6this concept is clear..ut sumplaces i cum across fractional o.n.wich is nt possible..how to get the correct value of o.n??
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.
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19Oxidation 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)
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1agree with debotosh. fractional O.N. is quite possible.e.g. superoxides like KO2 where o.n. of O is -1/2.
6yeah i kno its possible theoritically bt actually it does'nt happen ..........an element does'nt lose electron in fractions........
13Arrey, 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..!
1357You 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