Componds `(A)` and `B` are treated with dilute `HCl` separately. The gases liberated are `Y` and `Z` respectively. `Y` turns acidified `K_2 Cr_2 O_7` paper green while `Z` turns lead acetate paper black. The compounds `A` and `B` are respectively :

Options

A. `Na_2 S` and `Na_2 SO_3`
B. `Na_2 SO_3` and `Na_2 S`
C. `NaCl` and `Na_2 CO_3`
D. `Na_2 SO_3` and `Na_2 SO_4`

Answer

B

Explanation

`(A) + underset(dil.) (HCl) rarr underset(Gas) ((Y))`
`(B) + underset(dil. )(HCl) rarr underset(Gas) ((Z))`
`(Y)` turns acidified `K_(2) Cr_(2) O_(7)` paper green
`(Z)` turns lead acetate paper black.
`Na_(2) SO_(3)` on reaction with dil. `HCl`, liberated `SO_(2)` gas.
`Na_(2) SO_(3) + HCl rarr Na_(2) SO_(4) + SO_(2) + H_(2) O`
or `SO_(3) ^(2-) + overset("dil.") underset("dil.") (2 H^(oplus)) rarr underset((Y)) (SO_(2)) + H_(2) overset((Y))O`
`SO_(2)` has turns acidified `K_(2) Cr_(2) O_(7)` paper green
`SO_(2)` acts as reducing agent and reduces
`Cr_(2) O_(7)^(2-) to Cr^(3+)` (green)
`K_(2) Cr_(2) O_(7) + H_(2) SO_(4) + 3 SO_(2) rarr underset(("Green")) (Cr_(2)(SO_(4))_(3) + K_(2) SO_(4) + H_(2) O`
Ionic equation :
`Cr_(2) O_(7)^(2-) + 14^(oplus) + 6 e^(Ө) rarr 2 Cr^(3+) + 7 H_(2) O`
`[2 H_(2) O + SO_(2) rarr SO_(4)^(2-) + 4 H^(oplus)+ 2e^(Ө)] xx 3`.
`ulbar( Cr_(2) O_(7)^(2-) + 3SO_(2) + 2 H^(oplus) rarr 2 Cr^(3+)+ 3SO_(4)^(2-) + H_(2) O)`
`Na_(2) S+ 2 HCl rarr NaCl + H_(2) S uarr`
or `S^(2-) + overset("dil.") (2 H^(oplus) rarr H_(2) S`
`H_(2) S` gas turns lead acetate paper black due to formation of lead sulphide, `PbS`.
`S^(2-) + Pb^(2+) rarr PbS darr`
`H_(2) S Pb (CH_(3) COO)_(2) rarr PbS + 2CH_(3) COOH`
Hence `A = Na_(2) SO_(3)` and `B = Na_(2) S`
and option `(b)` is correct.