Class 12 Chemistry Chapter 1 The Solid State

Class 12 Chemistry Chapter 1 The Solid State

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State:

Section Name	Topic Name
1	The Solid State
1.1	General Characteristics of Solid State
1.2	Amorphous and Crystalline Solids
1.3	Classification of Crystalline Solids
1.4	Crystal Lattices and Unit Cells
1.5	Number of Atoms in a Unit Cell
1.6	Close Packed Structures
1.7	Packing Efficiency
1.8	Calculations Involving Unit Cell Dimensions
1.9	Imperfections in Solids
1.10	Electrical Properties
1.11	Magnetic Properties

NCERT Course book QUESTIONS Tackled

1.1. For what reason are solids inflexible?
Ans: The constituent particles in solids have fixed positions and can sway about their mean positions. Subsequently, they are unbending.

1.2. For what reason do solids have unmistakable volume?
Ans: Solids keep their volume on account of unbending nature in their construction. The interparticle powers are areas of strength for extremely. Also, the interparticle spaces are not very many and little too. Accordingly, their volumes can’t change by applying pressure.

1.3. Group the accompanying as nebulous or glasslike solids: Polyurethane, naphthalene, benzoic corrosive, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiberglass, copper
Ans: Translucent solids: Benzoic corrosive, potassium nitrate, copper Shapeless solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fiberglass

1.4. Why is glass thought of as super cooled fluid ? (C.B.S.E. Delhi 2013)
Ans: Glass is viewed as super cooled fluid since it shows a portion of the qualities of fluids, however it is a nebulous strong. For instance, it is somewhat thicker at the base. This can be conceivable provided that it has flown like fluid, however leisurely.

1.5. Refractive record of a strong is seen to have similar worth along all bearings. Remark on the idea of this strong. Could it show cleavage property?
Ans: As the strong has same worth of refractive list along all headings, it is isotropic in nature and consequently undefined. Being indistinct strong, it won’t show a spotless cleavage and when cut, it will break into pieces with unpredictable surfaces.

1.6. Group the accompanying solids in various classes in light of the idea of the intermolecular powers: sodium sulfate, copper, benzene, urea, smelling salts, water, zinc sulfide, jewel, rubedium, argon, silicon carbide.
Ans: Ionic, metallic, atomic, sub-atomic, sub-atomic (hydrogen-reinforced), atomic (hydrogen-fortified), ionic, covalent, metallic, sub-atomic, covalent (network).

1.7. Strong A will be an extremely hard electrical protector in. strong as well as in liquid state and melts at very high temperature. What sort of strong is it?
Ans: It is a covalent or network strong.

1.8. For what reason are ionic solids leading in the liquid state and not in the strong state?
Ans: In the ionic solids, the electrical conductivity is because of the development of the particles. Since the ionic versatility is immaterial in the strong express, these are non-directing in this state. After liquefying, the particles present get some portability. In this manner, the ionic solids become leading

1.9. What kind of solids are electrical channels, moldable and pliable?
Ans: Metallic solids

1.10. Give the meaning of a grid point.
Ans: The cross section point means the place of a specific constituent in the precious stone grid. It could be iota, particle or a particle. The game plan of the cross section focuses in space is liable for the state of a specific translucent strong.

1.11. Name the boundaries that describe a unit cell.
Ans: A unit cell is described by the accompanying boundaries:
(i)the aspects of unit cell along three edges: a, b and c.
(ii)the points between the edges: α (among b and c); β (among a and c) and γ (among a and b)

1.12. Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans:
(i) In a hexagonal unit cell :
a = b # c; α = β = 90° and γ = 120°
In a monoclinic unit cell :
a # b # c and α = γ = 90° and β # 90°

(ii) In a face-focused unit cell, constituent particles are situated at every one of the corners as well as at the focuses of the relative multitude of countenances.
In end-focused unit cell, constituent particles are situated at every one of the corners as well as at the focuses of two inverse countenances. (C.B.S.E Unfamiliar 2015)

1.13. Make sense of the number of parts of an iota that situated at
(i)corner and (ii)body focus of a cubic unit cell is important for its adjoining unit cell.
Ans: (I) A molecule at the comer is shared by eight adjoining unit cells. Consequently, piece of the iota at the comer that has a place with one unit cell=1/8.
(ii)An molecule at the body place isn’t shared by some other unit cell. Subsequently, it has a place completely with unit cell.

1.14. What is the two-layered coordination number of a particle in a square close-stuffed layer?
Ans: In the two-layered square close-stuffed layer, a specific particle is in touch with four atoms. Thus, the coordination number of the atom is four.

