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Coordination Compounds question

2019 · 10 Jan · Shift 1 · Q2
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Coordination Compounds question

2019 · 10 Jan · Shift 1 · Q2

JEE MainChemistryCoordination CompoundsMCQ+4 / −1
The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)][M(F)(Cl)(SCN)(NO_2)][M(F)(Cl)(SCN)(NO2​)] is
  1. A
    16
  2. B
    8
  3. C
    12
  4. D
    4
View written solutionFree

Correct answer: C

  1. Identify the type of complex

The complex is square planar: [M(F)(Cl)(SCN)(NO2)][M(F)(Cl)(SCN)(NO_2)][M(F)(Cl)(SCN)(NO2​)]

Here, all four ligands are different if we only consider their binding atoms as fixed. But two ligands are ambidentate:

  • SCN−SCN^-SCN− can bind through S or N.
  • NO2−NO_2^-NO2−​ can bind as nitro (M−NO2M-NO_2M−NO2​) or nitrito (M−ONOM-ONOM−ONO).

So we must count:

  • Geometrical isomers of the square planar arrangement
  • Linkage isomers due to ambidentate ligands
  1. Geometrical isomers for square planar complex with four different ligands

For a square planar complex of type [MABCD][MABCD][MABCD] with four different ligands, the number of geometrical isomers is: 333

This is because fixing one ligand, the second ligand can be:

  • trans to the first ligand in one way,
  • or in one of two distinct cis-relative arrangements.

Thus, for one fixed connectivity pattern: geometrical isomers=3\text{geometrical isomers} = 3geometrical isomers=3

  1. Count linkage possibilities

Now consider ambidentate ligands:

For SCN−SCN^-SCN−:

It can coordinate as:

  • thiocyanato-SSS : M−SCNM-SCNM−SCN
  • isothiocyanato-NNN : M−NCSM-NCSM−NCS

So number of possibilities from SCNSCNSCN is: 222

For NO2−NO_2^-NO2−​:

It can coordinate as:

  • nitro : M−NO2M-NO_2M−NO2​
  • nitrito : M−ONOM-ONOM−ONO

So number of possibilities from NO2NO_2NO2​ is: 222

Hence total linkage patterns: 2×2=42 \times 2 = 42×2=4

  1. Total number of isomers

For each linkage pattern, there are 333 geometrical isomers.

Therefore total isomers: 3×4=123 \times 4 = 123×4=12

  1. Check options
  • A: 161616 ❌
  • B: 888 ❌
  • C: 121212 ✅
  • D: 444 ❌
  1. Final answer

The total number of isomers is: 12\boxed{12}12​

So the correct option is C.

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