Answer

Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.

Answer

The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders.

Quartz can be converted into glass by heating and then cooling it rapidly.

Answer

Ionic $\rightarrow$ (ii)Ammonium phosphate $\left(\mathrm{NH_4}\right)_{3} \mathrm{PO_4}$, (x) LiBr

Metallic $\rightarrow$ (viii)Brass, (ix)Rb

Molecular $\rightarrow$ (i) Tetra phosphorus decoxide $\left(\mathrm{P_4} \mathrm{O_10}\right)$, (iv) $\mathrm{I_2},(\mathrm{v}) \mathrm{P_4}$.

Covalent (network) $\rightarrow$ (iii) SiC, (vii)Graphite, (xi) Si

Amorphous $\rightarrow$ (vi)Plastic

Answer

(i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number. (ii) The coordination number of atoms

(a)in a cubic close-packed structure is 12 , and

(b) in a body-centred cubic structure is 8

Answer

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

Let ’ $a$ ’ be the edge length of a unit cell of a crystal, ’ $d$ ’ be the density of the metal, ’ $m$ ’ be the mass of one atom of the metal and ’ $z$ ’ be the number of atoms in the unit cell.

Now, density of the unit cell

$$ \begin{equation*} =\frac{\text { Mass of the unit cell }}{\text { Volume of the unit cell }} \end{equation*} $$

$$\Rightarrow d=\frac{z m}{a^{3}}\tag{i}$$

[Since mass of the unit cell $=$ Number of atoms in the unit cell $\times$ mass of one atom]

[Volume of the unit cell $=(\text { Edge length of the cubic unit cell })^{3}$ ]

From equation ( $i$ ), we have:

$$ \begin{equation*} m=\frac{d a^{3}}{z} \tag{ii} \end{equation*} $$

Now, mass of one atom of metal $(m)$

$$ =\frac{\text { Atomic mass }(\mathrm{M})}{\text { Avogadro’s number }\left(\mathrm{N}_{\mathrm{A}}\right)} $$

Therefore,

$$ \begin{equation*} \mathrm{M}=\frac{d a^{3} \mathrm{~N}_{\mathrm{A}}}{z} \tag{iii} \end{equation*} $$

If the edge lengths are different (say $a, b$ and $c$ ), then equation (ii) becomes:

$M=d(a b c) N_{A}$

Answer

Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.

The melting points of the given substances are:

Solid water $\rightarrow 273 \mathrm{~K}$

Ethyl alcohol $\rightarrow 158.8 \mathrm{~K}$

Diethyl ether $\rightarrow 156.85 \mathrm{~K}$

Methane $\rightarrow 89.34 \mathrm{~K}$

Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.

Answer

i. 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 $\mathrm{d}$ is surrounded by 6 spheres and is called the octahedral void.

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 $\mathrm{C}$ occupy the tetrahedral voids $\mathbf{c}$.

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 $\mathrm{C}$ is present over the voids $\mathrm{C}$. In figure 4.2, layer B is present over the voids b and layer $\mathrm{C}$ is present over the voids $\mathrm{c}$. It can be observed from the figure that in this arrangement, the spheres present in layer $\mathrm{C}$ are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are arranged in an $A B A B . .$. pattern.

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 $\mathrm{C}$ is completely different from that in layers $\mathrm{A}$ or $\mathrm{B}$.

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.

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.

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.

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:

$a=2 r$

Now, volume of the cubic unit cell $=a^{3}$

$=(2 r)^{3}$

$=8 r^{3}$

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

Therefore, volume of the occupied unit cell $=\frac{4}{3} \pi r^{3}$

Hence, packing efficiency

$$ =\frac{\text { Volume of one particle }}{\text { Volume of cubic unit cell }} \times 100 \% $$

$$ \begin{aligned} & =\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100 \% \\ & =\frac{1}{6} \pi \times 100 \% \\ & =\frac{1}{6} \times \frac{22}{7} \times 100 \% \\ & =52.4 \% \end{aligned} $$

(ii) Body-centred cubic

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

From $\triangle \mathrm{FED}$, we have:

$b^{2}=a^{2}+a^{2}$

$\Rightarrow b^{2}=2 a^{2}$

$\Rightarrow b=\sqrt{2} a$

Again, from $\triangle A F D$, we have:

$c^{2}=a^{2}+b^{2}$

$\Rightarrow c^{2}=a^{2}+2 a^{2} \quad\left(\right.$ Since $\left.b^{2}=2 a^{2}\right)$

$\Rightarrow c^{2}=3 a^{2}$

$\Rightarrow c=\sqrt{3} a$

Let the radius of the atom be $r$.

