In how many of the following compounds of sulphur, the oxidation state of sulphur atom is + 6 ?
`H_(2)S_(2)O_(8), H_(2)SO_(5),H_(2)SO_(3), H_(2)SO_(4),H_(2)S_(2)O_(7), SO_(2) CL_(2), SOCL_(2)`

Options

A. 3
B. 5
C. 4
D. 6

Answer

C

Explanation

The oxidation state of sulphur in the given compounds can be found through their respective structures which are as follows :
(a) `H_(2) S_(2) O_(8)`

We can calculate oxidation state sulphur as :
`x + 2 (-2) +1 (-1) +1(-1) = 0 `
x = + 6
Thus, the sulphur is in + 6 oxidation state .
(b). `H_(2) SO_(3)`

`:. x + 2 (-2) +1 (-1) +1(-1) = 0 `
x = + 6
Thus, the sulphur is in + 6 oxidation state.
(c) H_(2) SO_(3)

x + 2 (-1) + 1 (-2) = 0
x = + 4
Thus, sulphur is present in + 4 oxidation state.
(d) `H_(2) SO_(4)`

x + 2 (-1) + 1 (-2) = 0
x = + 4
Thus, sulphur is present in + 4 oxidation state.
(e) `H_(2)S_(2)O_(7)`

x + 2 (-2) + 1 (-1) + 1 (-1) = 0
x = + 6
Thus, sulphur is present in + 6 oxidation state.
(f) `SO_(2) Cl_(2)`

x + 2 (-1) + 2(-2) = 0
x = + 6
Thus , sulphur has + 6 oxidation state.
(g) `SOCl_(2)`

x + 1 (-2) + 2 (-1) = 0
x = + 4
Thus, sulphur, has 4 oxidation state,
Compounds having+ 6 oxidation state of sulphur atom are `H_(2)S_(2)O_(2), H_(2)SO_(5), H_(2)S_(2)O_(7), SO_(2)Cl_(2)`
Hence, the total number of molecules having + 6 oxidation state of sulphur atom is 4 .