Question

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. $[Co(PPh_3)_3Cl]Cl_3$

Answer 1

Answer: (C)

$[Fe(PPh_3)_3NH_3ClBr]Cl$

Answer 2

$[Fe(PPh_3)_3NH_3ClBr]Cl$ contains different ligands ($NH_3$, $Cl$, and $Br$) and phosphine ligands ($PPh_3$). The different arrangements of these ligands around the central $Fe$ atom can result in various isomers. In this case, you can have geometric (cis-trans) isomers due to the presence of both cis and trans configurations of $Cl$ and $Br$ ligands with respect to each other, leading to two isomers. Additionally, you can have optical isomers (enantiomers) because of the chiral nature of the $PPh_3$ ligands, which can give you another two isomers.

So, the correct option is (C): $[Fe(PPh_3)_3NH_3ClBr]Cl$