Length of the body diagonal, $c=4 \pi$

$\Rightarrow \sqrt{3} a=4 r$

$\Rightarrow a=\frac{4 r}{\sqrt{3}}$

$r=\frac{\sqrt{3} a}{4}$

Volume of the cube,

$$ a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3} $$

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice $=2 \pi \frac{4}{3} r^{3}$

$=\frac{8}{3} \pi r^{3}$

$\therefore$ Packing efficiency $=\frac{\text { Volume occupied by two spheres in the unit cell }}{\text { Total volume of the unit cell }} \times 100 \%$

$=\frac{\frac{8}{3} \pi r^{3}}{\left(\frac{4}{\sqrt{3}} r\right)^{3}} \times 100 \%$

$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100 \%$

$=68 \%$

(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 $\triangle \mathrm{ABC}$, we have:

$\mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2}$

$\Rightarrow b^{2}=a^{2}+a^{2}$

$\Rightarrow b^{2}=2 a^{2}$

$\Rightarrow b=\sqrt{2 a}$

Answer

It is given that the edge length, $a=4.077 \times 10^{-8} \mathrm{~cm}$

Density, $d=10.5 \mathrm{~g} \mathrm{~cm}^{-3}$

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

We also know that, $\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$

Using the relation:

$$ \begin{aligned} & d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N_\mathrm{A}}} \\ & \begin{aligned} & \Rightarrow \mathrm{M}=\frac{d a^{3} \mathrm{~N_\mathrm{A}}}{z} \\ &=\frac{10.5 \mathrm{gcm}^{-3} \times\left(4.077 \times 10^{-8} \mathrm{~cm}\right)^{3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}}{4} \\ &=107.13 \mathrm{gmol}^{-1} \end{aligned} \end{aligned} $$

Therefore, atomic mass of silver $=107.13 \mathrm{u}$

Answer

It is given that the atoms of $\mathrm{Q}$ are present at the corners of the cube.

Therefore, number of atoms of $\mathrm{Q}$ in one unit cell

$$ =8 \times \frac{1}{8}=1 $$

It is also given that the atoms of $\mathrm{P}$ are present at the body-centre.

Therefore, number of atoms of $\mathrm{P}$ 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 $\mathrm{PQ}$.

The coordination number of both $\mathrm{P}$ and $\mathrm{Q}$ is 8.

Answer

It is given that the density of niobium, $d=8.55 \mathrm{~g} \mathrm{~cm}^{-3}$

Atomic mass, $\mathrm{M}=93 \mathrm{gmol}^{-1}$

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

We also know that, $\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$

Applying the relation:

$$ \begin{aligned} d= & \frac{z \mathrm{M}}{a^{3} \mathrm{~N_\mathrm{A}}} \\ \Rightarrow a^{3} & =\frac{z \mathrm{M}}{d \mathrm{~N_\mathrm{A}}} \\ & =\frac{2 \times 93 \mathrm{gmol}^{-1}}{8.55 \mathrm{gcm}^{-3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}} \end{aligned} $$

$=3.612 \times 10^{-23} \mathrm{~cm}^{3}$

So, $a=3.306 \times 10^{-8} \mathrm{~cm}$

For body-centred cubic unit cell:

$ \begin{aligned} r & =\frac{\sqrt{3}}{4} a \\ & =\frac{\sqrt{3}}{4} \times 3.306 \times 10^{-8} \mathrm{~cm} \end{aligned} $

$=1.432 \times 10^{-8} \mathrm{~cm}$

$=14.32 \times 10^{-9} \mathrm{~cm}$

$=14.32 \mathrm{~nm}$

Answer

A sphere with centre $\mathrm{O}$, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that $\triangle \mathrm{POQ}$ is right-angled

$\angle \mathrm{POQ}=90^{\circ}$

Now, applying Pythagoras theorem, we can write:

$ \begin{aligned} & \mathrm{PQ}^{2}=\mathrm{PO}^{2}+\mathrm{OQ}^{2} \\ & \Rightarrow(2 \mathrm{R})^{2}=(\mathrm{R}+\mathrm{r})^{2}+(\mathrm{R}+\mathrm{r})^{2} \\ & \Rightarrow(2 \mathrm{R})^{2}=2(\mathrm{R}+\mathrm{r})^{2} \\ & \Rightarrow 2 \mathrm{R}^{2}=(\mathrm{R}+\mathrm{r})^{2} \\ & \Rightarrow \sqrt{2} \mathrm{R}=\mathrm{R}+\mathrm{r} \\ & \Rightarrow \mathrm{r}=\sqrt{2} \mathrm{R}-\mathrm{R} \\ & \Rightarrow \mathrm{r}=(\sqrt{2}-1) \mathrm{R} \\ & \Rightarrow \mathrm{r}=0.414 \mathrm{R} \end{aligned} $

