UV-Vis spectra forCl2: Use the shade wheel and briefly comment on just how the oboffered color ofthis compound relates to your experimentally measured value ofλmax in the UV-Vis. A little table (name, shade watched,shade took in and λmax) would be beneficial.

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The product finished up a light purple color. Throughout the reactionit changed from gold/brvery own to dark brown/red. This is the table Ihave so far:

Does it seem right? And what are they looking for for the names?The intermediate stages?

Name Color SeenColor Absorbedλmax (nm)
orangeblue456.10 nm
orange/redblue/blue-green500.60 nm
red/purpleblue-green/green529.10 nm

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This is what we were given:

Initially Co(II) is oxidized to Co(III) utilizing 30%H2O2 (the formal charge of oxygenin hydrogen peroxide is -1).

2H+ + 2CoCl2·6H2O+ H2O2 ⟶⟶ 2Co3+ +4Cl− + 14H2O (1)

A complicated is formed which includes a water molecule as the sixthcoordinated ligand in the octahedral facility (considering that water isbound to cobalt by the oxygen the formula is composed "backwards",i.e., OH2)

4Cl− + 2Co3+ +8NH3 + 2NH4Cl +2H2O ⟶⟶ 2Cl3 +2H+ (2)

Combining reactions (1) and also (2) andeliminating widespread ions or molecules on opposite sides of theequation offers the complying with.

2CoCl2·6H2O +H2O2 + 8NH3 +2NH4Cl ⟶⟶ 2Cl3 +12H2O (3)

Finally one of the chlorides (present as an anion) exchangesareas through the water molecule that is bound to the cobalt in theoctahedral complicated providing the formula of the last compoundchloropentaaminecobalt(III) chloride.

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Cl3 ⟶⟶ Cl2 +H2O (4)

And this is the equation I came upwith: 2CoCl2*6H2O +H2O2 + 8NH3 +2NH4Cl -->2Cl2 +14H2O


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