How to solve question 1... Provide with references 1. Explain clearly the follow

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How to solve question 1... Provide with references

1. Explain clearly the following definitions/observations i) back-bonding in metal complexes ii) solutions of Mn2"dan Fe" have a very pale colour (pink and violet respectively) and exhibits a very weak absorption spectrum. ii) when a solution of hexaaquacuprum(II) is added to aqueous ammonia solution, the expected hexaamminecuprum(Il) complex is not obtained 2. When [ZrBr dppe)) is reacted with Mg(Me): ja product [2rt(Me)('dppe)] is formed NMR spectra showed that all the Me groups are equivalent. Draw clear diagrams to show the octahedral and trigonal prism structures for the product. Predict the correct structure for the product based on the NMR data and explain clearly how you arrive at your conclusions. dppe is 112-bis(diphenylphosphino lethane You are encouraged to refer to sources other than your textbook in order to abtain good marks for your answers. Include balanced chemical equations and clear diagrams where appropriate. Do not forget to cite references and acknowledge your sources. EN LG

Explanation / Answer

1)

i) Back-bonding in metal complexes:

  Back-bonding in metal complexes is the electron moves from the metal d-orbital to antibonding orbital of the ligand.

ii) Mn2+ is a high spin d5 and Fe3+ also having high spin d5. For high spin d5 complexes electronic transition is not only Laporte forbidden but also spin forbidden. Absorptions that are doubly forbidden transitions are extremely weak. Here electronic d-d transition is impossible unless the t2g electron being excited simultaneously flips its spin so that it can pair with the electron already in the eg orbital,this event is extremely improbable. Therefore the solutions of  Mn2+ and  Fe3+ have a very pale colour and exhibits a very weak absorption spectrum.

iii) The copper ion in the aqueous solution of exists predominantly as [Cu(H2O)6]2+.Since ammonia is a weak base, when it is added, hydroxide ion forms

NH3(aq) + H2O(l) <==> NH4+(aq) + OH-(aq)

The hydroxide ion reacts with the hexaaquacopper(II) ion to form the insoluble compound, copper(II) hydroxide dihydrate

[Cu(H2O)6]2+(aq) + 2 OH-(aq) <==> Cu(OH)2  + 2 H2O(s) + 2 H2O(l)

Continued addition of ammonia results the formation of a soluble deep blue complex copper ammonia ion

[Cu(H2O)6]2+(aq) + 4 NH3 (aq) <==> [Cu(NH3)4(H2O)2]2+(aq) + 4 H2O(l)

Here the resulting complex is not hexaammine copper(II) complex. Still 2 water molecules present in the complex.

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