Class 12 Engineering Chemistry - The Solid State

(i) Write the type of magnetism observed when the magnetic moment are oppositely aligned and cancel out each other?
(ii) Which stoichiometric defect does not change the density of the crystal?

(i) Anti-ferromagnetism is observed when the magnetic moments are oppositely aligned and cancel out each other.
(ii) The Frenkel defect does not change the density of the crystal.

(i) Write the type of magnetism observed when the magnetic moment are aligned in parallel and anti-parallel directions in unequal numbers?
(ii) Which stoichiometric defect decreases the density of the crystal?

(i) Ferrimagnetism: This effect is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti–parallel directions in unequal numbers.

(ii) Vacancy Defects: When some lattice sites left vacant while the formation of crystal, the defect is called vacancy defects. In vacancy defects, an atom is missing from its regular atomic site. Because of missing of atom the density of substance decreases, i.e., because of vacancy defects.

When 9.65 C of electricity is passed through a solution of silver nitrate (atomic mass of Ag =108 g $$mol^{-1}$$) , the amount of silver deposited is

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A unit cell of iron crystal has edge length $$288$$ pm and density$$7.86 g cm^{-3}$$. Find the number of atoms per unit cell and type of the crystal lattice.
Given: Molar mass of iron = $$56 g mol^{-1}$$, Avogadro's number $$N_A= 6.022\times 10^{23}mol^{-1}$$

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What is the crystallograph dimension of monoclinic unit cell?

In crystallography, the mono-clinic crystal system is one of the 7 crystal system. A crystal system is described by three vectors. In mono-clinic system, the crystal is described by vectors of unequal lengths, as in the orthorhombic system. They form a rectangular prism with a parallelogram as its base. Hence two vectors are perpendicular while the third vector meets the other two at an angle other than 90 degrees.

Lattice paths are paths consisting of one unit steps in the positive horizontal or positive vertical directions. Let distinct lattice paths from the point $$(-1,0)$$ to the point $$(3, 5)$$, if number of shortest path are $$\lambda$$. Let $$\mu$$ be the sum of all digits of $$\lambda$$. Then find the value of $$\mu$$.

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Zinc sulphide crystallises with zinc ions occupying half of the tetrahedral holes in a closed packed array of sulphide ions. What is the formula of zinc sulphide?

The zinc sulphide crystals are composed of equal number of $${Zn}^{+2}$$ and $$S^{2-}$$ ions.

Zinc ions occupy tetrahedral hole. Only half of the tetrahedral holes are occupied by $${Zn}^{+2}$$ so that the formula of the zinc sulphide is $$ZnS$$ i.e. the stoichiometry of the compound is 1:1 (Only alternate tetrahedral holes are occupied by $${Zn}^{+2}$$)

When atoms are placed at all the 8 corners of the cube, how many atoms are present per unit cell?

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In $$CaF_2$$ unit cell the co ordination number of $${ F }^{ - }$$ ion is?

Simple ionic structures are described by two coordination numbers, one for each type of ion.

Calcium fluoride $$(CaF_2)$$ is an (8, 4) structure,

each cation $$Ca^{2+} $$is surrounded by eight $$F^−$$ anion neighbors,

each anion $$F^−$$ by four $$Ca^{2+}$$.

So coordination number of $$Ca^{+2}=8$$ and $$F^−=4$$ ion in $$CaF_2$$ crystal

The density of iron crystal is 8.54 g $$cm^{-3}$$. If the edge length of unit cell is 2.8 Å and atomic mass is 56 g $$mol^{-1}$$.
Find the no. of atoms in the unit cell.
(given: Avagadro's number = $$6.022\times 10^{23}, 1 A=1\times 10^{-8}cm)$$

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The density of $$KBr$$ is $$2.75g\ cm^{-3}$$. The length of edge of unit cell is $$6.54\ $$Å$$ $$. Predict the type of cubic lattice to which unit cell of $$KBr$$ belongs. (Atomic mass $$K=39, Br=80$$)

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Calculate no. of unit cell present $$58.5\ g$$ of $$NaCl$$.

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Define the term (a) unit cell (b) space lattice?

a) A unit cell is the smallest building block of crystal structure, repetition of the unit cell forms a solid crystal. Example- Cubic unit cell, Hexagonal unit cell etc.
b) A space lattice is an array of points showing how particles (atoms, ions or molecules) are arranged at different sites in three dimensional spaces.

Match the Bravais lattices with their crystal systems.

Orthorhombic structure has 4 bravais lattices namely primitive, body centred, face centred and end centred.
Cubic structure has 3 bravais lattices namely primitive, body centred and face centred.
Hexagonal structure has only primitive bravais lattice.
Tetragonal structure has 2 bravais lattices namely primitive and body centred.

Out of seven possible crystal systems, the number of crystal systems having more than one Bravais lattice is ______.

The following are the crystal structures with more than one bravais lattice.

1. Cubic - 3
2. Tetragonal - 2
3. Rhombic - 4
4. Monoclinic - 2

In cubic system, how many arrangement of atoms exist in nature?

The type of lattice represented by this compound is simple cubic.

$$X\; atoms\; located\; at\; the\; corners\; \\ no.\; of\; corners\; =\; 8\\ Contribution\; of\; 1\; corners\; =\; \cfrac { 1 }{ 8 } \; atoms\\ no.\; of\; atoms\; of\; X\; =\; \cfrac { 1 }{ 8 } \times 8\; =\; 1\; atoms\\ 0\; atoms\; are\; located\; at\; the\; edge\; \\ no.\; of\; edges\; =\; 12\\ contribution\; of\; 1\; edge\; =\; \cfrac { 1 }{ 4 } atoms\; \\ no.\; of\; atoms\; of\; 0\; =\; \cfrac { 1 }{ 4 } \times 12\; =\; 3\; atoms\; \\ Na\; atoms\; are\; located\; at\; the\; centers\\ no.\; of\; center\; =\; 1\\ contribution\; of\; 1\; center\; =\; 1\; \\ no.\; of\; atoms\; of\; Na\; =\; 1\times 1\; =\; 1atom\\ Hence\; the\; formating\; Na\times { O }_{ 3 }\\ Hence\; it\; is\; a\; peroustuite\; crystal\; structure\; (eg.\; NaW{ O }_{ 3 })$$

Find the number of hexagonal faces that are present in a truncated octahedral.

If we chip all the corners, we get to get a truncated octahedral. Each chipped face will be a hexagon.
Thus, there are $$8$$ hexagonal faces that are present in a truncated octahedral.

A cubic structure of uniform spheres has the unit cell edge length of 0.8 mm. If the radius of the molecule in a simple cubic lattice is $$40\times10^{-x}$$, then what is the value of x?

As we know, for simple cubic lattice,

$$\displaystyle r=\dfrac{a}{2}=\dfrac{0.8}{2}=0.4\ mm$$

$$r= 40 \times 10^{-2}$$

By comparing, $$r= 40 \times 10^{-2}=40 \times 10^{-x}$$
we get, $$x=2$$
Hence, the correct answer is 2.

Calculate the efficiency of packing in case of a metal crystal for the following crystal structure (with the assumptions that atoms are touching each other):

(i) simple cubic

(ii) body-centered cubic

(iii) face-centered cubic

(i) The efficiency of packing in case of simple cubic unit cell is given below:
A simple cubic unit cell contains one atom per unit cell.
Also, $$a= 2r$$, where $$a$$ is the edge length and $$r$$ is the radius of atom.
Total volume of unit cell $$=\displaystyle a^3$$.
Packing efficiency $$ \displaystyle = \frac {\text {Volume of one sphere}}{\text {Total volume of unit cell}} \times 100 $$

Packing efficiency $$ \displaystyle = \frac {\displaystyle \frac {4}{3}\pi r^3}{8r^3} \times 100 = 52.4 \%$$

(ii) The efficiency of packing in case of body-centred cubic unit cell is given below:
A body-centred cubic unit cell contains two atoms per unit cell.
Also, $$\sqrt {3} a= 4r$$, where $$a$$ is the edge length and $$r$$ is the radius of atom.
Total volume of unit cell $$=\displaystyle a^3$$.
Packing efficiency $$ \displaystyle = \dfrac {\text {Volume of two spheres}}{\text {Total volume of unit cell}} \times 100 $$

Packing efficiency $$ \displaystyle = \dfrac {\displaystyle \frac {8}{3}\pi r^3}{\displaystyle \frac {64}{3\sqrt{3}}\times r^3} \times 100 = 68 \%$$

(iii) The efficiency of packing in case of face-centered cubic unit cell (with the assumptions that atoms are touching each other) is given below:
A face-centered cubic unit cell contains four atoms per unit cell.
Also, $$a= 2\sqrt {2} r$$, where $$a$$ is the edge length and $$r$$ is the radius of atom.
Total volume of unit cell $$=\displaystyle a^3$$

Packing efficiency $$ \displaystyle = \frac {\text {Volume of four spheres}}{\text {Total volume of unit cell}} \times 100 $$

Packing efficiency $$ \displaystyle = \frac {\displaystyle \frac {16}{3}\pi r^3}{16\sqrt {2}r^3} \times 100 = 74 \%$$

Copper crystallises into a fcc lattice with edge length $$3.61 \times 10^{-8} cm$$. Show that the calculated density is in agreement with its measured value of $$8.92\ g\ cm^{3}$$.

Copper crystallizes into a fcc lattice with edge length $$3.61×10^{−8}$$ cm.

Fcc unit cell has $$4$$ atoms per unit cell.

Cu has atomic mass of $$63.5$$ g/mol.

Avogadro's number is $$6.023 \times 10^{23}$$.

The calculated density is as below:

$$\displaystyle D = \frac {ZM}{a^3N_0} = \frac {4 \times 63.5}{(3.61 \times 10^{-8})^3 \times 6.022 \times 10^{23}} = 8.97 \: g\ cm^{-3}$$

This is close enough to the measured value of $$8.92\ g\ cm^{-3}$$.

Explain the following terms with suitable examples:
(i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres

(i) Schottky defect
Some atoms or ions are missing from their normal lattice sites. Unoccupied lattice sites are called vacancies or holes.
For example, in the crystal lattice of NaCl, equal number of cations and anions are missing to maintain electrical neutrality.
(ii) Frenkel defect
An ion is missing from its normal position and occupies an interstitial site.
Due to this, vacancy defect and interstitial defect are produced.
For example, in the crystal lattice of ZnS, a zinc cation is missing from its normal position and occupies an interstitial site.
(iii) Interstitials
They are sites between lattice points.
For example, in Frenkel defect, the occupancy of interstitials cause interstitial defect.
(iv) F-centres
The electrons trapped in anion vacancies are called F-centres. They impart colour to crystal.
For example, in NaCl crystal, F-centres impart yellow colour.

Define the term 'amorphous'. Give a few examples of amorphous solids.

$$Amorphous$$ $$solids$$ have short-range order with irregular shapes of constituent particles. They have isotropic nature and melt over a range of temperatures. They do not have a definite enthalpy of fusion.

Examples of amorphous solids are glass, rubber, plastic, etc.

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

$$Quartz$$ is crystalline solid with long range order and glass is amorphous solid (or psuedo solid or super cooled liquid) with short range order and has a tendency to flow. When quartz is heated, it can be converted into glass.

(i) What type of non-stoichiometric point defect is responsible for the pink colour of $$LiCl$$?
(ii) What type of stoichiometric defect is shown by $$NaCl$$ ?

(i) The metal excess defect caused by anionic vacancies is responsible for the pink colour of $$ LiCl$$.
(ii) The Schottky defect is found in $$NaCl$$. In this defect, an equal number of cations and anions are missing from their regular sites.

