Bond angles

Original post by MM2002

What are the bond angles for BrF3?
I am aware there are 3 bonding pairs and 2 lone pairs so the structure is based of a triangular bipyramidal. As there are two lone pairs is the bond angle going to be 85 and 115?

When there are 5bp, you get two bonds on the axis and three on the equator.

When there are 3bp and 2lp, where will the 2lp position themselves? i.e on the axis and/or on the equator?

Reply 2

Original post by MM2002

What are the bond angles for BrF3?
I am aware there are 3 bonding pairs and 2 lone pairs so the structure is based of a triangular bipyramidal. As there are two lone pairs is the bond angle going to be 85 and 115?

No. This is an exception to the rule. You're correct in saying that the shape is based on trigonol bipyramidal due to there being 5 electron pairs however, the shape is just trigonol planer as you'll soon find out why. The image below shows the general shape of a trigonal bipyramidal molecule. Knowing that it has 2 lone pairs, these lone pairs will repel the most, and therefore will want to be as far as possible in the molecule to create the largest bond angle. One of the lone pairs will be at the top of the molecule in the same plane, and one at the bottom in the same plane as this creates the largest bond angle and distance apart. This means there is no net repulsion, since the repulsion from lone pair at the top of the molecule cancels out the repulsion from the bottom, leaving a normal trigonal-planer shaped molecule. The bond angle is therefore 120 degrees as the bonding pairs don't undergo a net repulsion
https://p7.hiclipart.com/preview/623/474/363/trigonal-bipyramidal-molecular-geometry-vsepr-theory-pentagonal-bipyramidal-molecular-geometry-trigonal-planar-molecular-geometry-others.jpg

(edited 4 years ago)

Reply 3

Original post by Huckipity

the shape is just trigonol planer as you'll soon find out why.

Are you saying Wikipedia is wrong?

https://en.wikipedia.org/wiki/Bromine_trifluoride

How dare you?

Reply 4

username4500092

14

It would be 87.5, from what I'm reading from my notes

Reply 5

MM2002

OP

9

Original post by Huckipity

No. This is an exception to the rule. You're correct in saying that the shape is based on trigonol bipyramidal due to there being 5 electron pairs however, the shape is just trigonol planer as you'll soon find out why. The image below shows the general shape of a trigonal bipyramidal molecule. Knowing that it has 2 lone pairs, these lone pairs will repel the most, and therefore will want to be as far as possible in the molecule to create the largest bond angle. One of the lone pairs will be at the top of the molecule in the same plane, and one at the bottom in the same plane as this creates the largest bond angle and distance apart. This means there is no net repulsion, since the repulsion from lone pair at the top of the molecule cancels out the repulsion from the bottom, leaving a normal trigonal-planer shaped molecule. The bond angle is therefore 120 degrees as the bonding pairs don't undergo a net repulsion
https://p7.hiclipart.com/preview/623/474/363/trigonal-bipyramidal-molecular-geometry-vsepr-theory-pentagonal-bipyramidal-molecular-geometry-trigonal-planar-molecular-geometry-others.jpg

Thank you

Reply 6

Original post by sakthar23

It would be 87.5, from what I'm reading from my notes

Original post by Pigster

Are you saying Wikipedia is wrong?

https://en.wikipedia.org/wiki/Bromine_trifluoride

How dare you?

My teacher did a molecule with the exact same number of lone pairs and bonding pairs as the one here and said it will be trigonal planer for the reasons I explained. :confused:

Reply 7

Original post by Huckipity

My teacher did a molecule with the exact same number of lone pairs and bonding pairs as the one here and said it will be trigonal planer for the reasons I explained. :confused:

Teachers can get stuff wrong.

If you check an AQA MS (which I bet the OP is studying), then they allow either trigonal planar or T-shaped.

Reply 8

Original post by Pigster

Teachers can get stuff wrong.

If you check an AQA MS (which I bet the OP is studying), then they allow either trigonal planar or T-shaped.

So my teacher was right...

Reply 9

Original post by Huckipity

So my teacher was right...

BrF3 is T-shaped. It just is. What the exam board will accept and what is correct are not necessarily the same thing.

Reply 10