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 -3
C 4
C 4
N 5
3 H atom attached to first C atom.
1 H atom to second one
There is a double bound between second C atom and third one.
The third one has one H atom.
N atom has 2 dots underneath it and attached to 2 H atoms.
Note: I tried to display the diagram but it turned out to be like above. sorry about that.
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.
Note: I tried to display the diagram but it turned out to be like above. sorry about that.
This programing recognizes one space and ignores after that.
I would assign H as +1 each for a total of +7 for H.
I would assign N a +3 for a total of +3 for N
That leaves -10 for C (to make zero for the compound which it must be) which makes EACH H -3 1/3.
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???
thank you!!!
To calculate the oxidation number for each atom in the given molecule, you need to follow a few steps:
Step 1: Determine the number of valence electrons for each atom.
Step 2: Assign the shared electrons to the more electronegative atom in each bond.
Step 3: Assign the remaining electrons to the less electronegative atom in each bond.
Step 4: Determine the oxidation number by considering the electrons assigned to each atom.
Let's go through each atom and calculate their oxidation numbers:
1. Hydrogen (H): Hydrogen is in Group 1 of the periodic table, so it typically has an oxidation number of +1. However, in some cases, such as when it is bonded to a metal or boron, hydrogen can have an oxidation number of -1. In the given molecule, hydrogen is bonded to carbon, so its oxidation number is +1.
2. Carbon (C): Carbon has four valence electrons. In the molecule, one carbon is bonded to hydrogen and two other carbons. Since hydrogen has an oxidation number of +1, carbon has to have an oxidation number that balances out the charge. In this case, it is -3. You can use the equation 4 - (number of hydrogen atoms) = oxidation number of carbon. So, 4 - 1 = -3.
3. Nitrogen (N): Nitrogen has five valence electrons. In the molecule, it is bonded to carbon and has four remaining electrons. Since carbon has an oxidation number of -3, and there are no other atoms bonding with nitrogen, you can determine the oxidation number of nitrogen using the equation -3 + (number of remaining electrons) = oxidation number. So, -3 + 4 = +1. Therefore, the oxidation number of nitrogen in the given molecule is +1.
In summary, the oxidation numbers for each atom in the molecule are as follows:
H: +1
C: -3
C: Undetermined
C: Undetermined
N: +1