Question 10

NUMERICALHARD

The total number of all possible isomers for the square planar complex with formula K[M(NCS)(NO2)(gly)]K[M(NCS)(NO_2)(gly)] is ____. (MM = metal ion and gly=NH2CH2COO−gly = NH_2CH_2COO^-)

Correct Answer: 8

Detailed Solution

The complex is [M(NCS)(NO2)(gly)]−[M(NCS)(NO_2)(gly)]^-.

  1. Linkage Isomerism: Both NCS−NCS^- and NO2−NO_2^- are ambidentate ligands. Possible combinations of coordination sites:
  • (N−CS,N−O2)(N-CS, N-O_2)
  • (S−CN,N−O2)(S-CN, N-O_2)
  • (N−CS,O−NO)(N-CS, O-NO)
  • (S−CN,O−NO)(S-CN, O-NO) There are 4 such linkage combinations.
  1. Geometrical Isomerism: For each linkage combination, the complex is of the type [M(A)(B)(C−D)][M(A)(B)(C-D)], where C−DC-D is the unsymmetrical bidentate glycinate ligand (NH2−CH2−COO−NH_2-CH_2-COO^-). In a square planar geometry, the bidentate ligand occupies cis positions. The two monodentate ligands AA and BB can be arranged in 2 ways:
  • AA trans to NN (of gly), BB trans to OO (of gly)
  • BB trans to NN (of gly), AA trans to OO (of gly)

Total isomers = (4 linkage combinations) ×\times (2 geometrical isomers) = 8.

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