# Which molecule has maximum bond angle amongst BF₃, BCl₃ and BBr₃? [closed]

$$\ce{Br}$$ 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.

• 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. – Sameer Thakur Jun 27 '19 at 4:33
• Unfortunately, I still don't get the motive of this question. – William R. Ebenezer Jun 27 '19 at 12:57
• 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)? – Eric Towers Jun 27 '19 at 15:54
• @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). – Kyle_S-C Jun 27 '19 at 16:53
• You might find this related answer interesting chemistry.stackexchange.com/questions/14981/… – porphyrin Jun 28 '19 at 16:24

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.