EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393543971
Author: KARTY
Publisher: VST
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Chapter 1, Problem 1.31P
Interpretation Introduction
Interpretation:
It is to be explained if the chemical reactivity of the two given compounds will be similar or significantly different.
Concept introduction:
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C
This question shows how molecular orbital (MO) theory can be used to understand the chemical
properties of elemental oxygen O₂ and its anionic derivative superoxide Oz.
a)
Draw the MO energy diagram for both O2 and O2. Clearly label your diagram with atomic orbital names
and molecular orbital symmetry labels and include electrons.
Draw the Lewis structure of O2. How does the MO description of O2 differ from the Lewis structure, and
how does this difference relate to the high reactivity and magnetic properties of oxygen?
) Use the MO diagram in (a) to explain the difference in bond length and bond energy between superoxide
ion (Oz, 135 pm, 360 kJ/mol) and oxygen (O2, 120.8 pm, 494 kJ/mol).
Please draw
-Page: 8
nsition metal ions have high-spin aqua complexes except one: [Co(HO)₁]".
What is the d-configuration, oxidation state of the metal in [Co(H:O))"? Name and draw the geometry
of [Co(H2O)]?
b)
Draw energy diagrams showing the splitting of the five d orbitals of Co for the two possible electron
configurations of [Co(H2O)]:
Knowing that A = 16 750 cm and Пl. = 21 000 cm, calculate the configuration energy (.e., balance
or ligand-field stabilization energy and pairing energy) for both low spin and high spin configurations of
[Co(H2O)]. Which configuration seems more stable at this point of the analysis? (Note that 349.76 cm = 1
kJ/mol)
Exchange energy (IT) was not taken into account in part (d), but it plays a role. Assuming exchange
an occur within t29 and within eg (but not between tz, and ea), how many exchanges are possible in the low
in configuration vs in the high spin configuration? What can you say about the importance of exchange energy
07
Chapter 1 Solutions
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
Ch. 1 - Prob. 1.1PCh. 1 - Prob. 1.2PCh. 1 - Prob. 1.3PCh. 1 - Prob. 1.4PCh. 1 - Prob. 1.5PCh. 1 - Prob. 1.6PCh. 1 - Prob. 1.7PCh. 1 - Prob. 1.8PCh. 1 - Prob. 1.9PCh. 1 - Prob. 1.10P
Ch. 1 - Prob. 1.11PCh. 1 - Prob. 1.12PCh. 1 - Prob. 1.13PCh. 1 - Prob. 1.14PCh. 1 - Prob. 1.15PCh. 1 - Prob. 1.16PCh. 1 - Prob. 1.17PCh. 1 - Prob. 1.18PCh. 1 - Prob. 1.19PCh. 1 - Prob. 1.20PCh. 1 - Prob. 1.21PCh. 1 - Prob. 1.22PCh. 1 - Prob. 1.23PCh. 1 - Prob. 1.24PCh. 1 - Prob. 1.25PCh. 1 - Prob. 1.26PCh. 1 - Prob. 1.27PCh. 1 - Prob. 1.28PCh. 1 - Prob. 1.29PCh. 1 - Prob. 1.30PCh. 1 - Prob. 1.31PCh. 1 - Prob. 1.32PCh. 1 - Prob. 1.33PCh. 1 - Prob. 1.34PCh. 1 - Prob. 1.35PCh. 1 - Prob. 1.36PCh. 1 - Prob. 1.37PCh. 1 - Prob. 1.38PCh. 1 - Prob. 1.39PCh. 1 - Prob. 1.40PCh. 1 - Prob. 1.41PCh. 1 - Prob. 1.42PCh. 1 - Prob. 1.43PCh. 1 - Prob. 1.44PCh. 1 - Prob. 1.45PCh. 1 - Prob. 1.46PCh. 1 - Prob. 1.47PCh. 1 - Prob. 1.48PCh. 1 - Prob. 1.49PCh. 1 - Prob. 1.50PCh. 1 - Prob. 1.51PCh. 1 - Prob. 1.52PCh. 1 - Prob. 1.53PCh. 1 - Prob. 1.54PCh. 1 - Prob. 1.55PCh. 1 - Prob. 1.56PCh. 1 - Prob. 1.57PCh. 1 - Prob. 1.58PCh. 1 - Prob. 1.59PCh. 1 - Prob. 1.60PCh. 1 - Prob. 1.61PCh. 1 - Prob. 1.62PCh. 1 - Prob. 1.63PCh. 1 - Prob. 1.64PCh. 1 - Prob. 1.65PCh. 1 - Prob. 1.66PCh. 1 - Prob. 1.67PCh. 1 - Prob. 1.68PCh. 1 - Prob. 1.69PCh. 1 - Prob. 1.70PCh. 1 - Prob. 1.71PCh. 1 - Prob. 1.72PCh. 1 - Prob. 1.73PCh. 1 - Prob. 1.74PCh. 1 - Prob. 1.75PCh. 1 - Prob. 1.76PCh. 1 - Prob. 1.77PCh. 1 - Prob. 1.78PCh. 1 - Prob. 1.79PCh. 1 - Prob. 1.80PCh. 1 - Prob. 1.81PCh. 1 - Prob. 1.82PCh. 1 - Prob. 1.1YTCh. 1 - Prob. 1.2YTCh. 1 - Prob. 1.3YTCh. 1 - Prob. 1.4YTCh. 1 - Prob. 1.5YTCh. 1 - Prob. 1.6YTCh. 1 - Prob. 1.7YTCh. 1 - Prob. 1.8YTCh. 1 - Prob. 1.9YTCh. 1 - Prob. 1.10YTCh. 1 - Prob. 1.11YTCh. 1 - Prob. 1.12YTCh. 1 - Prob. 1.13YTCh. 1 - Prob. 1.14YTCh. 1 - Prob. 1.15YTCh. 1 - Prob. 1.16YTCh. 1 - Prob. 1.17YT
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