NCERT Solutions For Class 12 Chemistry Chapter 1 - The Solid State

1. Define the term ‘amorphous’. Give a few example of amorphous solids.

Ans. An amorphous solid consists of particles of irregular shape. In such solid, the arrangement of constituent particles has only short range order. In such as arrangement, a regular and periodically repeating pattern is observed over short distance only. Example: glass, rubber, plastics etc.

2. What makes a glass different from solid such as quartz? Under what condition could quartz be converted into glass?

Ans. Glass is an amorphous solid in which the constituent particles (SiO₄tetrahedra) have only a short range order. Quartz is a crystalline form of silica in which SiO₄ units are arranged in such a way that they have long ranged order. Quartz can be converted into glass by melting it and cooling it rapidly.

3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

(i) Tetra phosphorus decoxide (P₄O₁₀)

(ii) Ammonium phosphate (NH₄)₃PO

(iii) SiC

(iv) I₂

(v) P₄

(vi) Plastic

(vii) Graphite

(viii) Brass

(ix) Rb

(x) LiBr

(xi) Si

Ionic means contain ions, metallic means are metals, network solids are giant molecules, molecular solids contains covalent bond, amorphous solids have short range order, so classify the give solid on the basis of intermolecular forces.

Ans. (i) Molecular solid (as contains covalent bonds)

(ii) Ionic solid (as contains ions)

(iii) Network (covalent) solid (giant molecule)

(iv) Molecular solid (contains covalent bond)

(v) Molecular solid (contains covalent bond)

(vi) Amorphous solid (short range order)

(vii) Network (covalent) solid (giant molecule)

(viii) Metallic solid (metal)

(ix) Metallic solid (metal)

(x) Ionic solid (gives ions)

(xi) Network (covalent) solid (giant molecule)

4. (i) What is menat 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?

Ans. (i) Coordination number is defined as the number of nearest neighbours of any constituent a particle in a close packed structure.

(ii) (a) 12 (as each atom is attached with 12 other atoms).

(b) 8 (because each atom is attached with 8 other atoms).

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.

Ans. Atomic mass of element of unknown metal is determined by using formula:

$$(\text{M})=\frac{d×a^{3}×\text{N}_\text{A}}{\text{Z}}$$

where, d = density

a³ = volume of the unit cell

N_A = Avogadro’s number

Z = number of atoms present in one unit cell.

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 force between these molecules.

Ans. Stability of a crystal depends upon the magnitude of force of attraction between the constituent particles. Greater the attraction force, more will be the stability of the cyrstal and hence, the melting point of the solid will be higher.

The melting points of the given substances are as follows:

(a) Solid water = 273 K

(b) Ethyl alcohol = 158.8 K

(d) Methane = 90.5 K

The intermolecular forces in solid water and ethyl alcohol are mainly hydrogen bonding. The higher melting point of water than ethyl alcohol indicates that the hydrogen bonding in water is stronger than in ethyl alcohol. Diethyl ether is polar molecule and, therefore, the intermolecular forces in diethyl ether are dipole–dipole interactions. On the other hand, methane is a non polar molecule and the only force present in them are the weak van der Walls forces. These are weaker than dipole–dipole interactions and hence methane has very low melting point than diethyl ether.

7. How will you distinguish between the following pair of terms:

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

(ii) Crystal lattice and unit cell.

(iii) Tetrahedral void and octahedral void.

Ans. (i) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer. It means that tetrahedral voids of the second layer are covered by the spheres of the third layer. The AB AB AB ......... type.

In cubic close packing (ccp), the spheres of third layer cover the octahedral voids of second layer. But the spheres of the fourth layers are aligned with those of the first layer. The pattern is ABC ABC ........ type.

(ii) Crystal lattice is the three dimensional arrangement of identical point in the space which represent how the constituent particles (atoms, ions, molecules) are arranged in a crystal.

