a. Fe(CN)6³b. Fe(CN)64Which compound would have the largest value of Ao?the ligand CN ashere,andc. Co(NH3)62+B isis 2+,d. Ni(NH3)42+. Co(NH3)3C13since A F2+, A is large=

The question is asking which compound has the largest value of (Δ0), which is the crystal field…

Answered: a. Fe(CN)6³ b. Fe(CN)64 Which compound…

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter19: Complex Ions

Section: Chapter Questions

Problem 25QAP

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a. In the absorption spectrum of the complex ion Cr(NCS)63, there is a band corresponding to the absorption of a photon of light with an energy of 1.75 104 cm-1. Given 1 cm1 = 1.986 1023 J, what is the wavelength of this photon? b. The CrNC bond angle in Cr(NCS)63 is predicted to be 180. What is the hybridization of the N atom in the Ncs- ligand when a Lewis acid-base reaction occurs between Cr3+ and NCs- that would give a 180 CrNC bond angle? Cr(NCS)63 undergoes substitution by ethylenediamine (en) according to the equation Cr(NCS)63+2enCr(NCS)2(en)2++4NCS Does Cr(NCS)2(en)2+ exhibit geometric isomerism? Does Cr(NCS)2(en)2+ exhibit optical isomerism?
Trimethylphosphine, P(CH3)3, can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(Il) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH3)3 can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?
Give the number of unpaired electrons in octahedral complexes with strong-field ligands for (a) Rh3+ (b) Mn3+ (c) Ag+ (d) Pt4+ (e) Au3+
Four different octahedral chromium coordination compounds exist that all have the same oxidation state for chromium and have H2O and Cl as the ligands and counterions. When 1 mole of each of the four compounds is dissolved in water, how many moles of silver chloride will precipitate upon addition of excess AgNO3?
How many geometric isomers are possible for the square-planar complex ion [Pt(NH3)(CN)Cl2)?
Which of the following statement(s) is( are) true? a. The coordination number of a metal ion in an octahedral complex ion is 8. b. All tetrahedral complex ions are low-spin. c. The formula for triaquatrianuninechromiwn(III) sulfate is [Cr(H2O)3(NH3)](SO4)3. d. The electron configuration of Hf2+ is [Xe]4f126s2. e. Hemoglobin contains Fe3+.
An aqueous solution of [Rh(C2O4)3]3− is yellow. Predict the approximate wavelength and predominant color of light absorbed by the complex.
The square-planar complex Pt(en)Cl2 has chloride ligands in a cis configuration. No trans isomer is known. Based on the bond lengths and bond angles of carbon and nitrogen in the ethylenediamine ligand, explain why the trans compound is not possible.
Nickel and palladium both form complexes of the general formula M(PR3)2Cl2. (The ligand PR3 is a phosphine such as P(C6H5)3, triphenylphosphine. It is a Lewis base.) The nickel(II) compound is paramagnetic whereas the palladium(II) compound is diamagnetic. (a) Explain the magnetic properties of these compounds. (b) How many isomers of each compound are expected?
What types of isomers are possible for the following compounds or complex ions? (a) K[Co(NH3)2Cl4] (b) Pt(en)Cl2 (square-planar) (c) [Co(NH3)5Cl]2+ (d) [Ru(phen)3]Cl3 (e) Na2[MnCl4] (tetrahedral) (f) [Co(NH3)5NO2)2+
One of the following nitrogen compounds or ions is not capable of serving as a ligand: NH4+, NH3, NH2. Identify this species, and explain your answer.
Experiments show that K4[Cr(CN)6] is paramagnetic and has two unpaired electrons. The related complex K4[Cr(SCN)6] is paramagnetic and has four unpaired electrons. Account for the magnetism of each compound using the ligand field model. Predict where the SCN ion occurs in the spectro-chemical series relative to CN.

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a. Fe(CN)6³ b. Fe(CN)64 Which compound would have the largest value of Ao? the ligand CN as here, and c. Co(NH3)62+ B is is 2+, d. Ni(NH3)42+ . Co(NH3)3C13 since A F 2+, A is large=