Chapter 8: Problem 15
The electron configurations of ions derived from representative elements follow a common pattern. What is the pattern, and how does it relate to the stability of these ions?
Chapter 8: Problem 15
The electron configurations of ions derived from representative elements follow a common pattern. What is the pattern, and how does it relate to the stability of these ions?
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Define ionization energy. Ionization energy measurements are usually made when atoms are in the gaseous state. Why? Why is the second ionization energy always greater than the first ionization energy for any element?
The first four ionization energies of an element are approximately \(579 \mathrm{~kJ} / \mathrm{mol}, 1980 \mathrm{~kJ} / \mathrm{mol}, 2963 \mathrm{~kJ} /\) mol, and \(6180 \mathrm{~kJ} / \mathrm{mol}\). To which periodic group does this element belong?
In halogen displacement reactions a halogen element can be generated by oxidizing its anions with a halogen element that lies above it in the periodic table. This means that there is no way to prepare elemental fluorine, because it is the first member of Group 7A. Indeed, for years the only way to prepare elemental fluorine was to oxidize \(\mathrm{F}\) ions by electrolytic means. Then, in \(1986,\) a chemist reported that by reacting potassium hexafluoromanganate(IV) \(\left(\mathrm{K}_{2} \mathrm{MnF}_{6}\right)\) with antimony pentafluoride \(\left(\mathrm{SbF}_{5}\right)\) at \(150^{\circ} \mathrm{C},\) he had generated elemental fluorine. Balance the following equation representing the reaction: $$ \mathrm{K}_{2} \mathrm{MnF}_{6}+\mathrm{SbF}_{5} \longrightarrow \mathrm{KSbF}_{6}+\mathrm{MnF}_{3}+\mathrm{F}_{2} $$
A technique called photoelectron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with UV light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write $$ h \nu=I E+\frac{1}{2} m u^{2} $$ where \(\nu\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm} .\) Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?
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