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Which complex compound will give four isomers :
A
$[Fe(en)_3]Cl_3$
B
$[Co(en)_2Cl_2]Cl$
C
$[Fe(PPh_3)_3NH_3ClBr]Cl$
D
$[Co(PPh_3)_3Cl]Cl_3$
Detailed Solution
$[Fe(en)_3]^{3+}$: shows only optical isomerism (d and l), 2 isomers.
$[Co(en)_2Cl_2]^+$: cis (which is optically active, d and l) and trans: 3 isomers.
$[Fe(PPh_3)_3(NH_3)ClBr]^+$: an octahedral $[MA_3BCD]$ complex. The three $PPh_3$ groups can be facial (fac) or meridional (mer), and the positions of $NH_3$, Cl and Br relative to them give further arrangements, some of which are chiral. In total it gives four stereoisomers.
$[Co(PPh_3)_3Cl]^{3+}$ is a four-coordinate complex with no comparable isomerism.
So, as per the answer key, $[Fe(PPh_3)_3NH_3ClBr]Cl$ gives four isomers.
$[Co(en)_2Cl_2]^+$: cis (which is optically active, d and l) and trans: 3 isomers.
$[Fe(PPh_3)_3(NH_3)ClBr]^+$: an octahedral $[MA_3BCD]$ complex. The three $PPh_3$ groups can be facial (fac) or meridional (mer), and the positions of $NH_3$, Cl and Br relative to them give further arrangements, some of which are chiral. In total it gives four stereoisomers.
$[Co(PPh_3)_3Cl]^{3+}$ is a four-coordinate complex with no comparable isomerism.
So, as per the answer key, $[Fe(PPh_3)_3NH_3ClBr]Cl$ gives four isomers.