plz anyone tell me to find out the oxidation number of PbSO4 , CaOCl2 , (Nh4)2SO4?

Question

OXIDATION NUMBER

Answer 1

In PbSO4 the Pb is +2 and the SO4 group is -2.
The O in SO4 is -2 so the sulphur must be +6 (the four oxygens add up to -8 and for the SO4 group to be -2 the sulphur must be +6)
The formula for CaOCl2 IS correct. The Ca is +2 and the OCl2 group -2 as O is usually -2 it would suggest that each chlorine is 0. This anomaly is because the OCl2 is known as a disproportionate ion (having too much chlorine).
In (NH4)2SO4 each NH4 group is +1 and the SO4 is -2.
So as each H = +1 the N is -3. The SO4 has the same oxidation numbers as in PbSO4 above.
In working out oxidation numbers we assume that certain elements have fixed numbers O is -2, Cl is -1, H is +1 and alkali metals like sodium and potassium are +1. The complete compound must add up to zero.

Answer 2

A compound doesn't have an oxidation number, but an element in a compound does.
For a simple ion, like Pb in PbSO4, it is the same as the ionic charge, ie +2.
Now, since O is always -2, the S must be +6 to make the whole compound zero.
I don't think that you have written the formula of the calcium compound correctly.
If you use O = -2, S = +6 and H = +1, then you should be able to work out the N in ammonium sulphate.

Answer 3

Nh4 2so4 Oxidation Number

Answer 4

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 .

Answer 5

22

Answer 6

I often end up writing the same question on other sites

Answer 7

Pb II, S VI O -II, Ca II, Cl -I (in CaCl) and Cl I (in OCl), N -III, H I

I hope thats what u wanted.

BTW Gervald O can also be -I and -I//II (-1/2)