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Question

A and B are two ores of copper.
A undergoes calcination to form a solid, CO2(g) and O2(g)
B undergoes roasting to form a solid and a gas C which turns acidified K2Cr2O7 to green solution.
B also reacts with dil HCl to form a solid and gas D which turns lead acetate solution black. D also reacts with C to form colloidal sulphur in the presence of moisture.

The nature of the gas (D) and hybridization of the central atom is respectively:

A
Weakly acidic, sp3
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B
Strongly acidic, sp3
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C
Weakly acidic, sp2
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D
Strongly acidic, sp2
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Solution

The correct option is A Weakly acidic, sp3
A = Cu(CO3) and B = CuS

Reactions involved are:

Calcination of A:
Cu(CO3)(A)CuO(solid)+CO2+O2

Roasting of B:
CuS(B)CuO(solid)+SO2(C)

SO2(C)+K2Cr2O7K2(SO)4+Cr3+(green)

CuS(B)+dil.HClCuCl2(solid)+H2S(D)

'D' is H2S.

H2S is weakly acidic and according to VSEPR theory is tetrahedral in geometry and hence hybridisation is sp3.

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