Unit cell is the smallest portion of a crystal lattice which when repeated in different directions, generates the entire lattice.

(iii) Tetrahedral voids are surrounded by four spheres which lie at the vertices of a regular tetrahedron. There are 2 tetrahedral voids per atom in a crystal. Octahedral voids are surrounded by six spheres and formed by a combination of two triangular voids of the first and second layer. There is one octahedral void per atom in a crystal.

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

$$\textbf{Ans}.\text{(i)}\space\text{Number of corner atoms per unit cell = 8 corner atoms ×}\frac{1}{8}\text{atom per unit cell}\\=8×\frac{1}{8}=1\text{atom}\\\text{Number of face centred atoms per unit cell = 6 face centred atoms ×}\frac{1}{2}\text{atom per unit cell}\\=6×\frac{1}{2}=3\text{atoms}$$

∴ Total number of atoms of lattice points = 1 + 3 = 4

(ii) Face centered tetragonal:

8 lattice points at corners and 6 lattice points at face centres
∴ 8 × 1/8 + 6 × 1/2 = 4

(iii) In bcc unit cell, number of corner atoms per unit cell

$$=\text{8 corners×}\frac{1}{8}\text{per corner atom}\\=8×\frac{1}{8}=1\text{atom}$$

Number of atoms at body centre = 1 × 1 = 1 atom

∴ Total number of atoms (lattice point) = 1 + 1

= 2

9. Explain:

(i) The basis of similarities and differences between metallic and ionic crystals.

(ii) Ionic solids are hard and brittle.

Ans. (i) Similarities

(a) In ionic and metallic crystals, the constituent particles are held by electrostatic force of attraction. In ionic crystals, this electrostatic force acts between oppositely charged ions. Whereas in metallic crystals, this force acts between the valence electrons and the kernels (positive ions).

(b) Both metallic and ionic crystal have high melting point due to strong electrostatic force of attraction. And in both the crystals bond is non-directional.

Differences

(a) In ionic crystals, ions are not free to move. So, they don’t conduct electricity in solid state. They are conductors only in molten state or aqueous solution. In metallic crystals, valence electrons are free to move. Hence, they can conduct electricity in solid state.

(b) Ionic bond is strong due to strong electrostatic force of attraction whereas metallic bonds may be weak or strong depending upon the number of valence electrons and the size of kernel.

(ii) The ionic solids are hard because their ions are bound together by strong electrostatic force of attraction. They are brittle because bond is non-directional.

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).

$$\textbf{Ans.}\text{(i)}\space\text{In simple cubic, Packing efficiency =}\frac{\text{Volume of one atom}}{\text{Volume of cubic unit cell}}×100\\=\frac{\frac{4}{3}\pi r^{3}}{8 r^{3}}×100=\frac{\pi}{6}×100=52.4\%\\\text{Where, r = radius of one sphere or atom.}\\\text{(ii)}\space\text{In body centred cubic, Packing efficiency =}\frac{\text{Volume occupied by 2 atoms}}{\text{Volume of cubic unit cell}}×100\\=\frac{2×(4/3)\pi r^{3}×100}{[(4/\sqrt{3})r]^{3}}\\=\frac{(8/3)\pi r^{3}×100}{64/(3×\sqrt{3})r^{3}}=68\%\\\text{(iii)}\space\text{In face centred cubic, Packing efficiency =}\frac{\text{Volume occupied by 4 atoms}}{\text{Volume of cubic unit cell}}×100\\=\frac{4×(4/3)\pi r^{3}×100}{(2\sqrt{2r})^{3}} $$

$$=\frac{(16/3)\pi r^{3}×100}{16\sqrt{2} r^{3}}=74\%$$

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

Ans. Given, Density, d = 10.5 g cm^–3

Edge length, a = 4.07 × 10^–8 cm

As the lattice is fcc type, the number of atoms per unit cell, Z = 4 (fcc) atoms

N_A = 6.022 × 10²³ mol^–1 (Avogadro’s constant)

