# Bond angles in BrF5

If in an octahedral structure there is one lone pair placed on any two of the axial position then will the lone pair distort all the 90 degree angles due to repulsion from lone pair or the angles will remain same? I found one question regarding $\ce{BrF5}$.

You are right! You asked the question and you replied simultaneously. According to this website, the structure of bromine pentafluoride confirmed your claims.

Consider the figure given in the previous answer:

Due to a lone pair of $$\ce{e-}$$ and thus, due to lp-bp repulsions, all equatorial fluorine atoms will move upwards but still remain in a plane. Now, the $$\ce{Br}$$ will be out of that plane. This will cause the $$\ce{F_{axial}-Br-F_{equatorial}}$$ angle to reduce. I think $$\ce{F_{equatorial}-Br-F_{equatorial}}$$ angle would remain $$\mathrm{90^\circ}$$.

According to this website, the experimental data are:

$$\ce{F_{axial}-Br-F_{equatorial}}\mathrm{~=84.8^\circ}$$ $$\ce{F_{equatorial}-Br-F_{equatorial}}\mathrm{~=89.529^\circ}$$

I am wondering, where is the difference of $$~\mathrm{360^\circ -4\times89.529^\circ=1.884^\circ}~$$?

One more thing, why it is written equitorial not equatorial in this website?

Edit: I found another source$$^1$$ according to which the equatorial bond angles are at $$\mathrm{90^\circ}$$.

Reference:

1. Page 291, University Chemistry, 4th Edition, Bruce M. Mahan/ Rollie J. Meyers, Pearson (ISBN 978-81-317-2957-1)