Chapter 6

Cesium has the smallest ionization energy of all elements ( 376 \(\mathrm{kJ} / \mathrm{mol}\) ), and chlorine has the most negative electron affinity \((-349 \mathrm{~kJ} / \mathrm{mol}) .\) Will a cesium atom transfer an electron to a chlorine atom to form isolated \(\mathrm{Cs}^{+}(\mathrm{g})\) and \(\mathrm{Cl}^{-}(g)\) ions? Explain.

Born-Haber cycles, such as those shown in Figures \(6.6\) and \(6.7\), are called cycles because they form closed loops. If any five of the six energy changes in the cycle are known, the value of the sixth can be calculated. Use the following five values to calculate the lattice energy in kilojoules per mole for sodium hydride, \(\mathrm{NaH}(s)\) : \(E_{\mathrm{ea}}\) for \(\mathrm{H}(\mathrm{g})=-72.8 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Na}(g)=+495.8 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Na}(s)=+107.3 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{H}_{2}(g)=+435.9 \mathrm{~kJ} / \mathrm{mol}\) Net energy change for the formation of \(\mathrm{NaH}_{2}(s)\) from its elements \(=-60 \mathrm{~kJ} / \mathrm{mol}\).

Calculate a lattice energy for \(\mathrm{CaH}_{2}(s)\) in kilojoules per mole using the following information: \(E_{\mathrm{ea}}\) for \(\mathrm{H}(\mathrm{g})=-72.8 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Ca}(g)=+589.8 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Ca}(g)=+1145 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Ca}(s)=+178.2 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{H}_{2}(g)=+435.9 \mathrm{~kJ} / \mathrm{mol}\) Net energy change for the formation of \(\mathrm{CaH}_{2}(s)\) from its elements \(=-186.2 \mathrm{~kJ} / \mathrm{mol}\)

Calculate the overall energy change in kilojoules per mole for the formation of \(\mathrm{CsF}(s)\) from its elements using the following data: \(E_{\mathrm{ea}}\) for \(\mathrm{F}(\mathrm{g})=-328 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\operatorname{Cs}(g)=+375.7 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Cs}(g)=+2422 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Cs}(s)=+76.1 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{F}_{2}(\mathrm{~g})=+158 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{CsF}(s)=+740 \mathrm{~kJ} / \mathrm{mol}\)

Calculate the overall energy change in kilojoules per mole for the formation of \(\mathrm{CaCl}(s)\) from the elements. The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{Cl}(\mathrm{g})=-348.6 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Ca}(g)=+589.8 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Ca}(g)=+1145 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Ca}(s)=+178.2 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{Cl}_{2}(g)=+243 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{CaCl}_{2}(s)=+2258 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{CaCl}(s)=+717 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

\(\mathrm{Cu}^{+}\) has an ionic radius of \(77 \mathrm{pm}\), but \(\mathrm{Cu}^{2+}\) has an ionic radius of \(73 \mathrm{pm}\). Explain.

The following ions all have the same number of electrons: \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-} .\) Order them according to their expected sizes, and explain your answer.

The following ions all have the same number of electrons: \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-} .\) Order them according to their expected sizes, and explain your answer. 6.88 Calculate overall energy changes in kilojoules per mole for the formation of \(\mathrm{MgF}(s)\) and \(\mathrm{MgF}_{2}(s)\) from their elements. In light of your answers, which compound is more likely to form in the reaction of magnesium with fluorine, \(\mathrm{MgF}\) or \(\mathrm{MgF}_{2}\) ? The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{F}(\mathrm{g})=-328 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Mg}(g)=+737.7 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Mg}(g)=+1450.7 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Mg}(s)=+147.7 \mathrm{~kJ} / \mathrm{mol}\)Bond dissociation energy for \(\mathrm{F}_{2}(\mathrm{~g})=+158 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}_{2}(s)=+2952 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}(s)=930 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

Many early chemists noted a diagonal relationship among elements in the periodic table, whereby a given element is sometimes more similar to the element below and to the right than it is to the element directly below. Lithium is more similar to magnesium than to sodium, for example, and boron is more similar to silicon than to aluminum. Use your knowledge about the periodic trends of such properties as atomic radii and \(Z_{\text {eff }}\) to explain the existence of diagonal relationships.

