Bond angles (help please) [ENDORSED]

[AtreyiMitra2L]

Hello. I am still confused about certain bond angles. I posted this elsewhere but I want to say it again to highlight the discord. This what I feel it to be.

Sea-saw : <90, <120, <180
T-structure : <90, <120, <180
Square pyramidal : <90
Square planar : 90

Can someone please verify these all of these bond angles for me? I feel that the bond angle should be less than 120 because Lone pair atom repulsions are stronger. Therefore even in the equatorial position, the electrons should repulse the atoms. I feel that the existence of the lone pair should apply the same logic with why for sea saw it should be less than 90 and 180 for sea-saw.

[Cassidy 1G]

I believe the angle between the two equatorial ligands is exactly 120 degrees. The lone pairs repel the electrons just enough to keep them this position.

[Cassidy 1G]

In addition, the textbook on pages 114 to 115 talks about the repulsion of axial vs equatorial lone pairs. In the example given the axial bond takes more energy and the bond length is shorter than an equatorial bond, which the book says is 120 degrees. So i think it is always 120.

[Chem_Mod]

I just answered this question at the end of class.

Need to be careful on wording.

If there are three equatorial bonds as in AX₃ and AX₅ AND all the atoms (X) are the same, then the equatorial bond angles are 120^o.

Even with AX₃ and AX₅ some bond angles will be smaller and some larger than 120^o if the atoms are different.
Obviously the bond angle between the larger atoms will be slightly higher, and the bond angle involving a smaller atom will be slightly lower.

Here is the correct summary:

The angles are slightly less due to one or more lone pairs.

Seesaw : <90, <120, <180
T-structure : <90, <180
Square pyramidal : <90, <180
Square planar : 90 and 180 (here the lone pairs are symmetric, no net distortion)

Now I have 10 minutes to eat lunch before leaving for my next class. :-)