Which metal sulfide is not black?
A. $NiS$
B. $CoS$
C. $CuS$
D. $ZnS$
Answer
575.4k+ views
Hint: The metal sulfide which is not black among all the given options is transparent and occurs as a mineral sphalerite. Due to its transparent appearance, it is used as a luminescent material, optical material and as a pigment. It is also used in a semiconductor as it is an intrinsic wide-band gap semiconductor.
Complete answer:
The metal sulfides are the most important inorganic compound. They are charge forming materials. In the case of nickel sulfide, it is a black solid which is produced by reacting nickel(ll) salts with hydrogen sulfide. So, this is an incorrect option.
In the case of cobalt sulfide, it is black and used semiconducting. It is insoluble in water and nonstoichiometric. So, this is an incorrect option.
In the case of copper sulfide, it is black colloidal precipitate formed when hydrogen sulfide is bubbled through the solution of copper(ll) salts. It is a moderate conductor of electricity. So, this is an incorrect option.
Now, in the case of zinc sulfide, it is transparent white powder and prepared by igniting a mixture of zinc and sulphur. Zinc sulfide is insoluble in water and also prepared by treating zinc(ll) salt with hydrogen sulfide.
So, zinc sulfide is not black colored.
Hence, option D is correct.
Note:
Zinc sulfide exists in two main crystalline forms. And these two forms are the example of polymorphism. In each form, the geometry is tetrahedral. And the most stable cubic form is called zinc blende or sphalerite. And the hexagonal form is known as the mineral wurtzite. Wurtzite can also be produced synthetically. The transition from the sphalerite form to the wurtzite form occurs at high temperature around ${1020^0}C$ . A tetragonal form is called a polhemus site, with the formula \[(Zn,Hg)S\] .
.
Complete answer:
The metal sulfides are the most important inorganic compound. They are charge forming materials. In the case of nickel sulfide, it is a black solid which is produced by reacting nickel(ll) salts with hydrogen sulfide. So, this is an incorrect option.
In the case of cobalt sulfide, it is black and used semiconducting. It is insoluble in water and nonstoichiometric. So, this is an incorrect option.
In the case of copper sulfide, it is black colloidal precipitate formed when hydrogen sulfide is bubbled through the solution of copper(ll) salts. It is a moderate conductor of electricity. So, this is an incorrect option.
Now, in the case of zinc sulfide, it is transparent white powder and prepared by igniting a mixture of zinc and sulphur. Zinc sulfide is insoluble in water and also prepared by treating zinc(ll) salt with hydrogen sulfide.
So, zinc sulfide is not black colored.
Hence, option D is correct.
Note:
Zinc sulfide exists in two main crystalline forms. And these two forms are the example of polymorphism. In each form, the geometry is tetrahedral. And the most stable cubic form is called zinc blende or sphalerite. And the hexagonal form is known as the mineral wurtzite. Wurtzite can also be produced synthetically. The transition from the sphalerite form to the wurtzite form occurs at high temperature around ${1020^0}C$ . A tetragonal form is called a polhemus site, with the formula \[(Zn,Hg)S\] .
.
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