Coordination Compounds
The extent of crystal field splitting in octahedral complex of the given metal with particular weak field ligand are such that :
Geometrical shapes of the complexes formed by the reaction of $$Ni^{2+}$$ with $$Cl^{-},\ CN^{-}$$ and $$H_{2}O$$, respectively, are:
$$Ni^{2+}+4Cl^-\rightarrow [NiCl_4]^{2-}$$
$$[NiCl_4]^{2-}=3d^8;\ Cl^-$$ being weak ligand doesn't compel (configuration $$sp^3)$$ for the pairing of electrons.
Hence, the complex has a tetrahedral geometry.
$$[Ni(CN)_4]^{2-}\rightarrow 3d^8$$ configuration. $$(CN)^-$$ being strong ligand compel for the pairing of the electrons. Hence, square planar geometry.
$$Ni^{2+}+6H_2O\rightarrow [Ni(H_2O)_6]^{2+}$$
$$[Ni(H_2O)_6]^{2+}\rightarrow 3d^8$$ configuration. As with $$3d^8$$ configuration, two orbitals are not available for $$d^2sp^3$$ hybridisation.
So, it will have octahedral geometry.
Hence, option $$B$$ is correct.
The extent of crystal field splitting in octahedral complex of the given metal with particular weak field ligand are such that :
Define crystal field splitting energy.
On the basis of crystal field theory, write the electronic configuration for $$d^{4}$$ ion, if $$\Delta_0 > P$$
The complex that has the highest crystal splitting energy $$(\Delta)$$, is:
A metal ion $$M^{n+}$$ having $$d^4$$ valence electronic configuration combines with three bidentate ligands to form a compound. Assuming $$\Delta_0 > P$$: Explain orbital splitting during this complex formation.
Geometrical shapes of the complexes formed by the reaction of $$Ni^{2+}$$ with $$Cl^{-},\ CN^{-}$$ and $$H_{2}O$$, respectively, are:
Select the correct order of magnetic moment (in B.M.) from the following options :
The increasing order of the crystal field splitting power of some common ligands is?
In which of the following complexes of the Co (atomic no. $$=27$$), will the magnitude of $$\Delta_o$$ be the highest?