Please explain how to calculate the number of unpaired electrons in a coordination compound and how to estimate its geometry. Which elements in Spectrochemical Series form high spin and which ones from low spin complexes?

Asked by buttercup | 21st Mar, 2012, 02:57: PM

Expert Answer:

Suppose i take a complex [Cr(NH3)6]+3
Chromium (Z=24)
Electronic configuration is 3d54s1
But calculate the oxidation state of Cr =+3
Hence one 4s and two 3d electrons are removed
 
As NH3  is a strong spin complex thus the pairing does not take place in the electrons present in 3d.
 
Now the vacant orbitalsare:   two 3d  , one 4s and three 4p
Thus the ammonia forms occupy the six vacant orbitals and has octahedral geometry.
 
 
 
Spectrochemical series:
 
The ascending order of crysatl field splitting is :
I-,Br-,S-2,Cl-,NO3-,F-,OH-,C2O4-2,O-2,H2O,EDTA,NH3,Pyridine,en,dipyridyl,o-phenanthroline,NO2,CN,CO
 
The ascending order from high spin complex to low spin complex is as given above.

Answered by  | 27th Mar, 2012, 02:29: AM

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