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Why is the bond angle in BrF3, 86 degrees?

The shape is: trigonal bypyramidal, with 5 pairs of electrons in the outer shell. I dont understand why the bond angle changes, should'nt the angles remain the same ( 90 and 120 ) because there is an electron above and below? :confused:

(sorry cant really explain)

Kinda look like this:

( the one i drew was (a) )

(edited 11 years ago)

Medical-Intermediate-BrF3-Bromine-f-luoride-5-100.jpg1.5KB

Reply 1

EierVonSatan

You'd expect 90^o if there were 5 bonding pairs, but the lone pairs repel each other more, pushing the bonds closer together :smile:

Reply 2

Seatbelt

Original post by EierVonSatan

You'd expect 90^o if there were 5 bonding pairs, but the lone pairs repel each other more, pushing the bonds closer together :smile:

yes, but there are 2 lone pairs above and below, so they are repelling the same amount right?, so the question is why isnt it 90? :s-smilie:

Reply 3

EierVonSatan

Original post by Seatbelt

yes, but there are 2 lone pairs above and below, so they are repelling the same amount right?, so the question is why isnt it 90? :s-smilie:

Sorry, I didn't realise you had edited your post. The two lone pairs aren't above and below the BrF₃ plane as in (a). A molecule with 3 bonding pairs and 2 lone pairs adopts a T-shape (c).

Reply 4

MediterraneanX

Sorry to hijack this post but could it also be drawn as having two lone pairs at the top and then the T-shape as a normal T-shape so to speak as opposed to it being on its side. I never understand this part of mark schemes - I get the structure right but never facing the right way if that makes sense?

Reply 5

EierVonSatan

Original post by MediterraneanX

Yes, that's fine. It doesn't matter how you draw it - in the same way you can draw a square on it's side or standing on one of it's corners. It's still the same square :smile:

Reply 6

MediterraneanX

Original post by EierVonSatan

Yes, that's fine. It doesn't matter how you draw it - in the same way you can draw a square on it's side or standing on one of it's corners. It's still the same square :smile:

Ah thanks for that - I just wasn't sure as usually in the textbooks and revision books they tend to put the lone pair at the top but in mark schemes they tend to rotate the shapes around.

Reply 7

Seatbelt

I still dont understand :/, what do you mean T-shape?

Reply 8

EierVonSatan

Original post by Seatbelt

I still dont understand :/, what do you mean T-shape?

The molecule is in the shape of a T

http://en.wikipedia.org/wiki/T-shaped_molecular_geometry

Reply 9

Jambone2

My question is why was this included in the Jan 2012 module when no one covered this structure. To be honest it's still wasn't clear to me, until I actually looked at it indepth, how it was made up! http://sixthsense.osfc.ac.uk/chemistry/bonding/covalent.asp

Helpful site for anyone else confused. :smile:

(edited 11 years ago)

Reply 10

bloomingblossoms

is this expanding the octet?

Reply 11

EierVonSatan

Original post by bloomingblossoms

is this expanding the octet?

Indeed

Reply 12

username913907

Angles aren't perfect 120 and 90 etc as in VSEPR theory lone pairs take up 'more space' than bonding pairs. This can be rationalised by considering that lone pairs are localised on the central atom whereas bonding pairs are shared between the atoms. As the lone pairs are localised they are 'nearer' the central atom and have more influence upon the shape.
Axial positions (up and down) are more favorable to place large atoms than equatorial positions as the steric interactions are smaller than in the equatorial positions.