Co Ordination Compounds Question 285
Geometrical shapes of the complexes formed by the reaction of $ Ni^{2+} $ with $ Cl^{-} $ , $ CN^{-} $ and $ H_2O $ , respectively, are
Options:
A) octahedral, tetrahedral and square planar
B) tetrahedral, square planar and octahedral
C) square planar, tetrahedral and octahedral
D) octahedral, square planar, and octahedral
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Answer:
Correct Answer: B
Solution:
$ Ni^{2+}+4Cl^{-}\to {{[Ni{{(Cl)}_4}]}^{2-}} $ (Tetrahedral) $ Ni(Z=28)=3d^{8}4s^{2},Ni^{2+}=3d^{8} $ Since $ Cl^{-} $ ion is a weak ligand so pairing does not occur, thus $ sp^{3} $ hybridisation and is tetrahedral
(ii) $ N{i^{2+}}+4C{N^{-}}\to {{[Ni{{(CN)}_4}]}^{2-}} $ (Square planar) Since $ C{N^{-}} $ ion is a strong ligand so pairing occurs, thus $ dsp^{2} $ hybridisation and is square planar
(iii) $ N{i^{2+}}+H_2O\to {{[Ni{{(H_2O)}_6}]}^{2+}} $ (octahedral) Since $ H_2O $ is a weak ligand so pairing does not occur, thus $ sp^{3}d^{2} $ (outer complex) hybridisation and is octahedral
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