(i) What type of non - stoichiometric point defect is responsible for the pink color of $$LiCl$$?
(ii) What type of stoichiometric defect is shown by $$NaCl$$?
OR
How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell

$$\rightarrow \; Due\; to\; metal\; excess\; defect,\; anion\; IC\; vacancies\; are\; formed\; in\; \\ \; \; LiCl\; which\; are\; responsible\; for\; its\; pink\; colour.\; \\ \rightarrow \; NaCl\; shows\; schottly\; defect\; (stoichiometric\; defect\; )\; \\ \; \; in\; this,\; equal\; no.\; of\; cations\; and\; anions\; are\; missing\; from\\ \; \; there\; lattice\; \\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; OR\\ \Rightarrow \; Tetrahedral\; \\ \; \; (a)\; it\; is\; a\; void\; which\; is\; surrounded\; by\; 4\; spheres\; .i.e\; \; \\ \; coordination\; no.\; \; =\; 4\; \\ \; \; (b)\; they\; are\; observed\; on\; edges\; of\; the\; unit\; cell.\\ $$

$$\Rightarrow \; Octahedral\\ \; \; (a)\; it\; is\; a\; void\; which\; is\; surrounded\; by\; 6\; spheres\; i.e.\; \\ \; \; \; \; \; coordination\; no.\; =6\; \\ \; \; (b)\; they\; are\; observed\; in\; the\; center\; of\; the\; unit\; cell.\\ $$

$$\Rightarrow \; Physical\; Lattice\; \\ \; \; (a)\; it\; is\; a\; arrangement\; of\; point\; in\; sphere\; (3-g)\\ \; \; (b)\; lattice\; are\; complex\; and\; large\; \\ $$

$$\Rightarrow \; Unit\; cell\; \\ \; \; (a)\; it\; is\; a\; smallest\; part\; of\; a\; crystal\; lattice\; which\; when\; \\ repeated\; in\; different\; direction\; form\; the\; entire\; crystal\; lattice\\ \; \; (b)\; unit\; cell\; are\; simple\; and\; smaller.$$

What type of point defect is produced when $$AgCl$$ is doped with $$CdCl_{2}$$?

Explanation:

In $$AgCl$$, the cation is $$Ag^+$$ ion. When $$AgCl$$ is doped with $$CdCl_2$$, cation vacancy is created because two $$Ag^+$$ ions are replaced by one $$Cd^{2+}$$ ion. The $$Cd^{+2}$$ ion occupies the site of one of the $$Ag^+$$ ion and at another site of $$Ag^+$$ ion a vacancy is created. This defect is known cation vacancy defect.

Final Answer: The defect created is called cation vacancy defect.

Examine the given defective crystal.
$$\begin{matrix} { A }^{ + } & { B }^{ - } & { A }^{ + } & { B }^{ - } & { A }^{ + } \\ { B }^{ - } & O & { B }^{ - } & { A }^{ + } & { B }^{ - } \\ { A }^{ + } & { B }^{ - } & { A }^{ + } & O & { A }^{ + } \\ { B }^{ - } & { A }^{ + } & { B }^{ - } & { A }^{ + } & { B }^{ - } \end{matrix}$$
Answer the following questions:
(i) What type of stoichiometric defect is shown by crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect?

(i) Schottky defect is shown by the mentioned crystal, as equal number of cations and anions are missing in the crystal lattice.

(ii) This defect leads to decrease in density, as equal number of the cations and anions are missing from the crystal lattice.

(iii) This kind of defect is shown by those ionic substance in which the cations and anions are of almost similar sizes. Examples are $$KCl, NaCl, CsCl$$.

Examine the given defective crystal.

$$\displaystyle \begin{matrix}X^{+} & Y^{-} & X^{+} & Y^{-} &X^{+} \\ Y^{-} & O &Y^{-} & X^{+} & Y^{-}\\ X^{+}& Y^{-} & X^{+} & O & X^{+} \\ Y^{-}& X^{+} & Y^{-} & X^{+} & Y^{-}\end{matrix}$$

Answer the following questions :

(i) Is the above defect stoichiometric or non-stoichiometric?
(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect?
(iii) How does this defect affect the density of the crystal?

(i) Stoichiometric

(ii) Schottky defects, e.g, $$NaCl,CsCl$$

(iii) The presence of a large number of schottky defects in a crystal will lower its density markedly.

An element crystallizes in a f.c.c. lattice with a cell edge of 250 pm. Calculate the density, if 300 g of this element contain $$2 \times {10}^{24}$$ atoms.

Formula of density of the unit cell:

$$d = \dfrac {zM}{a^3N_A}$$

where,

$$z =$$ number of particles present per unit cell $$= 4$$

$$M =$$ Molecular mass $$= \dfrac {300}{2\times 10^{24}}\times 6.023 \times 10^{23} = 90.34 g$$ ,

$$N_A =$$ Avogadro’s number $$=6.023 \times 10^{23}$$

$$a =$$ Edge length $$= 250\ pm = 2.5 \times 10^{-8} cm$$.

Now, substituting the values we get,

$$d= \dfrac {4 \times 90.34}{(2.5)^3\times 10^{-24} \times 6.023 \times 10^{23}} =38.4\ g/cm^3$$

For a crystal of diamond, state the hybridization of the carbon atom.

Diamond is an allotrope of carbon in which carbon atom is present in $$sp^3$$ hybridization state.

Distinguish between crystalline solid and amorphous solid?

Crystalline solids	Amorphous solids
They have definite characteristic geometrical shape.	They have irregular shape.
They have long range order of regular pattern of arrangement of constituent particles.	They have short range order of regular pattern of arrangement of constituent particles.
They are true solids.	They are pseudo solids or super cooled liquids.
They have sharp melting points.	They do not have sharp melting points.
They are anisotropic in nature.	They are isotropic in nature.
They have definite heat of fusion.	They do not have definite heat of fusion.

Explain impurity defect in stainless steel with diagram.

The impurity defect found in stainless steel is interstitial impurity defect. In interstitial impurity defect, the impurity of cation is present in the interstitional position and make crystal defected. Stainless steel is an alloy of Iron and $$4\%$$ Chromium mixed with it. Stainless steel typically contains about $$1\%$$ Carbon, $$1-5\%$$ Manganese, $$0.05\%$$ Phosphorous, $$1-3\%$$ Silicon, $$5\% - 10\%$$ Nickel and $$15\% - 20\%$$ Chromium. Carbon is a second-period element that is non-metallic and much smaller tha iron. Carbon will therefore tend to occupy interstitial sites in the iron lattice.

For a crystal of diamond, state the type of lattice in which it crystallizes.

Diamond is a covalent solid with $$sp^3$$ hybridised C atoms and forms giant network type with fcc arrangement of C atoms. Each C atom has tetrahedral geometry.

On the basis of nature of ionic solids compare Frenkel defect with Schottky defect :

Frenkel defect	Schottky defect
Cation moves to an interstitial site.	Equal number of cationic and anionic vacancies are present.
It is observed in cations and anions having different size.	It is observed in cations and anions having similar size.
The density of crystal is not affected.	The density of crystal decreases.
It is observed in ZnS, AgBr, AgI, AgCl etc.	It is observed in alkali halides (NaCl, CsCl) etc.

Name the crystal structure of the copper metal.

The crystal structure of copper metal is Face Centered Cubic (fcc).

Calculate the packing efficiency in a Body Centered Cubic (BCC) lattice?

For Body Centered Cubic (BCC) lattice, the relationship between the edge length a and the radius r of the unit cell is $$\displaystyle a = \dfrac {4r}{\sqrt{3}}$$
The volume of the unit cell is $$\displaystyle a^3=(\dfrac {4r}{\sqrt{3}})^3=\dfrac {64r^3}{3\sqrt{3}}$$
The volume occupied by 1 atom is $$\displaystyle \dfrac {4}{3}\pi r^3$$
A BCC unit cell has 2 atoms per unit cell.
The volume occupied by 2 atoms is
$$\displaystyle 2 \times \dfrac {4}{3}\pi r^3= \dfrac {8}{3}\pi r^3$$
The packing efficiency $$\displaystyle = \dfrac {\text {Volume occupied by atoms in a unit cell }}{\text {total volume of unit cell}} \times 100 $$
The packing efficiency $$\displaystyle = \dfrac { \dfrac {8}{3}\pi r^3}{\dfrac {64r^3}{3\sqrt{3}}} \times 100=68.04$$ %

(a) Explain Schottky defect and Frenkel defect.
(b) Write briefly the adsorption theory of catalysis.

(a) Schottky Defect: This defect is caused if some of the lattice points are unoccupied. The points which are unoccupied. The points which are unoccupied are called lattice vacancies. The number of missing positive and negative ions is the same in this case and the crystal remains neutral. The existence of two vacancies, one due to a missing $${Na}^{+}$$ ion and the other due to a missing $${Cl}^{-}$$ ion in a crystal of $$NaCl$$. It appears in ionic crystals in which positive and negative ions do not differ much in size.
Frenkel Defects: This defect arise when an ion occupies an interstitial position between the lattice points. This defect occurs generally in ionic crystals in which the anion is much larger in size than the cation.
$$AgBr$$ is an example for this type of defect. One of the $${Ag}^{+}$$ ion occupies a position in the interstitial space rather than its own appropriate site in the lattice.
The crystal remains neutral since the number of positive ions is the same as the number of negative ions.
(b) Adsorption theory explains the mechanism of heterogenous catalysts. Here, the catalyst functions by adsorption of the reacting molecules on its surface.
In general there are four steps involved in the heterogenous catalysis.
$${ A }_{ (g) }+{ B }_{ (g) }\xrightarrow [ ]{ catalyst } { C }_{ (g) }+{ D }_{ (g) }\quad $$
Step 1: Adsorption of reactant molecules:
The reactant molecules $$A$$ and $$B$$ strike the surface of the catalyst. They are held up at the surface by weak Vanderwall's forces by partial chemical bonds.
Step 2: Formation of activated complex:
The particles of the reactants adjacent to one another join to form an intermediate complex $$(A-B)$$ . Tee activated complex is unstable.
Step 3: Decomposition of Activated Complex:
The activated complex breaks to form the products $$C$$ and $$D$$. The separated particles of the products hold to the catalyst surface by partial chemical bonds.
Step 4: Desorption of products:
The particles of the products are desorbed or released from the surface.

Sliver forms a ccp lattice. The edge length of its unit cell is 408.6 pm. Calculate the density of sliver?$$(N_A=6.022\times 10^{23}$$, Atomic mass of $$Ag=108 \,gmol^{-1})$$

The edge length of its unit cell is 408.6 pm or
$$\displaystyle 408.6 \times 10^{-10} $$ cm
The volume of the unit cell $$\displaystyle = a^3 =

(408.6 \times 10^{-10} \: cm)^3 = 68.27 \times 10^

{-24} \: cm^3$$
ccp unit cell has 4 atoms per unit cell.
Mass of one $$\displaystyle Ag$$ atom
$$\displaystyle = \dfrac {\text {molar mass of Ag}}{\text {Avogadro's number}}= \dfrac {108 \: g/mol}{6.022 \times 10^{23} /mol}=17.93 \times 10^{-23} \: g$$
Mass of unit cell $$\displaystyle = 4 \times 17.93 \times 10^{-23} \:g=71.72 \times 10^{-23} \: g$$
$$\displaystyle \text {Density of silver}=\dfrac {\text {mass of unit cell}}{\text {volume of unit cell}}$$

$$\displaystyle \text {Density of silver}= \dfrac {71.72 \times 10^{-23} \:g}{68.27 \times 10^{-24} \; cm^3}=10.51 \: g/cm^3 $$

a) Calculate the packing efficiency in a unit cell of Cubic Close Packing (CCP) structure?
b) Name the crystal defect which lowers the density in an ionic crystal?

a) For Cubic Close Packing (CCP) lattice, the relationship between the edge length a and the radius r of the unit cell is $$\displaystyle a = \sqrt{8}r$$
The volume of the unit cell is $$\displaystyle a^3= (\sqrt{8}r)^3= 8\sqrt{8}r^3$$
The volume occupied by 1 atom is $$\displaystyle \dfrac {4}{3}\pi r^3$$
A CCP unit cell has 4 atoms per unit cell.
The volume occupied by 4 atoms is
$$\displaystyle 4 \times \dfrac {4}{3}\pi r^3 = \dfrac {16}{3}\pi r^3 $$
The packing efficiency $$\displaystyle = \dfrac {\text {Volume occupied by atoms in a unit cell }}{\text {total volume of unit cell}} \times 100 $$
The packing efficiency $$\displaystyle = \dfrac { \dfrac {16}{3}\pi r^3} {8\sqrt{8}r^3}\times 100=74.0$$ %

b) Schottky defect (Vacancy defect) which lowers the density in an ionic crystal.

