Coordination Compounds 1 Question 6
####6. The total number of isomers for a square planar complex $\left[M(F)(Cl)(SCN)\left(NO _2\right)\right]$ is
(2019 Main, 10 Jan I)
(a) 12
(b) 16
(c) 4
(d) 8
Show Answer
Answer:
Correct Answer: 6. (c)
Solution:
- A square planar complex of general formula, $M _{a b c d}$ gives three geometrical isomers only.
Let, $a=F^{-}, b=Cl^{-}, c=S CN^{-}, d=NO _2^{-}$
$SCN^{-}$and $NO _2^{-}$are ambidentate ligands and they also show linkage isomerism (structural). Considering both linkage and geometrical isomerism.
Total number of possible isomers given by the complex,
$$ =3 \times(2+2)=12 $$
