For the synthesis of $\ce{KMnF3}$ I said the reaction equation was

$\ce{3KF + MnCl2 -> KMnF3 + 2KCl}$

Is it this simple or are am I missing something with these particular types of reactions?

  • $\begingroup$ scripts.iucr.org/cgi-bin/paper?S0365110X61001868 Also what types? $\endgroup$ – Mithoron Mar 10 '16 at 22:39
  • $\begingroup$ First row transition metals with KF $\endgroup$ – MrLuke370 Mar 10 '16 at 22:44
  • $\begingroup$ Maybe I misinterpreted the reaction, so would it be the HF formed in solution with the KF and nitric acid that would be reacting instead? The ratio of reagents in my lab manual would suggest 3:1 like the equation above. $\endgroup$ – MrLuke370 Mar 10 '16 at 22:50
  • $\begingroup$ Hmm, I guess it's OK. $\endgroup$ – Mithoron Mar 10 '16 at 23:03

Quoting from Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

$\ce{KMnF3}$ is prepared by the addition of $\ce{KF}$ to $\ce{Mn^2+}$ solution. [...] Fluorination of $\ce{MnF2}$ in presence of an alkali metal fluoride leads to the complex Mn(IV) fluorides of composition $\ce{M2MnF6 (M = K, Rb, Cs) }$.

So, the reaction is all right.


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