A) 16
B) 12
C) 8
D) 24
Correct Answer: B
Solution :
The total number of possible isomers for square-planar \[{{[Pt(Cl)(N{{O}_{2}})(N{{O}_{3}})(SCN)]}^{2-}}\]is \[12\] . THe square-planar complex of type \[|Mabcd{{|}^{n\pm }}\], where all four ligands are different, has 3 geometrical isomers. But if one of the ligands is ambidentate, then \[2\times 3=6\]geometrical isomers are possible. But if two ligands are ambidentate, then \[4\times 3=12\] geometrical isomers are possible. In our example, both \[NO_{2}^{-}\] and \[SC{{N}^{-}}\] are ambidentate ligands.
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