[AdiDex]

I know H₂CSF₄'s central atom is S . Having Hybridization sp₃d , Molecular structure is Triangular Bi-pyramid.
That's alright .
But Where its unhybridized orbital lies ?? How its π bond forms ??

My teacher told me that its unhybrid orbital ( Didn't told it is p or d , i think it should be d orbital ) lies in equatorial plane such that it can remain perpendicular atleast 3 orbitals i.e. from 2 axial and 1 equatorial orbital.
Ok it can be correct but the problem is that Unhybrid Orbital is d so it will have 4 lobes.  So what rubbish he told me i don't know. So now anybody can tell me ??
Second Problem Is he also told that CH₂= will lie on the equatorial plan such that 2 hydrogen atoms are same plane with the axial fluorine , He told this correct i checked in one book .

my problem is why CH₂= Can't be on axial position ??
In case of XeOF₂ , O= also lies in equatorial plane(as I watched Google Images of XeOF₂) ..!!
so there is something common that why Atoms having pi bond lie on the equatorial plane ...but i don't what is that common thing ..
so please help me.
Thanks in advance

[mjc123]

Why is the lone pair equatorial in SF₄?

[AdiDex]

See what i know is that -
'To minimize the repulsion between lone pair  and bond pair.'
Am i right or wrong ...??

[mjc123]

Yes, but you want to minimise repulsion between all the electron pairs. Why does the lone pair go equatorial? Can you see any parallel with a double bond?

[AdiDex]

because lone pair - bond pair repulsion is greater than bond pair bond pair ..
ok i got your point .
you are trying to say that  double bond has a pi cloud also , so the double bond - single bond repulsion is greater than single bond-single bond repulsion ??
Thank you very much .

i have another doubts also ..!!
Q.1 is there any mathematical proof that at equatorial position the repulsion is minimized. At equatorial position 2 bonds are with 120 degree angle and two with 90 degree angle .
if it is axial position then one bond is with 180 degree and another three with 90 degree .

i know there should be no simple maths behind this but i want to know, is there any formula ??

Q.2 in sp3d3 hybridization when there is one lone pair of electron we put it on equatorial position . and when there is two lone pairs then we put them on axial position . same question is there any mathematical proof ??

Q.3 i'm still confused with unhybrid orbitals ..when p orbitals are unhybrid then they tries to remain perpendicular to the hybrid orbitals .! is this correct or wrong ?? So what about d-orbitals ??

[mjc123]

Q.2 in sp3d3 hybridization when there is one lone pair of electron we put it on equatorial position . and when there is two lone pairs then we put them on axial position . same question is there any mathematical proof ??

Not true; ClF₃, for example, is T-shaped, with two equatorial lone pairs.

[AdiDex]

Not true; ClF₃, for example, is T-shaped, with two equatorial lone pairs.

bro you are right but i was talking about "SP3D3" not sp3d2 . ClF₃ is a sp3d2 hybridised atom .
XeF₅^- is an example of sp3d3 hybridised molecule .

[mjc123]

Oh I see. ClF₃ is actually sp³d - five electron pairs, five orbitals. IF₅ would be sp³d².
You are right, XeF₅^- is sp³d³ with two axial lone pairs. In this case, because of the 5 equatorial electron pairs rather than 3, the axial position is the one that minimises repulsion for the lone pairs. But do you know any examples with one lone pair where it goes equatorial? BrF₆^- and IF₆^- are octahedral; XeF₆ is - well, look it up, it's complicated, but it's not PB with an equatorial lone pair (or axial either).

[AdiDex]

I know that when there is one lone pair in sp3d3 it comes at equatorial and when there is 2 lone pairs they come at Axial position . That why i was asking for any mathematical formula so that by using it i can measure the repulsion ??
and what do yo mean by PB ??

[mjc123]

I know that when there is one lone pair in sp3d3 it comes at equatorial

Can you give any examples? I can't, offhand.
PB = pentagonal bipyramid

[AdiDex]

XeF₆

[mjc123]

No it's not:
"The molecule XeF₆ is an interesting case. As with IF₇, application of VSEPR rules suggests seven electron pairs. These are made up from six bonding pairs and one lone pair. In fact, the structure of XeF₆ is based upon a distorted octahedron, probably towards a monocapped octahedron. It is difficult to settle the geometry of the lowest energy configuration because the geometry of XeF₆ changes rapidly with time, that is, it is fluxional. The effect of this fluxional process is to average all the fluorine positions."
http://winter.group.shef.ac.uk/vsepr/high-cn.html