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???
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.
I'll be happy to answer follow up questions.
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.
To calculate the oxidation number for each atom in the molecule, we need to use the following guidelines:
1. The oxidation number of an atom in its elemental form (H2, O2, N2, etc.) is always zero.
2. The oxidation number of a Group 1 element (H, Li, Na, etc.) is always +1.
3. The oxidation number of a Group 2 element (Be, Mg, Ca, etc.) is always +2.
4. The oxidation number of oxygen (O) is usually -2, unless it is in a peroxide (O2^2-) where its oxidation number becomes -1.
5. The oxidation number of hydrogen (H) is usually +1, unless it is in a hydride (LiH, NaH, etc.) where its oxidation number becomes -1.
6. The sum of the oxidation numbers of all the atoms in a neutral molecule is always zero.
7. The sum of the oxidation numbers of all the atoms in a polyatomic ion is equal to the charge of the ion.
Now, let's go through the molecule step by step to calculate the oxidation numbers:
1. Hydrogen (H):
Since each hydrogen atom is bonded to a carbon or nitrogen atom, we can assume its oxidation number is +1.
2. Carbon (C):
The first carbon atom is bonded to four hydrogen atoms and no other atoms, so its oxidation number can be calculated as follows:
(+1 x 4 hydrogen atoms) + (0 other atoms) = +4
So the oxidation number of the first carbon atom is +4.
The second carbon atom is bonded to three hydrogen atoms and one carbon atom, so its oxidation number can be calculated as follows:
(+1 x 3 hydrogen atoms) + (+4 carbon atom) = +7
So the oxidation number of the second carbon atom is +7.
The third carbon atom is bonded to two carbon atoms (one through a double bond) and one hydrogen atom, so its oxidation number can be calculated as follows:
(+7 x 2 carbon atoms) + (+1 hydrogen atom) = +15
So the oxidation number of the third carbon atom is +15.
3. Nitrogen (N):
The nitrogen atom is bonded to three carbon atoms and two hydrogen atoms, so its oxidation number can be calculated as follows:
(+15 x 3 carbon atoms) + (+1 x 2 hydrogen atoms) = +47
So the oxidation number of the nitrogen atom is +47.
In summary, the oxidation numbers for each atom in the molecule are as follows:
H: +1
C1: +4
C2: +7
C3: +15
N: +47
Regarding your question about the number of electrons in a double bond between two carbon atoms, a double bond consists of a sigma bond and a pi bond. The sigma bond is formed by the overlapping of two atomic orbitals, and the pi bond is formed by the side-to-side overlap of p orbitals. Each bond (sigma or pi) consists of two electrons. In the case of a double bond, there are two pi bonds between the carbon atoms, so there are four electrons in total involved in the double bond.
Lastly, I apologize for any confusion caused by the assigned oxidation number of +3 for nitrogen (N). It was a mistake on my part. The correct oxidation number for nitrogen in this molecule should be +47, as explained earlier.