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. (c)

Solution:

  1. 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,

$$ =3 \times(2+2)=12 $$



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