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Find CFSE of $[Fe(H_2O)_6]^{2+}$ and $[NiCl_{4}^{-}]^{2-}$A. $-0.4\Delta_0$ and $-0.8\Delta_t$B. $-0.4\Delta_0$ and $-1.6\Delta_t$ C. $-0.8\Delta_0$ and $-0.4\Delta_t$D. $-1.2\Delta_0$ and $-1.2\Delta_t$

Last updated date: 02nd Aug 2024
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Complete step by step answer: $[Fe(H_2O)_6]^{2+}$ from this we can observe that $Fe^{+2}\longrightarrow[Ar]3d^3$ also $H_2O\longrightarrow$ weak field ligand, so pairing so not take place $t_{2g}^{2,1,1}$
Hence CFSE for $[Fe(H_2O)_6]^{2+}$ can be written as CFSE = $-0.4\times4\Delta_0+0.6\times2\Delta_0=-0.4\Delta_0$
In case of $[NiCl_{4}^{-}]^{2-}$ we can observe that $Ni^{+2}\longrightarrow[Ar]3d^8$ also $Cl^{-}$ is a weak field ligand, so pairing do not take place and have tetrahedral geometry $eg^{2,2}$
Hence CFSE for $[NiCl_{4}^{-}]^{2-}$ can be written as CFSE=$-0.6\times\Delta_t+0.4\times4\Delta_t=-2.4\Delta_t+1.6\Delta_t=-0.8\Delta_t$
Note: The main part of the above question is to identify the geometry of the desired ion as in the above case are $[Fe(H_2O)_6]^{2+}$ and $[NiCl_{4}^{-}]^{2-}$ 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.