Why is this not a geometric isomer?

[Declan McKeown]

While I'm fairly sure this question is pretty straight forward, I can't remember why (ii) is not considered a geometric isomer. (The answer is (c)) http://imgur.com/ObccNpv

Looking at ii I would have thought it to be trans-pent-2-ene due to the ethyl and methyl group either side of the double bond, however it is not and I am not sure why. Would anyone happen to know why this is?

Thanks :)

 

[Borek]

Are (i) and (ii) isomers at all?

 

[IcyRealm]

https://en.wikipedia.org/wiki/Cis–trans_isomerism

Well, (I) and (II) are not geometric isomers because they are the same compound just mirror images, besides both seem trans to me.
I II and III can not be geo. (cis-trans) isomers because that is three options.
III and IV don't make sense because IV is trans so III should be cis but it's weird because its not.

Sorry that's all I can help with my limited knowledge.

 

[epenguin]

https://en.wikipedia.org/wiki/Cis–trans_isomerism

Well, (I) and (II) are not geometric isomers because they are the same compound just mirror images, besides both seem trans to me.
I II and III can not be geo. (cis-trans) isomers because that is three options.
III and IV don't make sense because IV is trans so III should be cis but it's weird because its not.

Sorry that's all I can help with my limited knowledge.

Being mirror images does not necessarily make them the same compound! You probably remember now.

The essential is you can swivel it around a vertical axis and i coincides with ii - so it is the same thing. Any bodily movement leading one to coincide with another means it is the same thing. A reflection on the other hand is not a bodily movement.

I think that identity was the essential point of this exercise. Anything that has the same composition as another thing but is not the same thing is an isomer of it.