Determine oxidation number?
My question is, what oxidation number does the positively charged nitrogen atom have?
Normally I always look at the bonding partners and count the electrons depending on their electronegativity.
In this case, three electrons with nitrogen, since both pull equally strongly, 3 each and another 2 electrons with from the single bond with the carbon, so 5, then the oxidation number would be 0, but how can that be consistent with the positive charge?
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You have calculated correctly — oxidation numbers and formal charges have nothing to do with each other; they are different ways of assigning the electrons to the atomas. Therefore, their sum is the same over all atoms, namely the formation of the molecule (the division does not alter the number of electrons), but the similarities end.
Other example: In H3O+, O has the formal charge +1 and the oxidation number −II.
FALSCH! Since a positive atom has the greater electronegativity than a neutral atom of the same element. -3 would be formal.
You then use a modified method to assign electrons to atoms. This can be done, of course, but what comes out is no longer the commercially available oxidation number.
Electronegativity is a compromise between electron affinity and ionization energy. Both sizes can be measured, and both sizes are defined. Electronegativity and oxidation number are terms from the transition of alchemy to real chemistry as exact natural science.
Count from: +5 -3 (from adjacent N) -1 (from C) are +1
So oxidation number -4 and formal charge +1.
The nitrogen with the formal charge is more electronegative than the neutral nitrogen.
If you get this reckoned you get +5 -8 so -3.
This is the formal charge. For the oxidation number, the sticky gets 3 electrons from its N-neighborhood, but both electrons of the C-N bond (because N is more negative than C). Thus, five electrons are added, i.e. the number of oxidations 0.
That’s not true. According to your statement, for example, C in CH4 should have the oxidation number -8, but has -4.
I have the impression that you do not know how to determine oxidation numbers — try it for the N2 molecule |N=N|, you must get zero.
Of course, for both N 0 comes out. What oxidation number would you assign N and O to NO?
The nitrogen bound to the C has 4 electrons per se and is therefore simply positive because it normally belongs to 5 valence electrons. And he also has this excess charge, so his OZ = 0.
It is a discussion about Red Riding Hood’s hatch color, but formally 5 positive core charges have to be neutralized. The two electrons to the carbon atom draw the N atom. No discussion. And the three pairs of electrons to the terminal N atom also pulls the controversial N atom. So we have +5 – 2 – 6 = -3
So it is formal, and more than formalities are not here anyway.
Oxidation number -4 is not possible for nitrogen at all. He’d have to cross the octet. It doesn’t make nitrogen. In nitrides N^3 nitrogen has the oxidation number -3, more electrons do not go.
It is a discussion about Red Riding Hood’s hatch color, but formally 5 positive core charges have to be neutralized. The two electrons to the carbon atom draw the N atom. No discussion. And the three pairs of electrons to the terminal N atom also pulls the controversial N atom. So we have +5 – 2 – 6 = -3
So it is formal, and more than formalities are not here anyway.
You had the same number of oxidations and actual charges. That’s nonsense. The oxygen atoms in ozone have the oxidation number 0 and not -2 and 2x+1. In carbon monoxide, the carbon atom has the oxidation number +2, the oxygen atom -2, not vice versa. The rule, not exception, is that chemists no longer control the substance of basic studies after a while. That’s not bad anymore. But you have to realize that you can no longer calculate oxidation numbers if you do not have the formalities established for it and you do not have a desire or consider it unnecessary to refresh the knowledge.
You little flash marker 😉
Oxidation numbers have nothing to do with actual charges. First semester chemistry.
The decisive factor is that the positive N atom has a greater electron affinity (can be measured) and a greater ionization energy (can be measured) than the neutral N atom. ALSO also has a greater electronegativity.
Its oxidation number is actually negative and not zero, as the reactions of the diazonium salts show us.
https://www.chemie.de/lexikon/Diazonium salts.html
Why are you all always so hard to look beyond the edge of the plate?
I don’t understand the problem. According to formalism, the nitrogen atom has the formal charge +1 and the carbon atom to which the nitrogen atom is bonded has the oxidation number +1. The sum is thus +1 in both cases. After your calculation, the nitrogen atom in nitric acid has the oxidation number +3 and one of the oxygen atoms has the oxidation number 0.
You can’t let that stand. The sum of the oxidation numbers of all atoms in the molecule must be +1. I think we’re agreed, right? What oxidation numbers would you assign to the other atoms, so that the sum is +1? Otherwise, it should be inconceivable that the positive N atom has a stronger electronegativity than the neutral N atom.
N+IIO Ì„II
Yeah, looks like that.
The Fabian is a high-lifter. He’s lying like there’s no tomorrow and he’s always adding a degree in chemistry…
The formal rules for calculating the oxidation number in this case result in zero. Formal rules, according to which formal charges are first calculated and then oxidation numbers are determined on the basis of these charges, are not available in chemistry.
C is more electronegative than H, so all 8 binding electrons are added to it, four it has in the atom, i.e. oxidation number −IV.