Which molecule most likely to deviate from ideal values?

[LegitSci]

Hi,

Hoping I could get some help on this one.
How would I know what causes bond angles to deviate from the ideal values?

I greatly appreciate any help and advice on this.

Thanks

 

[Bystander]

Anything that breaks the symmetry; ³⁵Cl instead of ³⁶ in a CCl₄ molecule; you name it.

 

[LegitSci]

Is there any connection between polarity?

 

[Bystander]

polarity?

You name it.

 

[LegitSci]

If so, how would you compare between different molecules? Like if I were to compare which molecules deviated THE MOST from the "ideal" values...
for example BrCl3, XeCl4, or CO

 

[LegitSci]

does deviation correspond to the number of lone pairs on the central atom, but even then XeCl4 and BrCl3 both have 2 pairs so I'm super confused.
Or does it have to do with location of the lone pairs?

 

[Bystander]

bond angles

CO? A straight line is a straight line. What are the "ideal" values for the other two? Which isotopes/isotope combinations are you considering "ideal?"

XeCl4 and BrCl3 both have 2 pairs

Two pairs? What are you counting?

 

[LegitSci]

but arent there lone pairs of electrons for central atom (either C or O) CO or do they not affect the bond angles?

for XeCl4 and BrCl3, I am counting the number of lone pairs of electrons.
so there would be 2 pairs of lone pairs on the central atom (Xe and Br, respectively), right?

 

[LegitSci]

I am considering the VSEPR to be "ideal", sorry for any confusion

 

[LegitSci]

I think I got it. It has to do with the placement of lone pair of electrons on the central atom.

Thanks

 

[Bystander]

central atom (either C or O) CO

One bond can NOT make an angle.

would be 2 pairs of lone pairs on the central atom (Xe and Br, respectively)

Yes.

VSEPR

So, that's back in fashion --- two pairs --- one on one side and the other opposite --- "pair repulsion" --- what are your thoughts?

 

[Bystander]

I think I got it. It has to do with the placement of lone pair of electrons on the central atom.

Yes --- You're welcome.