question_answer

Choose the correct alternative for the compounds: \[[Ni{{(CO)}_{4}}]\]and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]

A)  Tetrahedral, Paramagnetic                         Tetrahedral, Diamagnetic

B)  Tetrahedral, Diamagnetic                            Square planar, Diamagnetic

C)  Square planar, Diamagnetic                        Square planar, Paramagnetic

D)  Trigonal pyramidal          Square pyramidal

Correct Answer: B

Solution :

                (I) In \[[Ni{{(CO)}_{4}}]\], Ni is present as Ni atom. Since, CO is a strong field ligand, therefore, it pairs up the unpaired electrons. \[_{28}Ni=[Ar]\,3{{d}^{8}}\,\,4{{s}^{2}}\]                 Due to \[S{{p}^{3}}\]-hybridisation, its shape is tetrahedral and due to absence of unpaired electron, it is diamagnetic in nature. II. In \[{{[Ni{{(CN)}_{4}}]}^{2-}}\], Ni is present as \[N{{i}^{2+}}\]-ion, since \[C{{N}^{-}}\] is also a strong field ligand, therefore, it also pairs up the unpaired electrons. \[_{28}Ni=[Ar]\,3{{d}^{8}}4{{s}^{2}}\], \[N{{i}^{2+}}=[Ar]\,3{{d}^{8}}\,\,4{{s}^{0}}\] Due to \[ds{{p}^{2}}\] -hybridisation, its shape is square planar and due to absence of unpaired electron, it is diamagnetic in nature.