
A certain sample of cuprous sulphide is found to have the composition \[C{{u}_{1.92}}{{S}_{1.00}}\] because of incorporation of \[C{{u}^{2+}}\] and \[C{{u}^{+}}\] ions in the crystal then ratio of \[C{{u}^{2+}}\] and \[C{{u}^{+}}\] ions is:
(a) 0.08: 1.00
(b) 1:23
(c) 1: 24
(d) 1:1
Answer
583.2k+ views
Hint: \[C{{u}_{2}}S\] can be prepared by heating copper strongly in presence of sulfur vapour or \[{{H}_{2}}S\]. The reaction of copper powder in molten sulfur rapidly produces \[C{{u}_{2}}S\]. 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 \[C{{u}_{2}}S\].
It exists in the form as given below,
\[C{{u}_{2}}S\to 2C{{u}^{+}}\,{{S}^{2-}}\]
Here we can see that, one mole of \[C{{u}^{2+}}\] replaces two moles of \[C{{u}^{+}}\].
Now due to incorporation of \[C{{u}^{2+}}\] and \[C{{u}^{+}}\] ions in the crystal the formula becomes \[C{{u}_{1.92}}{{S}_{1.00}}\].
Therefore, the total loss of Copper in the sample = 2 - 1.92 = 0.08
Percentage of \[C{{u}^{2+}}\] in the sample = \[\dfrac{0.08}{1.92}\times 100=4.1\]
Percentage of \[C{{u}^{+}}\] in the sample = 100 - 4.1 = 95.59
Ratio of \[C{{u}^{2+}}\]and \[C{{u}^{+}}\]= \[\dfrac{4.1}{95.59}=\dfrac{1}{23.31}\approx \dfrac{1}{23}\]
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 \[C{{u}_{1.96}}S\] which is non-stoichiometric, having a monoclinic structure with 248 copper and 128 sulfur atoms in the unit cell. It also ranges anywhere between \[C{{u}_{1.934}}S-C{{u}_{1.965}}S\]. Both \[C{{u}_{2}}S\] and \[C{{u}_{1.96}}S\] are similar in appearance and hard to distinguish one from another.
Complete step-by-step answer:
We know the molecular formula for Cuprous sulphide can be written as \[C{{u}_{2}}S\].
It exists in the form as given below,
\[C{{u}_{2}}S\to 2C{{u}^{+}}\,{{S}^{2-}}\]
Here we can see that, one mole of \[C{{u}^{2+}}\] replaces two moles of \[C{{u}^{+}}\].
Now due to incorporation of \[C{{u}^{2+}}\] and \[C{{u}^{+}}\] ions in the crystal the formula becomes \[C{{u}_{1.92}}{{S}_{1.00}}\].
Therefore, the total loss of Copper in the sample = 2 - 1.92 = 0.08
Percentage of \[C{{u}^{2+}}\] in the sample = \[\dfrac{0.08}{1.92}\times 100=4.1\]
Percentage of \[C{{u}^{+}}\] in the sample = 100 - 4.1 = 95.59
Ratio of \[C{{u}^{2+}}\]and \[C{{u}^{+}}\]= \[\dfrac{4.1}{95.59}=\dfrac{1}{23.31}\approx \dfrac{1}{23}\]
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 \[C{{u}_{1.96}}S\] which is non-stoichiometric, having a monoclinic structure with 248 copper and 128 sulfur atoms in the unit cell. It also ranges anywhere between \[C{{u}_{1.934}}S-C{{u}_{1.965}}S\]. Both \[C{{u}_{2}}S\] and \[C{{u}_{1.96}}S\] are similar in appearance and hard to distinguish one from another.
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