X is an ore of a metal M. X on calcination gives black precipitate (W) of metal oxide which belongs to group II of basic radicals in qualitative analysis.

X is an ore of a metal M. X is an ore of a metal M. X on calcination gives black precipitate (W) of metal oxide which belongs to group II of basic radical in qualitative analysis. X on roasting gives the metal (M) and gas as major byproducts. The gas when passed through an acidified K₂Cr₂O₇ solution turns green.

1. Identify the metal X. (1)
2. Write the reaction involved during calcination of X. (1)
3. Write the action of the gas on acidified K₂Cr₂O₇. (1)
4. Convert metal X into it’s vitriol. (2)

Solution:

(a) Copper (Note: Copper pyrite on calcination gives black sulphide. It does not give black oxide on calcination.)

(b) 2CuFeS₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

(c)

K₂Cr₂O₇ + H₂SO₄ + 3SO₂ → K₂SO₄ + Cr₂(SO₄)₃ +H₂O

(d)

$Cu + O\mathop  \to \limits^{upto\;{{110}^ \circ }} CuO$

CuO + H₂SO₄ à  Cu SO₄ + H₂O

$CuS{O_{4}}(aq.)\mathop  \to \limits^{crystallization} CuS{O_4}.5{H_2}O$