(a) Which element from each set has the largest atomic radius? Explain. (i) \(\mathrm{Ba}, \mathrm{Ti}, \mathrm{Ra}, \mathrm{Li}\) (ii) \(\mathrm{F}, \mathrm{Al}, \mathrm{In}, \mathrm{As}\) (b) Which element from each set has the smallest ionization energy? Explain. (i) \(\mathrm{Tl}, \mathrm{Po}, \mathrm{Se}, \mathrm{Ga}\) (ii) \(\mathrm{Cs}, \mathrm{Ga}, \mathrm{Bi}, \mathrm{Se}\)

a) Which of the elements \(\mathrm{Be}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) has the most negative electron affinity? Explain. (b) Which of the ions \(\mathrm{Se}^{2-}, \mathrm{F}, \mathrm{O}^{2-}\), and \(\mathrm{Rb}^{+}\) has the largest radius? Explain.

Given the following information, construct a Born-Haber cycle to calculate the lattice energy of \(\mathrm{CaC}_{2}(s)\). Net energy change for the formation of \(\mathrm{CaC}_{2}(s)=-60 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Ca}(s)=+178 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i]}\) for \(\mathrm{Ca}(g)=+590 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Ca}(g)=+1145 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{C}(s)=+717 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{C}_{2}(g)=+614 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\text {eal }}\) for \(\mathrm{C}_{2}(g)=-315 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{ea} 2}\) for \(\mathrm{C}_{2}(\mathrm{~g})=+410 \mathrm{~kJ} / \mathrm{mol}\)

Given the following information, construct a Born-Haber cycle to calculate the lattice energy of \(\mathrm{CrCl}_{2} \mathrm{I}(s)\) : Net energy change for the formation of \(\mathrm{CrCl}_{2} \mathrm{I}(s)=-420 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{Cl}_{2}(g)=+243 \mathrm{~kJ} / \mathrm{mol}\) Bond dissociation energy for \(\mathrm{I}_{2}(s)=+151 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{I}_{2}(s)=+62 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Cr}(s)=+397 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{il}}\) for \(\mathrm{Cr}(g)=652 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 2}\) for \(\mathrm{Cr}(g)=1588 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{i} 3}\) for \(\mathrm{Cr}(\mathrm{g})=2882 \mathrm{~kJ} / \mathrm{mol}\) \(E_{\mathrm{ca}}\) for \(\mathrm{Cl}(g)=-349 \mathrm{~kJ} / \mathrm{mol}\) \(E_{e a}\) for \(\mathrm{I}(g)=-295 \mathrm{~kJ} / \mathrm{mol}\)

Consider the electronic structure of the element bismuth. (a) The first ionization energy of bismuth is \(E_{i 1}=+703 \mathrm{~kJ} / \mathrm{mol}\). What is the longest possible wavelength of light that could ionize an atom of bismuth? (b) Write the electron configurations of neutral \(\mathrm{Bi}\) and the \(\mathrm{Bi}^{+}\) cation. (c) What are the \(n\) and \(l\) quantum numbers of the electron removed when \(\mathrm{Bi}\) is ionized to \(\mathrm{Bi}^{+} ?\) (d) Would you expect element 115 to have an ionization energy greater than, equal to, or less than that of bismuth? Explain.

Iron is commonly found as \(\mathrm{Fe}, \mathrm{Fe}^{2+}\), and \(\mathrm{Fe}^{3+}\). (a) Write electron configurations for each of the three. (b) What are the \(n\) and \(l\) quantum numbers of the electron removed on going from \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) ? (c) The third ionization energy of Fe is \(E_{i 3}=+2952 \mathrm{~kJ} / \mathrm{mol}\). What is the longest wavelength of light that could ionize \(\mathrm{Fe}^{2+}(g)\) to \(\mathrm{Fe}^{3+}(g) ?\) (d) The third ionization energy of \(\mathrm{Ru}\) is less than the third ionization energy of Fe. Explain.

The ionization energy of an atom can be measured by photoelectron spectroscopy, in which light of wavelength \(\lambda\) is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron \(\left(E_{\mathrm{K}}\right)\) is measured by determining its velocity, \(v_{y}\) since \(E_{\mathrm{K}}=1 / 2 m v^{2}\). The \(E_{\mathrm{i}}\) is then calculated using the relationship that the energy of the incident light equals the sum of \(E_{i}\) plus \(E_{\mathrm{K}}\) (a) What is the ionization energy of rubidium atoms in kilojoules per mole if light with \(\lambda=58.4 \mathrm{~nm}\) produces electrons with a velocity of \(2.450 \times 10^{6} \mathrm{~m} / \mathrm{s}\) ? (The mass of an electron is \(9.109 \times 10^{-31} \mathrm{~kg}\).) (b) What is the ionization energy of potassium in kilojoules per mole if light with \(\lambda=142 \mathrm{~nm}\) produces electrons with a velocity of \(1.240 \times 10^{6} \mathrm{~m} / \mathrm{s} ?\)