The complex compound [Co(SCN)_2(NH_3)_4]Cl do | vedclass

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(D) $1$. The complex $[Co(SCN)_2(NH_3)_4]Cl$ contains an ambidentate ligand $SCN^-$,which allows for linkage isomerism.$2$. It can exhibit ionisation

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Optical isomerism

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(D) $1$. The complex $[Co(SCN)_2(NH_3)_4]Cl$ contains an ambidentate ligand $SCN^-$,which allows for linkage isomerism.$2$. It can exhibit ionisation isomerism because the $Cl^-$ ion can exchange positions with the $SCN^-$ ligand.$3$. It exhibits geometrical isomerism due to the presence of two $SCN^-$ ligands in either $cis$ or $trans$ configurations.$4$. For optical isomerism in octahedral complexes of the type $[M(A)_4(B)_2]$,the complex must be in the $cis$ form. However,the $cis$ form of $[Co(SCN)_2(NH_3)_4]^+$ has a plane of symmetry,making it achiral. Therefore,it does not exhibit optical isomerism.

List-$I$	List-$II$
$A$. $[Fe(H_2O)_6]^{2+}$	$P$. $0$
$B$. $[Fe(H_2O)_6]^{3+}$	$Q$. $1$
$C$. $[Fe(CN)_6]^{4-}$	$R$. $2$
$D$. $[Fe(CN)_6]^{3-}$	$S$. $4$
$E$. $[Ni(H_2O)_4]^{2+}$	$T$. $5$

The complex compound $[Co(SCN)_2(NH_3)_4]Cl$ does not exhibit which of the following types of isomerism?

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What type of isomerism is exhibited by $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$?

Match the complex ion (in List-$I$) with the number of unpaired electrons (in List-$II$):

List-$I$ List-$II$

$A$. $[Fe(H_2O)_6]^{2+}$ $P$. $0$

$B$. $[Fe(H_2O)_6]^{3+}$ $Q$. $1$

$C$. $[Fe(CN)_6]^{4-}$ $R$. $2$

$D$. $[Fe(CN)_6]^{3-}$ $S$. $4$

$E$. $[Ni(H_2O)_4]^{2+}$ $T$. $5$

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