Homework Question 4.5 (6th edition)

(Polar molecules, Non-polar molecules, etc.)

Moderators: Chem_Mod, Chem_Admin

Dana Wilks 3I
Posts: 30
Joined: Fri Sep 28, 2018 12:23 am

Homework Question 4.5 (6th edition)

Postby Dana Wilks 3I » Thu Nov 15, 2018 11:03 am

For part b on question 4.5, it asks us to predict the OClO bond angle for a ClO2 + ion. I understand that the shape is angular because of the lone pair on the Cl atom, but I don't understand how we are supposed to know that the angle will be slightly less than 120 degrees.

AdityaGuru1H
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

Re: Homework Question 4.5 (6th edition)

Postby AdityaGuru1H » Thu Nov 15, 2018 11:16 am

This is because the overall structure if we include the lone pair is trigonal planar and in trigonal planar the bond angles are all 120 degrees. However this is if all 3 electron regions are bonding pairs, but in this case one of them is a lone pair which has more repulsion and will cause the bond angle between the two remaining bonding pairs to be less than 120 degrees. Hope this helps!

Schem_student
Posts: 29
Joined: Fri Sep 28, 2018 12:19 am

Re: Homework Question 4.5 (6th edition)

Postby Schem_student » Thu Nov 15, 2018 11:50 am

That is the hardest part when it comes to molecular shape because there is a large difference between elctron denisty domains that are lone pairs vs bonding pairs. The lone pair domain affects the elctrons involved in the bonding pairs significantly more than a bonding pair would. The electron electron repulsion that we learned about during electron config in the quantum section creates a similar repulsion in the molecular structure of a molecule.


Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 1 guest