lppm4411
Badges: 5
Rep:
?
#1
Report Thread starter 2 years ago
#1
I am stuck on this question, mark scheme says answer is C however I don't understand why
I thought it would be A as the 3 bidentate ligands are the same and wouldn't form a different arrangement in space? Can someone correct me I have uploaded the question, Thanks

https://imgur.com/a/LyeZMwe
0
reply
EierVonSatan
Badges: 21
#2
Report 2 years ago
#2
(Original post by lppm4411)
I am stuck on this question, mark scheme says answer is C however I don't understand why
I thought it would be A as the 3 bidentate ligands are the same and wouldn't form a different arrangement in space? Can someone correct me I have uploaded the question, Thanks

https://imgur.com/a/LyeZMwe
Optical isomerism occurs with these types of structures:

Image
You cannot rotate the left hand side one to become the right hand side one.
0
reply
Kian Stevens
Badges: 16
Rep:
?
#3
Report 2 years ago
#3
(Original post by EierVonSatan)
Optical isomerism occurs with these types of structures:

Image
You cannot rotate the left hand side one to become the right hand side one.
This is a message to the above answerer: It's more about non-superimposable mirror images rather than rotation; if I were to place the second enantiomer on top of the first one, in the position they're in now, they would never fit properly. But with the way those structures are drawn, I could easily rotate one of them to match the other. Of course, if they were drawn in 3D, then rotation would matter, but that's not what defines an enantiomer.


Now let's explain why it's C:

  • A, as mentioned, exhibits just optical isomerism. This is because it's a octahedral molecule with bidentate ligands, so you can form two non-superimposable mirror images with it,


  • B exhibits both optical and cis-trans isomerism. This is because it has two bidentate ligands and two monodentate ligands. In the image on the question it's the cis isomer, which can also form optical isomers, but you can place a hydroxyl ligand directly across from another one to form the trans isomer (trans = across),

  • Finally, D exhibits just cis-trans isomerism. This is because it's a octahedral molecule with two different ligands; two of one and four of another. In the image on the question it's the cis isomer, but you can place one of the chloride ligands directly across from another to form the trans isomer.

This leaves C, which exhibits no stereoisomerism. This is because no matter how you arrange the ligands, they will always have the same arrangement in space.
0
reply
EierVonSatan
Badges: 21
#4
Report 2 years ago
#4
(Original post by Kian Stevens)
This is a message to the above answerer: It's more about non-superimposable mirror images rather than rotation; if I were to place the second enantiomer on top of the first one, in the position they're in now, they would never fit properly. But with the way those structures are drawn, I could easily rotate one of them to match the other. Of course, if they were drawn in 3D, then rotation would matter, but that's not what defines an enantiomer.
I like you
0
reply
Kian Stevens
Badges: 16
Rep:
?
#5
Report 2 years ago
#5
(Original post by EierVonSatan)
I like you
Lol, you're flattering me bud
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (120)
28.1%
No - I have already returned home (57)
13.35%
No - I plan on travelling outside these dates (84)
19.67%
No - I'm staying at my term time address over Christmas (40)
9.37%
No - I live at home during term anyway (126)
29.51%

Watched Threads

View All