NCERT SOLUTION For THE SOLID STATE

THE SOLID STATE

   by    sagar kumar singh

Question – 1.1 - Define the term 'amorphous'. Give a few examples of amorphous solids.

Solution:

Solids having constituent particles with irregular shapes and short range order are called amorphous solids. Amorphous solids are isotropic in nature and melt over a range of temperature. Thus, amorphous solids are also referred as pseudo solids or super cooled liquids.

Amorphous solids do not have definite heat of fusion.

Amorphous solids give irregular surfaces, when cut with sharp tool.

Glass, rubber, plastic, etc. are some examples of amorphous solid.

Question – 1.2 - What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Solution:

It is the arrangement of constituent particles of glass which makes it different from quartz. The constituent particles of glass have short range order while quartz has constituent particles in long range order and short range order both.

By heating and cooling rapidly quartz can be converted into glass.

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

Solution:

(i) Tetra phosphorous decoxide (P₄O₁₀) - Molecular

(ii) Ammonium phosphate (NH₄)₃PO₄ – Ionic

(iii) SiC - Covalent (network)

(iv) I₂ - Molecular

(v) P₄ - Molecular

(vi) Plastic - Amorphous

(vii) Graphite – Covalent (network)

(viii) Brass - Metallic

(ix) Rb - Metallic

(x) LiBr - Ionic

(xi) Si – Covalent (network)

 

Question – 1.4 - (i) What is meant by the term 'coordination number'?(ii) What is the coordination number of atoms:(a) in a cubic close-packed structure?(b) in a body-centred cubic structure?

Solution:

(i) Coordination number is the number of nearest neighbours of any constituent particle present in the crystal lattice.

(ii) The coordination number of atoms

(a) In a cubic close-packed structure is 12

(b) In a body-centered cubic structure is 8

Question – 1.5 - How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Solution:

The atomic mass of an unknown metal can be determined by knowing its density and the dimension of unit cell.

Let ‘a’ be the edge length of a unit cell of a crystal.

‘d’ is the density of the metal

‘m’ is the atomic mass of the metal

‘z’ is the number of atoms in the unit cell

Now, the density of the unit cell

As we know that, mass of the unit cell = Number of atoms in the unit cell X Atomic mass

And Volume of the unit cell = (Edge length of the cubic unit cell) ³

Now, since mass of the metal (m)

If the edge lengths are different (say a, b and c), therefore, equation (iii) can be written as

Thus, using equation (iii) and the atomic mass of an unknown metal can be determined.

Question – 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?

Solution:

Stability of a crystal is reflected in the magnitude of its melting points because higher the melting point, greater is the intermolecular force and greater the intermolecular force greater is the stability. And hence, a substance with higher melting point would be more stable.

The melting points of the given substances are as follows:

Solid water - 273 K

Ethyl alcohol – 158.8 K

Diethyl ether – 156.85 K

Methane – 89.34 K

As we can see the melting point of solid water is highest and melting point of methane is lowest among the given substance. This says that intermolecular force in solid water is strongest and the intermolecular force in methane is weakest.

Question – 1.7 - How will you distinguish between the following pairs of terms:(i) Hexagonal close-packing and cubic close-packing?(ii) Crystal lattice and unit cell?(iii) Tetrahedral void and octahedral void?

Solution:

(i) Hexagonal close-packing and cubic close-packing

(ii) Crystal lattice and unit cell

(iii) Tetrahedral void and octahedral void

Question – 1. 8 - How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centred cubic
(ii) Face-centred tetragonal
(iii) Body-centred

Solution:

(i) One unit cell of a face-centered cubic has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

(ii) One unit cell of face-centered tetragonal has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

(iii) One unit cell of body centered has 8 lattice points are corners and 1 lattice points at faces, total 9 lattice points.

Question – 1.9 - Explain(i) The basis of similarities and differences between metallic and ionic crystals.(ii) Ionic solids are hard and brittle.

Solution:

(ii) In ionic solids, constituent particles are held together with strong electrostatic force of attraction along with their fixed position. The fixed position of ions and strong electrostatic force of attraction make ionic solids hard and brittle.

Question – 1.10 - Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).

Solution:

(i) Simple cubic:

Let the side of a simple cubic lattice is ‘a’ and radius of atom present in it is ‘r’.

Since, edges of atoms touch each other, therefore, a = 2r (for simple cubic lattice)

(ii) body-centred cubic

(iii) face-centered cubic

Question – 1.11 – Silver crystallizes 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.

Solution:

Question – 1.12 - A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Solution:

Given, Atoms of Q are at the corners of the cube and P at the body-centre.