(a) Write about the most common point defects.

Point defects are where an atom is missing or is in an irregular place in the lattice structure. Point defects include self interstitial atoms, interstitial impurity atoms, substitutional atoms and vacancies.

A substitutional impurity atom is an atom of a different type than the bulk atoms, which has replaced one of the bulk atoms in the lattice.

Interstitial impurity atoms are much smaller than the atoms in the bulk matrix. Interstitial impurity atoms fit into the open space between the bulk atoms of the lattice structure.

Explain Schottky and Frenkel defects in solids.

Schottky defect:
Same number of positive and negative ions are missing from their respective positions leaving behind a pair of holes. This defect is more common in ionic compounds with high coordination numbers with approximately equal sizes of cation and anion. The density of the crystal decreases. This defect is observed in $$NaCl, CsCl, KCl, KBr,$$ etc.

Frenkel defect:
An ion leaves its correct lattice site and occupies an interstitial site. It is common in ionic compounds with low coordination numbers and large differences in the size of cation and anion. This defect is observed in $$ZnS, AgCl, AgBr, AgI,$$ etc.

Explain Schottky and Frenkel defects.

a) Schottky Defects ;
This detect is caused if sonic of the lattice-points are unoccupied. The points which are unoccupied are called lattice vacancies. The number of missing positive end negative ions is, the same in this case and the crystal remains neutral. The existence of two vacancies, one due to a missing $$N^+$$ ion and the other due to a missing $$Cr^-$$ ion in a crystal of NaCl. It appears ionic crystals in which be positive and negative ions do not differ much in, size.
b) Frenkel defect:
This defect arise when an ion occupies interstitial position beteen the lattiec points.
This This defect occurs generally in ionic crystals in which much larger in size than the cation.
AgB is example for type of defect one of the Ag ion occupies a position in the interstilial space rather than its own appropriate site in the lattice.
The crystal remains neutral since the number of positive ions is the same as the number of negative ions.

The figures given below show the location of atoms in three crystallographic planes in a fcc lattice. Draw the unit cell for the corresponding structure and identify these planes in your diagram.

Refer to the image.

$$X\longrightarrow$$ anion

$$.\longrightarrow$$ cation

1175581_654639_ans_52c81fa1bc6f4c018f1a675e0f93c621.png

A unit cell of sodium chloride has four formula units. The edge length of unit cell is $$0.564\ nm$$. What is the density of sodium chloride?

$$Z=4, a=5.64\mathring A, M_A=58.5$$ $$\rho=\cfrac {Z\times M_A}{N_A\times a^{3}}$$

$$\rho=\cfrac {4\times 58.5}{6.023\times 10^{23}\times 179.4\times 10^{-24}}$$

$$\rho= 2.16$$ $$g/cm^{3}$$

Match the List-I with List-II and List-III:
List-I
(Solids)
List-II
(Unit cell)
List-III
(Coordination number)
(a) Rock salt (p) Face-centred cubic, anion in tetrahedral void (w) 6
(b) Fluorite (q) Face-centred cubic, cation in octahedral void (x) Cation (8), anion (4)
(c) Agl, ZnS (r) Face-centred cubic, cation in alternate tetrahedral void (y) Cation (4), anion (8)
(d) $$Na_2O$$ (s) Face-centred cubic, cation in tetrahedral void (z) Cation (4), anion (4)

(a) Rock Salt $$(NaCl)$$ Cation: All octahedral voids Coordination

Anion:FCC lattice $$6:6$$

(b) Fluorite Cation:FCC lattice, anion-tetrahedral void Coordination

Cation:$$8$$

Anion:$$4$$

(Zinc Blende) anion:FCC lattice $$4:4$$

(d) $$Na_2O$$ Cation:tetrahedral voids Coordination

(Antifluorite) anion:FCC lattice Cation:$$4$$

Anion:$$8$$

Lead sulphide has face centered cubic crystal structure. If the edge length of the unit cell of lead sulphide is $$495 pm$$, calculate the density of the crystal. (atomic weight $$Pb=207, S=32$$)

Atomic weight, $$ Pb $$=$$ 207, S $$=$$ 32$$

Molecular weight (M) of $$\displaystyle PbS = 207+32=239$$ g/mol.

An fcc unit cell has 4 formula units (Z) per unit cell.
The edge length (a) of the unit cell of lead sulphide is 495pm or $$\displaystyle 495 \times 10^{-10}$$ cm

The density $$\displaystyle d= \dfrac {ZM}{a^3N_A}$$

$$\displaystyle d = \dfrac {4 \times 239}{(495 \times 10^{-10})^3 \times 6.023 \times 10^{23}}$$
$$\displaystyle d=13.1 \: g/cm^3$$

A crystal of lead(II) sulphide has $$NaCl$$ structure. In this crystal the shortest distance between $$ Pb^{+2}$$ ion and $$ S^{2-} $$ ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? Also calculate the unit cell volume.

In $$NaCl$$ type structure shortest distance of the anion and cation=$$\cfrac {a}{2}$$ (where $$a$$ is edge length)

$$\Rightarrow \cfrac {a}{2}=297 \Rightarrow a=594$$ pm

Volume of unit cell=$$a^3=209.58\times 10^{-30}m^3$$

Distinguish between crystaline solids and amorphous solid.

There is some differences between these two solids.

$$\rightarrow$$ Crystalline solids have a definite shape with orderly arranged ions, molecules or atoms but in amorphous solid they are randomly arranged & not having definite shape.

$$\rightarrow$$ Crystalline components are held together by strong uniform intermolecular forces but they differ from atoms to atoms.

A cubic unit cell contains manganese ions at the corners and fluorides ions at the center of each edge. Calculate the density of the compound. (Given that atomic mass $$Mn=55$$amu and F is $$19$$amu).

$${Z_{Mn}}^{+3}=8\times \cfrac {1}{8}=1$$ Let the edge length be $$a\mathring A$$

$${Z_F}^{-}=12\times \cfrac {1}{4}=3$$

Formula:- $$MnF_3$$ $$M_A=112$$, $$Z=1$$ formula unit of $$MnF_3$$

$$\rho=\cfrac {Z\times M_A}{N_A\times a^{3}} \Rightarrow \rho= \cfrac {1\times 112}{6.02\times 10^{23}\times a^{3}\times 10^{-24}}$$

$$=\cfrac {186}{a^{3}}$$ $$g/cm^3$$

Determine the density of cesium chloride which cystallises in a bcc type structure with the edge length $$412.1 pm. $$ The atomic masses of $$Cs$$ and $$Cl$$ are 133 and $$35.5$$ respectively.

$$Z=1$$, $$a=4.121\mathring A$$, $$M_A=168.5$$

$$\rho=\cfrac {Z\times M_A}{N_A\times a^{3}}$$

$$\rho=\cfrac {1\times 168.5}{6.023\times 10^{23}\times 69.98\times 10^{-24}}$$

$$\rho=4$$ $$g/cm^3$$

Explain with the help of neat diagram $$AAAA$$ type of three.

1453979_711556_ans_6badfd52d9b944a4bbf0a0be34e08807.png

Write the value of axial distance and axial angles of triclinic crystal.

For triclinic crystal, $$a \neq b \neq c$$ and $$\alpha \neq \beta \neq \gamma \neq 90^0$$. Examples of tricilinic crystal include $$K_2Cr_2O_7$$ and $$H_3BO_3$$.

Give any one difference between anisotropy and isotropy nature of solid.

The ability of crystalline solids to change values of physical properties when measured in different directions is called anisotropy. It arises because the composition of solid changes with direction.
On the other hand, the values of the physical properties for amorphous solids do not change with change of direction. The properties remain same in all the directions. It is called isotropy.

Silver crystallizes in face-centered cubic structure. The edge length of the unit cell is found to be 408.7 pm. Calculate the density of silver. ($$Ag= 108g\ { mol }^{ -1 }$$)

Theoretical density of a crystal, $$\rho = \cfrac {nM}{N_oa^3} \ g/cm^3$$

For FCC, $$n=4 \ ; \ M= 100 \ g/mol \ ; \ a=408.7 \times 10^{-10} \ cm$$

$$\Rightarrow \rho = \cfrac {4 \times 108}{6.022 \times 10^{23} \times (408.7)^3 \times 10^{-30}}=10.5 \ g/cm^3$$

What type of stoichiometric defect is shown by:
(i) $$ZnS$$ (ii) $$AgBr$$

$$ZnS$$ shows Frenkel Defect.

$$AgBr$$ shows Frenkel Defect and Schottky Defect.

$$\rightarrow$$ Frenkel Defect: It is a kind of defect in crystalline solids in which atoms are displaced from their lattice position to interstitial site creating vacancy at the lattice point. It usually occurs in ionic solid with large difference in size of ions.

$$\rightarrow$$ Schottky Defect: This defect occurs when oppositely charged ions leave their lattice site creating vacancies in such a way that electrical neutrality of crystal is maintained. It is generally seen in highly ionic compounds where difference in size of cation and anion is small.

If the edge length of a $$NaH$$ unit cell is $$488\ pm$$, what is the length of $$Na-H$$ bond if it crystallizes in the fcc structure?

Given - Unit cell edge $$=a=488pm$$

Let $${ r }_{ + }$$ and $${ r }_{ - }$$ radius of cation and anion,

$$\sqrt { 2 } a=2\left( { r }_{ + }+{ r }_{ - } \right) $$

$$\left( { r }_{ + }+{ r }_{ - } \right) =\dfrac { a }{ \sqrt { 2 } } $$

$${ r }_{ + }+{ r }_{ - }=\dfrac { 488 }{ 1.414 } =345.12pm$$

Hence, Bond length between nuclei is $$345.12pm$$.

Calculate the number of unit cells in $$8.1$$g of aluminium if it crystallizes in a face-centred cubic (f.c.c.) structure. (Atomic mass of Al$$=27$$g $$mol^{-1}$$).

Face centered cubic unit cell has 4 atoms-

Atoms of aluminium $$= \dfrac{8.1}{27} \times N_a$$

NO. of unit cells $$= \dfrac {Total atoms of aluminium} {Atoms in one FCC unit}$$

No. of unit cells $$= \dfrac{.3N_a}{4} \approx .45 \times 10^{23}$$

A piece of chalk can be broken into small particles by hammering but a piece of iron cannot be broken into small particles. Which characteristic of the particles of matter is illustrated by these observations?

Here the intermolecular forces of attraction between the particles are identified or mentioned. Since in chalk the particles forces between the chalk particles are not so strong hence they can be easily broken up while the interparticle distance between the particles of iron is strong as there are strong metallic bonds between them. Hence they cannot be easily broken down.

Calculate packing efficiency in simple cubic lattice.

