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

NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State is the essential study material needed to perfect Chapter 1 The Solid State topics.

NCERT Class 12 Chemistry solutions provided here have correct answers to NCERT textbook questions. Solutions curated in a comprehensive manner will help students to understand the subtopics in this chapter in a better way.

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

Answer – The term “amorphous” comes from the Greek word amorphos which means “no form”. The amorphous solids are in short-range order i.e. periodically a repeating pattern is observed for short-range. The arrangement of constituent particles is irregular. They are isotropic in nature. They are also known as pseudo solids or super cooled liquids.
Example- rubber, glass.

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

Answer – The arrangement of constituent particles is different in glass as compared to in quartz. The glass is amorphous solids and possesses short-range order whereas the quartz is crystalline solids and possesses long-range order.

Quartz can be converted into glass by heating it and cooling it quickly. In the glass, SiO4 tetrahedra are joined in a random manner.

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

• (a) Tetra phosphorus decaoxide (P4O10)
• (b) Ammonium phosphate (NH4)3PO4
• (c) SiC
• (d) I2
• (e) P4
• (f) Plastic
• (g) Graphite
• (h) Brass
• (i) Rb
• (j) LiBr
• (k) Si

Answer –

• IONIC SOLIDS – (b) Ammonium phosphate (NH4)3PO4, (j) LiBr
• METALLIC SOLIDS – (h) Brass, (i) Rb
• MOLECULAR SOLIDS – (a) Tetra phosphorus decaoxide (P4O10), (e) P4, (d) I2
• COVALENT SOLIDS – (c) SiC (g) Graphite (k) Si
• AMORPHOUS SOLIDS – (f) Plastic

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?

Answer –

• (i) The number of adjacent neighbours of a particle is called the coordination number.
• (ii) The coordination number of atoms in
  • a cubic close-packed structure- 12
  • a body-centred cubic structure- 8

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.

Answer –

If you are given the density and dimension of the unit cell, then,
Let, the edge length of the unit cell be a, and,
The volume of the unit cell be a³,
The density of the unit cell be D and the atomic mass will be A,

Since,

the mass of unit cell=No. of atoms in unit cell x Mass of each atom = Z × m

Mass of an atom present in the unit cell, m = M/Na
where Na is Avogadro’s number.

As we know,

the mass of unit cell=No. of atoms in unit cell x Mass of each atom = Z × m

Mass of an atom present in the unit cell, m = M/N_a

where N_a is Avogadro’s number.

As we know,

therefore, Atomic mass,

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?

Answer – Substances with higher melting points are more stable than those with lower melting points. This is because a higher melting point implies strong intermolecular interactions between the molecules, thus giving greater stability.

The melting points of given substances from the data book are as follows:

• solid water= 273 K
• ethyl alcohol= 158.8 K
• diethyl ether=156.85 K
• methane=89.34 K

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?

Answer –

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

HEXAGONAL CLOSE PACKING	CUBIC CLOSE PACKING
Placing the third layer on top of the second layer so that the tetrahedral cavities are covered with spheres, produces a dense 3D packing, with every three or alternating layers aligned vertically. If you continue packing in this order, you will get the hexagonal close packing.	Placing the third layer on top of the second layer so that the octahedral cavity is covered with a sphere gives a different layer than the first (A) and second (B). Continue packing in this way to get the cubic close packing.

(ii) Crystal lattice and unit cell

CRYSTAL LATTICE	UNIT CELL
This is the smallest 3D dimensional part of a complete spatial lattice, which is repeated many times in a different direction than the crystalline lattice.	It is the regular orientation of the crystals in the 3D space.

(iii) Tetrahedral void and octahedral void

TETRAHEDRAL VOID	OCTAHEDRAL VOID
A tetrahedron is formed when the centres of four spheres are joined.	This void is surrounded by six spheres and a triangular shape in the centre.

Credit – NCERT

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

Answer –

• (i) Face-centred cubic= 14 lattice points are found, 8 from the sides and 6 from the faces.
• (ii) Face-centred tetragonal= 14 lattice points are found, 8 from the sides and 6 from the faces.
• (iii) Body-centred= 9 lattice points are found, 8 from the centres and 1 from the centre.

