Possible number of isomers of `[Pt(Cl)(NO_(2)) (NO_(3))(SCN)]^(2-)`, given it is a square planar complex, is _______.

Answer

12

Explanation

In the complex `Pt(Cl)(NO_(2))(NO_(3))(SCN)]^(2-)`, assume:
`M=Pt " " a=Cl^(-)" " b = NO_(2)^(-)" " c=NO_(3)^(-) " " d= SCN^(-)`
Now, it is (Mabcd] type complex, which exhibits three geometrical isomers as following:

Now, `NO_(2)^(-) SCN^(-)` are ambidentate ligands, hence, linkage isomerism is possible in each set of the three isomers.
The possible linkage combinations in each set of the three isomers are:
`1. NO_(2)^(-) , SCN^(-)`
`2. ONO^(-), SCN^(-)`
`3. NO_(2)^(-), NCS^(-)`
`4. ONO^(-) , NCS^(-)`
Total number of isomers`= 3 xx 4 = 12.`