Shapes of Molecules | A level Chemistry | Exam Question Walkthrough

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Shapes of Molecules.
A level Chemistry.
Exam question Walk-through.
VSEPR
Working out shapes.
Bond pairs and lone pairs.
Molecules with charge.
Explaining why a molecule has a particular shape.
Polar bonds and non-polar molecules.

[@umayyahosman2004]

Great video.. one of the best explanatory videos out there

[@chemistrytutor]

Excellent! Really pleased you found it helpful! 😀

[@pratyasha279]

Do you teach for the OCR exam board as well?

[@chemistrytutor]

The vast majority of exam boards have common content. The major differences between exam boards is simply how they group things together for teaching and for the exam

[@nadernassr6636]

in ur aqa a level chemistry paper 1 playlist i think u accidentally got a vid u didnt mean to put in there to do with speed paints or somethin

[@chemistrytutor]

Thanks, good spot!

[@oli6779]

how would you do something like XeO3 in which you would get an odd number when adding up

[@chemistrytutor]

This is a more complicated one. Since O is in group 6, it needs 2 more in the Valence energy level. So the connections with Xe would need to be double bonds. So in fact, the total would be 14 electrons. 7 pairs. 3 double bonds between Xe and O and one lone pair on the Xenon

[@Masowe.]

10:45 can i say bond angle is smaller?

[@chemistrytutor]

Yes definitely 😀

[@Masowe.]

@@chemistrytutor thank you a lot

[@meteorrjin]

For the KrF2 how come we dont minus 7.5 from 180 degress due to the 3 lone pairs.

[@chemistrytutor]

Good question! Due to the position of the Lone pairs the bonds are 180 degrees apart. The 'potential' repulsion would be up AND down. So the bonds would be pushed from 2 directions exactly equally and so the repulsion Up and Down cancel out 😀

[@jillian9098]

But as there’s 3 lone pairs, and there’s 2 of them at the top, wouldn’t that mean the forces pushing down would be stronger than the force pushing up by the one lone pair at the bottom?

[@chemistrytutor]

@jillian9098 the 3 lone pairs are in a triangle in the middle... the equatorial positions. The Bonding pairs make up the axial positions (the vertical). The additional repulsion for lone pairs relative to Bonding pairs that you've mentioned is correct. But their net effects cancel out since they're all at 120o angles to each other. It's not 2 on one side and 1 on the other, they are effectively at the vertices of an equilateral triangle