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Chapter 7: Problem 111

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. exo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm} .\) (a) What is the energy of a photon of this light, in joules? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(1.72 \times 10^{-18} \mathrm{~J}\). What is the first ionization energy of \(\mathrm{Hg}\), in $\mathrm{kJ} / \mathrm{mol} ?$ (d) Using Figure 7.10 , determine which of the halogen elements has a first ionization energy closest to that of mercury.

Short Answer

Expert verified
To find the energy of a photon of light, we use the energy formula: \(E = h \times v\), where h is Planck's constant, \(6.626\times 10^{-34}\, \mathrm{J\cdot s}\), and the frequency (v) is found using the formula: \(v = \frac{c}{\lambda}\), with c being the speed of light and \(\lambda\) the wavelength. For the ionization of Hg, the equation is: \(\mathrm{Hg} \xrightarrow{\mathrm{UV\,light}} \mathrm{Hg}^+ + e^-\). The first ionization energy of Hg is calculated using the formula: \(E_{\mathrm{photon}} - E_{\mathrm{kinetic}} = E_{\mathrm{ionization}}\). To convert the ionization energy to \(\mathrm{kJ/mol}\), we use the formula: \(E_{\mathrm{ionization\, kJ/mol}} = E_{\mathrm{ionization}} \times \frac{1000\, \mathrm{J}}{1\, \mathrm{kJ}} \times N_A\), with N_A being Avogadro's number. Finally, by referring to the ionization energy trends in Figure 7.10, we can find the closest halogen element with a first ionization energy similar to that of mercury.

Step by step solution

01

Find the energy of a photon of ultraviolet light

We are given the wavelength of the ultraviolet (UV) light: \(58.4\,\mathrm{nm}\). Let's first convert the wavelength to meters: s \[ 58.4\,\mathrm{nm}\times\frac{1\,\mathrm{m}}{10^9\,\mathrm{nm}}=5.84\times 10^{-8}\,\mathrm{m}. \] We will now find the frequency (v) of the light using the formula: speed of light (c) = frequency (v) * wavelength (λ) \[ v = \frac{c}{\lambda} \] where c is the speed of light, \(3.00\times 10^{8}\, \mathrm{m/s}\), and \(\lambda\) is the wavelength. Now, we will find the energy (E) of a photon of this light using the formula: \[ E = h \times v \] where h is Planck's constant, \(6.626\times 10^{-34}\, \mathrm{J\cdot s}\).
02

Write the ionization equation for Hg

Write an equation representing the first ionization energy of mercury (Hg): \[ \mathrm{Hg} \xrightarrow{\mathrm{UV\,light}} \mathrm{Hg}^+ + e^- \]
03

Find the first ionization energy of Hg

We are given the kinetic energy of the emitted electrons as \(1.72\times 10^{-18} \, \mathrm{J}\). Then, we will find the energy of the emitted photon using the measured kinetic energy. \[ E_{\mathrm{photon}} - E_{\mathrm{kinetic}} = E_{\mathrm{ionization}} \]
04

Convert the ionization energy to kJ/mol

To have the ionization energy in \(\mathrm{kJ/mol}\), we will first multiply the energy by \(\frac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\) and the Avogadro's number, \(N_A=6.022\times 10^{23}\, \mathrm{mol^{-1}}\): \[ E_{\mathrm{ionization\, kJ/mol}} = E_{\mathrm{ionization}} \times \frac{1000\, \mathrm{J}}{1\, \mathrm{kJ}} \times N_A \]
05

Determine the closest halogen element

Using the ionization energy trends for halogens given in Figure 7.10, we can determine which of the halogen elements has a first ionization energy closest to that of mercury.

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Most popular questions from this chapter

(a) Why does xenon react with fluorine, whereas neon does not? (b) Using appropriate reference sources, look up the bond lengths of Xe-F bonds in several molecules. How do these numbers compare to the bond lengths calculated from the atomic radii of the elements?

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, $S .(\mathbf{d})$ For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Which of the following is the expected product of the reaction of \(\mathrm{Mg}(s)\) and \(\mathrm{N}_{2}(g)\) under heat? (i) \(\mathrm{Mg}_{3} \mathrm{~N}(s)\) (ii) \(\mathrm{MgN}_{2}(s)\) (iii) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s),\) (iv) \(\mathrm{Mg}(s)\) and \(\mathrm{N}_{2}(g)\) will not react with one another.

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 forbitals at \(105 \mathrm{eV}\), 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 s$ orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for O. (a) Calculate the wavelength of the \(\mathrm{X}\) 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?

Consider the stable elements through lead \((Z=82) .\) In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?
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