# What is the oxidation state of SCN

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Samy

Registration date: 07/23/2013
Posts: 5

 Posted: Jul 23, 2013 21:37 Title: Oxidation number NaSCN My question: Hello everybody, I have a problem with determining the oxidation number of NaSCN. According to a sample solution for a test (which can of course also be wrong) the oxidation numbers would be: Na = +1 S = -2 N = -3 C = +4 I would be very grateful for your help! :) My ideas: It is clear to myself that Na must be +1 since it is an alkali metal. Unfortunately, I cannot determine the remaining atoms with the usual rules for determining the oxidation number. Is the CN cyanite here? According to some Google results, however, cyanite has an oxidation number of -1. Here, however, we would have +1 for CN (according to the sample solution).
ka

Registration date: 10/01/2012
Posts: 1832

Posted: Jul 23, 2013 22:30 Title: Re: Oxidation number NaSCN

 Samy wrote the following: It is clear to myself that Na must be +1 since it is an alkali metal. Unfortunately, I cannot determine the remaining atoms with the usual rules for determining the oxidation number.

Since nitrogen is more electronegative than carbon, all binding electrons from the triple bond are attributed to it. So formally he gets 3 electrons: OZ (N) = - 3

The carbon is involved in a total of 4 bonds. All ties to more electronegative partners. Formally, it gives off 3 electrons to nitrogen and one electron to sulfur: OZ (C) = + 4.

The sulfur receives an electron through the ionic bond to Na and formally a second electron from the bond to the carbon:
OZ (S) = - 2
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Friedrich Karl Schmidt
Thirsty for knowledge

Registration date: 04.08.2012
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 Posted: Jul 23, 2013 11:05 PM Subject: A little tip: To write a triple bond you can either use an equal sign - underlined (over "u"): HC = CH. Or in latex via "\ equiv": It will look like that: _________________If you made one mistake and don't correct it, you commit a second one.
Samy

Registration date: 07/23/2013
Posts: 5

Posted: Jul 24, 2013 00:10 AM Subject:

ka wrote the following:
 Samy wrote the following: It is clear to myself that Na must be +1 since it is an alkali metal. Unfortunately, I cannot determine the remaining atoms with the usual rules for determining the oxidation number.

Since nitrogen is more electronegative than carbon, all binding electrons from the triple bond are attributed to it. So formally he gets 3 electrons: OZ (N) = - 3

The carbon is involved in a total of 4 bonds. All ties to more electronegative partners. Formally, it gives off 3 electrons to nitrogen and one electron to sulfur: OZ (C) = + 4.

The sulfur receives an electron through the ionic bond to Na and formally a second electron from the bond to the carbon:
OZ (S) = - 2

Okay, thanks for the quick reply!
However, I have never heard of the fact that oxidation numbers are assigned using electronegativities.

So far I only knew the following rules:

 Quote: 1) Individual elements (elemental state of an atom) always have the oxidation number 0 (if ONLY S8: S = 0, ONLY Pt: Pt = 0, O2 = 0) 2) Sum of all oxidation numbers = total charge of the compound (ClO-: total charge is -1 (because of -): O = -2, Cl = 1 → -2 + 1 = -1) 3) For ions (charged atoms): Oxidation number = ion charge of the individual ion (NO3- = -1, OH- = -1, CO32- = -2, SO42- = -2; PO43- = -3, NH4 + = +1, MnO4- = -1) 4) Alkali metals (1st main group) except hydrogen (H): +1 5) Alkaline earth metals (2nd main group): +2 6) Fluorine always has -1 7) Halogens (7th main group): -1 [Exception (does NOT apply to fluorine): halogens with oxygen (O): +1] 8) Hydrogen (H): +1 [Exception: compound with alkali metal (1st main group): H = -1, alkali metal = +1] 9) Oxygen (O): -2 [Exceptions: connection with fluorine: O = +2 connection with O-O: -1 (H2O2: O = -1)] 10) adjust the remaining atoms so that rule 2) applies.

Is there a connection? Could I somehow solve it without the electronegativities?

And can you explain the electronegativities process to me in general?

LG, Samy
Thirsty for knowledge

Registration date: 04.08.2012
Posts: 537

Posted: Jul 24, 2013 12:44 AM Subject:

The auxiliary rules that you have learned are also all correct. For inorganic compounds, it is usually very simple (you rattle off the rules, in most cases), but for organic substances you have to record the structures in order to determine the oxidation numbers.

Without knowing the structure, it is m.M.n. not possible to find out the oxidation number of an organic compound (if you have not memorized it beforehand). There are some general rules like "in carboxylic acids the carbon has the oxidation number + III" and that means only one carbon atom, the one that is attached directly to the carboxy group - but I wouldn't rely on that too much. There is hardly a way around recording the structure: But it's not that difficult either, don't worry.

"Electronegativity" is nothing more than the ability to "attract" the binding electron in a chemical bond. Some examples:

1. Example: Methane: The carbon "pulls" more strongly on the bond - is therefore "more electronegative" (thus the binding electrons are formally assigned to it) and you can read that from the numbers: Electronegativity is (according to the Pauling scale) for Carbon 2.5 - for hydrogen 2.2. That means carbon gets the electrons: And that's how you look at the whole thing. You add up the electrons: 4 lines (4 bonds) make 8 electrons, since each bond means two electrons. Carbon normally has four outer electrons (because it is in the 4th main group); therefore it has four electrons too many (makes four negative (formal) charges): Oxidation number - IV. The hydrogen (1st main group) would normally have one electron; now it doesn't have any (in our simplified model). A negative charge is missing, if something negative is "missing", it has to be positive: Oxidation number + I. Just think about it for a moment. To show that the electrons are assigned to one of the partners (sometimes also assigned to both; then it would be a line exactly in the middle of the bond) it is "separated" by a curved line (see Fig. 1).