1.15. A compound forms hexagonal close-packed. structure. What is the total number of voids in 0. 5 mol of it? How many of these are tetrahedral voids?
Ans:
No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 10²³ =3.011 x 10²³
No. of octahedral voids = No. of atoms in packing =3.011 x 10²³
No. of tetrahedral voids = 2 x No. of atoms in packing
= 2 x 3.011 x 10²³ = 6.022 x 10²³
Total no. of voids = 3.011 x 10²³ + 6.022 x 10²³
= 9.033 x 10²³

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound? (C.B.S.E. Foreign 2015)
Ans: Let us suppose that,
the no. of atoms of N present in ccp = x
Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,
the no. of tetrahedral voids occupied = 2x/3
The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2
∴ The formula of the compound = N₃M₂ or M₂N₃

1.17. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centered cubic and (iii) hexagonal close-packed lattice?
Ans: Packing efficiency of:
Simple cubic = 52.4% bcc = 68% hcp = 74%
hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2:7 x 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m^-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)
Ans: 

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).

1.19. What sort of imperfection can emerge when a strong is warmed? Which actual property is impacted by it and how?
Ans: When a strong is warmed, opportunity imperfection is delivered in the gem. On warming, a few molecules or particles leave the grid site totally, i.e., cross section locales become empty. Because of this deformity, thickness of the substances diminishes.

1.20. What kinds of stoichiometric imperfections are shown by (C.B.S.E. Delhi 2013)
(I) ZnS
(ii) AgBr?
Ans:
(I) ZnS gems might show Frenkel surrenders since the cationic size is more modest when contrasted with anionic size.
(ii) AgBr gems might show both Frenkel and Schottky absconds.

1.21. Make sense of how opportunities are presented in an ionic strong when a cation of higher valence is added as a contamination in it.
Ans: Let us take a model NaCl doped with SrCl, contamination when SrCl2 is added to NaCl strong as a debasement, two Na+ particles will be supplanted and one of their destinations will be involved by Sr21-while the other will stay empty. Subsequently, we can say that when a cation of higher valence is added as a debasement to an ionic strong, at least two cations of lower valency are supplanted by a cation of higher valency to keep up with electrical impartiality. Consequently, a few cationic opportunities are made.

1.22. Ionic solids, which have anionic opening because of metal abundance imperfection, foster tone. Make sense of with the assistance of a reasonable model.
Ans: Let us take an illustration of NaCl. At the point when NaCl gem is warmed in presence of Na fume, a few Cl-particles leave their grid locales to join with Na to frame NaCl. The e-1 s lost by Na to frame Na+ (Na+ + Cl-— > NaCl) then, at that point, diffuse into the gem to involve the anion opportunities. These locales are called F-focuses. These e-s ingest energy from noticeable light, get eager to higher energy level and when they fall back to ground state, they give yellow tone to NaCl gem.

1.23. A gathering 14 component is to be changed over into n-type semiconductor by doping it with a reasonable pollutant. To which gathering should this contamination have a place?
Ans: Pollution from bunch 15 ought to be added to get n-type semiconductor.

1.24. What sort of substances would improve super durable magnets, ferromagnetic or ferrimagnetic. Legitimize your response.
Ans: Ferromagnetic substances improve super durable magnets. This is on the grounds that when put in attractive field, their spaces get situated in the headings of attractive field and a solid attractive field is delivered. This requesting of spaces endures in any event, when outside attractive field is taken out. Thus, the ferromagnetic substance turns into a long-lasting magnet.

NCERT Activities

1.1. Characterize the term ‘shapeless’. Give a couple of instances of shapeless solids.
Sol. Undefined solids are those substances, wherein there is no standard plan of its constituent particles, (i.e., particles, iotas or atoms). The game plan of the comprising particles has just short-range request, i.e., a customary and occasionally rehashing design is seen over brief distances just, e.g., glass, elastic, and plastics.

1.2. What makes glass not the same as a strong like quartz? Under what conditions might quartz at any point be changed over into glass?
Sol. Glass is a supercooled fluid and a shapeless substance. Quartz is the glasslike type of silica (SiO2) in which tetrahedral units SiO4 are connected with one another so that the oxygen particle of one tetrahedron is imparted to another Si iota. Quartz can be changed over into glass by dissolving it and cooling the soften quickly. In the glass, SiO4 tetrahedra are participated in an irregular way.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent), or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate, (NH₄)₃PO₄ (iii) SiC (iv) I₂ (v) P₄ (vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si
Sol.

1.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atom
(a) in a cubic close-packed structure?
(b) in a body centred cubic structure?
Sol. (i) The number of nearest neighbours of a particle are called its coordination number.
(ii) (a) 12 (b) 8

1.5. How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)
Sol.

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7. How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Sol.
(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).
(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.
Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.

Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.

1.8 How many lattice points are there is one unit cell of each of the following lattices?
(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic
Sol.

