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Shape of molecule HELP!

So how would you describe the shape of KrF2 (krypton difluoride) and state it's bond angle?

What I have tried to do is draw the shape of the molecule itself. I know that in KrF2, the total number of valence electrons is:

8 + 2(7) = 8 + 14, hence it has 22 valence electrons

From that I found the number of lone pairs available in krypton by doing the following:

22 - 16 = 6, but for every lone pair has 2 electrons in it, so 6/2 = 3, so in fact the molecule has 3 lone pairs on the central krypton atom as well as being bonded to 2 fluorine atom; where there is one fluorine atom covalently bonded to either side of the krypton atom.

This means that in total the molecule has 2 bonding pairs and 3 lone pairs.
Now then apparently the molecule is said to have a linear shape with a corresponding bond angle of 180 degrees.
This was actually a question that I had gone through in class with my teacher, but I didn't really understand why it was.
How can the molecule have a linear shape and bond angle of 180 degrees if it has 3 lone pairs and 2 bonding pairs?

I initially thought that molecules can only have a linear shape when they have 2 bond pairs and zero lone pairs. Because if it did have lone pairs, then the bond angle would be less than 180 since lone pairs can repel more to bond pairs (approximately reducing bond angle by 2.5 degrees)

Can someone please help explain the logic why KrF2 has a linear shape and a bond angle of 180 degrees?

Any help is really appreciated! Thanks!

(edited 5 years ago)

Reply 1

Yatayyat

I'm thinking along the lines that it has something to do with the molecule being non-polar, because regardless of having lone pairs on the central krypton atom, it maybe has a symmetrical arrangment that cancels out the bond dipoles when they are summed together. Although I don't know if that is correct to say and how it helps to explain the shape and bond angle?

Reply 2

Pigster

Do you know what shape a molecule with 5 bp 0 lp would be?

When swapping a bp for an lp, do you know whether the equatorial bps are swapped first or polar bps? (i.e. do lps end up equatorial or polar?)

What about the next and then next lps? Where to they go?

Reply 3

Yatayyat

Original post by Pigster

The shape of a molecule with 5 bp and 0 lp is trigonal bipyramidal.

You had me lost about equatorial bps, never heard of this before. What is the difference between a equatorial bp and polar bp?
The molecule I'm dealing with has only has 3 lp, but I'm not sure about whether it is equatorial or polar and what effect it has in changing the shape.

But I think I see why you are talking about a molecule with 5 bp because the 3 lone pairs and 2 bond pairs can all be treated as 5 bp in essence maybe? I'm probaby just guessing at this point.

Reply 4

Yatayyat

Could this mean the shape of KrF2 is a slightly modified version of the trigonal bipyrimadal arrangement if we are dealing with 5 electron pairs?

Reply 5

Pigster

Original post by Yatayyat

The shape of a molecule with 5 bp and 0 lp is trigonal bipyramidal.

3 of the bp in trig bipyrimidal arange themself just like a trigonal plannar set of 3 bp would. These are known as equatorial. The other 2 bp are at 90o to that plane (making the points of the pyramids). These two are the polar bp.

If you swap one of those bp and convert to an lp, where would the lp be to reach maximum separation from the 4 bps? On the equator or pole?

Reply 6

username3028808

Original post by Yatayyat

Could this mean the shape of KrF2 is a slightly modified version of the trigonal bipyrimadal arrangement if we are dealing with 5 electron pairs?

In terms of electron geometry, KrF2 would show a trigonal bipryamidal arrangement. However, when describing the molecule's geometry, you ignore the lone pairs, meaning the molecules shape is linear, you got the right idea as this is because the molecule is symmetrical so bond dipoles cancel.

Reply 7

Yatayyat

Original post by Pigster

I'm thinking for the max separation it might be that the one of the bp's at the 'equator' of the molecule is swapped places to be a lone pair instead. You wouldn't need to swap the 2 polar bp's right?

(edited 5 years ago)

Reply 8

Yatayyat

Original post by naem071

But why do you ignore the lone pairs? Doesn't lone pairs repel much more than bond pairs which essentially distorts the shape?

Reply 9

username3028808

Original post by Yatayyat

But why do you ignore the lone pairs? Doesn't lone pairs repel much more than bond pairs which essentially distorts the shape?

The lone pairs in this case repel from one another by occupying the equatorial positions on the Kr atom. Think of the two F atoms as occupying the Z axis, whilst the three lone pairs are spread across the X and Y axes to maximise repulsion.

Reply 10

Yatayyat

Original post by naem071

I see where you are coming from now, so in simpler terms even though there are 3 lp's in the equatorial positions, spread apart equally to how a trigonal planar shape would arrange it self, it still results in showing 2 fluorine atoms directly opposite to the Kr atom on either side which is a linear shape. So do the polar bp's for fluorine at the top and bottom Kr stay in that position (i.e. poles of the molecule), as this is the only way to get maximum distance from all the electron pairs at the equatorial position?

(edited 5 years ago)

Reply 11

Pigster

Original post by Yatayyat

It is more a case of the 3 lps getting maximum separation from each other being more important than the 2 bps (which are less repulsive than the lps).

The 3 bps are 120o apart, the 2 lps have to make do with only being 90o from other e- pairs.

Reply 12

Yatayyat

Original post by Pigster

From your last sentence, is it not meant to be 3 lps 120 degrees apart and the 2 bps 90 degrees apart from the other electron pairs below?

Reply 13

Pigster

If you consider going from 5bp to 4bp and 1lp... where should the lp go?

It has a choice of 1. going polar, now it has 3 bp at 90o (the 3 equatorial bp) and 1bp at 180o (the other polar one). It doesn't really like having bp so close at 90o.
OR 2. going equatorial, now it has 2bp at 90o (the 2 polar bp) and 2bp at 120o. This one it prefers

Comparing the two, both involve 2bp at 90o, but the first has a third at 90o, whereas the second has ones at 120o.

I shouldn't say 'likes' etc., I mean the bp replusions lead to lowest energetic states, but people like explainations in anthropomorphic language.