# Calculate the oxidation # for each atom in the molecule below. Can you help me with the rest. Thank you!!!

H H H H
/ / / /
H -- C ---C ===C ----N
..
/ /
H H

Group # # of E Oxidation #

H 1 0 1 x 7 = 7

C 4 7
C 4

C 4

N 5

3 H atom attached to first C atom.

1 H atom to second C atom.

There is a double bound between second C atom and the third C atom.

The third C atom has one H atom.

N atom has 2 dots underneath it and attached to 2 H atoms.

How many for each Carbon though!

That was the part I was confused!

Another question?

If we have double bond How many electrons we will have between each Carbon atoms?

Why did you assign +3 for N???

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1. If you go with what I did the oxidation number for each C is 3 1/3 just as I posted. This whole process is just a matter of book keeping. H can have +1 or -1 and we stay out of trouble if we assign +1 in most cases unless we know the compound is a hydride such as NaH or CaH2. N is in group V (or 15) depending upon what system you are using so it has usual oxidation states of -3 or +3 or +5. I arbitrarily assigned it +3. I noticed you had +5 and that is ok too; of course that will change C from 3 1/3 to some other number.

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2. For C=C there are four electrons between each C atom. But don't count the bonds or electrons as a way of calculating the oxidation state. Counting the electrons is a way to calculate the formal charge on the atom but that isn't the same thing as oxidation state.

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