plz tell me how to find the oxidation number...?

first of all you need to know what is oxidation:-

oxidation is a process which involves either of the following:

1. addition of oxygen

2. removal of hydrogen

3. addition of an electronegative element or group

4. removal of an electropositive element or group

5. increase in the valency of an electropositive element

6. loss of one or more electrons by an atom or an ion or a molecule

for the oxidation number following rules are followed in any type of element:

1. the oxidation number of an atom in free elements is zero no matter how complicated the molecule is.

2. the oxidation number of FLUORINE is always -1.

3. the oxidation number of OXYGEN is -2 in all compounds EXCEPT in PEROXIDE, SUPER OXIDES and OXYGEN FLUORIDES.

4. the oxidation number of HYDROGEN is +1 in all of its compounds EXCEPT in METALLIC HYDRIDES. in the METALLIC HYDRIDES, oxidation number og HYDROGEN is -1.

5. the oxidation number of an ION is equal to the ELECTRICAL CHARGE present.

6. the oxidation number of ALKALI METALS is +1 and that of ALKALINE EARTHLY METALS is +2.

7. for complex ions, the algebraic sum of oxidation number of all atoms is equal to the net charge on the ion.

8. in case of neutral molecules, the algebraic sum of the oxidation number of all the atoms present in the molecules is zero.

group no. - charge

IA elements +1

IIA elements +2

IIIA elements +3,+1

IVA elements -4 to +4

VA elements -3 to +5

VIA elements -2 to +6

VIIA elements -1 to +7

i hope this might help you......:)

let we have find the O. N> of Mn in KMnO4

(1) let O.N. of Mn is x

now there are some points { rule} which should be kept

in mind while calculating O. N.--------

(a) O.N.of Ois always taken -2 except in peroxide where it is -1

& in OF2 it is +2

(b) O,N. of H is always taken as -1 except in hydride it is +1

(c)K, Na . Li [ ist gp metal have O.N. +1

(d) Ba , Ca. Mg have O. N. +2

(e) F, Cl , Br, I i.e. halogen have O. N. -- 1

IMPORTANT--[--1] IN . NEUTRAL COMPOUNDS TOTAL OF

O.N. OF ALL THE ELEMENNTS IS EQUAL TO ZERO ( vvvvvvvv important )

[2] the total of O.N. of all the atoms in a radical is equal the charge

now in KMnO4 we have ----

1 + x +4(-2) =0 [ as above rule (1) ]

so x= +7 [ans [ sign + Or -- is necessary ]

ANOTHER EXAMPLE ------ to find O. N.of Cr in Cr2O7 --2[dichromate ion]

let O. N.of Cr is x

so 2x + 7(-2) = --2 [ charge on Cr2 O7 is --2 ]

x = + 6 ans

ANOTHER EXAMPLE------ find O.N.of Fe in K4 Fe (CN)6

let it be x

so 4(+1) + x + 6(-1) =0 [ CN-1 cynide has -1 ]

x = +2 [ so potassium ferro cynaide ]

note ---in K3 Fe (CN)6 O. N. of Fe = +3 [ pot. ferri cynide ]

NOTE ---O.N. OF ANY RADICLE IS EQUAL TO THE CHARGE

IT BEARS ON e.g. ( SO4--2) , (CO3--2), has --2

N-3 . , PO4-3 has --3

NH4+1 has +1

LAST find O.N. of Mn in MnO4-1 & Mno4--2

ans in MnO4-1 , x+ 4(-2) =-1 or x= +7

ans, in MnO4--2 x+ 4(-2) = -2 so x= +6

while in MnO2 , x+ 2(-2) =0 sox= +4

i think clear .

Every element is trying to get tn neutral state.if hydrogen gets one more electron its stable.by inert gases as ref u can get easy way to remember oxy number and many secrets by knowing how periodic table is arranged.

finding oxidation no. does not involve only rule. Different problems are dealt with diff. Rules