Coordination Compounds 1 Question 13
13. Geometrical shapes of the complexes formed by the reaction of $Ni^{2+}$ with $Cl^{-}, CN^{-}$and $H _2 O$, respectively, are
(2011)
(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: 13. (b)
Solution:
- $Ni^{2+}+4 CN^{-} \rightarrow\left[Ni(CN) _4\right]^{2-}$
Here, $Ni^{2+}$ has $d^{8}$ -configuration with $CN^{-}$ as strong ligand.
$d^{8}$ -configuration in strong ligand field gives $d s p^{2}$ -hybridisation, hence square planar geometry.
$ Ni^{2+}+4 Cl^{-} \longrightarrow\left[NiCl _4\right]^{2-} $
Here, $Ni^{2+}$ has $d^{8}$ -configuration with $Cl^{-}$ as weak ligand.
$d^{8}$ -configuration in weak ligand field gives $s p^{3}$ -hybridisation, hence tetrahedral geometry. $Ni^{2+}$ with $H _2 O$ forms $\left[Ni\left(H _2 O\right) _6\right]^{2+}$ complex and $H _2 O$ is a weak ligand.
Therefore, $\left[Ni\left(H _2 O\right) _6\right]^{2+}$ has octahedral geometry.
BETA
