(a) [Ni(CO)4] possesses tetrahedral geometry while [Ni(CN)4] is square planar. Why?
(b) Write the IUPAC name of K 2[Ni(CN)4l.
(c) Write the hybridization and shape of [Fe(CN)6]3-.
(a) Ni has the atomic no 28. And it has d8 configuration. As, CO is weak ligand, it is unable to cause internal pairing has make bonds with next 4 available orbital that is 1 5s and 3 4p orbital. And have sp3 hybridization and tetrahedral shade. Whereas CN is a strong field ligand and hence cause pairing and use the left vacant d orbital and take next 1 5s and 2 4p orbital. And hence has dsp2 hybridization and square planer shape.
(b) As it is an anionic complex, and have 4 cyano group, the name will be Potassium Tetracyanonickelate(II).
(c) As CN is a strong field ligand, it has the ability to cause internal pairing and use inner d orbital. So it will make inner orbital compound and the hybridization of [Fe (CN)6]3- will be d2sp3. And the shape will be octahedral.
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