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for [Co(NH3)6]^(+3) in many books hybridisation is given as d2sp3. this is correct. but for compound [Fe(NH3)6]^(+2) the hybridisation is given as sp3d2. WHY? Both Co3+ and Fe 2+ are d^6 confguration. and nh3 is strong field ligand. so it should also form inner spin complex in fe2+ compound. PLEASE HELP. I WILL BE VERY GRATEFUL OF YOU.....


for [Co(NH3)6]^(+3) in many books hybridisation is given as d2sp3. this is correct.



 but for compound [Fe(NH3)6]^(+2) the hybridisation is given as sp3d2. WHY?  


Both Co3+ and Fe 2+ are d^6 confguration. and nh3 is strong field ligand. so it should also form inner spin complex in fe2+ compound.


PLEASE HELP. I WILL BE VERY GRATEFUL OF YOU.....


Grade:

2 Answers

Parth Malik
9 Points
10 years ago

the shapes of the compounds are predicted according to the valence bond theory. one of the limitations of valence bond theory is that it does not explain why some compounds form outer orbital complexes while some form inner orbital complexes. the shapes and nature of the compounds are first found by spectroscopy and then valence bond theory is applied. in this case [Co(NH3)6](+2) is DIAMAGNETIC while [Fe(NH3)](+2) is PARAMAGNETIC. it is just to explain their MAGNETIC NATURE THEIR HYBRIDISATION IS DIFFERENT. this is one of the limitations of valence bond theory. just to explain their nature found by spectroscopy they have different hybridisations.

ruchi yadav
askIITians Faculty 27 Points
8 years ago
for hybridisation it does not depend fully on the ligand tendency. it also depends the behaviour of the compound i.e paramag. and diamag. actually considering properties like this the series is formed for ligands.
as fe[NH6](+2) is paramagnetic in nature thus sp3d2 is most appropriate for it. the series is experimental.
eg. check for [MnBr4](2-) mag. moment=5.59 (5 unpaired electrons) as dsp2 or sp3 ans- sp3
and
for [MnBr6](4-) u= 5.59 it is sp3d2


Thank You
Ruchi
Askiitians Faculty

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