what is the oxidation state of sulphur in H₂S₂O_8,

Na₂S₄O₈ and Na₂S₂O_3.

@ bhavika 
in H2S2O8, since 2 oxygen is forming the peroxide linkage here , so, the o.n of sulpher will be +6 . and in the 2^nd case NA2S4O8 , in thaat case take sulpher o.n to be x 
after that , 2(1) +4(x) + 8 (-2) =0 
so, by calculating the value of x u will get sulpher will present in a oxidation state of 3.5 
coming to the 3^rd case in na2s2o3, 
take that in pieces and calculate the oxidation number , 
u know , na hv a o.n of +1 , as there are two , so, it will be +2 , 
now , oxygen is -2 , there are 3 atoms , so, that will be -6 , 
so, u need to balance a +4 in it , thatswhat u left with , and there are 2 sulpher left , 
so, each one hv a +2 charge . 
SO, THE O.N OF SULPHER HERE IS +2 . 
HOPE IT CLEARS YOUR DOUBT 
ALL THE BEST ...
 

but hou do i know that give compound has peroxy linkage.

i think that is concluded by drawing structure only . i am not able to draw thw structure.

u need to remember that which structure have peroxide linkage and which structure have not , 
there are only some exception with peroxide linkage , 
go through the ncert redox reaction chapter solved example , then start solving exercise . 
ALL THE BEST /..

First calculate the oxidation number 

Case

1- if oxidation number is equal to group number then must have C.A-B.A-C A(That means central atom bond with bonded atom then with central atom)

2-if oxidation number is greater than group number than peroxy linkage is must

3-if oxidation number is less than group number then C.A-C.A

 

3-if oxidation number is less than group number then C.A-C.A