A manganese(Il) cyanide complex has been isolated and its six line epr spectrum

Question

A manganese(Il) cyanide complex has been isolated and its six line epr spectrum due to hyperfine coupling with the 100%abundant 55Mn nucleus with I-5/2 is shown below: a=71 G This complex is believed to have four cyanide ligands, [Mn(CN)4]2. One way to establish this coordination number, is to use a labeled CN, which we could enrich in either 13C or 15N, and both these nuclei are spin ½. Ignore any differences in the cost of the enriched cyanide since you have an unlimited budget. Which isotope would you suggest enriching all the cyanide ligands with and explain your choice. (a) (b) In this thought experiment, what results would you expect to observe after enriching to confirm a coordination number of four? Assuming you actually were to observe what you expect, would it establish a tetrahedral or square planar geometry for this complex?

Explanation / Answer

a) We would use cyanide ligand enriched in 13C isotope in place of 15N isotope. This is because it's mentioned that we are given a cyanide complex of Mn. Had it been isocyanide complex then we would have used 15N. In cyanide complex Mn is directly attached to 4 Carbons of CN- ligand and we know hyperfine interaction takes place between 2 directly attached nucleus.

b) After enriching all the CN- ligands with 13C , we will observe hyperfine interaction between unpaired electron of Mn and 13C.

No. Of Hyperfine lines = (2nI1+1)(2nI2+1) (this formula used since 2 different type of nucleus)

= (2*1*5/2 +1)(2*4*1/2+1) = 30 hypersplitting lines will be observed

If the observed reading is 30 then it confirms the coordination number.

Now if we use [Mn(CN)4]2- with 12C , i.e, no more coupling between Mn and C then no. of hyperfine lines we are supposed to observe for Mn2+ are:

i) TETRAHEDRAL :

No. of fine lines = no. of unpaired electrons = 5 (from sp3 hybridization)

then , no. of hyperfine lines = (2nI+1) * (no. of fine lines) = (2*1*5/2+1)(5) = 30

ii) SQUARE PLANAR :

No. of fine lines = no. of unpaired electrons = 1 ( from d2sp hybridization)

then, no. of hyperfine lines = (2nI+1) * (no. of fine lines) = (2*1*5/2+1)(1) = 6

Thus we can confirm the structure of the complex too. Since CN is strong ligand and pairing of e-s takes place square planar structure is most likely to be observed.

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