IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

thegreateye

What will the be the oxidation states of the marked nitrogen atoms in the question?

pls help

thnx

svj29

looking at the figure, we see that 2 and 3 form a double bond, which will be non polar, and so will be that of 1-2, 1-3. therefore the Oxidation number of 2,3 will be 0. as per 1 is concerned, it will be -1. (bcoz of H)
1. -1

2. 0

3. 0

am i correct dude...???

mukundmadhav

Yeah I think -1, 0, 0 is right..

thegreateye

i am not sure of the answer, as this was asked in a test

one pt. there is electronegativity difference b/w 1 & 3 and 2&3 (because of s character). so will that affect the answer in any way?

mukundmadhav

That actually is a bone of contention. I haven't seen this question, but there's another question, in which you have to find the oxidation state of Sulphur in H₂S₂O_3..

There one sulphur atom is doubly bonded to the another. Basically structure is same as H₂SO₄with one S replacing one O. Now some books list the oxidation states as 0 and 4, while others do it as +6 and -2. The second by drawing analogy with sulphuric acid..

Let me check what JD Lee says, and get back to you..

mukundmadhav

JD Lee says +6, -2..

VISHUG

1] +1
2] -1
3] -1
reason-
[2] and [3] N r sp2 hybridized
more electronegative than [1].

svj29

hybridization doesn't count that much in find the oxidation state.
check any book for theory.

svj29

EDITED

mukundmadhav

J.D. Lee is never wrong :P.. They use the text to teach in IIT, man.. Give some credit to the guy..
Regarding your doubt, You're forgetting that the molecule is not strictly covalent, the OH will exist as O-H+.. So the O- will have an electron donating effect on S, increasing negative charge on it, and thus pushing the S=S shared pairs of electrons away from the central atom. Hence +6, -2..

thegreateye

hybridisation does account for the electronegativity difference. that is what is confusing me. I gave the same answer as Vishug.

svj29

@mukundmadhav

everyone can make mistakes,
just like i had commited one in the above answer :P
dude your explanation is totally wrong.
and you too made a big mistake, O can never donate its electrons to S ( i mean shifting too). and the oxygen hydrogen bond is a covalent bond, check any book...hope ya knw H2O is covalent.

so ur explanation is wrong.
and my previous one too was wrong.
Here is the correct one:

dude see, talking about the central atom (S):

it will form one single bond with OH (2 in total)
one coordinate bond with O

and one coordinate bond with S

now the two OH will induce an charge of +2 on total
the O atom +2
and as per the above sulphur atom is concerned, remember the bond is COORDINATE, NOT COVALENT...both the electrons involved are from the central atom. Hence the other S will get a net charge of -2 on it and the central atom will get another +2, therefore amounting the total charge as 2(+1) + 2 + 2 = +6

this is the correct explanation, and JD Lee is correct, sorry for the above answer...

if you smhw still find any mistake here, do tell..
although i think my answer is perfect. :)

svj29

@thegreateye.

sorry dude, for deviating from the topic, but here is your answer:

HYBRIDIZATION DOESN'T MAKE DIFFERENCE IN PREDICTING THE OXIDATION NUMBER.
ITS ASSURED, AND my 1st answer is correct.

CHANGE OF HYBRIDIZATION, really doesn't make a HUGE DIFFERENCE, hence it is not included. :)

hope its clear now :)

pulsar

Oxidation State Question

svj29

@mukundmadhav

is my explanation correct...??? or still smdng wrong..??

@ thegreateye

is the answer satisfactory...???

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