Page No 31: - Chapter 1 The Solid State Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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?

Answer:

1. A 2-D hexagonal close-packing contains two types of triangular voids (a and b) as shown in figure 1. Let us call this 2-D structure as layer A. Now, particles are kept in the voids present in layer A (it can be easily observed from figures 2 and 3 that only one of the voids will be occupied in the process, i.e., either a or b). Let us call the particles or spheres present in the voids of layer A as layer B. Now, two types of voids are present in layer B (c and d). Unlike the voids present in layer A, the two types of voids present in layer B are not similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is surrounded by 6 spheres and is called the octahedral void.

Figure 2

Figure 3

Now, the next layer can be placed over layer B in 2 ways.

Case 1: When the third layer (layer C) is placed over the second one (layer B) in such a manner that the spheres of layer C occupy the tetrahedral voids c.

In this case we get hexagonal close-packing. This is shown in figure 4. In figure 4.1, layer B is present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is present over the voids b and layer C is present over the voids c. It can be observed from the figure that in this arrangement, the spheres present in layer C are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are arranged in an ABAB….. pattern.

Figure 4.1                 

  Figure 4.2

Case 2: When the third layer (layer C) is placed over layer B in such a manner that the spheres of layer C occupy the octahedral voids d.

In this case we get cubic close-packing. In figure 5.1, layer B is present over the voids a and layer C is present over the voids d. In figure 5.2, layer B is present over the voids b and layer C is present over the voids d. It can be observed from the figure that the arrangement of particles in layer C is completely different from that in layers A or B. When the fourth layer is kept over the third layer, the arrangement of particles in this layer is similar to that in layer A. Hence, we can say that the layers in cubic close-packing are arranged in an ABCABC….. pattern.

Figure 5.1                                                                            

                                                                      Figure 5.2

The side views of hcp and ccp are given in figures 6.1 and 6.2 respectively.

Figure 6.1                                                                          Figure 6.2

(ii) The diagrammatic representation of the constituent particles (atoms, ions, or molecules) present in a crystal in a regular three-dimensional arrangement is called crystal lattice.

A unit cell is the smallest three-dimensional portion of a crystal lattice. When repeated again and again in different directions, it generates the entire crystal lattice.

(iii) A void surrounded by 4 spheres is called a tetrahedral void and a void surrounded by 6 spheres is called an octahedral void. Figure 1 represents a tetrahedral void and figure 2 represents an octahedral void.

Figure 1                                                                          Figure 2

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

Answer:

(i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.

(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.

(iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.

Question 1.9:

Explain

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

(ii) Ionic solids are hard and brittle.

Answer:

(i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points.

The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.

(ii) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are 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).

Answer:

(i) Simple cubic

In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

Let the edge length of the cube be ‘a’ and the radius of each particle be r.

So, we can write:

a = 2r

Now, volume of the cubic unit cell = a³

= (2r)³

= 8r³

We know that the number of particles per unit cell is 1.

Therefore, volume of the occupied unit cell 

Hence, packing efficiency 

(ii) Body-centred cubic

It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.

From ΔFED, we have:

Again, from ΔAFD, we have:

Let the radius of the atom be r.

Length of the body diagonal, c = 4π

or,

Volume of the cube, 

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice 

(iii) Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.

From ΔABC, we have:

Let r be the radius of the atom.

Now, from the figure, it can be observed that:

Now, volume of the cube, 

We know that the number of atoms per unit cell is 4.

So, volume of the occupied unit cell 

= 74%

Question 1.11:

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.

Answer:

It is given that the edge length, a = 4.077 × 10⁻⁸ cm

Density, d = 10.5 g cm⁻³

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that, N_A = 6.022 × 10²³ mol⁻¹

Using the relation:

= 107.13 gmol⁻¹

Therefore, atomic mass of silver = 107.13 u

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?

Answer:

It is given that the atoms of Q are present at the corners of the cube.

Therefore, number of atoms of Q in one unit cell

It is also given that the atoms of P are present at the body-centre.

Therefore, number of atoms of P in one unit cell = 1

This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1

Hence, the formula of the compound is PQ.

The coordination number of both P and Q is 8.

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.

Answer:

It is given that the density of niobium, d = 8.55 g cm⁻³

Atomic mass, M = 93 gmol⁻¹

As the lattice is bcc type, the number of atoms per unit cell, z = 2

We also know that, N_A = 6.022 × 10²³ mol⁻¹

Applying the relation:

= 3.612 × 10⁻²³ cm³

So, a = 3.306 × 10⁻⁸ cm

For body-centred cubic unit cell:

= 1.432 × 10⁻⁸ cm

= 14.32 × 10⁻⁹ cm

= 14.32 nm

Question 1.14:

If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Answer:

A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled

∠POQ = 90⁰

Now, applying Pythagoras theorem, we can write:

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.

Answer:

Edge length, a = 3.61 × 10⁻⁸ cm

As the lattice is fcc type, the number of atoms per unit cell, z = 4

Atomic mass, M = 63.5 g mol⁻¹

We also know that, N_A = 6.022 × 10²³ mol⁻¹

Applying the relation:

= 8.97 g cm⁻³

The measured value of density is given as 8.92 g cm⁻³. Hence, the calculated density 8.97 g cm⁻³ is in agreement with its measured value.

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?

Answer:

The formula of nickel oxide is Ni_0.98O_1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100

Now, total charge on 100 O²⁻ ions = 100 × (−2)

= −200

Let the number of Ni²⁺ ions be x.

So, the number of Ni³⁺ ions is 98 − x.

Now, total charge on Ni²⁺ ions = x(+2)

= +2x

And, total charge on Ni³⁺ ions = (98 − x)(+3)

= 294 − 3x

Since, the compound is neutral, we can write:

2x + (294 − 3x) + (−200) = 0

⇒ −x + 94 = 0

⇒ x = 94

Therefore, number of Ni²⁺ ions = 94

And, number of Ni³⁺ ions = 98 − 94 = 4

Hence, fraction of nickel that exists as Ni²⁺

= 0.959

And, fraction of nickel that exists as 

= 0.041

Alternatively, fraction of nickel that exists as Ni³⁺ = 1 − 0.959

= 0.041

Question 1.17:

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

Answer:

Semiconductors are substances having conductance in the intermediate range of 10^-6 to 10⁴ ohm⁻¹m⁻¹.

The two main types of semiconductors are:

(i) n-type semiconductor

(ii) p-type semiconductor

n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.

Si and Ge have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, P and As contain five valence electrons each. When Si or Ge is doped with P or As, the latter occupies some of the lattice sites in the crystal. Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Si or Ge atoms. The remaining fifth electron becomes delocalised and increases the conductivity of the doped Si or Ge.

p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p-type semiconductor. When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated.

When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the 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?

Answer:

In the cuprous oxide (Cu₂O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu⁺ ions is slightly less than twice the number of O²⁻ ions. This is because some Cu⁺ ions have been replaced by Cu²⁺ ions. Every Cu²⁺ ion replaces two Cu⁺ ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.

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.

Answer:

Let the number of oxide (O²⁻) ions be x.

So, number of octahedral voids = x

It is given that two out of every three octahedral holes are occupied by ferric ions.

So, number of ferric (Fe³⁺) ions

Therefore, ratio of the number of Fe³⁺ ions to the number of O²⁻ ions, Fe³⁺ : O²⁻ 

= 2 : 3

Hence, the formula of the ferric oxide is Fe₂O₃.

Question 1.20:

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

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

Answer:

(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.

(ii) B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be a p-type semiconductor.