Paradox Creating tension

There is a section in Redox reactions Paradox of fractional oxidation no.I have understood the concept but i could not understand one thing that how do we calculate the real oxidation numbers of elements in a compoud.could anybody please make me understand how to do it and please give a good way to solve for the oxidn. no.Thank u.

8 Answers

19
Debotosh.. ·

Oxidation number and Oxidation state

The o.n. of an atom in a molecule or in a polyatom ion is a hypothetical charge the atom would have if the electrons in each bond were located on the more electronegative atom .
→ oxidation state is the oxidation number per atom
→ an unreacted element has o.n. zero
→ a monoatomic ion has o.n. equal to its charge

Rules for assigning o.n
o.n. of one particular element in a covalent compound or ion is determined by taking o.n. of other elements to be fixed . these are summarized below :
Group I (alkali metals) : +1
Group II (alkaline earth metals) : +2
Group III (boron family) : +3
Hydrogen : +1 , -1(in metal hydrides)
Oxygen : -2 , -1(in peroxides) , +2(in OF2)
Nitrogen : -3(in ammonia and nitrides), varies when in combination with oxygen
Halogens : -1 (when in direct combination with metals) and varies when in combination with oxygen and other halogens ( as in ICl)

6
Kalyan IIT-K Beware I'm coming ·

this concept is clear..ut sumplaces i cum across fractional o.n.wich is nt possible..how to get the correct value of o.n??

3
msp ·

u need to know atleast their complete structures.

19
Debotosh.. ·

who said frac o.n is impossible ? look at ferric oxide !

1
Arka Halder ·

agree with debotosh. fractional O.N. is quite possible.e.g. superoxides like KO2 where o.n. of O is -1/2.

6
Kalyan IIT-K Beware I'm coming ·

yeah i kno its possible theoritically bt actually it does'nt happen ..........an element does'nt lose electron in fractions........

13
Avik ·

Arrey, matlab it is actually impossible, but it had to be made possible in a way (koi aur chaara nahi bachaa hogaa) :P

Kyunki the structure of some compounds, like S2O32- mein ek S is in 0 n the other is in +4, n in reactions v need to see just "Sulphur", wahaan par ye thodi kahenge ki this particular alpha-sulphur is in 0 n now it is brking off n being further attached with pqr....n this beta-sulphur atom which was in +4 oxidation is losing so many electrons n going to falaana oxid.n state........i mean just imagine!!

Isliye Average O.S. had to be defined, to explain reactions clearly on the basis of elements n not individual atoms..!

1357
Manish Shankar ·

You need to know the structures of the compounds for calculating their individual oxidation numbers.

e.g. in Fe3O4 it exists as FeO.Fe2O3

The first one has ON of 2 and second has that of 3. and they give average of 8/3

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