
Concept explainers
Using orbital potential energies, show that group orbital 4 is more likely than group orbital 2 to interactstrongly with the
The
orbitals of oxygen. Therefore, group orbital 6 is nonbonding.

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Chapter 5 Solutions
Inorganic Chemistry
- An electrode process takes place at a metal-solution interface. Indicate the current condition that must be met for Faradaic rectification to occur.arrow_forwardAt a metal-solution interface, an electron is exchanged, and the symmetry factor beta < 0.5 is found in the Butler-Volmer equation. What does this indicate?arrow_forwardTopic: Photochemistry and Photophysics of Supramoleculesarrow_forward
- When two solutions, one of 0.1 M KCl (I) and the other of 0.1 M MCl (II), are brought into contact by a membrane. The cation M cannot cross the membrane. At equilibrium, x moles of K+ will have passed from solution (I) to (II). To maintain the neutrality of the two solutions, x moles of Cl- will also have to pass from I to II. Explain this equality: (0.1 - x)/x = (0.1 + x)/(0.1 - x)arrow_forwardCalculate the variation in the potential of the Pt/MnO4-, Mn2+ pair with pH, indicating the value of the standard potential. Data: E0 = 1.12.arrow_forwardGiven the cell: Pt l H2(g) l dis X:KCl (sat) l Hg2Cl2(s) l Hg l Pt. Calculate the emf of the cell as a function of pH.arrow_forward
- The decimolar calomel electrode has a potential of 0.3335 V at 25°C compared to the standard hydrogen electrode. If the standard reduction potential of Hg22+ is 0.7973 V and the solubility product of Hg2Cl2 is 1.2x 10-18, find the activity of the chlorine ion at this electrode.Data: R = 8.314 J K-1 mol-1, F = 96485 C mol-1, T = 298.15 K.arrow_forward2. Add the following group of numbers using the correct number of significant figures for the answer. Show work to earn full credit such as rounding off the answer to the correct number of significant figures. Replace the question marks with the calculated answers or write the calculated answers near the question marks. 10916.345 37.40832 5.4043 3.94 + 0.0426 ? (7 significant figures)arrow_forwardThe emf at 25°C of the cell: Pt l H2(g) l dis X:KCl (sat) l Hg2Cl2(s) l Hg l Pt was 612 mV. When solution X was replaced by normal phosphate buffer solution with a pH of 6.86, the emf was 741 mV. Calculate the pH of solution X.arrow_forward
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