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
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
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 😀
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?
@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
Great video.. one of the best explanatory videos out there
Excellent! Really pleased you found it helpful! 😀
how would you do something like XeO3 in which you would get an odd number when adding up
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
Do you teach for the OCR exam board as well?
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
For the KrF2 how come we dont minus 7.5 from 180 degress due to the 3 lone pairs.
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 😀
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?
@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
10:45 can i say bond angle is smaller?
Yes definitely 😀
@@chemistrytutor thank you a lot
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
Thanks, good spot!