Bundle: Physical Chemistry, 2nd + Student Solutions Manual
2nd Edition
ISBN: 9781285257594
Author: David W. Ball
Publisher: 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
Bundle: Physical Chemistry, 2nd + Student Solutions Manual
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_forward10. Consider two hydrogen atoms. The electron in the first one is in n=1 state, whereas in the second the electron is in the n=3 state. (a) which atom is in the ground state configuration? Why? (b) Which orbital has a larger radius? (c) Which electron is moving faster and why? (d) Which electron has a lower potential energy? (e) Which atom has higher ionization energy? Hint: assume that the radius of the n=3 orbital is =5 rBarrow_forward
- Using 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_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_forwardUse 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_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-1 Use the bond energies listed in the table below to estimate the standard enthalpy of the reaction: CCl3CHCl2(g) + 2HF(g) → CCl3CHF2(g) + 2HCl(g) C – H 412 kJ mol-1 C – C 348 kJ mol-1 C – Cl 338 kJ mol-1 H – Cl 431 kJ mol-1 C – F 484 kJ mol-1 H – F 565 kJ mol-1arrow_forwardPhotoelectron spectroscopy applies the principle of the pho-toelectric effect to study orbital energies of atoms and mol-ecules. High-energy radiation (usually UV or x-ray) is absorbedby a sample and an electron is ejected. The orbital energy can becalculated from the known energy of the radiation and the mea-sured energy of the electron lost. The following energy differ-ences were determined for several electron transitions:ΔE 2→1=4.098X10⁻¹⁷J, ΔE 3→1=4.854X10⁻¹⁷J, ΔE 5→1=5.242X10⁻¹⁷J, ΔE 4→2=1.024X10⁻¹⁷J Calculate the energy change and the wavelength of a photon emitted in the following transitions:(a) Level 3→2 (b) Level4→1 (c) Level5→4arrow_forward
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