compounds having the respective symmetries C, D, and C. Without consulting refer
ID: 542946 • Letter: C
Question
compounds having the respective symmetries C, D, and C. Without consulting reference material, which of these should display the greatest number of CO stretching bands in the IR spectrum? Check your answer and give the number of expected bands for each by nsulting Table 22.7. 22.8 Provide plausible reasons for the differences in IR wavenumbers between each of the following pairs: (a) Mo(CO),(PF,), 2040, 1991 cm versus Mo(CO),(PMe,), 1945, 1851 cm-1, (b) MnCp(CO), 2023, 1939 cm1 versus MnCp*(CO), 2017, 1928 cm. 22.9 The compound Ni,(C,H.),(CO), has a single CO stretching bsorption at 1761 cm-1. The IR data indicate that all C,H, ligands are pentahapto and probably in identical environments. (a) On the basis of these data, propose a structure. (b) for each metal in your structure agree with the 18-electron rule? If not, is nickel in a region of the periodic table where deviations from the 18-electron rule are common? Does the electron countExplanation / Answer
22.8 a) The phosphine ligands bound to Mo in [Mo(CO)3(PF3)3] are substituted by electron withdrawing fluorides, hence the phosphine ligands become electron defincient, as a result, the metal Mo will also become electron defincient. Therefore, the carbonyl stretching frequency is higher for this complex, when compared to [Mo(CO)3(PMe3)3], which is bearing methyl substituted electron rich phosphine ligands and thereby decreasing the C-O bond order then corresponding stretching frequency decreases.
b) The Cp ligand bound to Mn in [MnCp(CO)3] are substituted by neutral hydrogen atoms, as a result, the metal Mn will become electronically neutral. Therefore, the carbonyl stretching frequency is higher for this complex, when compared to [MoCp*(CO)3], which is bearing electron rich methyl substituted Cp* ligand and thereby decreasing the C-O bond order then corresponding stretching frequency decreases.
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