To find: Atomic mass of silver, M

$$\text{Formula}:\space d=\frac{\text{Z×M}}{a^{3}×\text{N}_{\text{A}}}\\\text{M}=\frac{d×a^{3}×\text{N}_\text{A}}{\text{Z}}\\=\frac{10.5 g\space\text{cm}^{\normalsize-3}×(4.077×10^{8}\text{cm})^{3}×(6.022×10^{23}\text{mol}^{\normalsize-1})}{4}\\=\frac{10.5×67.767×10^{-24}×6.022×10^{23}}{4}\text{g mol}^{\normalsize-1}$$

= 107.09 mol^–1

Hence, the atomic mass of silver is 107.09 mol^–1.

12. A cube solid is made up 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 number of P and Q?

$$\textbf{Ans.}\space\text{Contribution of atom Q present at 8 corners of the cube}=\frac{1}{8}×8=1$$

Contribution of atoms P present at the body centre = 1

Therefore, ratio of P and Q = 1 : 1

So, the formula is PQ.

Hence, the coordination number of atom P and Q = 8.

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.

Ans. Given: density d = 8.55 g cm^–3

Atomic mass of the element (M) = 93 g mol^–1

Number of particles in bcc type unit cell (Z) = 2

To find: Atomic radius, r =?

$$\text{Mass of unit cell =}\frac{\text{Z×M}}{\text{N}_\text{A}}\\=\frac{2×(93 \text{g mol}^{\normalsize-1})}{(6.022×10^{23}\text{mol}^{\normalsize-1})}\\=30.89×10^{\normalsize-23}g\\\text{Density of unit cell (d)} = 8.55 g \space\text{cm}^{\normalsize–3}\\\text{Volume of unit cell (a}^{3}) =\frac{\text{Mass of unit cell}}{\text{Density of unit cell}}\\=\frac{(30.89×10^{\normalsize-23}g)}{(8.55 \text{g cm}^{\normalsize-3})}$$

= 36.16 × 10–24 cm3

Edge length of unit cell (a) = (36.13 × 10^–24 cm³)^1/3

= 3.31 × 10^–8 cm

$$\text{For bcc structure, r =}\frac{\sqrt{3a}}{4}\\=\frac{\sqrt{3}×(3.31×10^{-8}\text{cm})}{4}=1.43×10^{\normalsize-8}\text{cm}$$

= 143 pm

Hence, the atomic radius of niobium is 143 pm.

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.

Ans. Derivation of relation between r and R

A sphere fitted into the octahedral void is shown by shaded circle. The spheres present in other layers are not shown in the figure.

∴ ΔABC is a right angled triangle.

∴ We apply Pythagoras theorem.

AC² = AB² + BC²

(2R)² = (R + r)² + (R + r)² = 2 (R + r)²

4R² = 2(R + r)²

2R² = (R + r)²

$$(\sqrt{2\text{R}})^{2}=\sqrt{(\text{R+r})^{2}}\\\sqrt{2}\text{R}=\text{R+r}\\r=\sqrt{2}\text{R}-\text{R}\\r=(\sqrt{2}-1)\text{R}$$

r = (1.414 – 1)R

r = 0.414R

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.

Ans. Given: a = 3.61 × 10^–8 cm

d = 8.92 g cm^–3

Z = 4

To determine: Calculated density

$$ \text{Formula: d =}\frac{\text{Z×M}}{a^{3}×\text{N}_{\text{A}}}\\\text{For fcc lattice of copper, Z = 4}\\\text{Atomic mass of copper, M = 63.5 g mol}^{\normalsize–1}\\\therefore\space d=\frac{4×63.5\text{g mol}^{\normalsize-1}}{(3.61×10^{\normalsize-8}\text{cm})^{3}×(6.022×10^{23}\text{mol}^{\normalsize-1})}$$

= 8.97 g cm^–3

Hence, the above calculated value of density is approximately in agreement with its measured value of 8.92 cm^–3.

