I had already asked this question and recieved the answer [(n-C4H9)4]TcOBr4 whic
ID: 577227 • Letter: I
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I had already asked this question and recieved the answer [(n-C4H9)4]TcOBr4 which is C2V point group
but HOW do you get this answer???
The reaction of KTcO4 with 48% aqueous HBr leads to a brown solution from which yellow green plates of tetra-n-butylammonium, (n-C4H9)4N+, salt may be obtained. The product has the following chemical analysis: C, 28.45%; H, 5.35%; N, 2.16%; Br, 47.03%. A strong infrared spectral band occurs at 1011 cm1 that is conspicuously absent in [(n-C4H9)4N]2TcBr6. What is the product? Draw the two most likely structures and give their point group symmetryExplanation / Answer
KTcO4 + 8 HBr -----------------------> KTcBr6 + 4 H2O + 2Br + [(n-C4H9)4 N+ ] ---------------> [(n-C4H9)4]TcOBr4
When we react KTcO4 with 48 % HBr then we get the product KTcBr6 and when we further do its reaction with (n-C4H9)4 N+ then we obtain the product [(n-C4H9)4]TcOBr4. This oxohallide is insoluble in aqueous acids but soluble in organic solvents. Raman spectra is obtained for this salt. When we compare the IR spectra of oxochloride and oxoiodide with oxobromide then we get the same peaks.
When we obtained all these information then we gather the infomation that when technitium oxohallides react with bulkier cations then they yeild TcOX-4 where X = Cl, Br, I and when it reacts with smaller or medium cations they they form TcOX-5. The TcOBr-4 has distorted square pyramidal structure and C2v symmetry detected by X-ray crystallography.
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