4.1

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

Moderators: Chem_Mod, Chem_Admin

304922790
Posts: 25
Joined: Fri Sep 29, 2017 7:04 am

4.1

Postby 304922790 » Sun Nov 12, 2017 9:20 pm

For 4.1, how does the bond angle determine if the molecule "must have lone pairs" or "may have lone pairs?" If there is a lone pair for part B, will the shape of the molecule still be linear?

Lourick Bustamante 1B
Posts: 50
Joined: Wed Nov 16, 2016 3:02 am

Re: 4.1

Postby Lourick Bustamante 1B » Sun Nov 12, 2017 9:29 pm

Bond angle determines if a molecule has lone pairs because lone pairs have strong repulsive factors that cause it to displace the other atoms, pushing other atoms closer together. This causes the bond angle between the other two atoms to decrease. Therefore, if there is a lone pair for part B, the shape will be bent or v-shaped and no longer linear.

Timothy Kim 1B
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: 4.1

Postby Timothy Kim 1B » Sun Nov 12, 2017 11:17 pm

Lone pairs are the reasons for the bond angles being closer together. We learned in class that the lone-lone pair repulsion strength is the greatest, followed by the lone-bonding pairs. We know that bonding pairs want to be as far apart as possible. Thus, the explanation for the bond angles being close together must be the existence of lone pairs.

Cooper1C
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

Re: 4.1

Postby Cooper1C » Mon Nov 13, 2017 11:24 am

It is possible for (b) to have lone pairs if it has 3 lone pairs around the central atom. It will still be a linear shape. See page 118 of the book for a graphic.

Chem_Mod
Posts: 17828
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 406 times

Re: 4.1

Postby Chem_Mod » Mon Nov 13, 2017 2:31 pm

Please post/state what the problem is rather than simply referring to it.

Lindsay H 2B
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: 4.1

Postby Lindsay H 2B » Mon Nov 13, 2017 3:03 pm

the bond angle can determine whether an atom can, must, or cannot have lone pairs because the given bond angle in this exercise suggests the number of regions of e- density around the central atom. for part (a), the 120º bond angle means 3 regions of e- density around the central atom but since there are only 2 atoms bonded to the central atom, there must be a lone pair on the central atom as well.


Return to “Determining Molecular Shape (VSEPR)”

Who is online

Users browsing this forum: No registered users and 2 guests