A complex with the composition [MA_3B]^{n pm} | vedclass

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(C) For a complex with the composition $[MA_3B]$,we analyze the possibility of geometrical isomerism in different geometries:$1$. Tetrahedral complexe

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

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(C) For a complex with the composition $[MA_3B]$,we analyze the possibility of geometrical isomerism in different geometries:$1$. Tetrahedral complexes: In a tetrahedral geometry,all four positions are equivalent relative to each other. Therefore,any substitution of ligands does not lead to different geometrical arrangements. Thus,$[MA_3B]$ in a tetrahedral geometry does not show geometrical isomerism.$2$. Square planar complexes: In a square planar geometry,for a complex of the type $[MA_3B]$,all positions are equivalent. Any arrangement of the three $A$ ligands and one $B$ ligand results in the same structure. Thus,$[MA_3B]$ in a square planar geometry also does not show geometrical isomerism.Since both geometries do not exhibit geometrical isomerism for this specific composition,the correct answer is $(c)$.

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$

$A$ complex with the composition $[MA_3B]^{n \pm}$ is found to have no geometrical isomers. The possible structure$(s)$ of the complex is/are (where $A$ and $B$ are monodentate ligands):

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How many unpaired electrons are present in the cobalt ion in the $+3$ oxidation state prior to hybridization in the $[CoF_6]^{3-}$ complex ion?

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

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

$D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$

The geometry and the number of unpaired electron$(s)$ of $[MnBr_4]^{2-}$,respectively,are

In $[NiCl_4]^{2-}$,the number of unpaired electrons is

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