Question

An element occupies group No.16 and period number 2. This element
on passing through silence electric discharge forms (A). (A) also reacts
with lead sulphide and forms (B). (A) also reacts with BaO 2 and forms
(C). It reacts with H 2 O 2 and forms (D). Identify the dement (A), (B),
(C) and (D).

Answer 1

the element of group no. 16 is O2
3O2  -----------------------------> 2O3
          electric discharge      (A)
4O3  +  PbS  --------------->   PbSO4  + 4O2
 (A)                                     (B)
H2O2  + O3  ------------------------> H2O  + 2O2
             (A)                             (C)
BaO2  + O3  -------------------------->  BaO  +   2O2
            (A)                                 (D)
A = O3
B = PbSO4
C = H2O 
D = BaO

Answer 2

the element is oxygen O2
A=O3(O2 will get broken into two nascent oxygen atoms[O],which will then combine with another oxygen molecule)
B=PbO2(oxygen on reaction with PbSO3 will displace sulphide radical(SO3) and combine with Pb) 
C=BaO(when oxygen reacts with BaO2 it forms BaO)
D=H2O(2O2+H2O2------->2H2O + O2)