In the complex $[SbF_5]^{2-}$,$sp^3d$ hybridization is present. The geometry of the complex is

  • A
    Square
  • B
    Square pyramidal
  • C
    Square bipyramidal
  • D
    Tetrahedral

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Similar Questions

Match List-$I$ with List-$II$:
List-$I$ (Tetrahedral Complex) List-$II$ (Electronic configuration)
$A$. $TiCl_4$ $I$. $e^2, t_2^0$
$B$. $[FeO_4]^{2-}$ $II$. $e^4, t_2^3$
$C$. $[FeCl_4]^{-}$ $III$. $e^0, t_2^0$
$D$. $[CoCl_4]^{2-}$ $IV$. $e^2, t_2^3$

Choose the correct answer from the options given below:

An aqueous solution of metal ion $M1$ reacts separately with reagents $Q$ and $R$ in excess to give tetrahedral and square planar complexes,respectively. An aqueous solution of another metal ion $M2$ always forms tetrahedral complexes with these reagents. Aqueous solution of $M2$ on reaction with reagent $S$ gives a white precipitate which dissolves in excess of $S$. The reactions are summarized in the scheme given below:
$1.$ $M1$,$Q$ and $R$,respectively are :
$(A)$ $Zn^{2+}, KCN$ and $HCl$
$(B)$ $Ni^{2+}, HCl$ and $KCN$
$(C)$ $Cd^{2+}, KCN$ and $HCl$
$(D)$ $Co^{2+}, HCl$ and $KCN$
$2.$ Reagent $S$ is :
$(A)$ $K_4[Fe(CN)_6]$
$(B)$ $Na_2HPO_4$
$(C)$ $K_2CrO_4$
$(D)$ $KOH$
Give the answer for question $1$ and $2$.

$EDTA^{4-}$ is the ethylenediaminetetraacetate ion. The total number of $N-Co-O$ bond angles in the $[Co(EDTA)]^{-}$ complex ion is:

Square-planar geometry is shown by

Which of the following is not a square planar complex?

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