Question

[Pt(NH3)(H2O)(SCN)(NO2)]3 how many isomers are possible?
10
11
12
3​

Answer 1

Answer:

3 geometrical isomers are possible

Explanation:

in square planar complex of formula [Mabcd](where M is the central atom and a,b,c,d are ligands) three geometrical isomers are possible.

another example--[Pt(NH3)(NH2OH)(NO2)(Py)]+

hope it helps you.

Answer 2

Solution:

Total number of isomers are 12 for given complex.

Explanation:

• $[Pt(NH_{3} )(H_{2} O)(SCN)(NO_{2} )]^{3}$ is a square planar complex of type $[Mabcd]^{n}$3 geometrical isomers.

• But, if one of the ligands is ambidentate, then  $2*3=6$ geometrical isomers are possible.

• In this complex, there is 2 ambidentate ligands i.e. $(SCN)^{-}$ and $(NO_{2} )^{-}$  .

• Hence, there will be $2*2*3= 12$ isomers in given complex.

• The total number of possible isomers for square-planar   $[Pt(NH_{3} )(H_{2} O)(SCN)(NO_{2} )]^{3}$12.  

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