How to find possible hybridization of complex compounds?
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Lets say I have to find hybridisation of [Ni(CO)4][Ni(CO)X4]and [Ni(CN)4]2−[Ni(CN)X4]X2−.
The metal atom/ion in these compounds are NiNiand Ni2+Ni2+ respectively.
So the outer shell configuration for NiNi in [Ni(CO)4][Ni(CO)X4] would be
4s2 3d2xy3d1yz3d1zx3d2z23d2x2−y2 4p0 4d04s2 3dxy23dyz13dzx13dz223dx2−y22 4p0 4d0
The electrons from ss oribital will jump to dd orbital and so I expect COCO to donate electron pairs in 4p4pand 4s4s orbitals and form sp3sp3 hybridisation.
The outer shell configuration for Ni2+Ni2+ in [Ni(CN)4]2−[Ni(CN)X4]2− would be
4s2 3d0xy3d1yz3d1zx3d2z23d2x2−y2 4p0 4d04s2 3dxy03dyz13dzx13dz223dx2−y22 4p0 4d0
Here also I expect the electrons to jump from ss to ddand hybridsation to be sp3sp3, but the answer given here is dsp2dsp2. Why ?
I am heavily confused with this hybridsation stuff, is there a systematic and logical way of knowing hybridsation of coordination compounds ?
I tried to search this topic on internet and I came across H=12(V+M−I)H=12(V+M−I) where VV is number of valence electron, MM is number of single bonds and II is charge on the atom. Here if H=2H=2 then it is spsp , H=3H=3 then sp2sp2 and so on.
For [Ni(CO)4][Ni(CO)X4], here V=10V=10, M=4M=4 and I=0I=0, so H=7H=7 hence sp3d3sp3d3.
Also for [Ni(CN)4]2−[Ni(CN)X4]2−, here V=8V=8, M=4M=4 and I=−2I=−2, so H=7H=7 hence sp3d3sp3d3.
As it can be seen, this formula is complete failure.
Why this formula does not work for coordination compounds whereas it works for compounds like PCl5,PCl3PCl5,PCl3 and SF6SF6 ?
The metal atom/ion in these compounds are NiNiand Ni2+Ni2+ respectively.
So the outer shell configuration for NiNi in [Ni(CO)4][Ni(CO)X4] would be
4s2 3d2xy3d1yz3d1zx3d2z23d2x2−y2 4p0 4d04s2 3dxy23dyz13dzx13dz223dx2−y22 4p0 4d0
The electrons from ss oribital will jump to dd orbital and so I expect COCO to donate electron pairs in 4p4pand 4s4s orbitals and form sp3sp3 hybridisation.
The outer shell configuration for Ni2+Ni2+ in [Ni(CN)4]2−[Ni(CN)X4]2− would be
4s2 3d0xy3d1yz3d1zx3d2z23d2x2−y2 4p0 4d04s2 3dxy03dyz13dzx13dz223dx2−y22 4p0 4d0
Here also I expect the electrons to jump from ss to ddand hybridsation to be sp3sp3, but the answer given here is dsp2dsp2. Why ?
I am heavily confused with this hybridsation stuff, is there a systematic and logical way of knowing hybridsation of coordination compounds ?
I tried to search this topic on internet and I came across H=12(V+M−I)H=12(V+M−I) where VV is number of valence electron, MM is number of single bonds and II is charge on the atom. Here if H=2H=2 then it is spsp , H=3H=3 then sp2sp2 and so on.
For [Ni(CO)4][Ni(CO)X4], here V=10V=10, M=4M=4 and I=0I=0, so H=7H=7 hence sp3d3sp3d3.
Also for [Ni(CN)4]2−[Ni(CN)X4]2−, here V=8V=8, M=4M=4 and I=−2I=−2, so H=7H=7 hence sp3d3sp3d3.
As it can be seen, this formula is complete failure.
Why this formula does not work for coordination compounds whereas it works for compounds like PCl5,PCl3PCl5,PCl3 and SF6SF6 ?
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