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a
Fe(CN)64− and Fe(CN)63− - both are octahedral and diamagnetic with d2sp3- hybridisation
b
Ni(CO)4 and Ni(CN)42−both are tetrahedral and diamagnetic with sp3-hybridisation
c
Ni(CO)4 and Co(CO)4−-both are tetrahedral and diamagnetic
d
CoH2O63+ and CrH2O63+- both are paramagnetic and metal is d2sp3 -hybridised
answer is C.
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Detailed Solution
Ni(CO)4 is tetrahedral sp3 and diamagnetic. Ni(CN)42− is square planar dsp2 and diamagnetic. Fe(CN)64− (0 unpaired electron) and Fe(CN)63− (1 unpaired electron). Both have d2sp3hybridisation and octahedralCoH2O63+→0 unpaired electron (diamagnetic) CrH2O6d2sp33+→3 unpaired electron (paramagnetic)