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$
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Answer:
Correct Answer: 6. (a)
Solution:
A square planar complex of general formula, $[Mabcd]$ 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 for the complex,
$ =3 \times(2+2)=12 $
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