9. Explain:

• (i)The basis of similarities and differences between metallic and ionic crystals.
• (ii) Ionic solids are hard and brittle.

Answer –

Metallic and Ionic crystals:

SIMILARITIES:-

• The force of attraction is electrostatic in both crystals.
• They both have high melting points.
• They both are non-directional bonds in nature.

DIFFERENCES:-

• Ionic crystals are poor conductors of electricity in the solid state because ions are not free to move. Therefore, they can only conduct electricity in the molten state or in an aqueous solution. Whereas, metallic crystals are good conductors of electricity in the solid state because the electrons are free to move around.
• The ionic bonds are strong in nature due to strong electrostatic attraction. Whereas, metallic bonds can be strong or weak depending on the number of valence electrons and the size of the nuclei.
• Ionic solids are hard due to strong electrostatic attraction. The brittleness of ionic crystals is due to the non-directional bonds they contain.

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 assumptions that atoms are touching each other).

Answer –

(i) simple cubic

(ii) body-centred cubic

(iii) face-centred cubic (with assumptions that atoms are touching each other).

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.

Answer –

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?

Answer –

Atoms Q are present at the eight corners of the cube,
Number of Q atoms per unit cell = 1/8 x 8 = 1
Atom P is present at the body centre,
Number of P atoms per unit cell = 1
Ratio of P and Q is 1:1
∴ The formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

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.

Answer –

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.

Answer –

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 g cm-3.

Answer –

16. Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Answer –

From the given formula of nickel oxide, we can see that 98 Ni-atoms are associated with 100 O-atoms. Out of 98 Ni-atoms, let Ni present be,
Ni²⁺ = x
Ni³⁺ = (98 – x)

The total charge on ions xNi²⁺ and (90-x)Ni³⁺ should be equal to the charge on 100 O^2- ions.
Therefore,
= x×2 + (98–x)×3 = 100×2
= 2x + 294–3x = 200
= x = 94
The fraction of Ni present as Ni²⁺ = 94/98 × 100 = 96%
The fraction of Ni present as Ni³⁺ = 4/98 × 100 = 4%

17. What are semi-conductors? Describe the two main types of semiconductors and contrast their conduction mechanisms.

Answer – Semiconductors are substances with electrical conductivities with intermediate-range, i.e between conductors and insulators.

Two main types of semiconductors are as follows-

• n-type semiconductor– When a silicon or germanium crystal is doped with a group of 15 elements such as P or As the doped atom forms four covalent bonds like a-Si or Ge atom but the fifth electron remains unused and delocalised. This unused electron increases the conductivity of the conductor. Hence silicon or germanium doped with P or As is called an n-type semiconductor, a sign of negative because it is an electrically conductive electron.
• p-type conductor– When silicon or germanium is doped with a group of 13 elements such as B or Al, the doped atom has only three valence electrons. An electron void or hole is created at the position of the missing fourth electron. Here the hole moves through the crystal lattice as a positive charge giving rise to electrical conductivity. So, Si or Ge doped with B or Al is called a p-type semiconductor, p stands for a positive hole because it is a positive hole that conducts electricity.

Credit – NCERT

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?

Answer – In this oxide, the copper to oxygen ratio is slightly less than 2:1 showing that Cu+ (cuprous) ions are replaced by Cu2+ (cupric) ions. In order to maintain electrical neutrality and stability, for one Cu2+ ion, two Cu+ ions will be replaced. This will leave positive holes. Thus, the conduction will be because of the positive holes making non-stoichiometric cuprous oxide (Cu2O) a p-type semiconductor.

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.

Answer –

Let the number of oxide ions (O2-) present in the packing be 90
∴ Number of octahedral voids = 90
Given,
2/3rd of the octahedral voids are occupied by ferric ions (Fe3+),
Then,
the number of ferric ions present = 2/3 × 90 = 60
∴ Ratio of Fe3+: O2- = 60:90 = 2:3
Hence, the formula of ferric oxide is Fe2O3.