The edge length or side of the cube be 'a', and the radius of each particle be 'r'.
$$a=2r$$
Volume of cube $$=a^3$$
$$(2r)^3=8r^3$$
In simple cube, each unit cell has only one sphere,
Volume of sphere $$=\dfrac{4}{3}\pi r^3$$
Packing efficiency $$\dfrac{\text{Volume if a sphere}}{\text{Volume of the cube}}\times 100\%$$
$$=\dfrac{4_{/3}\pi r^3}{8r^3}\times 100=52.4\%$$.

Give the differences between crystalline and amorphous solids with respect to shape and melting point.

Crystalline solids	Amorphous solids
They have definite characteristic geometrical shape as constituent particles are arranged in orderly regular long range manner.	They do not have definite geometrical shape. This is due to short range order of constituent particles.
They have sharp melting point. They melt completely at one constant temperature.	They do not have sharp melting point. As temperature is raised, they gradually soften, becomes less viscous and melt over a temperature range.

What is lattice enthalpy of an ionic crystals? Give the relation.

Lattice energy is the energy released when on mole of an ionic crystal is formed from one mole of their constituent ions . Greater the value of lattice energy, more stable is the ionic compound. The lattice energy of sodium chloride is -185 kcal/mole. The negative sign indicates that the process is exothermic process and that the energy of the system is lowered as the ionic solid is formed.

On heating crystals of $$KCl$$ in potassium vapours, the crystals start exhibiting violet colour. Why?

Space lattice is a regular, indefinitely repeated array of points in three dimensions in which the points lie at the intersections of three sets of parallel equidistant planes.

This happens due to the creation of F-centers, as a result of Chloride anion vacancy created upon heating KCl in excess of K vapors. The hole created by the chloride anion is filled by an electron from K, giving it violet appearance.

Preparation of models of crystal (cube).

Primitive (Simple) Cubic Structure-

Placing a second square array layer directly over a first square array layer forms a "simple cubic" structure. The simple “cube” appearance of the resulting unit cell is the basis for the name of this three-dimensional structure.
This packing arrangement is often symbolized as $$"AA"$$, the letters refer to the repeating order of the layers, starting with the bottom layer. The coordination number of each lattice point is six. This becomes apparent when inspecting part of an adjacent unit cell.

The unit cell in appears to contain eight corner spheres, however, the total number of spheres within the unit cell is 1 (the only$$\dfrac{ 1}{8}^{th}$$ of each sphere is actually inside the unit cell). The remaining$$\dfrac{ 7}{8}^th$$ of each corner sphere resides in $$7$$ adjacent unit cells.

Reapplication of the theorem to another right triangle created by an edge, a face diagonal, and the body diagonal allows for the determination of the body diagonal as a function of $$r$$.
Few metals adopt the simple cubic structure because of inefficient use of space. The density of a crystalline solid is related to its "percent packing efficiency".
The packing efficiency of a simple cubic structure is only about $$52\%. (48\%$$ is empty space!)

$$\% packing efficiency = \dfrac{Volume of atoms in a unit cell}{Volume of unit cell}\times 100$$

977264_1018676_ans_b04a9d48e7f1493e9f45d460d584790f.PNG

In a face-centered unit cell with all the positions occupied by A atoms, the body-centered octahedral holes in it are occupied by an atom B of an appropriate size.
Calculate the void fraction per unit volume of a unit cell for such crystal.

$$Volume\quad =\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\\ \sqrt { 3 } a=4r\\ a=\cfrac { 4r }{ \sqrt { 3 } } \\ So,\\ P.F.=\cfrac { 4\times \cfrac { 4 }{ 3 } \times \cfrac { 22 }{ 7 } \times { r }^{ 3 } }{ { a }^{ 3 } } \\ =\cfrac { 4\times \cfrac { 4 }{ 3 } \times \cfrac { 22 }{ 7 } \times { r }^{ 3 } }{ { \left( \cfrac { 4r }{ \sqrt { 3 } } \right) }^{ 3 } } \\ =\cfrac { 4\times \cfrac { 4 }{ 3 } \times \cfrac { 22 }{ 7 } \times { r }^{ 3 }\times 3\sqrt { 3 } }{ { \left( 64 \right) }^{ 3 } } \\ =0.74\\ Void\quad fration=1-0.74\\ =0.26$$

Which type of stoichiometric defect is shown by AgCl?

Frenkel defect is shown by those in which there is large difference in the size of cations and anions. Hence, Frenkel defect is shown by $$AgCl$$ due to small size of $$Ag^+$$ ion.

A metal has bcc structure and the edge length of its unit cell is $$3.04\overset{o}{A}$$. Find the volume of the unit cell in $$cm^3$$.

Volume of unit cell =$$a^{3}$$

= $$(3.04\times 10^{-8}cm)^{3}$$

=$$2.81\times 10^{-23}cm^{3}$$

The density of a metal is $$10.8$$ x $$10^3$$ kg/$$m^3$$. Find the relative density of the metal. (Density of water is $$10^3\ Kg/m^3$$)

Relative density$$=\cfrac{density\quad of\quad given\quad substance}{Density\quad of \quad standard\quad substance(water)}$$

Relative density$$=\cfrac{10.8\times 10^{3}}{10^{3}(density\quad of\quad water)}=10.8$$

Calculate the packing efficiency in primitive unit cell.

In primitive unit cell edge length $$=1=2\times $$ radius of element.

$$a=2r$$

$$\implies r=\cfrac{a}{2}$$

Volume occupied by element $$=8\times \cfrac{1}{8}\times \cfrac{4}{8}\times \pi\times \cfrac{a^3}{8}$$

Volume of unit cell $$=a^3$$

Packing efficiency $$=\cfrac{4}{3}\cfrac{\pi a^3}{8}\times \cfrac{1}{a^3}=\cfrac{\pi}{6}=0.5234$$

Calculate packaging efficiency in simple cubic lattice.

Volume of unit cell $$=a^3$$

Volume of atoms present at body corners $$=8\times \cfrac{1}{8}\times \cfrac{4}{3}\times \pi\times r^3$$

In simple cubic $$a=2r$$

$$\implies r=\cfrac{a}{2}$$

$$\therefore$$ volume of the atoms present at body corners$$=\dfrac{\pi}{6}a^3$$

$$Packaging\quad Fraction =\cfrac{Volume\quad of\quad atoms\quad present}{Volume\quad of\quad unit\quad cell}=\cfrac{\dfrac{\pi}{6}a^3}{a^3}=\cfrac{\pi}{6}=0.523$$

An element crystallizes in fcc structure with a unit cell edge length of 200 pm. Calculate its density if 200g of this element contain $$24\times10^{23}$$ atoms.

For Fcc

$$Z=4$$

$$N=24 \times 10^{23}$$ atoms

$$a = 200$$ $$pm$$

$$= 2 \times 10^{-8}$$

$$ d= \cfrac {2 \times M}{a^3 \times N} = \cfrac {4 \times 200}{8 \times 10^{-24} \times 24 \times 10^{23}}$$

$$= 41.67$$ $$g/cm^3$$

The density of $$200$$ $$g$$ of this element is $$= 41.67$$ $$g/cm^3$$

Refractive index of a solid is observed to have the same value along all direction. comment on the nature of this solid. would it show cleavage property?

Isotropic solid has the same value of the $$\text{refractive index}$$ in all the direction, all isotropic solid and amorphous solid.

Amorphous solid cut into two pieces with irregular surfaces, therefore amorphous solid does not show cleavage property.

What is the distance between $${ Na }^{ + }$$ and $${ Cl }^{ - }$$ ions in $$NaCl$$ crystal if its density is $$2.165 g { cm }^{ -3 }$$ .[Atomic mass of $$Na-23 u, Cl = 35.5 u$$, Avogadro,s Number=$$6.023 \times 10^{23}$$]

$$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$

$$Z=4$$

$$M_A=58.5$$

$$\rho=2.165$$ $$g/cm^3$$

$$a=?$$

$$2.165=\cfrac{4\times 58.5}{6.023\times 10^{23}\times a^3}$$

$$\implies a^3=\cfrac{4\times 58.5}{6.023\times 2.165\times 10^{23}}$$

$$\implies a=5.64\times 10^{-8}$$ $$cm$$

$$\implies a=5.64 \mathring A$$

Distance between $$Na^{+}$$and $$Cl^{-}=\cfrac{a}{2}=2.82\mathring A$$

Iron has a body-centered cubic unit cell with a cell edge of $$286.65 pm.$$. The density of iron is $$7.874 g { cm }^{ -3 }$$. Use this information to calculate Avogadro's number. $$(At.mass\ of\ iron=55.845 u)$$

$$a=2.8665 \mathring A$$

$$\rho= 7.874$$ $$gm/cm^3$$

$$M_A=55.845u$$

$$Z=2$$

$$N_A=?$$

$$\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$

$$\implies 7.874=\cfrac{2\times 55.845}{N_A\times 23.55\times 10^{-24}}$$

$$\implies N_A=6.0232\times 10^{23}$$

$$\circleddash \oplus \circleddash O\circleddash \oplus \circleddash \oplus $$
$$\oplus \circleddash \oplus \circleddash \oplus \circleddash \oplus \circleddash $$
$$\circleddash \oplus O\oplus \circleddash \oplus \circleddash \oplus $$
$$\oplus \circleddash \oplus O\oplus \circleddash \oplus \circleddash $$
$$\circleddash \oplus \circleddash \oplus \circleddash O\circleddash \oplus $$

(a) What are these types of vacancy defects called?
(b) How is the density of a crystal affected by these defects?
(c) Name one ionic compound which can show this type of defect in the crystalline state.
(d) How is the stoichiometry of the compound affected?

(a)Schottky defect:If in an ionic crystal of the type $$A+B^-$$ equal number of cations and anoins are missing from the the lattice site.so electrical neutrality is remain same.

(b)Density decreases due to vacancy of ions from the lattice.

(c)NaCl,KCl

(d) stoichiometry of the compound is not affected.

What type of change will occur when AgCl is dropped with $$CdCl_2$$.

When AgCl is doped with $$CdCl_2$$ a cation vacancy defect is created i.e. an $$Ag^+$$ ion from the lattice is absent in position due to the presence of $$Cd^{2+}$$ ions in its adjacent.

Ammonium chloride cystallise in a body-centred cubic lattice with a unit distance of $$387pm$$. Calculate
(a) the distance between oppositely charged ion in the lattice and
(b) the radius of the $${NH}_{4}^{+}$$ ion if the radius of $${Cl}^{-}$$ ion is $$181pm$$

$$1.$$ $$NH_4^+$$ is at body centre and $$Cl^-$$ at body corners. Distance between opposite charged ion $$=\cfrac{a\sqrt{3}}{2}=\cfrac{387\sqrt{3}}{2}=335.15$$ $$pm$$

$$2.$$ For BCC, Ideal ration of $$\cfrac{r_+}{r_-}=0.732$$

$$\cfrac{r_{NH_4^+}}{r_{Cl^-}}=0.732$$

$$\implies r_{NH_4^+}=0.732\times 181=132.49\ pm$$

Ice crystallizes in a hexagonal lattice. At the low temperature at which the structure was determined, the lattice constants were $$a=34\mathring{A}$$ and $$c=7.41\mathring{A}$$. How many $${H}_{2}O$$ molecules are contained in a unit cell? (The density of ice is $$0.92g/cc$$ at $${0}^{o}C$$ A unit cell of $${H}_{2}O$$ is shown.)

