7th edition 2E. 21

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

Moderators: Chem_Mod, Chem_Admin

Alyssa Bryan 3F
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am
Been upvoted: 1 time

7th edition 2E. 21

Postby Alyssa Bryan 3F » Sun Nov 18, 2018 6:16 pm

For part d of the question where it asks for the Lewis structure and bond angle N2H4, the book gives the exact bond angle of 107. I know in class Dr. Lavelle said for trigonal pyramidal structures we only have to know that the angle is less than 109.5, so for our answer would it be okay to say less than 109.5 and not give the exact angle even though that is what the book says?

Michael Torres 4I
Posts: 92
Joined: Thu May 10, 2018 3:00 am
Been upvoted: 1 time

Re: 7th edition 2E. 21

Postby Michael Torres 4I » Sun Nov 18, 2018 6:39 pm

I think it would be good to know the exact angles of the trigonal pyramidal structure. However, saying less than 109.5 should probably suffice if you don't know it since Lavelle already said we don't need to know the specific value.

Jessica Helfond 2F
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

Re: 7th edition 2E. 21

Postby Jessica Helfond 2F » Sun Nov 18, 2018 6:54 pm

It's not necessary for us to know the exact angle, I think it was just a specific example to show us the proof that the bond length was less. As long as you know it's less than 109.5, you should be good.

Anna O 2C
Posts: 98
Joined: Fri Sep 28, 2018 12:19 am

Re: 7th edition 2E. 21

Postby Anna O 2C » Sun Nov 18, 2018 7:47 pm

Test questions, unless given experimental data to permit exact bond angles, will generally accept "less than 109.5" because, without an experiment, one can't really know the extent to which the lone pairs can alter the position of the bonded atoms because there isn't a calculation to tell us that.

Anjali 4A
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

Re: 7th edition 2E. 21

Postby Anjali 4A » Sun Nov 18, 2018 9:17 pm

Dr. Lavelle said he does not expect us to know the specific bond angles, and on exams he will accept answers 'less than'.


Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 2 guests