Co-Ordination Compounds Question 164
Question: Geometrical shapes of the complexes formed by the reaction of $ N{i^{2+}} $ with $ C{I^{-}} $ , $ C{N^{-}} $ 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
Show Answer
Answer:
Correct Answer: B
Solution:
(i) $ N{i^{2+}}+4C{l^{-}}\to {{[Ni{{(Cl)}_4}]}^{2-}} $ (Tetrahedral) $ Ni(Z=28)=3d^{8}4s^{2},N{i^{2+}}=3d^{8} $ Since $ C{l^{-}} $ 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 weak 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