Number of atoms of P in one unit cell = 1

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

Solution:

Thus atomic radius of niobium = 14.31 nm

Question – 1.14 - If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Solution:

In the given figure, let an sphere having centre ‘O’ is fitted in the octahedral void.

As given, radius of the sphere fitted in the octahedral void = r

And radius of the atoms in close packing = R

This is the required relation between r and R.

Question – 1.15 - Copper crystallises into a fcc lattice with edge length 3.61 × 10^–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^–3.

Solution:

Question – 1.16 - Analysis shows that nickel oxide has the formula Ni_0.98O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?

Solution:

Question – 1.17 - What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Solution:

Semiconductor: - Solids having intermediate range of conductivity, i.e. from 10^–6 to 10⁴ ohm^–1 m^–1 are called semiconductors. Semiconductors are of following two types:

(i) n – type of semiconductors

(ii) p – type of semiconductors

(i) n – type semiconductors – Semiconductors formed after doping with electron rich impurities to increase their conductivity are called n-type of semiconductors.

Example –

Silicon and germanium, each has four valence electrons as they belong to 14th group of periodic table. Arsenic and phosphorous belong to 15th group of periodic table and they have valence electrons equal to 5. When silicon or germanium is doped with phosphorous or arsenic, four electrons of phosphorous or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon or germanium.

Since the electrical conductivity of silicon or phosphorous is increased because of negatively charged particle (electron), thus this is known as n-type of semiconductor.

(ii) p – type of semiconductor - Semiconductors formed by the doping with electron deficient impurities; to increase their conductivity; are called p-type semiconductors. In p - type of semiconductors, conductivity increase because of formation of electron holes.

Example - Electrical conductivity of silicon or germanium is doped with elements, such as Boron, Aluminium or Gallium having valence electrons equal to 3. Three valence electrons present in these elements make covalent bonds with three electrons present in valence shell out of four of silicon or germanium leaving one electron delocalized. The place from where one electron is missing is called electron hole or electron vacancy.

When the silicon or germanium is placed under electrical field, electron from neighbouring atom fill the electron hole, but in doing so another electron hole is created at the place of movement of electron. In the influence of electrical filed electron moves toward positively charge plate through electron hole as appearing the electron hole as positively charged and are moving towards negatively charged plate.

Question – 1.18 - 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:

When cuprous oxide is prepared in laboratory; the ratio of copper to the oxygen in the compound becomes slightly less than 2:1. This happens because some of the Cu⁺ ions are replaced by the Cu²⁺ ions. In this process, one Cu2+ ions replaces two Cu⁺ ions. As two Cu⁺ ions are replaced by one Cu2⁺ ion, this creates defects because of creating vacant space, i.e. positive holes.

Because of creation of holes due to this defect; this compound conducts electricity through these positive holes.

As semiconductors which are formed by electron deficient impurities are called p-type of semiconductors; thus, cuprous oxides so formed in laboratory are the p – type of semiconductors.

Question – 1.19 - Ferric oxide crystallises in a hexagonal close-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:

Given,

Ferric oxide crystallizes in a hexagonal close packed array of oxide ions.

Two out of every three octahedral holes are occupied by ferric ions.

Let the number of oxide ions = x

Therefore number of octahedral voids = x

Since, two out of every three octahedral holes are filled by ferric ions,

Thus, voids filled by ferric ions

Therefore, number of ferric ions

Now, ratio of ferric ions to the oxide ions

Thus, formula of ferric oxide is (Fe)₂2 O₃

Question – 1.20 – Classify each of the following as being either a p-type or a n-type semiconductor:

(i) Ge doped with In (ii) Si doped with B.

Solution:

(i) Ge doped with In – ‘Ge’ belongs to group 14 in periodic table and ‘In’ belongs to group 13. Thus, when ‘Ge’ is doped with ‘In’, it makes hole or electron vacancy and acts as p-type of conductor.

(ii) Si doped with B – ‘Si’ belongs to group 14 in periodic table and ‘B’ belongs to group 13. Thus, when ‘Si’ is doped with ‘B’, it makes hole or electron vacancy and acts as p-type of conductor.

Question – 1.21 – Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Solution:

Given, atomic radius = 0.144 nm

Type of unit cell = face centered

Thus side of the given cell = 407 nm

Question – 1.22 - In terms of band theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and a semiconductor?

Solution:

Molecular orbirals of metals are formed by atomic orbitals. These orbitals are so close to each other as they form band or valence band.

(i) Difference between conductor and insulator - In conductors there is no energy gap between the valence band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

While in insulators there is large energy gap between the valence band and electrons cannot jump to it i.e. large energy gap prevents the flow of electricity.

