General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
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Textbook Question
Chapter 21, Problem 21.87QP
Sketch a diagram showing the formation of energy levels from the valence orbitals for K, K2, K3, and Kn. On the diagram, place arrows indicating how the electrons fill these energy levels.
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Use the electronic configuration of zinc to explain why it forms only a +2. From which orbital(s) are the electrons removed?
Mixing SbCl 3 and GaCl 3 in a one-to-one ratio gives a solid ionic compound of empirical formula GaSbCl 6 . A controversyerupts over whether this compound is [SbCl 2 ] + [GaCl 4 ] ─ or [GaCl 2 ] + [SbCl 4 ] ─ . It is learned that the cation in the compound has abent structure. Based on this fact, which combination is more likely to be correct? Explain.
4. The common oxidation number for an alkaline earth metal is +2.
(a) Using the Born-Mayer equation (for determining the lattice enthalpy) and a Born-Haber
cycle (draw it), show that CaCl is an exothermic compound (negative AHf). Make a
reasonable prediction to estimate the ionic radius of Ca (explain your reasoning). The
sublimation (atomization) enthalpy for Ca(s) is 178 kJ/mol.
(b) Show that an explanation for the non-existence of CaCl can be found in the enthalpy
change for the reaction below. The AHf for CaCl2(s) is -190.2 kcal/mol.
2 CaCl(s) → Ca(s) + CaCl2(s)
Chapter 21 Solutions
General Chemistry - Standalone book (MindTap Course List)
Ch. 21.9 - Considering the fact that N2 makes up about 80% of...Ch. 21.10 - Prob. 21.2CCCh. 21 - Prob. 21.1QPCh. 21 - Prob. 21.2QPCh. 21 - Prob. 21.3QPCh. 21 - Prob. 21.4QPCh. 21 - Prob. 21.5QPCh. 21 - Prob. 21.6QPCh. 21 - Prob. 21.7QPCh. 21 - Prob. 21.8QP
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- Using the following thermodynamic data, calculate the lattice enthalpy of lithium oxide: Li(g) → Li*(g) + e* AH₁= 540 kJ /mol Li(s) → Li(g) AHS= +146 kJ/mol O₂(s) → 20(g) AH₂= +488 kJ /mol O(g) + 1e →→ O(g) AHA1= -142 kJ /mol O` (g) + 1e¯ → 0²-(g) AHÃ₂= +844 kJ /mol 2Li(s) + 1/2O₂(g) → Li₂O(s) AH₁= -586 kJ /mol NOTE: Give your answer in kJ mol-¹arrow_forwardUsing the following data, estimate the overall enthalpy of formation (in kJ/mol) for potassium chloride: K(s) + ½ Cl₂(g) → KCI(s). Process Lattice energy of KCI lonization energy of K Electron affinity of Cl Bond dissociation energy of Cl, Enthalpy of sublimation for K Question 21 of 28 Change in Energy (AHO) -690 kJ/mol 419 kJ/mol -349 kJ/mol 239 kJ/mol 90 kJ/molarrow_forwardUse the Born-Haber cycle to calculate the lattice energy of KF. [The heat of sublimation of K is 91.6 kJ·mol−1 and ΔfH(KF) = −567.3 kJ·mol−1. Bond enthalpy for F2 is 158.8 kJ·mol−1. Other data may be found in the Ionization Energies Table and the Electron Affinities Table.]arrow_forward
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