Volume of unit cell, $$V=$$ Area of Base $$\times$$ Height

$$=2\times \cfrac{\sqrt{3}}{4}a^2\times c$$

$$=2\times \cfrac{\sqrt{3}}{4}\times 1156\times 10^{-16}\times 7.41\times 10^{-8}$$ $$cm^3$$

Mass of unit cell $$=V\times S$$

$$=7.409\times 10^{-21}\times 0.92$$

$$=6.816\times 10^{-21}$$ $$gm$$

Let there be $$N$$ water molecules in the given cell

$$\implies 6.816\times 10^{-21}=n\times \cfrac{18}{6.023\times 10^{23}}$$

$$\implies n=227.9\approx 228$$ $$H_2O$$ molecules

Caesium bromide has $$CsCl$$ structure $$(BCC$$ type of lattice$$)$$. Its density is $$4.49gm.cc$$. Calculate the side of the unit cell. $$[$$Molar mass of $$Cs=132.09 g/mol]$$

For $$bcc$$, $$z=2$$ & $$M=132.09+80=212g{ mol }^{ -1 }$$

$$S=4.49g/cc$$

$$\therefore \quad { a }^{ 3 }=\cfrac { Z\times M }{ S\times NA } $$

$$=\cfrac { 2\times 212 }{ 4.49\times 6.022\times { 10 }^{ 23 } } $$

$$\therefore \quad { a }^{ 3 }=1.568\times { 10 }^{ -22 }=0.1568\times { 10 }^{ -21 }$$

$$\therefore \quad a=0.539\times { 10 }^{ -7 }m$$

$$\therefore $$ Side of the unit cell $$=0.539\times { 10 }^{ -7 }m$$

The void fraction of ..... and HCP is same.

Packing fraction of FCC and HCP= $$74$$%= $$0.74$$

Void Fraction= $$1-0.74=0.26$$

Void Fraction of FCC and HCP is same.

Calculate the efficiency of packing in case of a metal crystal for face-centered cubic.

Packing efficiency of face center cubic :-

$$P.E=\cfrac{Z\times \cfrac{4}{3}\pi r^3}{a^3}\times 100=\cfrac{4\times \cfrac{4}{3}\pi r^3}{{(2\sqrt{2}r)}^3}\times 100=74\%$$ $$Z\longrightarrow atoms/unit\text{ }shell=4;\quad a=2\sqrt{2}r$$

How many $$Na^+$$ ions in each unit cell is in a cubic ideal crystal of $$NaCl$$?

$$Na^+$$ ions are present in the edge centre and at the body centre.

The number of $$Na^+$$ ions present in one unit cell $$=12\times\dfrac{1}{4}+1=4\ atoms$$

$$Cl^-$$ ions are present at the corner and at the centre of the faces.

The number of $$Cl^-$$ ions present in one unit cell $$=6\times\dfrac{1}{4}+8\times\dfrac{1}{8}=4\ atoms$$

So, each unit cell has 4 atoms of each $$Na^+$$ and $$Cl^-$$.

Hence, there are total $$4$$ formula units of $$NaCl$$ in each unit cell.

Name the types of point defects in solids.

Types of Point defect

1. Stoichiometric Defect

Vacancy defect
Intersttial defect
Frenkel defect
Schottky defect.

2. Impurity defect
3. Non Stoichiometric defect

Metal deficiency defect
Metal excess defect

Give one example of amorphous substance.

Amorphous, or non-crystalline, solids lack this long-range order. Accordingly, they lack the elasticity, distinct melting points, and other properties of crystalline solids.
Amorphous solids lack a characteristic geometry, have identical properties along all axes, have wide ranges over which they melt, and break to form curved or irregular shapes.
Example of an amorphous solid is glass. However, amorphous solids are common to all subsets of solids. Additional examples include thin film lubricants, metallic glasses, polymers, and gels.

Iron has an edge length $$288$$pm. Its density is $$7.86 gm cm^{-1}$$. Find the type of cubic lattice to which the crystal belongs. (Atomic mass of iron = $$56$$).

Edge length, $$a= 288\,pm = 2.88\times {10}^{-10}\,cm$$

Mass of one atom, $$M=$$Molar mass of iron $${N_A}= \dfrac{56}{6.022\times{10}^{23}}$$

$$= 9.30\times {10}^{23}$$

Now,

Density, $$D=\dfrac{n\times M}{a^3}$$

where $$n=$$ no. of atoms in unit cell

$$n=\dfrac{7.86\times 23.89\times {10}^{23}}{6.90\times{10}^{-23}}$$

$$n=2$$

Thus unit cell contains 2 atoms in it. But bcc crystal structure contains 2 atoms in its unit cell. Hence the crystal structure for iron is body centred cubic structure.

An elements crystallizers in a cubic closed packed structure having FCC unit of an edge length $$200$$ pm. calculate the density if $$200$$g this elements contains $$24 \times 10^{23}$$ atoms.

For, FCC, given M$$=200g$$

z$$=4$$

N$$=24\times 10^{23}$$ atoms

a$$=200pm=200\times 10^{-10}m=2\times 10^{-8}cm$$

Density is given by the equation,

$$d=\cfrac{z\times N}{a^3\times N}=\cfrac{1000}{24}$$

$$\therefore d=41.67 gcm^{-3}$$

In the solid, oxide ions are arranged in hcp. one third of $$ O_{v} $$ are occupied by A and one sixth $$ T_{v} $$ are occupied by B. Find the formula ?

Total number of oxide ions in hcp $$=6$$

$$\therefore $$ $$ 6$$ oxygen atoms.

No. of octa hedral voids $$=6$$

$$\therefore$$ No. of $$A$$ atoms $$=\cfrac{1}{3}\times 6=2$$

No. of tetrahedral voids $$=2\times 6=12$$

$$\therefore$$ No. of $$B$$ atoms $$=\cfrac{1}{6}\times 12=2$$

$$\therefore$$ Formula of compound is $$A_2B_2O_6$$ or simplyfied formula is $$ABO_3$$.

Molybdenum has a atomic mass $$96 gmol^{-1}$$ with density $$10.3gcm^{-3}$$.The edge length of unit cell is $$314$$pm.Determine its lattice structure.

Density of element, $$\rho$$ = 10.3 g $$ {cm}^{-3}$$

Cell edge a=314 pm

$$ N_A = 6.023\times{10}^{23} {mol}^{-1}$$

Atomic mass = 96 g $$ {mol}^{-1}$$

$$ P = \frac{{Z}\times{M}} {{a}^{3}\times{}N_A}$$

$$ Z = \frac{{P}\times{N_A}\times{a}^{3}} {M}$$

$$= \frac{10 g {cm}^{-3}\times{10}^{-30}{cm}^{3}\times6.023\times{10}^{23} {mol}^{-1}} {96 g {mol}^{-1}}$$ = 2

The structure of the crystal lattic is B.C.C

An element of atomic mass $$98.5gmol^{-1}$$ occurs in FCC structure. If its density is $$5.22gcm^{-3}$$, determine the edge length of unit cell.

For FCC, density is given by

$$d=\cfrac{Z\times M}{a^3\times N}$$ where $$z$$ is $$4$$, i.e. Number of atoms per unit cell.

$$a^3=\cfrac{4\times 98.5}{5.22}$$ given: $$d=5.22g cm^{-3}$$and atomic mass is $$98.5g\text{ }mol^{-1}$$

$$a^3=75.48\\\therefore a=25.16 cm$$

Thallium (I) chloride crystallises in either a SCC or FCC of $$Cl^-$$ ion. The density of the solid is $$8.9gcm^{-3}$$ and edge of the unit cell is $$3.95 \times 10^{-8}cm^{-3}$$. Predict the type of unit cell.
[Molar mass of thallium chloride $$=239.8gmol^{-1}]$$

$$d= \dfrac{A.m}{a^3N_o}$$

$$9=Z\times239.8/(9×10^{−8})3\times6.023\times10^{23}$$

$$Z=8.9\times3.95\times3.95\times3.95\times10^{−24}\times6.023\times10^{23}/239.8$$

$$Z=1.178$$ or $$ 1$$

So, $$TiCl$$ is simple cubic.

Write an example of elements in which atoms occupy the position of lattice points.

Lattic points are the points on a crystal lattic where atoms or ions can be placed lattic points are the position where you can place on atom.

In covalent solids atoms occupy the position of lattic points eg : graphite, diamond etc.

Lowest Lattice energy will be possessed by which molecule?

The bond between ions of opposite charge is strongest when the ions are small. The lattice energies for the alkali metal halides is, therefore, largest for $$LiF$$ and smallest for $$CsCl$$.

What is meant by voids in crystal structure?

Voids literally mean $$gaps$$ between the constituent particles. Voids in solid states mean the vacant space between the constituent particles in a closed packed structure. Close packing in solids can be generally done in three ways: 1D close packing, 2D close packing and 3D close packing. In 2 dimensional structures when the atoms are arranged in square close packing and hexagonal close packing, we see empty spaces left over between the atoms. These empty spaces are called voids and in the case of hexagonal packing, these voids are in triangular shapes and are known as the triangular voids.

The size and distribution of voids in materials plays a role in determining In the close packed crystals (FCC, HCP) there are two types of voids - $$tetrahedral$$ and $$octahedral$$ voids (identical in both the structures as the voids are formed between two layers of atoms).

Write two examples of molecular solids.

$$Molecular \ solids$$—Made up of atoms or molecules held together by London dispersion forces, dipole-dipole forces, or hydrogen bonds. Characterized by low melting points and flexibility and are poor conductors. An example of a molecular solid are $$sucrose$$ and $$ solid $$ $$CO_2$$.

Calculate number of unit cells in $$58.5$$ of a compound which forms $$BCC$$. [Molar mass of a compound $$=58.5$$]

We will apply the concept of Avogadro's constant in this question.

Avogadros constant :

1 mole = $$6.022 \times {10}^{23}$$

The number of moles in the substance is :

Moles $$= \dfrac{mass}{molar\,mass}$$

Moles $$= \dfrac{58.5}{58.5}$$ = 1 moles

Number of cells $$= 1 \times 6.022 \times {10}^{23} = 6.022 \times {10}^{23}$$

Define line defect.

A line defect is a line along which whole rows of atoms in a solid are arranged anomalously. The resulting irregularity in spacing is most severe along a line called the line of dislocation. Line defects can either weaken or strengthen solids.

A line defect acting as a boundary between the slipped and non-slipped region in the slip plane is called a dislocation.

Write an example of soild in which the force between the particles having electro-staic force.

The force between the particles is an electrostatic force, hence these are ionic solids.

Define crystal lattice.

Crystal lattice is a regular arrangement of the constituent atoms or ions or molecules in three-dimensional space. A crystal lattice also called a space lattice or simply, a lattice. There are only 14 possible three-dimensional lattices. These are called Bravais’ lattices.
Characteristics of a crystal lattice:

Each point in a lattice is called a lattice point or lattice site.
Each point in a crystal lattice represents one constituent particle, which may be an atom, a molecule or an ion.
Lattice points are joined by straight lines to bring out the geometry of the lattice.

The radius of cation and anion is $$0.95\ A^{0}$$ and $$1.81\ A^{0}$$. Name of structure of ionic crystal.

We know,

$$Radius\,ratio=\dfrac{Radius\,of\,cation}{Radius\,of\,anion}$$

$$Radius\,ratio=\dfrac{0.95}{1.81}= 0.525$$

For the ratio of 0.414 to 0.732, the ionic crystal has octahedral structure

So, structure is octahedral.

What is the packing efficiency of $$BCC$$ unit cell?

The packaging efficiency of Body Centred Cubic cell is 68%. Thus 32% volume is empty space.

A metal crystallizes with a face-centered cubic lattice. The edge length of the unit cell is 408 pm. The diameter of the metal atom is:

For FCC,

$$r =\dfrac {a}{{2}\sqrt{2}}$$

Hence Diameter = $$\dfrac {a}{\sqrt{2}}$$

$$ =\dfrac {408} {1.414} = 288.5pm$$

Which crystal lattice has most efficient packing efficiency?

