I1. How many unpaired electrons are there in the low spin complex, (Colenklbnt y
ID: 967635 • Letter: I
Question
I1. How many unpaired electrons are there in the low spin complex, (Colenklbnt y unpaired electrons are there in the low spin complex, [Colen) a) 0 b) 1 c) 2 d) 3 e) 4 12. Determine the color and the crystal field splitting energy, A, for the complex bt 90 he complex IFetCNl which [Fe(CN)6 3 which forms an octahedral complex. The aqueous complex ion absorbs at 500 nm. 13. A solution of [CoCk], is green (hex-736nm). A solution of [Co(NH3)6]3t is orange 462nm). Based on this data, which is the stronger field ligand? Explain your answer. Does this agree with the spectrochemical series?Explanation / Answer
12. Data
[Fe(CN)6]3-
= 500 nm
1 m = 1 x 109 nm
= hc/ = ??
Where,
h = planck's constant = 6.625×1034 Js
c = speed of light = 2.998×108 m/s
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Answer.
= (6.625×1034 Js)(2.998×108 m/s)/(500 nm)(1m/1 x109 m)
= 3.97×10-19 J
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There's no need to determine the color of coordination with a formula, you have to use the color wheel (a experimental study, you can find on your textbook and guides). A factor to consider: If a complex absorbs any color, it will have the color appereance that is directly opposite of it on the color wheel.
In this case, 500nm is in the range which absorbs Blue color, it can be concluded that the solution will look Orange.
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13. Data
[CoCl6]3- (green)
max = 736 nm
[Co(NH3)6]3+ (orange)
max = 462 nm
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Answer.
= (6.625×1034 Js)(2.998×108 m/s)/(736 nm)(1m/1 x109 m)
= 2.69×10-19 J
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= (6.625×1034 Js)(2.998×108 m/s)/(462 nm)(1m/1 x109 m)
= 4.29×10-19 J
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By experience it is known, the greater the wavelength of complex, the lower the energy, that means, a strong field ligand results in a large energy (). As you can look, [CoCl6]3- < [Co(NH3)6]3+, also as reference you can look that these results are in tune with the visible spectrum series.
It can be conclude that [Co(NH3)6]3+ will have a stronger ligand field.