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 7.107 ), 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 forbitals at \(105 \mathrm{eV},\) from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the 1 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 \(4 f\) electrons in mercury and the 1 s 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? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full + \(1 .\)

Step by step solution

01

(a) Calculate the wavelength of the X-rays used in this experiment

To find the wavelength of the X-rays used in the experiment, we can use the relation between energy and wavelength of an electromagnetic wave, given by Planck's equation: \(E = h \cdot c / \lambda\) where: E = energy (1253.6 eV = 1253.6 * 1.6 × 10^(-19) J) h = Planck's constant (\(6.626 \times 10^{-34} \text{J s}\)) c = speed of light (\(3.00 \times 10^8 \text{m s}^{-1}\)) \(\lambda\) = wavelength Solving for \(\lambda\), we have: \(\lambda = \frac{h \cdot c}{E}\)

02

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

The energies of the 4f electrons in mercury and the 1s electrons in oxygen can be found from the given code in the problem: 105 eV for Mercury and 531 eV for Oxygen. The first ionization energies of Mercury and Oxygen can be found from provided data or literature values: - First ionization energy of Mercury: 10.44 eV - First ionization energy of Oxygen: 13.61 eV Now we can compare the energies: Mercury: 105 eV (4f electrons) compared to 10.44 eV (first ionization) Oxygen: 531 eV (1s electrons) compared to 13.61 eV (first ionization)

03

(c) Ground-state electron configurations and valence electrons of Hg^(2+) and O^(2-)

The ground-state electron configurations for Hg^(2+) and O^(2-) are: - Hg^(2+): [Xe] 4f^14 5d^8 - O^(2-): [He] 2s^2 2p^6 In each case, the valence electrons are: - Hg^(2+): 5d - O^(2-): 2s and 2p

04

(d) Z_eff estimation for the 4f and valence electrons of Hg^(2+) and O^(2-)

Using Slater's rules, we can estimate the effective nuclear charge (Z_eff) for the 4f and valence electrons of each ion: Hg^(2+): Z_eff (4f) = Z - S Z = atomic number of Hg = 80 S = total shielding factor due to other electrons: S(4f) = screening effect from 1s + 2s + 2p + ... + 3d = 1 * 75 = 75 Z_eff (4f) = 80 - 75 = 5 O^(2-): Z_eff (valence) = Z - S Z = atomic number of O = 8 S = total shielding factor due to other electrons: S(2s) = screening effect from 1s = 2 * 0.35 = 0.7 S(2p) = screening effect from 1s = 2 * 0.35 = 0.7 Z_eff (valence) = 8 - 0.7 = 7.3 Estimated Z_eff values: - Hg^(2+) 4f electrons: 5 - O^(2-) valence electrons: 7.3

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