The packing efficiency of both types of close packed structure is 74%, i.e. 74% of the space in hcp and ccp is filled. The hcp and ccp structure are equally efficient; in terms of packing. The packing efficiency of simple cubic lattice is 52.4%. And the packing efficiency of body centered cubic lattice (bcc) is 68%.

The density of $$CaF_2$$ (fluorite structure) is 3.18 $$g/cm^3$$. The length of the side of the unit cell is:

Density of $$CaF_2 = 3.18$$

Molar Mass of $$CaF_2 = 78 g/mol.$$

Using the Formula,

$$Density = \frac{Mass}{Volume}$$.

$$Volume = \frac{Mass}{Density}$$

$$Volume = \frac{78}{3.18}$$

$$Volume = 24.53$$ cubic units.

Now, Volume of one cube of $$CaF_2 = (Side)^3$$

$$Side = \sqrt[3]{Volume}$$

$$Side = \sqrt[3]{24.53}$$

$$Side = 2.905 units.$$

Hence, the Length of the side of the unit cell of the $$CaF_2 is 2.905 units$$.

Calculate the packing efficiency of particles in a simple cube.

In a simple cubic lattice the atoms are located only on the corners of the cube.

Let take edge length or side of the cube = a,

Let take radius of each particles = r

The relation between radius and edge a

a = 2r

The volume of the cubic unit cell $$= side^3$$

$$= a^3$$

$$= (2r)^3$$

$$= 8r^3$$

Number of atoms in unit cell = 8 x 1 /8

= 1

The volume of the occupied space $$= (4/3)πr^3$$

Paking efficiency $$= \dfrac{\text{volume of one atom }}{\text{volume of the cubic unit cell}} \times 100$$ %

$$= \dfrac{4/3 \pi r^3}{8r^3} \times 100 $$%

$$= \pi /6 \times 100$$ %

$$= 52.36$$ % or $$52.4 $$%

Lithium metal has a body centred cubic structure. Its density is $$0.53 g \, cm^{-3}$$ and its molar mass is $$6.94g \, mol^{-1}$$. Calculate the volume of a unit cell of lithium metal.

Density of unit cell is given as

Density$$-\cfrac{no.\quad of\quad atom\quad per\quad unit\quad cell\times Molar\quad mass\quad of\quad atom}{MA\times Volume\quad of\quad unit\quad cell}$$

In BCC structure

at corner $$=\cfrac{1}{8}\times8=1$$

at Body centre$$+1=1 atom$$

Total atom in BCC $$=2 atom$$
Molar mass of lithium $$=6.94$$

Given density $$=0.53g cm^{-3}$$
On putting all values we have

$$0.53gcm^{-3}=\cfrac{2\times 6.94}{6.022\times10^{23}mol^{-1}\times0.53gcm^{-3}}\\V=4.34\times10^{23}cm^3\\ V=4.34\times10^{-23}ml$$

Thus volume of unit cell is $$4.34\times 10^{-23} cm^3$$

Aluminium forms face centered cubic crystals .The density of $$AI$$ is $$2.7 g/cm^3$$ .Calculate the length of the side of the unit cell $$AI$$. (At.wt. of $$AI=27$$)

For FCC, number of unit cells= $$4$$

Density= $$2.7 g/cm^3$$

$$d=\cfrac {Z \times M}{a^3\times N_A}$$

$$\Rightarrow 2.7=\cfrac {4 \times 27}{a^3\times 6.023\times 10^{23}}$$

$$\Rightarrow a^3=\cfrac {4\times 27}{2.7 \times 6.023 \times 10^{23}}$$

$$\Rightarrow a= 4.05 \times 10^{-8} cm$$

The unit cell cube length for $$LiCl$$ is $$5.14 \overset{o}A $$. Assuming anion-cation contact, calculate the ionic radius for chloride ion.

In $$LiCl$$ structure, anions have face centered cubic arrangements. In such a unit cell face diagonal is four times the radius of anions.

Face diagonal $$=\sqrt { 2 } .a=\sqrt { 2 } \times 5.14{ A }^{ 0 }$$

Face diagonal $$=4r$$

$$\therefore$$ $$4r=\sqrt { 2 } \times 5.14{ A }^{ 0 }$$

$$\Rightarrow r=1.82{ A }^{ 0 }$$

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An element having atomic mass $$65.1 g/mol$$ has face centred cubic unit cell with edge length $$3.608 \times 10^{-8} cm$$. Calculate the density of the element. [Given $$N_A$$=$$6.022 \times 10^{23}atoms/mol$$]

Unit-cell edge length $$=360.8pm$$

Volume of unit cell $$=(360.8×10{−12}cm)3=4.7×10−23cm^3$$

Mass of the unit cell =Number of atoms in the unit cell $$\times$$ Mass of each atom

Number of atoms in the fcc unit cell $$=8\times18+6\times12=4$$

Mass of one atom =Atomic mass/Avogadro number

$$=61.5/6.023\times10^{23}$$

Mass of the unit cell $$=\dfrac{4\times60}{6.023\times10^{23}}$$

Density of unit cell =Mass of unit cell/Volume of unit cell

$$=\dfrac{4\times65.1}{6.023\times10^{23}}\times \dfrac {1}{4.7\times10^{−23}}$$

$$=9.2gcm^{−3}$$

A compound $$AB$$ has (rock salt type structure) molecular weight of $$AB$$ is $$6.023 Y \, amu$$ and closest $$A-B$$ distance is $$Y^{1/3}$$ nm, where $$Y$$ is an arbitrary number (in amu). Find the density of lattice, if the density of lattice is found to be $$20 \, kg \, m^{-3}$$.

Fcc structure (n=4) and formula weight of AB is 6.023y g

closest distance $$A-B=y\dfrac{1}{3}nm$$

Therefore , edge length of unit cell,a $$= 2(A^+ + B^-) = 2\times y/3\times 10^{-9}m$$

Therefore density of AB $$= \dfrac{n\times mol.wt}{N_AV}$$

$$= \dfrac{4\times 6.023\times y\times10^{-3}}{6.023\times 10^{23}(2\times \dfrac{y}{3}10^{-9})^3}$$

$$=5.0kgm^{-3}$$

The density of gold is $$419.3 g per cm^3$$.Calculate the diameter of solid gold sphere having a mass of 4239

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Ions replace Na+ ions at the edge centers of NaCI lattice then the number of vacancies in $$1$$ mole of NaCl be:

$$Na^+$$ the number of atoms $$= 12\times \dfrac{1}{4} = 3$$
And

$$3 Na^+ = Al^{3+}$$
Then for 1 mole that is $$1\times 6\times 10^{23}$$ atoms of $$Na^+$$ would have $$= \dfrac{6\times 10^{23}}{3 Al^{3+}}$$ atoms =$$2\times 10^{23}atoms$$.

A Metallic element crystallises into a lattice containing a sequence of layers of $$ABABAB....$$. Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?

For a hcp unit cell, there are 6 atoms per unit cell. If r is the radius of the metal atoms, volume occupied by the metallic-

Atoms $$6 \times \dfrac{4}{3} \times \pi r^3 = 6 \times 1.33 \times \dfrac{22}{7} \times r^3 = 25.08 \times r^3$$

Geometrically it has been shown that the base area of hcp unit cell

$$= 6 \times √3/4 \times 4r^2$$ and the height $$= 4r \times √2/3$$

Volume of the unit cell

$$= Area \times height = 6 \times √3/4 \times 4r^2 \times 4r \times \sqrt2/3= 33.94 r^3$$

Volume of the empty space of one unit cell

$$= 33.94 r^3 – 25.08 r^3 = 8.86 r^3$$

$$=\dfrac{8.816 r^3}{33.94 r^3} \times 100=26.1$$%

Name the type of solid in $$S_8$$ molecule.

REF.Image.

Crown-shaped

(structure of $$S_{8}$$ molecules)

This crown structure is perfectly stable thus

$$S_{8}$$ is Molecular solid.

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What are mass defect and binding energy?

Mass Defect-

The nuclear binding energy holds a significant difference between the nucleus actual mass and its expected mass depending on the sum of the masses of isolated components.

Since energy and mass are related based on the following equation-

$$E=mc^2$$

Where c is the speed of light. In nuclei, the binding energy is so high that it holds the considerable amount of mass.

The actual mass is less than the sum of individual masses of the constituent neutrons and protons in every situation because energy is ejected when the nucleus is created. This energy consists of mass which is ejected from the total mass of the original components and called as mass defect. This the mass is missing in the final nucleus and describes the energy liberated when the nucleus is made.

Mass defect is determined as the difference between the atomic mass observed (Mo) and expected by the combined masses of its protons (mp, every proton has a mass of 1.00728 AMU) and neutrons (mn, 1.00867 AMU).

Nuclear Binding Energy-

Once the mass defect is calculated, nuclear binding energy can be determined by converting mass to energy by applying E=mc2. When this energy is calculated which is of joules for a nucleus, you can scale it into per-mole quantities and per-nucleon. You need to multiply by Avogrado's number to convert into joules/mole and divide by the number of nucleons to convert to joules per nucleon.

Nuclear binding energy is also applied to situations where the nucleus splits into fragments that consist of more than one nucleon wherein, the binding energies of the fragments can be either negative or positive based on the position of the parent nucleus on the nuclear binding energy curve. When heavy nuclei split or if the new binding energy is known when the light nuclei fuses, either of these processes results in liberation of binding energy.

Mention one consequence of metal excess defect.

Metal excess defect may be of two type :
due to excess cation
- zn zno , $$zn^{+2}$$ are found in roids along with lattice sides, solid is metal due to interstitial dectrons and due to such $$e^-s$$ zinc oxide is found as yellow coloured solid.

The density of copper metal is $$8.95$$ gm $$cm^{ -3 }$$. If the radius of copper atom is $$127.8$$ pm, Will the copper unit cell be a simple cubic, a body-centred cubic or a face centred cubic? (At mass of Cu=$$63.54$$g $$mol^{ -1 }$$ and N$$_A$$ =$$6.02\times 10^{ 23 }\quad mol^{ -1 }$$)

Density of metal $$(\rho)=\cfrac{Z\times M}{N_A\times a^3}$$

For, simple cubic, $$a=2r, Z=1$$

$$\therefore \rho=\cfrac{1\times 63.54}{(2\times 127.8\times 10^{-10})^3\times 6.02\times 10^{23}}=6.31$$ $$g/cm^3$$

For BCC, $$Z=2, a=\cfrac {4}{\sqrt{3}}r$$

$$\therefore \rho=\cfrac{2\times 63.54}{\left(\cfrac {4}{\sqrt{3}}\times 127.8\times 10^{-10}\right)^3\times 6.02\times 10^{23}}=8.2$$ $$g/cm^3$$

For FCC, $$Z=4, a=\cfrac {4}{\sqrt{2}}r$$

$$\therefore \rho=\cfrac{4\times 63.54}{\left(\cfrac {4}{\sqrt{2}}\times 127.8\times 10^{-10}\right)^3\times 6.02\times 10^{23}}=8.92$$ $$g/cm^3$$

The density of copper metal is $$8.95$$ $$g/cm^3$$ which is nearest to FCC. Therefore, the copper unit cell is face centre cubic.