1.6 ‘Soundness of a precious stone is reflected in the extent of its liquefying focuses’. Remark. Gather dissolving points of strong water, ethyl liquor, diethyl ether and methane from an information book. What might you at any point say regarding the intermolecular powers between these particles?
Sol. Higher the liquefying point, more noteworthy are the powers keeping the constituent particles intact and subsequently more noteworthy is the dependability of a gem. It are following to Dissolve points of given substances. Water = 273 K, Ethyl liquor = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar powers present in the event of water and ethyl liquor are fundamentally because of the hydrogen holding which is answerable for their high softening places. Hydrogen holding is more grounded if there should be an occurrence of water than ethyl liquor and consequently water has higher liquefying point then ethyl liquor. Dipole cooperations are available if there should be an occurrence of diethylether. The main powers present in the event of methane is the frail van der Waal’s powers (or London scattering powers).

1.7. How might you recognize the accompanying sets of terms :
(a) Hexagonal close pressing and cubic close pressing
(b) Gem cross section and unit cell
(c) Tetrahedral void and octahedral void.
Sol.
(a) In hexagonal close pressing (hcp), the circles of the third layer are upward over the circles of the principal layer
(ABABAB… … . type). Then again, in cubic close pressing (ccp), the circles of the fourth layer are available over the circles of the principal layer (ABCABC… ..type).
(b) Gem cross section: It deplicts the real shape as well as size of the constituent particles in the gem. It is along these lines, called space grid or gem cross section.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).
Sol. Packing efficiency: It is the percentage of total space filled by the particles.

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Sol.

1.12. A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?
Sol. Contribution by atoms Q present at the eight corners of the cube = \(\frac { 1 }{ 8 } \)= x 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

1.13 Niobium crystallises in a body centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium, using its atomic mass 93u.
Sol.

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.
Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Sol.
This calculated value of density is closely in agreement with its measured value of 8.92 g cm³.

Question 16.
Analysis shows that nickel oxide has the formula Ni_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni²⁺ = x
Then Ni present as Ni³⁺ = 98 – x
Total charge on x Ni²⁺ and (98 – x) Ni³⁺ should
be equal to charge on 100 O^2- ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94
∴ Fraction of Ni present as Ni²⁺ = \(\frac { 94 }{ 98 } \) × 100 = 96%
Fraction of Ni present as Ni³⁺ = \(\frac { 4 }{ 98 } \) × 100 = 4%

(I) n-type semiconductor: When a silicon or germanium gem is doped with bunch 15 component like P or As, the dopant molecule structures four covalent bonds like Si or Ge iota however the fifth electron, not utilized in holding, becomes delocalised and proceeds with its portion towards electrical conduction. Consequently silicon or germanium doped with P or As is called H-type semiconductor, a-demonstrative of negative since the electron conducts power.

(ii) p-type semiconductor: When a silicon or germanium is doped with bunch 13 component like B or Al, the dopant is available just with three valence electrons. An electron opportunity or an opening is made at the spot of missing fourth electron. Here, this opening moves all through the precious stone like a positive charge leading to electrical conductivity. Hence Si or Ge doped with B or Al is called p-type semiconductor, p represents positive opening, since the positive opening is answerable for conduction.

Question 18.
Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Solution:
The ratio less than 2 : 1 in Cu₂0 shows cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. For maintaining electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal dose- packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Solution:
Suppose the number of oxide ions (O^2-) in the packing = 90
∴ Number of octahedral voids = 90
As 2/3^rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present = \(\frac { 2 }{ 3 } \) × 90 = 60
∴ Ratio of Fe³⁺ : O^2- = 60 : 90 = 2 : 3
Hence, the formula of ferric oxide is Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or n-type semiconductor :

1. Ge doped with In
2. B doped with Si.

Solution:

1. Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
2. B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Solution:
For a face centred cubic unit cell (fcc)
Edge length (a) = \(2\sqrt { 2 } r\) = 2 x 1.4142 x 0.144 mm = 0.407 nm

Question 22.
As far as band hypothesis, what is the distinction

between a guide and an encasing

between a guide and a semiconductor?

Arrangement:
In the vast majority of the solids and in many protecting solids conduction happens because of relocation of electrons affected by electric field. Notwithstanding, in ionic solids, the particles are answerable for the directing way of behaving because of their development.

(I) In metals, conductivity firmly relies on the quantity of valence electrons accessible in a particle. The nuclear orbitals of metal particles structure sub-atomic orbitals which are so close in energy to one another, as to shape a band. In the event that this band is to some extent filled or it covers with the higher energy vacant conduction band, then electrons can stream effectively under an applied electric field and the metal acts as a director.

In the event that the hole between valence band and next higher empty conduction band is enormous, electrons can’t bounce into it and such a substance acts as protector.

