Question
Question: Find CFSE of \([Fe(H_2O)_6]^{2+}\) and \([NiCl_{4}^{-}]^{2-}\) A. \(-0.4\Delta_0\) and \(-0.8\Delt...
Find CFSE of [Fe(H2O)6]2+ and [NiCl4−]2−
A. −0.4Δ0 and −0.8Δt
B. −0.4Δ0 and −1.6Δt
C. −0.8Δ0 and −0.4Δt
D. −1.2Δ0 and −1.2Δt
Solution
In such questions one should first identify the type of ligand that is being included in the reaction and according to that the type of geometry should be defined for the ions.
Complete step by step answer: [Fe(H2O)6]2+ from this we can observe that Fe+2⟶[Ar]3d3 also H2O⟶ weak field ligand, so pairing so not take place t2g2,1,1
Hence CFSE for [Fe(H2O)6]2+ can be written as CFSE = −0.4×4Δ0+0.6×2Δ0=−0.4Δ0
In case of [NiCl4−]2− we can observe that Ni+2⟶[Ar]3d8 also Cl− is a weak field ligand, so pairing do not take place and have tetrahedral geometry eg2,2
Hence CFSE for [NiCl4−]2− can be written as CFSE=−0.6×Δt+0.4×4Δt=−2.4Δt+1.6Δt=−0.8Δt
Therefore A is the correct option.
Note: The main part of the above question is to identify the geometry of the desired ion as in the above case are [Fe(H2O)6]2+ and [NiCl4−]2− as after that the only thing left is to apply the well defined formula of CFSE so one should concentrate on defining the geometry precisely other than anything else also the precise knowledge of configuration is also required in order to determine the geometry so it should also be kept in mind.