A white solid forms Rinmann’s green in the charcoal cavity test in an oxidising flame. On treatment with dilute \[{H_2}S{O_4}\], this solid produces a gas that turns an acidified dichromate paper green and lead acetate paper black. The white solid is:
a) \[PbS\]
b) \[ZnS{O_3}\]
c) \[ZnS\]
d) \[N{a_2}S\]

Hint: Find the compound which gives the color Rinmann’s green in oxidizing flame in the charcoal cavity. Then try to find the compounds which turn acidified dichromate paper green and lead acetate paper black.

Complete answer:
Given, a white solid forms Rinmann’s green in oxidizing flame, in the charcoal cavity. On treatment with dilute \[{H_2}S{O_4}\], the solid produces a gas that turns an acidified dichromate paper green and lead acetate paper black.
From the given question , it is evident the compound is ZnS, since zinc salts give green color in oxidizing flame in charcoal cavity test, and \[{H_2}S\] and \[PbS\] turn acidified dichromate paper green and lead acetate paper black respectively .
$ZnS + {H_2}S{O_4} \to ZnS{O_4} + {H_2}S$
$8KMn{O_4} + 7{H_2}S{O_4} + 5{H_2}S \to 4{K_2}S{O_4} + 8MnS{O_4} + 12{H_2}O $
Thus, the given compound in ZnS.

Hence, the required option for the given question is c).

Additional Information: In the last part of the given question, it is given that the reducing gas turns lead acetate paper green. Hence it must be forming the product PbS which is black in color. had it been the salt given in option d) which is \[ZnS{O_3}\], it would have turned white , in the last step of the question .
Thus, the given white compound is ZnS.

Note: In the charcoal cavity test, the elements which give the color green in oxidizing flame is zinc. The zinc salts give the color in oxidizing flame. Now, the gas which would turn acidified dichromate paper green must be a reducing gas \[({H_2}S)\]. Hence the white solid is either ZnS or \[ZnS{O_3}\] out of the given compounds. Students need to keep in mind the above chemical properties of the given elements to be able to successfully solve the sum.