16. Analysis shown that nickel oxide has the formula Ni_0.98 O_1.00. What fraction of nickel exist as Ni²⁺ and Ni³⁺ ions?

Ans. The ratio of Ni and O atoms in pure nickel oxide (NiO) = 1 : 1.

Let x be number of Ni (II) atoms replaced by Ni (III) atoms in the oxide.

∴ Number of Ni (II) atoms present = (0.98 – x)

Since, the oxide is neutral in nature,

Total charge on Ni atoms = Charge on oxygen atoms

2(0.98 – x) + 3x = 2

1.96 – 2x + 3x = 2

x = 2 – 1.96 = 0.04

$$\text{\%}\space\text{of Ni (III) atoms in nickel oxide =}\frac{\text{Number of Ni (III) atoms}}{\text{Total number of Ni atoms}}×100\\=\frac{0.04}{0.98}×100=4.01\%$$

% of Ni (II) atoms in nickel oxide = 100 – 4.01

= 95.99% or 96%

Hence, the fractions of nickel that exists as Ni²⁺ and Ni³⁺ are 96% and 4% respectively.

17. What is semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Ans. Semiconductors: Semiconductor is a solid which is perfect insulator at 0 K but conduct some electricity when impurities added into it such as Si and Ge.

Type of Semiconductors

(i) n-type semiconductor: When silicon or germanium crystal is doped with group 15 element like P or As, the dopant forms four covalent bonds like a Si or Ge atom but the fifth electron not used in bonding, becomes delocalised and contribute it share towards electrical conduction. Thus, silicon or germanium doped with P or As is called n-type semiconductor (negative-type).

(ii) p-type semiconductor: When silicon or germanium is doped with group 13 element like B or Al. The dopant atom forms three covalent bond like a B or Al atom, but at the place or fourth electron a hole is created. Here, this hole moves through the crystal like a positive charge giving rise to electrical conductivity. Thus, Si or Ge doped with B or Al is called p-type of semiconductor, (p stands for positive hole) since, it is the positive hole that is responsible for conduction.

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?

Ans. The ratio of copper to oxygen is slightly less than 2 : 1. This shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. In order to maintain the electrical neutrality of the molecule, every two Cu⁺ ions will be replaced by one Cu⁺ ion which results in creating cation vacancies leading to positive holes. Since, the conduction is due to positive holes, it is a p-type semiconductor.

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.

Ans. Let x represents the number of oxide ions.

Therefore, the number of octahedral voids = x

$$\text{Number of ferric ions =}\frac{2}{3}x\\\text{The ratio of the number of ferric ions to the number of oxide ion is}\space\frac{2}{3}x:x=\frac{2}{3}:1=2:3\\\text{Hence, ferric oxide has formula Fe}_2\text{O}_3.$$

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

(i) Ge doped with In

(ii) B doped with Si

Ans. (i) p-type semiconductor.

(ii) n-type semiconductor.

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

Ans. For a face-centred cubic unit cell (fcc)

$$\text{Edge length (a) =}2\sqrt{2}r\\=2×1.4142 × 0.144\space\text{nm}\\=0.407\space\text{nm}$$

22. In terms of band theory, what is the difference:

(i) Between a conductor and an insulator?

(ii) Between a conductor and a semiconductor?

Ans. (i) The energy gap between the valence band and conduction band in an insulator is very large while in a conductor, the energy gap is very small or there is overlapping, between valence band and conduction band.

(ii) In a conductor, the valence band is partially filled or there is overlapping between valence band and conduction band while in semiconductor, there is always a small energy gap between them.

23. Explain the following terms with suitable examples:

(i) Scottky defect

(ii) Frenkel defect

(iii) Interstitials, and

(iv) F-centers

Ans. (i) Schottky defect: This defect arises when equal number of cations and anions are missing from the lattice. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect: This defect arises when some of the ions of the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation. e.g., AgBr, ZnS etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall composition of the crystal.

