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According to the formula it should be +5 but according to its structure shouldn't it be +4? How is it +5 then?

enter image description here

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    $\begingroup$ $5\cdot(-2) + 2\cdot(+5) = 0$ $\endgroup$ – Klaus-Dieter Warzecha Feb 18 '16 at 7:44
  • $\begingroup$ But according to the structure it should be + 4 $\endgroup$ – rishabh gupta Feb 18 '16 at 8:32
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    $\begingroup$ No. Why do you think so? $\endgroup$ – aventurin Jul 2 '16 at 17:46
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    $\begingroup$ A coordinate bond gives a formal charge of +1 to N, implying that it effectively gets a raise in oxidation state by +2 $\endgroup$ – Supernova Mar 25 '17 at 17:59
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It is +5 .Even by the structure.

enter image description here

Notice that it is a resonance structure of:

enter image description here

Now one of the bonds are coordinate bonds where nitrogen donates BOTH its electrons to oxygen (otherwise, oxygen cannot have a single bond without a negative charge and no peroxy type linkage).

So, while determining oxidation state from the structure :

O.N=+2+2+1=+5

References: https://en.wikipedia.org/wiki/Dinitrogen_pentoxide

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    $\begingroup$ Your second image is better since it shows formal charges. $\endgroup$ – Jan Jul 2 '16 at 18:10
  • $\begingroup$ True, but that's not much of an improvement. Evidently, I picked it up for representation purpose. $\endgroup$ – Varun Jul 4 '16 at 2:37
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    $\begingroup$ Oh, I always advocate properly assigning formal charges to all atoms that need them, which is why I wouldn’t have upvoted your answer if only the first structure were shown ;) $\endgroup$ – Jan Jul 5 '16 at 0:11
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O. S is a theoretical concept. Electrons shared between two unlike atoms are counted towards more electronegative element.

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Nitrogen cannot exhibit a +5 oxidation state due to the absence of d-orbitals.

PROOF: Quantum Mechanics basics

As Nitrogen is in 2nd period in the modern periodic table, principal quantum number being n=2, the Azimuthual Quantum numbers = 0 to n-1 for Nitrogen it ranges from 0 to 1.

0 (zero) represents an s orbital and 1 represents p orbital hence no D orbitals are there.

... which clearly tells us that due to absence of d orbitals, nitrogen cannot expand its oxidation state greater than 4. Hence by the basic principle of quantum mechanics we can say that in N2O5 nitrogen has +4 oxidation state.

NOTE: Formulae are made to simplify the things. But they have some limitations But in order to overcome those limitations one must use the basics of the subject.

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    $\begingroup$ Sorry ,I think you don't even know a basic difference between covalency and oxidation state.It may sound same but they are entirely different. $\endgroup$ – Satyajeet Mar 25 '17 at 16:33
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    $\begingroup$ Don't count bonds. Count electrons. Nitrogen comes in with five valence electrons but all five are used to form bonds with more electronegative atoms (oxygen). That means all five valence electrons are considered lost from nitrogen in the oxidation state calculation. Thus +5. $\endgroup$ – Oscar Lanzi Mar 25 '17 at 18:11
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    $\begingroup$ Although this is a really crappy answer, it's an attempt at answering the question and thus I would downvote, not flag. $\endgroup$ – M.A.R. Mar 25 '17 at 22:17
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    $\begingroup$ @AbirPal Throw that book away. If they make such a bold and painfully obvious wrong statement, nothing, absolutely nothing in it is trustworthy. $\endgroup$ – Martin - マーチン Apr 24 '17 at 9:07
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    $\begingroup$ @Martin A lot many people, myself included would have to throw that book out! It is the standard textbook for 12 grade in India. But don't worry, I checked it and it clearly says that the covalency is limited to 4, not oxidation state! Abir probably conflated them both. $\endgroup$ – FreezingFire May 16 '17 at 17:16

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