In a ideal unit cell of perovskite $\ce{KMnF3}$ (closed packed with purely ionic bondings), calculate the unit cell length:

radius: $\ce{K}=1.7~$ $\ce{F}=1.2~$ $\ce{Mn}=0.8$

The structure enter image description here

Green $\ce{K}$, red $\ce{F}$, blue $\ce{Mn}$. (ignore the sizes in the picture)

Am I going right by using the $\ce{F-K-F}$ diagonal length to calculate the cell length using pytagroas, as the Mn diameter is to small and therefore the $\ce{Mn-F-Mn}$ will have some spacing on it and are not fully adjacent.

I would get unit cell length =4.101. can somebody confirm this?

Just a bit confused as I have the radius of $\ce{Mn}$ but don't use it for calculation

Many Thanks

  • 1
    $\begingroup$ Presumably the radius of Mn is there for you to figure out which set of ions is close-packed. $\endgroup$ – orthocresol Mar 19 '17 at 22:16

Experimental X-ray crystallography data yields a value of 4.182 for the lattice constant for $\ce{KMnF3}$. See this reference for further details. Given that the paper was published in 1961, there are likely more recent refinements experimentally. That said, your number is within the error of observed values.

Reference: K. Knox. Perovskite-like Fluorides. I. Structures of $\ce{KMnF3}$, $\ce{KFeF3}$, $\ce{KCoF3}$, $\ce{KNiF3}$, and $\ce{KZnF3}$. Crystal Field Effects in the Series and in $\ce{KCrF3}$ and $\ce{KCuF3}$. $\it{Acta\,Cryst.}$ (1961) $\bf{14}$, 583.


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