Which of the following will have two stereois | vedclass

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(B) To determine the number of stereoisomeric forms:$I. [Cr(NO_3)_3(NH_3)_3]$ is of the type $MA_3B_3$. It exhibits facial $(fac)$ and meridional $(me

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$I$ and $II$

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(B) To determine the number of stereoisomeric forms:$I. [Cr(NO_3)_3(NH_3)_3]$ is of the type $MA_3B_3$. It exhibits facial $(fac)$ and meridional $(mer)$ isomers. Thus,it has $2$ stereoisomers.$II. K_3[Fe(C_2O_4)_3]$ is of the type $M(AA)_3$. It exists as a pair of enantiomers ($d$ and $l$ forms). Thus,it has $2$ stereoisomers.$III. [CoCl_2(en)_2]^+$ is of the type $M(AA)_2B_2$. It exhibits $cis$ and $trans$ isomers. The $cis$ form is optically active ($d$ and $l$ forms),while the $trans$ form is optically inactive. Thus,it has $3$ stereoisomers.$IV. [CoBrCl(ox)_2]^{3-}$ is of the type $M(AA)_2BC$. It exhibits $cis$ and $trans$ isomers. The $cis$ form is optically active ($d$ and $l$ forms),while the $trans$ form is optically inactive. Thus,it has $3$ stereoisomers.Therefore,only $I$ and $II$ have exactly $2$ stereoisomeric forms.

Complex	Structure and magnetic property
$i$. $NiCl_4^{2-}$	$p$. tetrahedral and diamagnetic
$ii$. $Ni(CO)_4$	$q$. tetrahedral and paramagnetic
$iii$. $PtCl_4^{2-}$	$r$. square planar and diamagnetic
$iv$. $Ni(CN)_4^{2-}$	$s$. square planar and paramagnetic

Which of the following will have two stereoisomeric forms?
$I. [Cr(NO_3)_3(NH_3)_3]$
$II. K_3[Fe(C_2O_4)_3]$
$III. [CoCl_2(en)_2]^+$
$IV. [CoBrCl(ox)_2]^{3-}$

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Which of the following will have two stereoisomeric forms?$I. [Cr(NO_3)_3(NH_3)_3]$$II. K_3[Fe(C_2O_4)_3]$$III. [CoCl_2(en)_2]^+$$IV. [CoBrCl(ox)_2]^{3-}$