It represents the interatomic distance in a unit cell. It is calculated by using following formula
$$a = \root 3 \of {{{Z \times M} \over {\rho \times {N_A}}}} $$
For simple cubic, body-centred cubic and face-centred cubic structures value of Z is 1, 2 and 4 respectively.

density $$ = \frac{{mass}}{{volume}}$$

vol. of unit cell $$ = {a^3}$$

Total mass = mass of one unit cell $$ \times $$ no of unit cell

i.e $$d = \frac{{ZM.}}{{{a^3}}} = \frac{{Z.M/{N_A}}}{{{a^3}}}$$

$$ \Rightarrow d = \frac{{ZM.}}{{{a^3}{N_A}}} \Rightarrow {a^3} = \frac{{ZM}}{{d \times {N_A}}}$$

$$a = \root 3 \of {\frac{{ZM}}{{d \times {N_A}}}} $$

simple cubic has $$1$$ atom

body-centred has $$2$$ atom/unit cell

face centred has $$4$$ atom/unit cell.

How many unit cells are present in
a] $$2.5$$ cubic crystal of KCl &
b] With each edge of crystal. [ given KCl=$$ 74.5$$, Cell const=4]

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Name the parameters which are characteristics of unit cell.

A unit cell is the smallest part of a crystal lattice.

The characteristics of unit cells are-

1. Its dimensions are along the 3 edges i.e $$a$$,$$b$$ and $$c$$. These edges may or may not be equal.

2. The angles between these edges $$\alpha, \beta$$ and $$\gamma$$.

What is two-dimensional coordination number in square close packing layer of a molecule?

A coordination number is the number of ions and atoms around a central neutron or the total number of attachment locations to the central element in a molecule. The packing of constituent particles inside lattice in such a way that they occupy maximum available space in the lattice is known as Close Packing. The packing in which one sphere touches two spheres placed in two different rows one above and one below is known as square closed packaging. The coordination number here is 4.

Perovskite a mineral containing calcium, oxygen and titanium. crystallises in the unit cell as shown below.
On the basis of above information write the formula of perocskite?

A unit cell is the most basic and least volume consuming repeating structure of any solid.

The chemical formula of a perovskite structure is $$ABO_3$$. So the formula of given structure is $$T:CO_3$$

How many unit cells are there in $$1.00\ cm^{3}$$ of aluminium?

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An element $$'X'$$ crystallizes in $$bcc$$. Find volume of unit cell in $$(\mathring {A})^{3}$$, if atomic radius is $$\sqrt {3}\mathring {A}$$.

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Find the distance between $$K^+$$ and $$F^-$$ ion in potassium fluoride, which has sodium chloride type of crystal lattice. The density of KF is $$2.4\,g\, cm^{-3}$$. ($$M=58, N_A=6.022 \times 10^{23}$$)

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Classify the following solids into different types
Graphite

Graphite is a covalent solid as the constituent atoms in the graphite are carbon atoms and are held together by a strong covalent bond.In graphite each carbon bonds with three oter carbon atoms with strong covalent bond.So graphite called as covalent solid.

In the cubic crystal of $$CsCl(d=3.97 g\quad cm^{-3})$$ the eight corners are occupied by $$Cl^{-}$$ with a $$Cs^{+}$$ at the centre and vice-versa. Calculate the distance between the neighbouring $$Cs^{+}$$ and $$Cl^{-}$$ ions. What is the radius ratio of the two ions? (At. wt. of Cs$$=132.91$$ and Cl=$$35.45$$)

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Classify the following solids into different types
Plastic

Plastic is an amorphous solid substance .It has irregular arrangement of atoms or particles and do not have a definite structure. Thus it does not satisfy the definition of crystalline solid substance . So It is an amorphous solid.

Classify the following solids into different types
Ammonium Phosphate

$$(NH_4)_3PO_4$$(Ammonium phosphate ) is a ionic solid.it is a ionic solid as the constituent particles in the solid are ions and the constituent ions are $$NH_4^{+},PO_4^{3-}$$ ions.

Give the significance of a lattice point

The significance of a lattice point is that each lattice point represents one constituent particle of a solid which may be an atom, a molecule (group of atom), or an ion

Which of the following lattices has the highest packing efficiency?
(i) simple cubic
(ii) body-centred cubic
(iii) hexagonal close-packed lattice.

The packing efficiency in lattice is as follows:

(i) Simple cubic $$ = 52.4 $$ %

(ii) Body-centerd cubic $$ = 68$$ %

(iii) Hexagonal close packing $$ = 74 $$ %

Hence the maximum packing efficiency is of hexagonal packing lattice.

Potassium crystallizes in a body centred cubic lattice. How many unit cells are present in $$2g$$ potassium? (At. mass of $$K=39$$)

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What makes the crystal of $$KCl$$ appear sometimes violet?

An excess of potassium ions make $$KCl$$ crystals appear violet.

The density of $$KBr$$ is $$2.38\ g\ cm^{-3}$$. The length of the edge of the unit cell is $$700\ pm$$. Find the number of formula unit of $$KBr$$ present in the single unit cell.
$$(N_{A} = 6\times 10^{23} mol^{-1}$$, At. mass : $$K = 39, Br = 80)$$.

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Unit cell or iron crystal has edge length of 288 pm and density of 7.86 g $$cm^{-3}$$. Determine the type of crystal lattice [Fe = 56]

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Name the parameters that characterize a unit cell

The six parameters that characterize a unit cell are as follows.

(i) Its dimensions along the three edges, a, b, and c These edges may or may not be equal.

(ii) Angles between the edges These are the angles (between edges b and c), (between edges a and c), and (between edges a and b).

Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

(i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore,1/8th portion of the atom is shared by one unit cell. (ii) An atom located at the body centre of a cubic unit cell is not shared by its neighbouring unit cell. Therefore, the atom belongs only to the unit cell in which it is present i.e.,its contribution to the unit cell is 1.

$$AgCl$$ has the same structure as the $$NaCl$$. The edge length of unit cell of $$AgCl$$ is found to be $$555$$ pm and the density of $$AgCl$$ is $$5.561$$ g $$cm^{-3}$$. Find the percentage of sites that are unoccupied.

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A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

The ccp lattice is formed by the atoms of the element N. Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N. According to the question, the atoms of element M occupy 1/3rd of the tetrahedral voids. Therefore, the number of atoms of M is equal to 2 1/3 = 2/3rd of the number of atoms of N. Therefore, ratio of the number of atoms of M to that of N is M : N = (2/3):1 = 2:3 Thus, the formula of the compound is M2 N3

Analysis shows nickel oxide has the formula $$Ni_{0.98} O_{1.00}$$. What fraction of $$Ni$$ exists as $$Ni^{2+}$$ and $$Ni^{3+}$$ ions in given oxide?

Formula is $${Ni}_{0.98}{O}_{1.00}$$

Ration of $$Ni$$ to $$O = 0.98 : 1$$, i.e.,

$$0.98$$ atom of $$Ni$$ combines with $$1$$ atom of $$O$$.

Let $$x$$ be the number of atoms of $${Ni}^{+2}$$

Therefore,

Number of atoms of $${Ni}^{+3} = 0.98 - x$$

Total charge on $$Ni = -$$ Total charge on $$O$$

$$\because 1$$ oxygen atom has charge $$-2$$.

$$\therefore 3 \left( 0.98 - x \right) + 2x = 2$$

$$2.94 - 3x + 2x = 2$$

$$- x = - 0.94$$

$$\Rightarrow x = 0.94$$

Therefore,

Number of atoms of $${Ni}^{+2} = x = 0.94$$

Number of atoms of $${Ni}^{+3} = 0.98 - x = 0.98 - 0.94 = 0.04$$

Now

Percentage of $${Ni}^{+2} = \cfrac{94}{98} \times 100 \approx 95.92 \%$$

Percentage of $${Ni}^{+3} = 100 - 95.92 = 4.08 \%$$

Assertion (A): The packing efficiency is maximum for the fee structure. Reason (R): The coordination number is 12 in fee structures

In fee unit cell, there is cep arrangement with packing efficiency of 74.01% which is maximum. In cep arrangement, coordination number is 12.

Under which situations can an amorphous substance change to crystalline form?

An amorphous solid on heating at some temperature may become crystalline. Slow heating and cooling over a long period makes an amorphous solid acquires some crystalline character.

In spite of long range order in the arrangement of particles why are the crystals usually not perfect?

Crystals have long range in the arrangement of particles but usually the crystals are not perfect this is because when crystallisation occurs at a fast rate or moderate rate, the constituent particles may not get sufficient time to arrange themselves in a perfect order

Define Anisotropy. Distingusih between crystalline solids and amorphous solids.

Anistropy - It is the property of being directionally dependent means it has different properties in different directions.
Example - hight coming through a polarizer

	Crystalline solids		Amorphous solids
1.	Regular pattern and repeating arrangement of components in solid	1.	The irregular pattern of ions, molecules or atoms.
2.	Have a sharp melting point	2.	Melt over a range of a Temperature
3.	Definite heat of fusion	3.	No definite heat of fusion
4.	Eg.- Diamond	4.	Eg.- Glass

Name and draw the three common crystal structures adopted by metals.

The following are some statements related with solid state of matter:
(i) All crystalline solids are isotropic.
(ii) Amorphous solids are super cooled liquids with high viscosity
(iii) In crystals, short range order exists.
(iv) A unit cell is the smallest building unit in the crystals
(v) Every object possesses an identity element
(vi) Each lattice point in a crystal has the same environment
(vii) The basic unit must have the same stoichiometric composition as the entire crystal
(viii) A crystal space lattice cannot have a positive ion at one lattice point and a negative ion at another lattice point
The number of true statements is

$$5$$

A metal (atomic mass $$=307.2$$) exist in BCC structure. If the uncovered distance between the atoms along the edge is equal to $$120pm$$, the density of crystal (in $$g/{cm}^{3}$$) is $$\left( { N }_{ A }=6\times { 10 }^{ 23 },\sqrt { 3 } =1.732 \right) $$

For a BCC structure;

Radius of metal atom (r) is given by $$\mathrm{\cfrac{a\sqrt3}{4}}$$, where r is the radius of metal atom.

Distance left uncovered along the edge = $$\mathrm{a- 2r}$$ = $$\mathrm{a- \cfrac{a\sqrt3}{2}}$$

$$\implies$$ Distance left uncovered along the edge = $$\mathrm{a \times \cfrac{2-\sqrt3}{2} = 0.134 \ a}$$

$$\implies$$ 120 = 0.134 a

$$\implies$$ a = 895.52 pm

$$\implies$$ r = $$\mathrm{\cfrac{a\sqrt3}{4}}$$ = 387.76 pm

Density of a metal crystal is given by $$\mathrm{d = \cfrac{Z\times M}{N_A \times a^3}}$$

$$\mathrm{\implies d = \cfrac{2\times 307.2}{6 \times 10^{23} \times (895.52 \times 10^{-10})^3}}$$

$$\mathrm{\implies d = 1.65 g \ cm^{-3}}$$

List the type of defect that occur in the solid state and given an example of each. Explain in each case if any electrical conduction is possible and by what mechanism.

What is the approximate percentage of vacant space in a Silicon cubic cell having crystal structure similar to diamond?

Vacant space in diamond structure is 66%. It would be same for silicon in diamond structure as well.

What is peculiarity in electrical property of quartz?

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Find type of defect.

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Calculate the Avogadro constant from the following data:
Density of solid $$NaCl = 2.165\ g/cc$$.
Distance between centres of adjacent $$Na^{+}$$ and $$Cl^{-} = 0.2819\ nm$$.
Also, calculate the edge length of a cube containing $$1\ mole$$ of $$NaCl$$ and the number of ions $$(Na^{+} + Cl^{-})$$ along one edge of the cube.

$$density = \dfrac {Z\times \text {Formula wt of NaCl}}{N_{A}\times \text {volume of unit cell}}$$

No. of atom of $$NaCl$$ structure $$= 4$$

Formula unit $$wt = 23 + 35.5 = 58.5\ g$$

Distance between $$Na^{+}$$ and $$Cl^{-} = 0.2819\ nm$$.

