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(A) For tetrahedral complexes,the crystal field splitting results in $e$ orbitals at lower energy and $t_2$ orbitals at higher energy.$TiCl_4$: $Ti^{4

Vedclass · Accepted Answer

$3$

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(A) For tetrahedral complexes,the crystal field splitting results in $e$ orbitals at lower energy and $t_2$ orbitals at higher energy.$TiCl_4$: $Ti^{4+}$ $(d^0)$,configuration $e^0 t_2^0$. (No electrons in $t_2$)$[MnO_4]^{-}$: $Mn^{7+}$ $(d^0)$,configuration $e^0 t_2^0$. (No electrons in $t_2$)$[FeO_4]^{2-}$: $Fe^{6+}$ $(d^2)$,configuration $e^2 t_2^0$. (No electrons in $t_2$)$[FeCl_4]^{-}$: $Fe^{3+}$ $(d^5)$,configuration $e^2 t_2^3$. (Has electrons in $t_2$)$[CoCl_4]^{2-}$: $Co^{2+}$ $(d^7)$,configuration $e^4 t_2^3$. (Has electrons in $t_2$)The complexes with no electrons in the $t_2$ orbital are $TiCl_4$,$[MnO_4]^{-}$,and $[FeO_4]^{2-}$.Thus,the total number is $3$.

The number of complexes from the following with no electrons in the $t_{2g}$ orbital is: $TiCl_4, [MnO_4]^{-}, [FeO_4]^{2-}, [FeCl_4]^{-}, [CoCl_4]^{2-}$

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