First, draw the 2-hexene. Th is is a molecule of six carbons with a double bond in the second carbon:
CH3 - CH = CH2 - CH2 - CH2 - CH3
Secong, put one Br on the second carbon and one Br on the third carbon:
CH3 - CBr = CBr - CH2 - CH2 - CH3
Third, cis means that the two Br are placed in opposed positions, this is drawn with one Br up and the other down. So, you need to represent the position of the Br in the space:
H Br H H H
| | | | |
H - C - C = C - C - C - C - H
| | | | |
H Br H H H
The important fact to realize is that the two Br are in opposed sides of the molecule.
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Not sure what you are asking. I have two possible answers though...
It could either be more negatively charged, or valence electrons.
The more away from the nucleus a electron is, the more negatively charged it is.
The electrons on the outermost electron shell is valence electrons.
Again, I don't know what you were asking, but one of these answers may be correct.
Answer:
Zn(s) → Zn⁺²(aq) + 2e⁻
Explanation:
Let us consider the complete redox reaction:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
This is a redox reaction because, both oxidation and reduction is simultaneously taking place.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.
Here Zn(s) is undergoing oxidation from OS 0 to +2
And H in HCl (aq) is undergoing reduction from OS +1 to 0.
Therefore, for this reaction;
Oxidation Half equation is:
Zn(s) → Zn⁺²(aq) + 2e⁻
Reduction Half equation is:
2H⁺ + 2e⁻ → H₂(g)