Answer

Edge length, $a=3.61 \times 10^{-8} \mathrm{~cm}$

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

Atomic mass, $\mathrm{M}=63.5 \mathrm{~g} \mathrm{~mol}^{-1}$

We also know that, $\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$

Applying the relation:

$ \begin{aligned} d & =\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}} \\ & =\frac{4 \times 63.5 \mathrm{~g} \mathrm{~mol}^{-1}}{\left(3.61 \times 10^{-8} \mathrm{~cm}\right)^{3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}} \end{aligned} $

$=8.97 \mathrm{~g} \mathrm{~cm}^{-3}$

The measured value of density is given as $8.92 \mathrm{~g} \mathrm{~cm}^{-3}$. Hence, the calculated density $8.97 \mathrm{~g} \mathrm{~cm}^{-3}$ is in agreement with its measured value.

Answer

The formula of nickel oxide is $\mathrm{Ni_{0.98}} \mathrm{O_{1.00}}$.

Therefore, the ratio of the number of $\mathrm{Ni}$ atoms to the number of $\mathrm{O}$ atoms,

$\mathrm{Ni}: \mathrm{O}=0.98: 1.00=98: 100$

Now, total charge on $100 \mathrm{O}^{2}$-ions $=100 \times(-2)$

$=-200$

Let the number of $\mathrm{Ni}^{2+}$ ions be $x$.

So, the number of $\mathrm{Ni}^{3+}$ ions is $98-x$.

Now, total charge on $\mathrm{Ni}^{2+}$ ions $=x(+2)$

$=+2 x$

And, total charge on $\mathrm{Ni}^{3+}$ ions $=(98-x)(+3)$

$=294-3 x$

Since, the compound is neutral, we can write:

$2 x+(294-3 x)+(-200)=0$

$\Rightarrow-x+94=0$

$\Rightarrow x=94$

Therefore, number of $\mathrm{Ni}^{2+}$ ions $=94$

And, number of $\mathrm{Ni}^{3+}$ ions $=98-94=4$

Hence, fraction of nickel that exists as $\mathrm{Ni}^{2+}=\frac{94}{98}$

$=0.959$

And, fraction of nickel that exists as $\mathrm{Ni}^{3+}=\frac{4}{98}$ $=0.041$

Alternatively, fraction of nickel that exists as $\mathrm{Ni}^{3+}=1-0.959$

$=0.041$

Answer

Semiconductors are substances having conductance in the intermediate range of $10^{-6} \mathrm{to} 10^{4} \mathrm{ohm}^{-1} \mathrm{~m}^{-1}$.

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 $\mathrm{Si}$ or $\mathrm{Ge}$ is doped with a group 15 element such as $\mathrm{P}$ or As, an $n$-type semiconductor is generated.

$\mathrm{Si}$ and $\mathrm{Ge}$ have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, $\mathrm{P}$ 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 $\mathrm{Si}$ or $\mathrm{Ge}$ atoms. The remaining fifth electron becomes delocalised and increases the conductivity of the doped Si or Ge.

Perfect crystal

$n$ - type

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 $\mathrm{Si}$ or $\mathrm{Ge}$ is doped with a group 13 element such as $\mathrm{B}, \mathrm{Al}$, or $\mathrm{Ga}$ (which contains only three valence electrons), a p-type of semiconductor is generated.

When a crystal of $\mathrm{Si}$ is doped with $\mathrm{B}$, the three electrons of $\mathrm{B}$ 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.

Perfect crystal

$p$ - type

Answer

In the cuprous oxide $\left(\mathrm{Cu}_{2} \mathrm{O}\right)$ 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 $\mathrm{O}^{2}$ ions. This is because some Cu+ions have been replaced by $\mathrm{Cu}^{2+}$ ions. Every $\mathrm{Cu}^{2+}$ ion replaces two $\mathrm{Cu}^{+}$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.

Answer

Let the number of oxide $\left(\mathrm{O}^{2 \cdot}\right)$ 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 $\left(\mathrm{Fe}^{3+}\right)$ ions $=\frac{2}{3} x$

Therefore, ratio of the number of $\mathrm{Fe}^{3+}$ ions to the number of $\mathrm{O}^{2}$ ions,

$\mathrm{Fe}^{3+}: \mathrm{O}^{2-}=\frac{2}{3} x: x$

$=\frac{2}{3}: 1$

$=2: 3$

Hence, the formula of the ferric oxide is $\mathrm{Fe} _{2} \mathrm{O} _{3}$.

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.