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.

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

Case 1: When the third layer(layer C) is placed over the second one (layer B) in such a manner that thespheres 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 aand layer C is present over the voids c. In figure 4.2, layer B is present overthe voids b and layer C is present over the voids c. It can be observed fromthe figure that in this arrangement, the spheres present in layer C are presentdirectly above the spheres of layer A. Hence, we can say that the layers inhexagonal close-packing are arranged in an ABAB….. pattern.

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

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

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

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

A unit cell is the smallest three-dimensional portionof a crystal lattice. When repeated again and again in different directions, itgenerates the entire crystal lattice.

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

Question - 32 : - How many lattice points are there in one unit cell of each of the following lattice?

Answer - 32 : -

(i) Face-centred cubic

(ii) Face-centred tetragonal

(iii) Body-centred

Answer

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

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

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

Question - 33 : -
Explain
(i) The basis of similarities and differences between metallicand ionic crystals.

Answer - 33 : - (ii) Ionic solids are hard and brittle.

Answer

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

The basis of differences between metallic andionic crystals is that in metallic crystals, the electrons are free to move andso, metallic crystals can conduct electricity. However, in ionic crystals, theions 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 particlesof ionic crystals are ions. These ions are held together in three-dimensionalarrangements by the electrostatic force of attraction. Since the electrostaticforce of attraction is very strong, the charged ions are held in fixedpositions. This is the reason why ionic crystals are hard and brittle.

Question - 34 : - Calculate the efficiency of packing in case of a metal crystal for

Answer - 34 : -

(i) simple cubic

(ii) body-centred cubic

(iii) face-centred cubic (with the assumptions that atoms aretouching each other).

Answer

(i) Simple cubic

In a simple cubic lattice, the particles arelocated 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

Therefore, volume of the occupied unit cell

Hence, packing efficiency

(ii) Body-centredcubic

It can be observed from the above figurethat the atom at the centre is in contact with the other two atoms diagonallyarranged.

From ΔFED, we have:

Again, from ΔAFD, we have:

Let the radius of theatom be r.

Length of the bodydiagonal, c = 4π

Volume of the cube,

A body-centred cubiclattice contains 2 atoms.

So, volume of the occupied cubic lattice

(iii) Face-centredcubic

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

From ΔABC, we have:

Let r bethe radius of the atom.

Now, from the figure, itcan be observed that:

Now, volume of the cube,

We know that the numberof atoms per unit cell is 4.

So, volume of the occupied unit cell

Question - 35 : -
Silver crystallises infcc 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 - 35 : -

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

Density, d =10.5 g cm⁻³

As the lattice isfcc 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 massof silver = 107.13 u

Question - 36 : -
A cubic solid is made oftwo elements P and Q. Atoms of Q are at the corners of the cube and P at thebody-centre. What is the formula of the compound? What are the coordinationnumbers of P and Q?

Answer - 36 : -

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

Therefore, number of atomsof Q in one unit cell

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

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

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

Hence, the formula ofthe compound is PQ.

The coordination numberof both P and Q is 8.

Question - 37 : -
Niobium crystallises inbody-centred cubic structure. If density is 8.55 g cm⁻³, calculate atomic radius of niobium using its atomic mass 93 u.

Answer - 37 : -

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

Atomic mass, M = 93 gmol⁻¹

As the lattice is bcctype, 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 cubicunit cell:

= 1.432 × 10⁻⁸ cm

= 14.32 × 10⁻⁹ cm

= 14.32 nm

Question - 38 : -
If the radius of theoctachedral void is r and radius of the atoms in close packing is R, deriverelation between r and R.

Answer - 38 : -

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

∠POQ = 90⁰

Now, applying Pythagorastheorem, we can write:

Question - 39 : -
Copper crystallises intoa fcc lattice with edge length 3.61 × 10⁻⁸ cm. Show that thecalculated density is in agreement with its measured value of 8.92 g cm⁻³.

Answer - 39 : -

Edge length, a =3.61 × 10⁻⁸ cm

As the lattice is fcctype, 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 ofdensity is given as 8.92 g cm⁻³. Hence, the calculateddensity 8.97 g cm⁻³ is in agreement with its measured value.

Question - 40 : -
Analysis shows thatnickel oxide has the formula Ni_0.98O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?

Answer - 40 : -

The formula of nickeloxide is Ni_0.98O_1.00.

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

Now, total charge on 100O²⁻ 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 isneutral, 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 nickelthat exists as Ni²⁺

= 0.959

And, fraction of nickelthat exists as

= 0.041

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

= 0.041

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