2nd example:-S-C = N. Nitrogen is more electronegative than carbon and carbon is (at least on Pauling's electronegativity scale) more electronegative than sulfur. Now you just have to "assign" the electrons. One line = one bond = two electrons. Do not forget that there are still free electron pairs there, so it looks like in Figure 2.

Each free electron pair also means two electrons again. When sulfur gets 6 (external) electrons it is "happy" (because it is in the 6th main group) - i.e. no charge, carbon needs 4 electrons and nitrogen 5 electrons (5th main group). For every electron that is too much, that atom gets a negative charge. Of course, for every missing electron it gets a positive charge.

As an exercise, you can now try it for the thiocyanate, analogous to example 1 (as it was done in the first picture). I've already explained everything you need to know. And if you can't do it, it could also be because of the time; it makes more sense if you start this tomorrow. We are still available if you have any questions.

I can reassure you: there are usually not that many elements in organic chemistry. Always carbon and hydrogen, rarely sulfur, more often oxygen and a few, few others. But once you have understood the principle, you automatically assign the electrons to the more "electronegative" element; Especially in school chemistry, there are not too complicated examples.

Another note: please omit full-text quotations, something like this only disturbs the flow of reading and is usually unnecessary because the quoted is directly above it.

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Samy

Registration date: 07/23/2013
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Posted: Jul 24, 2013 11:10 AM Subject:

That has already helped me a lot! So basically I have now understood the electronegativities (as far as I can tell).

However, I still can't quite come up with the right solution with the SCN.

Here are my thoughts on this:

1) Electronegativities: S> Na. So sulfur gets the electron pair to Na. So we come to the conclusion that Na has +1 and S has first got the bond to sodium for itself.

This brings us to your starting formula for the SCN with the 1-fold negatively charged S.

2) Electronegativities of N> C. Hence, N gets the complete 3-fold bond. So N has 8 electrons, but only needs 5. So the charge of N = -3

3) Electronegativity from C> S (can one assume that the electronegativity is greater? Because the difference is only 0.1. Or, with such small differences, do we have to assume that it is approximately the same?): In any case, C > S means that C gets the bond to S. As a result, C would have the 2 electrons of the bond from S and none otherwise. However, C needs 4 electrons and would therefore have a double positive charge.

4) It follows that S is awarded 6 electrons: The two who are on S anyway, 0 from C, and 2 from the bond to Na.
Since S also strives to have 6 electrons, it would have the oxidation number 0.

So there must be a mistake here somewhere?!?

And is it actually generally possible that an atom has an oxidation number of 0? Or do I then automatically know that something must be wrong?

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Daniel35

Registration date: 09/10/2012
Posts: 1127

 Posted: Jul 24, 2013 12:14 PM Subject: On the Pauli8ng scale, sulfur has a higher electronegativity than carbon (lower in Mulliken and Allred-Rochow). To determine the oxidation number, which is only a formalism anyway, it makes more sense to consider sulfur to be more electronegative.
Thirsty for knowledge

Registration date: 04.08.2012
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 Posted: Jul 24, 2013 12:24 PM Subject: Draw the sulfur rather like in picture 2 (i.e. with three free electron pairs), also with a negative charge and the sodium rather than Na+ Besides. This is namely an ionic bond there (see also the link). You don't have to worry about sodium anyway. Great: Rule No. 4: Alkali metals always have the oxidation number + I. To 2): Nitrogen is fine! Otherwise; when asked "is sulfur or carbon more electronegative" -> I agree with Daniel / ka. Feel free to assume that sulfur is more electronegative here and gets all of the electrons (so draw again!). In principle, the oxidation number zero is also possible; E.g. in the elementary state: O2, F2, etc. have the oxidation number zero, but also a carbon that formally has exactly four (external) electrons, for example. It just doesn't appear in our example. Then revise that again; you have to come to the same result as "ka" at the beginning (see the oxidation numbers in his article)._________________If you made one mistake and don't correct it, you commit a second one.
Samy

Registration date: 07/23/2013
Posts: 5

 Posted: Jul 24, 2013 2:11 PM Subject: All right, if I assume that sulfur has a higher electronegativity than carbon then it works out wonderfully Can I make this assumption in principle? Or is that just by chance the case here with the SCN? And are there any other such assumptions where one should assume that the electronegativity of one atom is higher than that of the other, although it would not be the case according to the periodic table?
Thirsty for knowledge

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Posted: Jul 24, 2013 9:55 PM Subject:

For now, assume that sulfur is always more electronegative than carbon.

All you have to do is use the electronegativity scale (the "table" below) as a guide; It doesn't matter whether there is a difference of 2.0 or 0.1: These elements are then simply more electronegative and get the electrons.

And sulfur is a bit exotic, usually when it comes to organic compounds, you get oxygen, carbon and hydrogen (although the former does not have to be given). You don't need to check the electronegativity every time, the following applies: oxygen> carbon> hydrogen. If you know that you don't need to look any more, except when you have something like "sulfur" again (which is rather rare).

The trend in the periodic table would be: Electronegativity increases "towards the top" and "towards the right". The most electronegative element is at the top right (fluorine) [noble gases excluded] and the most electropositive element is at the bottom left. See table in the appendix.

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Samy

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