The number of complexes from the following wi | vedclass

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Question

(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

$3$

Vedclass · Answer

(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 $3$.

List-$I$ (Complex)	List-$II$ ($CFSE$ in $\Delta_0$)
$A$. $[Ti(H_2O)_6]^{2+}$	$I$. $-1.2$
$B$. $[V(H_2O)_6]^{2+}$	$II$. $-0.8$
$C$. $[Mn(H_2O)_6]^{3+}$	$III$. $-0.6$
$D$. $[Fe(H_2O)_6]^{3+}$	$IV$. $0$

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

Similar Questions

The metal $d-$orbitals that directly face the ligands in $K_3[Co(CN)_6]$ are:

Match List-$I$ with List-$II$:

List-$I$ (Complex) List-$II$ ($CFSE$ in $\Delta_0$)

$A$. $[Ti(H_2O)_6]^{2+}$ $I$. $-1.2$

$B$. $[V(H_2O)_6]^{2+}$ $II$. $-0.8$

$C$. $[Mn(H_2O)_6]^{3+}$ $III$. $-0.6$

$D$. $[Fe(H_2O)_6]^{3+}$ $IV$. $0$

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What is the correct order of increasing field strength for the ligands $NH_3$,$en$,$CN^-$,and $CO$?

Why are low spin tetrahedral complexes not formed?

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