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A certain sample of cuprous sulphide is found to have the composition Cu1.92S1.00 because of incorporation of Cu2+ and Cu+ ions in the crystal then ratio of Cu2+ and Cu+ ions is:
(a) 0.08: 1.00
(b) 1:23
(c) 1: 24
(d) 1:1

Answer
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Hint: Cu2S can be prepared by heating copper strongly in presence of sulfur vapour or H2S. The reaction of copper powder in molten sulfur rapidly produces Cu2S. Same can be done with pellets of copper but they will require much higher temperature.

Complete step-by-step answer:
We know the molecular formula for Cuprous sulphide can be written as Cu2S.
It exists in the form as given below,

Cu2S2Cu+S2

Here we can see that, one mole of Cu2+ replaces two moles of Cu+.
Now due to incorporation of Cu2+ and Cu+ ions in the crystal the formula becomes Cu1.92S1.00.
Therefore, the total loss of Copper in the sample = 2 - 1.92 = 0.08
Percentage of Cu2+ in the sample = 0.081.92×100=4.1
Percentage of Cu+ in the sample = 100 - 4.1 = 95.59

Ratio of Cu2+and Cu+= 4.195.59=123.31123
So, the correct option is (b).

Additional Information:
Naturally, Cuprous sulfide is found in the form of black powder or lumps and is as the mineral chalcocite. Large quantities of the compound can be obtained by heating cupric sulfide (CuS) in a stream of hydrogen. It is insoluble in water but readily soluble in ammonium hydroxide and nitric acid. It has various applications including use in solar cells, luminous paints, electrodes, and certain varieties of solid lubricants.

Note: There is also a crystallographically-distinct phase with stoichiometry Cu1.96S which is non-stoichiometric, having a monoclinic structure with 248 copper and 128 sulfur atoms in the unit cell. It also ranges anywhere between Cu1.934SCu1.965S. Both Cu2S and Cu1.96S are similar in appearance and hard to distinguish one from another.