[ 5 Marks Questions ]

Question 15 Marks

Explain briefly seven types of unit cells.

Answer

Seven types of unit cells:

Cubic
$a=b=c$
$\alpha=\beta=\gamma=90^{\circ}$

Rhombohedral
$a=b=c$
$\alpha=\beta=\gamma \neq 90^{\circ}$

Hexagonal
$a=b \neq c$
$\alpha=\beta=90^{\circ}, \gamma=120^{\circ}$

Tetragonal
$a=b \neq c$
$\alpha=\beta=\gamma=90^{\circ}$

Orthorhombic
$a \neq b \neq c$
$\alpha=\beta=\gamma=90^{\circ}$

Monoclinic
$a \neq b \neq c$
$\alpha=\gamma=90^{\circ}, \beta \neq 90^{\circ}$

Triclinic
$a \neq b \neq c$
$\alpha \neq \beta \neq \gamma \neq 90^{\circ}$
$\text { 1. } \text { Cubic }-NaCl$
$\text { 2. Rhombohedral - Cinnabar Cubic }$
$\text { 3. Hexagonal }-ZnO$
$\text { 4. Tetragonal }-TiO_2$
$\text { 5. Orthorhombic }-BaSO_4$
$\text { 6. Monoclinic }-PbCrO_4$
$\text { 7. Triclinic }-H_3 BO_3$
They differ in the arrangements of their crystallographic axes and angles.
Corresponding to the above seven, Bravis defined 14 possible crystal systems as shown in the figure.

View full question & answer→

Question 25 Marks

Calculate the percentage efficiency of packing in the case of body centered cubic crystal.

Answer

In body centered cubic arrangement the spheres are touching along the leading diagonal of the cube as shown in the figure.

In $\triangle ABC$
$
\begin{aligned}
& A C^2=A B^2+B C^2 \\
& A C=\sqrt{A B^2+B C^2} \\
& A C=\sqrt{a^2+a^2} \\
& =\sqrt{2 a^2} \\
& =\sqrt{2} a
\end{aligned}
$
In $\triangle A C G$
$
A G^2=A C^2+C G^2
$
$
\begin{aligned}
& A G=\sqrt{A^2+C^2} \\
& A G=\sqrt{(\sqrt{2} a)^2+a^2} \\
& A G=\sqrt{2 a^2+a^2} \\
& =\sqrt{3 a^2} \\
& =\sqrt{3} a
\end{aligned}
$
i.e., $\sqrt{3} a=4 r$
$
r=\frac{\sqrt{3}}{4} a
$
∴ Volume of the sphere with radius ‘r’
$\begin{aligned} & =\frac{4}{3} \pi r ^3 \\ & =\frac{4}{3} \pi\left(\frac{\sqrt{3}}{4} a \right)^3 \\ & =\frac{\sqrt{3}}{16} \pi a ^3\end{aligned}$
Number of spheres belong to a unit cell in bcc arrangement is equal to two and hence the total volume of all spheres
$
\begin{aligned}
& =2 \times\left(\frac{\sqrt{3} \pi a ^3}{16}\right) \\
& =\frac{\sqrt{3} \pi a ^3}{8}
\end{aligned}
$
Packing fraction $=\frac{\text { Total volume occupied by spheres in a unit cell }}{\text { Volume of the unit cell }} \times 100$
Packing fraction $=\frac{\left(\frac{\sqrt{3} \pi a ^3}{8}\right)}{\left( a ^3\right)} \times 100$
$
\begin{aligned}
& =\frac{\sqrt{3} \pi}{8} \times 100 \\
& =\sqrt{3} \pi \times 12.5 \\
& =1.732 \times 3.14 \times 12.5 \\
& =68 \%
\end{aligned}
$
i.e., 68 % of the available volume is occupied. The available space is used more efficiently than in simple cubic packing.

View full question & answer→

Question 35 Marks

Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.

Answer

1. AAAA type of three-dimensional packing: This type of three-dimensional packing arrangement can be obtained by repeating the AAAA type two-dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical.
All spheres of different layers of crystal are perfectly aligned horizontally and also vertically so that any unit cell of such arrangement as simple cubic structure as shown in fig.

Simple Cubic (SC)
In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in a simple cubic arrangement is 6.
2. ABABA type of three-dimensional packing: In this arrangement, the spheres in the first layer (A-type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in the figure.
The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

aba arrangement - hcp structure
3. ABCABC type of three-dimensional packing: In this arrangement (FCC) second layer spheres are arranged at the dips of the first layer. Third layer spheres are arranged in a manner such that they cover the octahedral void. Then no longer the third layer is similar to the first or second layer.
Third layer gives a different arrangement. Fourth layer spheres are similar to the first layer. If the first, second and third layers are represented as A, B, C then this type of packing gives the arrangement of layers as ABCABC… (i.e.,), the first three layers do not resemble the first, second and third layers respectively and the sequence is repeated.
with the addition of more layers. In this arrangement, atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12. Voids – The empty spaces between the three-dimensional layers are known as voids. There are two types of common voids possible. They are tetrahedral and octahedral voids.
Tetrahedral void – A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom layer is a hole between four spheres. The spheres are arranged at the vertices of a regular tetrahedron such as a hole or void is called a tetrahedral void.

abc arrangement – ccp structure
Octahedral void: A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

View full question & answer→

Question 45 Marks

Write short note on metal excess and metal deficiency defect with an example.