(ii) On the off chance that the hole between the valence band and conduction band is little, a few electrons might bounce from valence band to the conduction band. Such a substance shows some conductivity and it acts as a semiconductor. Electrical conductivity of semiconductors increments with expansion in temperature, since additional electrons can leap to the conduction band. Silicon and germanium show this sort of conduct and are called characteristic semiconductors. Directors have no prohibited band.

Question 23.
Make sense of the accompanying terms with appropriate models :

Schottky imperfection

Frenkel imperfection

Interstitial imperfection

F-focuses.

Arrangement:
(I) Schottky deformity : In Schottky imperfection a couple of opening or openings exist in the gem grid because of the shortfall of equivalent number of cations and anions from their cross section focuses. It is a typical imperfection in ionic mixtures of high coordination number where the two cations and anions are of a similar size, e.g., KCl, NaCl, KBr, and so forth. Because of this deformity thickness of gem diminishes and it starts to lead power less significantly.

(ii) Frenkel imperfection : This deformity emerges when a portion of the particles in the grid possess interstitial locales leaving cross section destinations empty. This imperfection is for the most part found in ionic gems where anion is a lot bigger in size than the cation, e.g., AgBr, ZnS, and so on. Because of this deformity thickness doesn’t change, electrical conductivity increments to a little degree and there is no adjustment of in general synthetic creation of the precious stone.

(iii) Interstitial imperfection : When a few constituent particles (iotas or atoms) involve an interstitial site of the precious stone, having interstitial defect is said. Because of this deformity the thickness of the substance increments.

(iv) F-Focuses : These are the anionic destinations involved by unpaired electrons. F-focuses bestow variety to gems. The variety results by the excitation of electrons when they assimilate energy from the apparent light falling on the gem.

Question 24.
Aluminum takes shape in a cubic close stuffed structure. Its metallic span is 125 p

1. What is the length of the side of the unit cell?
2. How many unit cells are there in 1.00 cm³ of aluminium?

Solution:
(i) For an fee unit cell, r = \(\frac{a}{2 \sqrt{2}}\) (given, r = 125 pm)
a = 2√2 r = 2√2 × 125 pm
= 353.55 pm
≅354 pm

(ii) Volume of one unit cell = a³ = (354 pm)³
= 4.4 × 10⁷ pm³
= 4.4 × 10⁷ × 10^-30cm³
= 4.4 × 10^-23 cm³
Therefore, number of unit cells in 1.00 cm³ = \(\frac{1.00 \mathrm{~cm}^{3}}{4.4 \times 10^{-23} \mathrm{~cm}^{3}}\)
= 2.27 × 10²²

Question 25.
If NaCI is doped with 10^-3 mol % SrCl₂, what is the concentration of cation vacancies?
Solution:
Let moles of NaCI = 100
∴ Moles of SrCl₂ doped = 10^-3
Each Sr²⁺ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.
∴ Moles of cation vacancy in 100 moles NaCI = 10^-3
Moles of cation vacancy in one mole
NaCI = 10^-3 × 10^-2 = 10^-5
∴ Number of cation vacancies
= 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸ mol^-1

Question 26.
Explain the following with suitable example:

1. Ferromagnetism
2. Paramagnetism
3. Ferrimagnetism
4. Antiferromagnetism
5. 12-16 and 13-15 group compounds.

Solution:
(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO₂ show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO₂, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are feebly drawn in by the outside attractive field are called paramagnetic substances. The property in this way shown is called paramagnetism. They are polarized in the very bearing as that of the applied field. This property is shown by those substances whose iotas, particles or atoms contain unpaired electrons, e.g., O2, Cu2+, Fe3+, and so forth. These substances, in any case, lose their attraction without even a trace of the attractive field.

(iii) Ferrimagnetic substances : Substances which are supposed to have huge attraction based on the unpaired electrons yet really have little net attractive second are called ferrimagnetic substances, e.g., Fe3O4, ferrites of the recipe M2+Fe2O4 where M = Mg, Cu, Zn, and so forth. Ferrimagnetism emerges because of the inconsistent number of attractive minutes in inverse heading bringing about some net attractive second.

(iv) Antiferromagnetic substances : Substances which are supposed to have paramagnetism or ferromagnetism based on unpaired electrons however they have zero net attractive second are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is because of the presence of equivalent number of attractive minutes in the contrary bearings

(v) 13-15 gathering compounds : When the strong state materials are created by mix of components of gatherings 13 and 15, the mixtures in this way acquired are called 13-15 mixtures. For instance, InSb, High mountain, GaAs, and so on.

12-16 gathering compounds : Blend of components of gatherings 12 and 16 yield a few strong mixtures which are alluded to as 12-16 mixtures. For instance, ZnS, Cds, CdSe, HgTe, and so on. In these mixtures, the bonds have ionic person.