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

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

Answer –

• (i) Ge doped with In – Ge is doped with In which is Group 13 element, which makes positive holes making it a p-type semiconductor.
• (ii) B doped with Si– As B is a 13 Group element and Si is a 14 Group element, therefore, one electron will be unused and delocalised. This is an n-type conductor.

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?

Answer –

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

• (i) between a conductor and an insulator
• (ii) between a conductor and a semiconductor?

Answer – In the majority of solids, conduction takes place due to the movement of electrons under the influence of an electric field. The atomic orbitals of metal atoms form molecular orbitals which are close in energy to each other to form a band.

(i) In a conductor, this band is either filled or partially overlaps with the higher energy unoccupied band, and then electrons flow easily under the influence of the electric field. Whereas, in insulators, the gap between the valence band and the conduction band is quite large, which makes it difficult for electrons to jump and conduct electricity.

(ii) If the gap between the valence band and the conduction band is small, some electrons can jump from the valence band to the conduction band. Such substances exhibit conductivity and behave like semiconductors. As more electrons can jump into the conduction band, the electrical conductivity of a semiconductor increases with increasing temperature. Silicon and germanium exhibit this behaviour and are called intrinsic semiconductors. Whereas the conductor has no banned bands.

Credit – NCERT

23. Explain the following terms with suitable examples :

• (i)Schottky defect
• (ii) Frenkel defect
• (iii) Interstitial defect
• (iv) F-centres.

Answer –

(i) Schottky defect

In the Schottky defect, a pair of vacancies or holes exist in the lattice because there are not equal numbers of cations and anions from their lattice points. This is a common defect in ionic compounds with high coordination numbers where the cations and anions have the same size, e.g. KCl, NaCl, KBr, etc. Due to this defect, the density of the crystal decreases and it begins to conduct electricity to a lesser extent.

Credits – NCERT

(ii) Frenkel defect

This defect arises when some ions in the lattice occupy the interstitial sites leaving the lattice sites vacant. This defect is often found in ionic crystals where the anion is much larger than the cation, for example, AgBr, ZnS, etc. Due to this defect, the density does not change, the conductivity increases to a small extent, and there is no change in the overall chemical composition of the crystal.

Credits – NCERT

(iii) Interstitial Defect

When some constituent particles (atoms or molecules) occupy the interstitial position of the crystal, it is said to have an interstitial defect. The density of the substance increases, due to this defect.

(iv) F-Centers

These are anionic sites occupied by unpaired electrons. The F centres give colour to the crystals. The colour is the result of the excitation of electrons as they absorb energy from visible light falling on the crystal.

Credits – NCERT

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?

Answer –

25. If NaCI is doped with 10-3 mol % SrCl2, what is the concentration of cation vacancies?

Answer –

26. Explain the following with suitable example:

• (i) Ferromagnetism
• (ii) Paramagnetism
• (iii) Ferrimagnetism
• (iv) Antiferromagnetism
• (v) 12-16 and 13-15 group compounds.

Answer –

(i) Ferromagnetism

Substances like Fe, Ni, Co and CrO2 are attracted very strongly by a magnetic field. These substances are called ferromagnetic substances and show ferromagnetism. Ferromagnetic substances can remain permanently magnetized. Ferromagnetism is due to the orientation of unpaired electrons in the same direction.

(ii) Paramagnetism

Paramagnetic substances such as O2, Cu2+, Fe3+, etc are weakly attracted by the magnetic fields. Paramagnetism is due to the presence of one or more unpaired electrons which are attracted by the magnetic field. These substances are magnetised only in the presence of a magnetic field.

(iii) Ferrimagnetism

Ferrimagnetism can be seen when the magnetic moment of domains is arranged in parallel and anti-parallel orientation in unequal numbers. They are weakly attracted as compared to ferromagnetic substances. They lose ferromagnetism on heating and become paramagnetic. Ferrimagnetic substances are magnetite Fe3O4 and ferrites.

(iv) Antiferromagnetism

Substances like MnO are domain structures similar to ferromagnetism, but they have their domains oriented opposite in direction and cancel each other’s magnetic moment.

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

When the solid-state materials are produced by a combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc. whereas the combination of elements of groups 12 and 16 returns some solid compounds which are referred to like 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have an ionic character.

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