Answer

Metal excess defect arises due to the presence of more metal ions as compared to anions. Alkali metal halides NaCl, KCl show this type of defect. The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electrons present in the interstitial position.

Metal Excess Defect
For example, when NaCl crystals are heated in the presence of sodium vapour, $Na ^{+}$ions are formed and are deposited on the surface of the crystal. Chloride ions $\left( Cl ^{-}\right)$diffuse to the surface from the lattice point and combines with $Na ^{+}$ions.
The electron lost by the sodium vapour diffuses into the crystal lattice and occupies the vacancy created by the $Cl ^{-}$ions. Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl which contains excess $Na ^{+}$ions can be written as $Na _{1+x} Cl$.
Metal deficiency defect: Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states. For example, in FeO crystal, some of the $Fe ^{2+}$ ions are missing from the crystal lattice.
To maintain electrical neutrality, twice the number of other $Fe ^{2+}$ ions in the crystal is oxidized to $Fe ^{3+}$ ions. In such cases, the overall number of $Fe ^{2+}$ and $Fe ^{3+}$ ions is less than the $O ^{2-}$ ions. It was experimentally found that the general formula of ferrous oxide is $Fe _{ x } O$, where x ranges from 0.93 to 0.98 .

Metal Deficiency Defect

View full question & answer→

Question 55 Marks

An element crystallizes in a fcc lattice with cell edge of 400 pm . The density of the element is $7 g / cm ^3$. How many atoms one present in 280 g of the element?

Answer

$ \text { Volume of unit cell } a ^3 \text { ( } a =\text { edge length) }$
$=400 pm$
$=\left(400 \times 10^{-12} m \right)^3$
$=\left(400 \times 100^{-10} cm \right)^3=64 \times 10^{-24} cm ^3$
$\text { Volume of } 208 g \text { of the element }=\frac{208}{7 g / cm ^3}=29.71 cm ^3$
$\text { Number of unit cells in this volume }=\frac{\text { vol.of given amount }}{\text { vol of one unit cell }}$
$\qquad=\frac{29.71}{64 \times 10^{-24}}=0.46 \times 10^{24}$
$\text { Since each f.c.c. unit cell contains } 4 \text { atoms therefore, }$
$\text { Total number }=4 \times 0.46 \times 1024,$
$=1.84 \times 1024 \text { atoms. } $
Common Errors

sc, bcc, fee structures may get confused if they ask in different order.
Calculation of atoms at each cube may get confused.
Sharing of atoms may get con¬fused.

Rectifications

In sc, only simple cube can be drawn, bcc structure is with one dot at centre, fee structure is with sc with six dots at each face.
sc – only one atom, bcc – two atoms, fee – four atoms (we can remember as f and f).
Atom at comer is shard by 8 unit cells. Atom at centre is not shared. Atom at face is shared by 2 unit cells. Atom at edge is shared by 4 unit cells.

View full question & answer→

Question 65 Marks

What are molecular solids?Explain their classification with suitable examples.

Answer

In molecular solids, the constituents are neutral molecules. They are held together by weak vander Waals forces. Generally molecular solids are soft and they do not conduct electricity. These molecular solids are further classified into three types.

Non polar molecular solids.
Polar molecular solids.
Hydrogen bonded molecular solids.

1. Non polar molecular solids

In this type, molecules are held together by weak dispersion forces or London forces.
They have low melting points and are usually in liquids or gaseous state at room temperature. Examples – Naphthalene, Anthracene etc.

2. Polar molecular solids

In this type, molecules formed by polar covalent bonds.
They are held together by strong dipole-dipole interactions.
They have higher melting points than the non – polar molecular solids. Examples., solid $CO _2$, solid $NH_3$

3. Hydrogen bonded molecular solids

The constituents are held together by hydrogen bonds.
The constituents are held together by hydrogen bonds.
They are soft solids under room temperature Examples: Solid ice, glucose, urea.

View full question & answer→

Question 75 Marks

What are ionic solids? Give their characteristics.

Answer

The structural units of an ionic crystals are cations and anions. They are bound together by strong Na electrostatic attractive forces.
To maximize the attractive force, cations are surrounded by as many anions as possible and vice versa.
Ionic crystals possess definite crystals structure.
Many solids are cubic close packed.

Characteristics

Ionic solids have high melting points.
These solids do not conduct electricity, because the ions are fixed in their lattice positions.
They do not conduct electricity in molten state (or) when dissolved in water, because the ions are free to move in the molten state or solution.
They are hard as only strong external force can change the relative positions of ions.
Example – The arrangement of Na and Cl ions in NaCI crystal.

View full question & answer→

Question 85 Marks

Write a note about classification of solids with suitable examples.

View full question & answer→

Question 95 Marks

What are general characteristics of solids?

Answer

Solids have definite volume and shape.
Solids are rigid and incompressible.
Solids have strong cohesive forces.
Solids have short interatomic, ionic (or) molecular distances.
Their constituents (atoms, ions or molecules) have fixed positions and can only oscillate about their mean positions.
Unlike gases, in solids, the atoms, ions (or) molecules are held together by strong force of attraction.

View full question & answer→

Chemistry [ 5 Marks Questions ]

Test yourself on this topic