in which of the following pairs, same geometry but different hybridisation ?a. [NiCl4]2-, [Ni(CN)4]2- b. [CoF6]3-, [Co(NH3)6]3+

c. [Ni(CO)4], [Ni(CN)4]2- d. [Cu(NH3)4]2+, [Ni(NH3)6]2+

experts please help.

I am not clear with your question. In complexes, for each hybridization there would be specific geometry. Any way I have given you details of each complex pairs geometry and hybridization. Please go through.

[NiCl4]2- = tetrahedral geometry (sp3), [Ni(CN)4]2-  = square planar geometry (dsp2) both has different geometry and hybridization.

[Ni(CN)4]2- is square planar because CN is strong ligand it in the presence of this ligand electrons are pair up in 3d orbitals. so inner orbital comples it is undergo dsp2 hybridization so it is in square panar geometry while [Ni(Cl)4] it is undergo sp3 hybridization so it has tetrahydral geometry .

[CoF6]3- = Octahedral geometry (d2sp3), F - is not a strong enough base to cause spin pairing, so the metal must use its 4d orbitals in bonding to the Co, This complex is called a high spin (spin free, outer orbital) complex.

[Co(NH3)6]3+ = Octahedral geometry (d2sp3), NH3 is a strong enough base to cause the d electrons to pair. Such complexes are called low spin, spin paired, or inner orbital complexes. [CoF6]3-, [Co(NH3)6]3+ both have same geometry same hybridization

[Ni(CO)4] = tetrahedral geometry (sp3),  [Ni(CN)4]2- = square planar geometry (dsp2), both have different geometry and hybridization.

[Ni(CN)4]2- is square planar because CN is strong ligand it in the presence of this ligand electrons are pair up in 3d orbitals. so inner orbital comples it is undergo dsp2 hybridization so it is in square panar geometry while [Ni(co)4] it is undergo sp3 hybridization so it has tetrahydral geometry .

[Cu(NH3)4]2+ = Square planar geometry (dsp2)

[Ni(NH3)6]2+ = octahedral geometry (d2sp3), both have different geometry and hybridization.