# $XeF_6$ and the ways it is.

1. May 4, 2015

### mooncrater

1. The problem statement, all variables and given/known data
I am confused about the structure of $XeF_6$.

2. Relevant equations

3. The attempt at a solution
Is it octahedral or a pyramid with a pentagonal base? $Xe$ has $sp^3d^3$ hybridization. Then I think the latter one is correct. Since the first one has nonarrangement for the lone pair over $Xe$.

Last edited: May 4, 2015
2. May 4, 2015

### Raghav Gupta

I think there is no thing as IF6.
It should be [IF6]+ as that is the stable specie.
So it has then a octahedral structure.

3. May 4, 2015

### mooncrater

Sorry I meant to say $XeF_6$.... did that in hurry.

4. May 4, 2015

### Raghav Gupta

Hey how you edited the title?
Did you reported it to staff?
XeF6 does not have perfect octahedral or the other structure you have showed in attempt.

5. May 4, 2015

### mooncrater

It has $sp^3d^3$ hybridization for sure. But it's structure is still not clear to me.

6. May 4, 2015

7. May 4, 2015

### mooncrater

But when we search about $sp^3d^3$ in images section of Google then we can see the structure of $IF_7$ which has $sp^3d^3$ structure. Why it has a different structure?

8. May 4, 2015

### Raghav Gupta

They are saying that it has an unusual pentagonal bi pyramidal structure .
And remember one has lone pair and IF7 does not have that. So some differences will be there.

9. May 4, 2015

### mooncrater

Ok then.... I assume my doubt cleared.
Thank you. ($R\longrightarrow R$ help)

10. May 16, 2015

### James Pelezo

Good call RG... The Pentagonal Bipyramid is correct. Now, can you apply the VSEPR theory and show => AX6E geometry for :XeF6? and then the Valence Bond theory to show how the ground state electron configuration of Xe hybridizes at the valence level to give seven (7) sp3d3 hybrid orbitals? I'll bet you already know how to do it. Go for it!