A) \[Ni{{(CO)}_{4}}\]-Tetrahedral, paramagnetic
B) \[Ni(CN)_{4}^{2-}\]-Square planar, diamagnetic
C) \[Ni{{(CO)}_{4}}\]-Tetrahedral, diamagnetic
D) \[{{[Ni{{(Cl)}_{4}}]}^{2-}}\] Tetrahedral, paramagnetic
Correct Answer: A
Solution :
In \[Ni{{(CO)}_{4}},\,Ni\] has zero oxidation number So \[_{28}Ni\,=1{{s}^{2}},\,2{{s}^{2}}\,2{{p}^{6}},\,3{{s}^{2}}3{{p}^{6}}3{{d}^{8}},\,4{{s}^{2}}\] In excited state and during the formation of \[Ni{{(CO)}_{4}}\] \[\to \] Hence in it, no unpaired electron is present. So it shows the property of diamagnetism and tetrahedral structure.You need to login to perform this action.
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