(iii) Interstitials: Atoms or ions which occupy normally vacant interstitials (voids) in a crystal are called interstitials.

(iv) F-centres: These are the anionic sites occupied by unpaired electrons. F-centers impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal. F-centers are produced when an alkali halide is heated with excess alkali metal. e.g., when NaCl crystals are heated in presence of sodium vapour, yellow colour is produced due to F-centers.

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 cell are there in 1.00 cm³ of aluminium?

Ans. (i) Given: r = 125 pm

To determine = a

$$\text{For an fcc unit cell}\space r=\frac{a}{2\sqrt{2}}\space[\text{Given , r = 125 \text{pm}}]\\ a=2\sqrt{2}×r=2×1.414×125$$

= 353.5 pm

≅ 354 pm

Hence, the length of the side of the unit cell is 354 pm.

(ii) Volume of unit cell = a³ = (353.5 × 10^–10 cm)³= 442 × 10^–25 cm³

$$\text{Number of unit cell =}\frac{1.00\text{cm}^{3}}{442×10^{\normalsize-25}\text{cm}^{3}}$$

= 2.26 × 10²² unit cells

Hence, there are 2.26 × 10²² unit cells.

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

Ans. It is given that NaCl is doped with 10^–3mol% of SrCl₂. This means that 100 mol of NaCl is doped with 10^–3 mol of SrCl₂.

$$\text{Therefore, 1 mol of NaCl is doped with}\space\frac{10^{\normalsize-3}}{100}\text{mol of SrCl}_2 = 10^{\normalsize–5} \text{mol of SrCl}_2$$

Cation vacancies produced by one Sr⁺² ion = 1

Total number of cation vacancies = 10^–5 × NA

= 10^–5 mol × 6.022 × 10²³ mol^–1

= 6.022 × 10¹⁸

26. Explain the following with suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

(iv) Antiferromagnetism

(v) 12-16 and 13-15 group compounds.

Ans. (i) Ferromagnetism: The substance like iron, cobalt, nickel, gadolinium, CrO₂ etc are strongly attracted by magnetic field. Such substances are called ferromagnetic substances. In solid state, the metal ions are grouped into small regions (domains), which act as small magnet. When such substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced. If the magnetic field is removed the magnetic property persists. Thus, a permanent magnet is formed. This property is called ferromagnetism.

(ii) Paramagnetism Some substances are weakly attracted by a magnetic field. Atoms of these substances have one or more unpaired electrons, which are attracted by magnetic field. They are magnetised in the direction of magnetic field. When the external magnetic field is removed, the magnetism is also lost. Thus, the paramagnetism is temporary. Examples of paramagnetic substance are O₂, Cu²⁺, Fe³⁺, Cr³⁺ etc.

(iii) Ferrimagnetism In ferrimagnetic substances, the net magnetic moment is small due to unequal number of magnetic moment of domains aligned in parallel and antiparallel directions. These substances lose the ferrimagnetism on heating and become paramagnetic. Examples: Fe₃O₄, MgFe₂O₄, ZnFe₂O₄ etc.

(iv) Antiferromagnetism These substances have domain structure similar to those of ferromagnetic substances but their domains are oppositely oriented and cancel out each other’s magnetic moment. MnO is an antiferromagnetic substance.

(v) 12-16 and 13-15 group compounds A large variety of solid state materials have been prepared by combination of group 13 and 15 or 12 and 16 to stimulate average valence of four as in Ge or Si. Typical compounds of groups 13-15 are InSb, AIP and GaAs. Gallium arsenide (GaAs) semiconductors have very fast response and have revolutionised the design of semiconductor devices. Zns, Cds, CdSe and HgTe are examples of groups 12-16 compounds. In these compounds, bonds are having same ionic character along with covalent. The ionic character depends on the electronegativities of the two elements.