A substance on treatment with dilute `H_(2)SO_(4)` liberates a colourless gas which produces `(I)` turbidity with baryta water and `(ii)` turns acidified dichromate solution green. The reaction indicates the presence of `:`

Options

A. `CO_3^(2-)`
B. `S^(2-)`
C. `SO_3^(2-)`
D. `NO_2`

Answer

C

Explanation

Baryta water is a solution of `Ba(OH)_2` in water. `SO_2` gives white ppt. (or turbidity ) with `Ba(OH)_2` solution. It is due to the formation of insoluble `BaSO_3`.
`SO_2+Ba(OH)_2 to BaSO_3 darr + H_2O`
`SO_2` also turns acidified `K_2Cr_2O_7` solution green. It is due to the formation of green coloured `Cr_2(SO_4)_3`.
`({:(K_2Cr_2O_7+4H_2SO_4to,K_2SO_4+Cr_2(SO_4)_3+4H_2O+3O),(SO_2+O+H_2Oto,H_2SO_4"]"xx3):})/ul(K_2Cr_2O_7+3SO_2+H_2SO_4toK_2SO_4+Cr_2(SO_4)_3+H_2O)`
`SO_2` is obtained by the action of dil. `H_2SO_4` on a sulphite, `SO_3^(2-)` (or thiosulphate).