(ii)Difference between conductors and semiconductor - In conductors there is no energy gap between the valence band and conduction band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

While in semi conductors, there is small energy gap between valence bond and conduction band. The small gap between band facilitates some electrons to jump to the conduction band by acquiring extra energy.

Question – 1.23 - Explain the following terms with suitable examples:
(i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres.

Solution:

(i) Schottky defects : When cations and anions both are missing from regular sites, the defect is called Schottky Defect. In Schottky Defects, the number of missing cations is equal to the number of missing anions in order to maintain the electrical neutrality of the ionic compound.

Schottky Defect is type of simple vacancy defect and shown by ionic solids having cations and anions; almost similar in size, such as NaCl, KCl, CsCl, etc. AgBr shows both types of defects, i.e. Schottky and Frenkel Defects.

Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.

(ii) It is a type of vacancy defect. In ionic compounds, some of the ions (usually smaller in size) get dislocated from their original site and create defect. This defect is known as Frenkel Defects. Since this defect arises because of dislocation of ions, thus it is also known as Dislocation Defects. As there are a number of cations and anions (which remain equal even because of defect); the density of the substance does not increase or decrease.

Ionic compounds; having large difference in the size between their cations and anions; show Frenkel Defects, such as ZnS, AgCl, AgBr, AgI, etc. These compounds have smaller size of cations compared to anions.

(iii) Interstitials – Sometime in the formation of lattice structure some of the atoms or ions occupy vacant interstitial site, and are known as interstitials. These interstitials are generally small size non-metals, such as H, B, C, etc. Defect arises because of interstitials is called interstitial defect.

(iv) F-centres – This is type of defect and called metal excess defect. These type of defects seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F-centre and responsible for showing colour by the compound.

Question – 1.24 - Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm³ of aluminium?

Solution:

Given, radius of atom (r) = 125 pm

(i) For ccp structure, we know that

(ii) Volume of 1 unit cell = a³

Thus, number of unit cell of aluminium in 1 cm³

Question – 1.25 - If NaCl is doped with 10^–3 mol % of SrCl₂, what is the concentration of cation vacancies?

Solution:

We know that two Na⁺ ions is replaced by each of the Sr⁺⁺ ion while SrCl₂ is doped with NaCl. But in this case only one lattice point is occupied by each of the Sr⁺⁺ ion and produce one cation vacancy.

Here 10 ^{– 3} mole of SrCl₂ is dopped with 100 moles of NaCl

Thus, cation vacancies produced by NaCl = 10 ^{– 3} mol

Since, 100 mole of NaCl produce cation vacancies after doping = 10 ^{– 3} mol

Therefore, 1 mole of NaCl will produce cation vacancies after doping

Question – 1.26 - Explain the following with suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Antiferromagnetism
(v) 12-16 and 13-15 group compounds.

Solution:

(i) Ferromagnetism - Substances that are attracted strongly with magnetic field are called ferromagnetic substances, such as cobalt, nickel, iron, gadolinium, chromium oxide, etc. Ferromagnetic substances can be permanently magnetized also.

Metal ions of ferromagnetic substances are randomly oriented in normal condition and substances do not act as a magnet. But when metal ions are grouped together in small regions, called domains, each domains act like a tiny magnet and produce strong magnetic field, in such condition ferromagnetic substance act like a magnet. When the ordering of domains in group persists even after removal of magnetic field a ferromagnetic substance becomes a permanent magnet.

(ii) Paramagnetism - Substances which are attracted slightly by magnetic field and do not retain the magnetic property after removal of magnetic field are called paramagnetic substances. For example O², Cu²⁺, Fe³⁺, Cr³⁺, Magnesium, molybdenum, lithium, etc.

Substances show paramagnetism because of presence of unpaired electrons. These unpaired electrons are attracted by magnetic field.

(iii) Ferrimagnetism - Substances which are slightly attracted in magnetic field and in which domains are grouped in parallel and anti-parallel direction but in unequal number, are called ferromagnetic substances and this property is called ferrimagnetism.

For example, magnetite (Fe³O⁴), ferrite (MgFe²O⁴), ZnFe²O⁴, etc.

Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic.

(iv) Antiferromagnetism - Substances in which domain structure are similar to ferromagnetic substances but are oriented oppositely, which cancel the magnetic property are called antiferromagnetic substances and this property is called antiferromagnetism. For example; MnO.

(v) 12-16 and 13-15 group compounds –

Compounds belong to 12 – 16 group are formed by the combination of elements of 12 and 16 groups. For example – ZnS, Cds, etc.

Compounds belong to 13 – 15 group are formed by the combination of the elements of 13 and 15 groups. For example – InSb, GaAs, etc.

Bonds in these compounds are not perfectly covalent.

The ionic characters of the bonds in these compounds depends on the electronegativities of the two elements.

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