The pair(s) of complexes wherein both exhibit | vedclass

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(A) $[FeCl_4]^-$: $Fe^{3+}$ is $d^5$ system. $Cl^-$ is a weak field ligand,so hybridization is $sp^3$ (Tetrahedral). $[Fe(CO)_4]^{2-}$: $Fe^{2-}$ is $

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(A) $[FeCl_4]^-$: $Fe^{3+}$ is $d^5$ system. $Cl^-$ is a weak field ligand,so hybridization is $sp^3$ (Tetrahedral). $[Fe(CO)_4]^{2-}$: $Fe^{2-}$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral). $(B)$ $[Co(CO)_4]^-$: $Co^-$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral). $[CoCl_4]^{2-}$: $Co^{2+}$ is $d^7$ system. $Cl^-$ is a weak field ligand,so hybridization is $sp^3$ (Tetrahedral). $(C)$ $[Ni(CO)_4]$: $Ni^0$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral). $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $d^8$ system. $CN^-$ is a strong field ligand,so hybridization is $dsp^2$ (Square planar). $(D)$ $[Cu(py)_4]^+$: $Cu^+$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral). $[Cu(CN)_4]^{3-}$: $Cu^+$ is $d^{10}$ system. Hybridization is $sp^3$ (Tetrahedral). Therefore,pairs $(A), (B),$ and $(D)$ exhibit tetrahedral geometry.

Similar Questions

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