{1}{\text{.68×10}}^{-17}\text{J}\end{array}$

Energy required for 1 mole of mercury is calculated as,

$\begin{array}{c}\text{Energy}\text{reguired}\text{by}\text{1}\text{Hg}\text{atom}=\text{1}{\text{.68×10}}^{\text{-17}}\text{J}\\ \text{Energy}\text{reguired}\text{by}\text{6}\text{.02}\times {\text{10}}^{23}\text{Hg}\text{atom}=\text{1}{\text{.68×10}}^{\text{-17}}\text{J}\times \text{6}\text{.02}\times {\text{10}}^{23}\\ =\text{1}\text{0,113}\text{.6}\text{kJ/mol}\end{array}$

The first ionization energy of mercury is $\text{1007}\text{kJ/mol}$ and the XPS energy to remove electron from $\text{4f}$ orbital is $\text{1}\text{0,113}\text{.6}\text{kJ/mol}$ which is $10$ times greater than the first ionization energy. Hence, $10$ times more energy is required to remove electron from $\text{4f}$ orbital as compared to the $\text{6s}$ orbital.

The XPS energy for the removal of $\text{1s}$ electron of oxygen is $\text{531}\text{eV}$ .

The conversion of $\text{eV}$ into joules $\left(\text{J}\right)$ is done as,

$\text{1}\text{eV}=\text{1}{\text{.602×10}}^{\text{-19}}\text{J}$

Hence, the conversion of $\text{531}\text{eV}$ into joules $\left(\text{J}\right)$ is,

$\begin{array}{c}\text{531}\text{eV}=\frac{\text{1}{\text{.602×10}}^{-19}\text{J}}{\text{1}\text{eV}}\times \text{531}\text{eV}\\ =\text{8}{\text{.51×10}}^{-17}\text{J}\end{array}$

Energy required for 1 mole of oxygen is calculated as,

$\begin{array}{c}\text{Energy}\text{reguired}\text{by}\text{1}\text{O}\text{atom}=\text{8}{\text{.51×10}}^{-17}\text{J}\\ \text{Energy}\text{reguired}\text{by}\text{6}\text{.02}\times {\text{10}}^{23}\text{O}\text{atom}=\text{8}{\text{.51×10}}^{-17}\text{J}\times \text{6}\text{.02}\times {\text{10}}^{23}\\ =\text{51,230}\text{kJ/mol}\end{array}$

The first ionization energy of oxygen is $\text{1314}\text{kJ/mol}$ and the XPS energy to remove electron from $\text{1s}$ orbital is $\text{51,230}\text{kJ/mol}$ which is $40$ times greater than the first ionization energy. Hence, $40$ times more energy is required to remove electron from $\text{1s}$ orbital as compared to the $\text{2p}$ orbital of oxygen.

Conclusion

For mercury, $10$ times more energy is required to remove electron from $\text{4f}$ orbital as compared to the $\text{6s}$ orbital.

For oxygen, $40$ times more energy is required to remove electron from $\text{1s}$ orbital as compared to the $\text{2p}$ orbital of oxygen.

Interpretation Introduction

To determine: The ground state electronic configuration of ${\text{Hg}}^{\text{2+}}$ and ${\text{O}}^{2-}$ ion and the valence electron present in them.

Answer to Problem 1DE

Solution: The ground state electronic configuration of ${\text{Hg}}^{\text{2+}}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}$ and the valence electrons in ${\text{Hg}}^{\text{2+}}$ are $\text{5d}$ orbital electrons.

The ground state electronic configuration of ${\text{O}}^{2-}$ is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^6$ and the valence electrons in ${\text{O}}^{2-}$ are $\text{2p}$ orbital electrons.

Explanation of Solution

The atomic number of $\text{Hg}$ is $80$ and the electronic configuration is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}$ . If $\text{Hg}$ undergoes ionization to ${\text{Hg}}^{\text{2+}}$ , the removal of 2 electrons will take place from outermost shell which is $\text{6s}$ in mercury. So, the ground state electronic configuration of ${\text{Hg}}^{\text{2+}}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}$ . The valence electrons are those that present in outermost orbital of the atom or ion. Hence, valence electrons in ${\text{Hg}}^{\text{2+}}$ are $\text{5d}$ orbital electrons.

The atomic number of $\text{O}$ is $8$ and the electronic configuration is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^{\text{4}}$ . If $\text{O}$ gains electrons to form ${\text{O}}^{2-}$ , the 2 electrons will add to outermost shell which is $\text{2p}$ in oxygen. So, the ground state electronic configuration of ${\text{O}}^{2-}$ is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^6$ . The valence electrons are those that present in outermost orbital of the atom or ion. Hence, valence electrons in ${\text{O}}^{2-}$ are $\text{2p}$ orbital electrons.

Conclusion

The ground state electronic configuration of ${\text{Hg}}^{\text{2+}}$ is $\left[\text{Xe}\right]{\text{4f}}^{\text{14}}{\text{5d}}^{\text{10}}$ and the valence electrons in ${\text{Hg}}^{\text{2+}}$ are $\text{5d}$ orbital electrons.

The ground state electronic configuration of ${\text{O}}^{2-}$ is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^6$ and the valence electrons in ${\text{O}}^{2-}$ are $\text{2p}$ orbital electrons.