Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Textbook Question
Chapter 12, Problem 12.20E
Why isn’t the electron configuration of beryllium, given as
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Chapter 12 Solutions
Physical Chemistry
Ch. 12 - In the Stern-Gerlach experiment, silver atoms were...Ch. 12 - Prob. 12.2ECh. 12 - Prob. 12.3ECh. 12 - Suppose s=12 for an electron. Into how many parts...Ch. 12 - Using and labels, write two possible...Ch. 12 - List all possible combinations of all four quantum...Ch. 12 - What are the degeneracies of the H atom...Ch. 12 - Prob. 12.8ECh. 12 - a Differentiate between the quantum numbers s and...Ch. 12 - Is the spin orbital 1s for the H atom still...
Ch. 12 - Draw a diagram analogous to Figure 11.15, but now...Ch. 12 - Are mathematical expressions for the following...Ch. 12 - Prob. 12.13ECh. 12 - Prob. 12.14ECh. 12 - a Assume that the electronic energy of Li was a...Ch. 12 - Spin orbitals are products of spatial and spin...Ch. 12 - If 1 and 2 are the individual wavefunctions for...Ch. 12 - Show that the correct behavior of a wavefunction...Ch. 12 - Prob. 12.19ECh. 12 - Why isnt the electron configuration of beryllium,...Ch. 12 - Prob. 12.21ECh. 12 - Write a Slater determinant for the lithide ion,...Ch. 12 - Why does the concept of antisymmetric...Ch. 12 - a Construct Slater determinant wavefunctions for...Ch. 12 - Prob. 12.25ECh. 12 - Prob. 12.26ECh. 12 - Prob. 12.27ECh. 12 - Suppose an electron had three possible values of...Ch. 12 - Using a periodic table or Table 12.1, find the...Ch. 12 - Write an acceptable electron configuration for...Ch. 12 - Prob. 12.31ECh. 12 - Prob. 12.32ECh. 12 - Prob. 12.33ECh. 12 - An anharmonic oscillator has the potential...Ch. 12 - Prob. 12.35ECh. 12 - In a particle-in-a-box having length a, the...Ch. 12 - Prob. 12.37ECh. 12 - Prob. 12.38ECh. 12 - Prob. 12.39ECh. 12 - The Stark effect is the change in energy of a...Ch. 12 - Prob. 12.41ECh. 12 - Prob. 12.42ECh. 12 - Prob. 12.43ECh. 12 - Show that a variation theory treatment of H using...Ch. 12 - Prob. 12.45ECh. 12 - Explain why assuming an effective nuclear charge,...Ch. 12 - Prob. 12.47ECh. 12 - Consider a real system. Assume that a real...Ch. 12 - Prob. 12.49ECh. 12 - Prob. 12.50ECh. 12 - Prob. 12.51ECh. 12 - Prob. 12.52ECh. 12 - State the Born-Oppenheimer approximation in words...Ch. 12 - Prob. 12.54ECh. 12 - Spectroscopy deals with differences in energy...Ch. 12 - Prob. 12.56ECh. 12 - What is the bond order for the lowest excited...Ch. 12 - The helium atom was defined as two electrons and a...Ch. 12 - Explain how we know that the first in equation...Ch. 12 - Prob. 12.60ECh. 12 - Prob. 12.61ECh. 12 - Use molecular orbital arguments to decide whether...Ch. 12 - Prob. 12.63ECh. 12 - Prob. 12.65ECh. 12 - Prob. 12.67ECh. 12 - Prob. 12.68E
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- What is the electron configuration of the Ba3+ ion? Suggest a reason why this ion is not normally found in nature.arrow_forwardPalladium, with an electron configuration of [Kr] 4d10, is an exception to the aufbau principle. Write the electron configuration of the 2+ cation of palladium. Does the fact that palladium is an exception influence the electron configuration of Pd2+?arrow_forwardUsing the following data, draw the Born Haber cycle for the formation of hypothetical compound of MX(s) from its elements. Using the Born Haber cycle, calculate the electron affinity of X. M (s) → M (g) ∆Hº = 77 kJ mol-1M (g) → M+ (g) + e- ∆Hº = 433 kJ mol-1X2 (g) → 2X (g) ∆Hº = 129 kJ mol-1M (s) + ½ X2(g) → MX(s) ∆Hº = -530 kJ mol-1M+ (g) + X-(g) → MX (s) ∆Hº = -760 kJ mol-1arrow_forward
- Use the following data and the Born-Haber cycle to calculate the first ionization energy (ΔH IE1) of K(g) K(s) → K(g) 89 kJ mol-1 Cl(g) + e– → Cl–(g) -349 kJ mol-1 K(s) + ½ Cl2(g) → KCl(s) -437 kJ mol-1 K(g) → K+(g) + e– ΔH IE1 Cl2(g) → 2Cl(g) 244 kJ mol-1 K+(g) + Cl–(g) → KCl(s) -717 kJ mol-1arrow_forwardWhat is the best reason for the observation that carbon can form four bonds and not two as expected from its electron configuration? * a) Carbon is in main group 4 of the periodic table. b) Carbon has four valence electrons in the n = 2 energy level. c) An electron is transferred from the filled 2s orbital to an unoccupied 2p orbital. d) The energy gap between the n = 1 and n = 2 is large and no further electrons can, therefore, be promoted from lower energy levels to facilitate more bonds. e) All of the reasons given above explain the observationarrow_forwardAll the noble gases have stable configuration ns2, np6 except He.arrow_forward
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