Question

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 core electrons 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 4\(f\) orbitals at 105 eV, from an X-ray source that provided 1253.6 \(\mathrm{eV}\) of energy \(\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{J}\right)\) The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV},\) corresponding to the 1 \(\mathrm{s}\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\) (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 \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-} ;\) which electrons are the valence electrons in each case?

Short answer

The wavelength of the X-rays used in the experiment is 0.0998 nm. The energies from XPS are higher than the first ionization energies for both Mercury (105 eV vs 10.43 eV) and Oxygen (531 eV vs 13.62 eV). The ground-state electron configurations for \(\mathrm{Hg}^{2+}\) is [Xe] \(4f^{14} 5d^{10}\) with valence electrons \(5d^{10}\), and for \(\mathrm{O}^{2-}\) is [He] \(2s^2 2p^6\) with valence electrons \(2s^2 2p^6\).

Step-by-step solution

(a) Calculate the wavelength of X-rays

To find the wavelength of X-rays used in the experiment, we'll make use of Planck's equation, which relates energy and wavelength: \(E = h \cdot \nu\) where E is the energy, h is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and \(\nu\) is the frequency of the X-rays. From the problem, we know the X-ray energy, E = 1253.6 eV = \(1253.6 \cdot 1.602 \times 10^{-19} \mathrm{J}\). We also know that the speed of light, c = \(\nu \cdot \lambda\), where \(\lambda\) is the wavelength. We'll first find the frequency, \(\nu = \frac{E}{h}\), and then find the wavelength, \(\lambda = \frac{c}{\nu}\). Frequency: \(\nu = \frac{1253.6 \cdot 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}} = 3.004 \times 10^{18} \mathrm{s}^{-1}\) Wavelength: \(\lambda = \frac{3 \times 10^{8} m/s}{3.004 \times 10^{18} \mathrm{s}^{-1}} = 9.98 \times 10^{-11} \mathrm{m} = 0.0998 \mathrm{nm} \) The wavelength of the X-rays used in this experiment is 0.0998 nm.

(b) Compare the energies of the 4f electrons in mercury and the 1s electrons in oxygen

From the problem, we know the energy of the 4f electrons of mercury (Hg) is 105 eV and the energy of the 1s electrons of oxygen (O) is 531 eV. The first ionization energies for Mercury and Oxygen can be found in tables. They are 1007.1 kJ/mol for Mercury (Hg) and 1313.9 kJ/mol for Oxygen (O). We need to convert these ionization energies to electron volts (eV) in order to compare them with the given values in the problem. \(1 \mathrm{ev} = 1.602 \times 10^{-19} \mathrm{J}\) \(1 \mathrm{mol} = 6.022 \times 10^{23} atoms\) Ionization energy of Mercury in eV: (1007.1 kJ/mol × \(10^{3} \mathrm{J/1 \mathrm{kJ}\) × \(1 \mathrm{eV/1.602 \times 10^{-19} \mathrm{J}\)) / \(6.022 \times 10^{23} atoms/mol\) = 10.43 eV/atom Ionization energy of Oxygen in eV: (1313.9 kJ/mol × \(10^{3} \mathrm{J/1 \mathrm{kJ}\) × \(1 \mathrm{eV/1.602 \times 10^{-19} \mathrm{J}\)) / \(6.022 \times 10^{23} atoms/mol\) = 13.62 eV/atom Now, let's compare the energies: 4f electrons in Mercury (Hg): 105 eV (from XPS) vs 10.43 eV (first ionization energy) 1s electrons in Oxygen (O): 531 eV (from XPS) vs 13.62 eV (first ionization energy) The energies from XPS are higher than the first ionization energies for both Mercury and Oxygen.

(c) Write the ground-state electron configurations for Hg(2+) and O(2-); identify the valence electrons

To write the ground-state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\), we'll remove or add electrons to the ground-state electron configuration of the neutral atoms to reflect the change in charge. Ground-state electron configuration of Mercury (Hg): [Xe] \(4f^{14} 5d^{10} 6s^2\) \(\mathrm{Hg}^{2+}\) configuration: Remove 2 electrons from the outermost subshell (6s): [Xe] \(4f^{14} 5d^{10}\) Ground-state electron configuration of Oxygen (O): [He] \(2s^2 2p^4\) \(\mathrm{O}^{2-}\) configuration: Add 2 electrons to the outermost subshell (2p): [He] \(2s^2 2p^6\) Valence electrons are the electrons in the outermost orbitals of an atom. For \(\mathrm{Hg}^{2+}\), valence electrons: \(5d^{10}\) For \(\mathrm{O}^{2-}\), valence electrons: \(2s^2 2p^6\)

Key concepts

Oxidation States

Oxidation states indicate the degree of oxidation or reduction of an element within a chemical compound. They are useful for understanding the electron transfer process.
In the case of mercury, it can exist in oxidation states of 0, +1, and +2. This is important in environmental chemistry, where mercury can change states depending on its surroundings.
X-ray Photoelectron Spectroscopy (XPS) is a powerful tool used to determine these oxidation states by examining the energies needed to eject core electrons. Different oxidation states affect core electron energies, providing insights into the chemical behavior of the element.

• The oxidative state impacts the stability and reactivity of elements.
• Mercury’s ability to switch states complicates its environmental impact.
• XPS helps in identifying these states by measuring the energy levels of core electrons.

Understanding oxidation states is crucial for predicting chemical reactions and properties.

Electron Configurations

Electron configurations describe the distribution of electrons in an atom's orbitals. They follow a specific order based on energy levels, starting with the lowest.
For mercury and oxygen ions, understanding changes in electron configurations is key:

• Mercury (Hg) loses two electrons to form Hg²⁺, resulting in the configuration \([Xe] 4f^{14} 5d^{10}\).
• Oxygen (O) gains two electrons to form O^2-, resulting in the configuration \([He] 2s^2 2p^6\).

The valence electrons, or the outermost electrons, play a critical role in chemical bonding and reactions. They determine how an atom will interact with others.
Knowing these configurations helps predict and explain chemical behavior and reactivity.

Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom or ion. It reflects an element's tendency to hold onto its electrons.
For metals and non-metals, ionization energies differ significantly, impacting their reactivity:

• Mercury has a first ionization energy of 10.43 eV per atom, showing a lower tendency to lose an electron compared to oxygen.
• Oxygen’s ionization energy is 13.62 eV per atom, indicating a stronger hold on its electrons.

XPS measures similar energy levels, providing context for these comparisons. By comparing ionization energies to results from XPS, researchers can validate their understanding of electron interactions within materials.
Understanding ionization energies is essential for predicting chemical behavior, determining bonding potential, and explaining reactivity patterns in various environments.