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$\ce{Br}$ is the least electronegative from the given compounds. So, in $\ce{BBr3}$ the electrons will be the closest towards boron and thus due to electron-electron repulsions $\ce{BBr3}$ will have maximum bond angle. Steric hindrance due to large size of bromine atom can be a factor too.

But the bond angles are same according to most of the websites.

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    $\begingroup$ I think for judging bond angles you should take a look at VSEPR structures for them. For all practical purposes, the changes due to electronegativity are negligible in most cases. $\endgroup$ – Sameer Thakur Jun 27 at 4:33
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    $\begingroup$ Unfortunately, I still don't get the motive of this question. $\endgroup$ – William R. Ebenezer Jun 27 at 12:57
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    $\begingroup$ How would the bond angles be different? If you have a ball and stick model, how do you move the bonds around so that you don't have an equilateral triangle (due to symmetry)? $\endgroup$ – Eric Towers Jun 27 at 15:54
  • $\begingroup$ @EricTowers I think the angles they're comparing between are the X-A-X vs X'-A-X' angles. Each of the X-A-X angles will necessarily be identical, but for different identities of A and X, you can change the angles (e.g. ammonia being pyramidal). $\endgroup$ – Kyle_S-C Jun 27 at 16:53
  • $\begingroup$ You might find this related answer interesting chemistry.stackexchange.com/questions/14981/… $\endgroup$ – porphyrin Jun 28 at 16:24
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With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for σ-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can deduce that the bond angle is $\angle(\ce{X-B-X}) = 120^\circ$ for all of them.

Bent's rule won't have any effect on that.

There is also considerable π-backdonation in these molecules, but that only has an effect on bond lengths, not bond angles.

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