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(C) To determine the number of stereoisomeric forms,we analyze each complex:$(i) \ [Cr(NO_3)_3(NH_3)_3]$ is of the type $[MA_3B_3]$. It exhibits facia

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$(iii)$ and $(iv)$

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(C) To determine the number of stereoisomeric forms,we analyze each complex:$(i) \ [Cr(NO_3)_3(NH_3)_3]$ is of the type $[MA_3B_3]$. It exhibits facial $(fac)$ and meridional $(mer)$ geometric isomerism. It does not show optical isomerism. Thus,it has $2$ stereoisomers.$(ii) \ K_3[Co(C_2O_4)_3]$ is of the type $[M(AA)_3]$. It shows optical isomerism ($d$ and $l$ forms). It has $2$ stereoisomers.$(iii) \ K_3[CoCl_2(C_2O_4)_2]$ is of the type $[M(AA)_2B_2]$. The cis-isomer is optically active ($d$ and $l$ forms),while the trans-isomer is optically inactive (meso-form). Thus,it has $3$ stereoisomers $(d, l, 	ext{and } trans)$.$(iv) \ [CoBrCl(en)_2]$ is of the type $[M(AA)_2BC]$. The cis-isomer is optically active ($d$ and $l$ forms),while the trans-isomer is optically inactive (meso-form). Thus,it has $3$ stereoisomers $(d, l, 	ext{and } trans)$.Therefore,$(iii)$ and $(iv)$ have three stereoisomeric forms.

Which of the following will have three stereoisomeric forms?
$(i) \ [Cr(NO_3)_3(NH_3)_3]$
$(ii) \ K_3[Co(C_2O_4)_3]$
$(iii) \ K_3[CoCl_2(C_2O_4)_2]$
$(iv) \ [CoBrCl(en)_2]$

Similar Questions

Which of the following complexes or complex ions can have an optically active isomer?

Arrange the following complexes in the increasing order of their spin only magnetic moment (in $B.M$):
$I$. $[Fe(CN)_6]^{4-}$
$II$. $[MnCl_4]^{2-}$
$III$. $[Mn(CN)_6]^{4-}$
$IV$. $[Cr(NH_3)_6]^{3+}$

Arrange the following in order of increasing number of unpaired electrons in them:
$i$. $[Fe(CN)_6]^{3-}$
$ii$. $[MnCl_6]^{3-}$
$iii$. $[FeF_6]^{3-}$
$iv$. $[Co(NH_3)_6]^{3+}$

$Cis-trans$ isomerism is found in square planar complexes of the molecular formula: ($A$ and $B$ are monodentate ligands):

In coordination compounds,hydrate isomers differ in:

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Which of the following will have three stereoisomeric forms?$(i) \ [Cr(NO_3)_3(NH_3)_3]$$(ii) \ K_3[Co(C_2O_4)_3]$$(iii) \ K_3[CoCl_2(C_2O_4)_2]$$(iv) \ [CoBrCl(en)_2]$