In $$NaCl$$ structure $$Cl^{-}$$ form Fcc structure and $$Na^{+}$$ will be present in each octahedral void.

$$a = 2 (Dist. between\ Na^{+}$$ and $$Cl^{-}) = 2\times 0.2819\ nm$$.

$$\therefore$$ volume of cube $$= a^{3}$$

$$Density = \dfrac {4 \times 58.5\ g}{N_{A}\times (0.5638\times 10^{-7} cm)^{3}}$$

$$2.165\ g/cc = \dfrac {4\times 58.5}{N_{A} \times 0.179\times 10^{-21}}$$

$$N_{A} = \dfrac {4\times 58.5}{0.179\times 2.165}\times 10^{21} = 6.03\times 10^{23}$$

For side length of $$1$$ mole of cube

$$\therefore density = \dfrac {mass\ of\ 1\ moles}{volume\ of\ 1\ mole}$$

Volume of $$1\ mole = \dfrac {58.5\ g}{2.165} = 27.02\ cm^{3}$$

$$\therefore$$ If the side length of cube containing $$1$$ mole of substance.
$$a^{3} = 27.02\ cm^{3}\Rightarrow a = 3\ cm$$

No. of molecules in $$1$$ mole of $$NaCl = 6.023\times 10^{23}$$

No. of ions in whole volume $$= 2\times 6.023\times 10^{23} = 1.2046\times 10^{24}$$

$$\therefore$$ No of ions along are side $$= (1.2046\times 10^{24}) = 1.064\times 10^{8} ions$$.

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Total number of faces in a truncated octahedron $$= x$$
Total number of faces in a truncated tetrahedron $$= y$$
Then $$x + y$$ is :

In truncated octahedron: Hexagonal faces $$= 8$$

Square faces $$= 6$$

In truncated tetrahedron: Hexagonal faces $$= 4$$

Triangular faces $$= 4$$

Total faces $$= 8+6+4+4=22$$

Calculate the number (n) of atoms contained with:
(a)cubic cell,
(b)a body-centred cubic cell,
(c)a face-cenred cubic cell.

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Consider the arrangement of circles of equal radii with their centres arranged as per the 2-dimensional lattice defined by $$a = b,\ \theta =60^{\circ }$$ such that each circle is touching all its nearest neighbour. If all the void areas present are additionally occupied by smaller circles of relevant size so that the void circles are just contacting their neighbours find the packing efficiency of the configuration in percent.

Packing fraction $$=\dfrac{\dfrac{1}{2}\pi R^{2} + \pi (0.155R)^{2}}{\dfrac{1}{2}\left ( 2R \right )^{2}\dfrac{\sqrt{3}}{2}}\times 100 = 95\%.$$
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Why does the window glass of the old buildings look milky ?

It is due to heating during the day and cooling at night , i.e., due to annealing over a number of years , glass acquires crystalline character .

Which point defect in its crystal units increases the density of a solid ?

Interstitial defect, because in this defect a foreign atom/ions occupy interstitial positions. Hence the overall density of a solid increases.

How can a material be made amorphous ?

By melting the material and then cooling it rapidly .

What is the meaning of the term imperfection in solids ?

Imperfection refers to the irregularities in the arrangement of atoms , ions or molecules in the structure of crystalline substances.

Why is the window glass of old building thick at the bottom ?

Glass is not a true solid but a supercooled liquid of high viscosity ( called pseudo solid ). It has the property to flow. Due to the incident of sunlight for a long period of time the flows towards the bottom and thus gets thick.

Complete the statement given below by selecting the correct word/s.

___________ is an example of a crystalline substance .
[wax, sugar, tea]

Wax, sugar is an example of a crystalline substance.

Answer the following questions .
Why does presence of excess of lithium makes LiCl crystals pink ?

Excess of lithium leads to metal excess defect . Lithium atoms lose electrons to form $$ Li^+ $$ ions . These electrons diffuse in to the crystal and form F-centres. Therefore , $$ LiCl $$ crystals become pink.

Tearing of paper is said to be a change that cannot be reversed. What about paper recycling?

Recycling of paper is the same as the tearing of paper.
As on recycling properties of paper change, like the texture,
colour, quality etc. therefore, it is an irreversible change.

What is the difference between sugar and sand ?

Sugar is a substance where as sand is a mixture.

Examine the given crystal :

$$ X^+ $$ $$ Y^- $$ $$ X^+$$ $$ Y^-$$ $$X^+$$
$$ Y^- $$ $$ X^+$$ $$Y^- $$ $$X^+ $$ $$Y^- $$
$$ X^+ $$ $$ Y^- $$ $$ X^+$$ $$ e^-$$ $$X^+$$
$$ Y^- $$ $$ X^+$$ $$Y^- $$ $$X^+ $$ $$Y^- $$

Answer the following questions :
Is the above defect stoichiometric or non-stoichiometric ?

Non-stoichiometric defect

Classify the following solid as ionic, metallic, molecular, network (covalent) or amorphous.
$$I_2$$

$$I_2$$ is a Molecular solid.

Answer the following questions .
Write any two differences between amorphous solids and crystalline solids .

$$ { Distinction \ between \ Crystalline \ and \ Amorphous \ Solids } $$

Property	Crystalline Solids	Amorphous Solids
Shape	Definite characteristics and geometric shape .	Irregular shape .
Melting point	Melt at a sharp and characteristics temperature .	Gradually soften over a range of temperature .
Cleavage property	When cut with a sharp edged tool , they split into two pieces and then newly generated surfaces are plain and smooth	When cut with a sharp edged tool , they cut into two pieces with irregular surfaces .
Heat of fusion	They have definite and characteristics heat of fusion .	They do not have a definite heat of fusion.
Isotropy	Anisotropic in nature.	Isotropic in nature.
Nature	True solids .	Pseudo solids or super cooled liquids .
Order in arrangement of constituents particles	Long range order .	Only short range order .

Calculate the efficiency of packing in case of metal crystal for body-centred cubic.

spheres at the corners, body diagonal, AD $$= 4r$$ further, face diagonal,

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{a^2 + a^2} = \sqrt{2a}$$

and body diagonal,

$$AD = \sqrt{AC^2 + CD^2} = \sqrt{2a^2 + a^2} = \sqrt{3a}$$

$$\sqrt{3a} = 4r$$ or $$a = \dfrac{4r}{\sqrt 3}$$

$$\therefore$$ volume of the unit cell

$$= a^3 = \left [ \dfrac{4r}{\sqrt 3} \right ]^3 = \dfrac{64 r^3}{3\sqrt 3}$$

Number of spheres per unit cell $$= 8 \times \dfrac{1}{8} + 1 = 2$$

volume of $$2$$ spheres $$= 2 \times \dfrac{4}{3} \pi r^3 = \dfrac{8}{3} \pi r^3$$

$$\therefore$$ Fractional occupied i.e. packing fraction

$$= \dfrac{ \dfrac{8}{3} \pi r^3}{\dfrac{64r^3}{3 \sqrt 3}} = \dfrac{\pi \sqrt 3}{8} = 0.68$$ or % occupied $$= 68\%.$$

1740466_1827523_ans_55f3967b63974bb0a416ff32c352e9fb.png

What do you understand by 'lattice point' ?

Each $$\text{lattice point}$$ represents one constituent particle of solid. This constituent particle may be an atom, a molecule (group of atoms) or an ion.

Why solids are hard ?

In solids the positions of constituent particles like atoms or molecules or ions are fixed i.e., they are not free to move. They are stationary in their fixed position. They have strong intermolecular or interionic forces of attraction. Hence solids are hard.

What type of stoichiometric defect is shown by ZnS

Frenkel defect type of stoichiometric defect is shown by ZnS.

How does amorphous silica differ from quartz ?

In amorphous Silica, $$SiO_4$$ tetrahedra are randomly joined to each other whereas in quartz, they are linked in a regular manner.

Calculate the efficiency of packing in case of metal crystal for simple cubic.

Top/Bottom/side view of simple cubic unit cell As spheres are touching each other,

evidently $$a = 2r$$

Number of spheres per unit cell $$= \dfrac{1}{2} \times 8 = 1$$

Volume of the sphere $$= \dfrac{4}{3} \pi r^3$$

Volume of the cube $$= a^3 = (2r)^3 = 8r^3$$

$$\therefore$$ Fraction occupied, i.e. packing fraction $$(\dfrac{4}{3} \pi r^3) / 8r^3$$

or % occupied $$= 52.4\%$$

1740414_1827499_ans_3da4c5c43b624e598e6d6bd05b17eea6.png

What type of stoichiometric defect is shown by AgBr ?

AgBr shows both $$\text{Frenkel as well as Schottky defects.}$$

Classify the following solid as ionic, metallic, molecular, network (covalent) or amorphous.
Brass.

Brass is an alloy of Cu and Zn. It is a Metallic solid.

Calculate the efficiency of packing in case of metal crystal for face-centred cubic.

A, B, C are the centres of the sphere. As sphere on the face centre is touching the spheres at the corners evidently $$AC = 4r$$ But from right angled triangle ABC

$$AC = \sqrt{AB^2 + BC^2} + \sqrt{a^2 + a^2} = \sqrt{2}a$$

$$\therefore \sqrt{2} a = 4r$$

or $$a = \dfrac{4}{\sqrt 2^r}$$

$$\therefore$$ volume of the unit cell

$$= a^3 = \left [ \dfrac{4}{\sqrt 2}r \right ]^3 = \dfrac{32}{\sqrt 2}r^3$$

Number of spheres in the unit cell

$$= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = 4$$

volume of 4 spheres $$= 4 \times \dfrac{4}{3} \pi r^3 = \dfrac{16}{3} \pi r^3$$

$$\therefore$$ fraction occupied, i.e. packing fraction

$$= \dfrac{16 \pi r^3 / 3}{32 r^3 \sqrt 2} = \dfrac{\pi \sqrt 2}{6} = 0.74$$

or % occupied $$= 74\%$$

1740488_1827541_ans_fa56c1d08699471f8777215c37c585cc.png

Ionic solids which have anionic vacancies due to metal excess defect , develop colour. Explain with the help of a suitable example.

When $$NaCl$$ crystals are heated in the presence of sodium vapour some chloride ions left the lattice sites and combine with sodium to form NaCl. For this combination sodium atoms lose electrons to form $$Na^+$$ ions. The released electrons diffuse into crystal to occupy the anion vacancies which are created by $$Cl^-$$ ions. In this condition the crystal have excess of sodium. The sites occupied by unpaired electrons are called F-centres or colour centres. F-centre imparts yellow colour to $$NaCl$$ crystal because they absorb energy from the visible light and get excited.

How will you distinguish between the following pairs of terms
Hexagonal and monoclinic unit cell

$$\text{Hexagonal and monoclinic unit cell}$$

Properties	Hexagonal unit cell	Monoclinic unit cell
(i) No. of three dimensional lattices.	$$1$$	$$2$$
(ii) Possible variations	Primitive	Primitive and end centered
(iii) Edge lengths	$$ a = b \neq c $$	$$ a \neq b \neq c $$
(iv) Axial angles	$$ \alpha = \beta = 90^{\circ} , \gamma =120^{\circ} $$	$$ \alpha = \gamma =90^{\circ}, \beta \neq 120^{\circ} $$
(v) Examples	Graphite , ZnO , CdS Sulphur	$$ Na_2SO_4.10H_2O$$

Explain the following term with suitable examples:
Interstitial defect

In this case, the extra anions may be occupying interstitial positions, The extra negative charge is balanced by the extra charges (oxidation of equal number of cations to higher oxidation states) on the adjacent metal ions. Such type of defect is not common because the negative ions are usually very large and they can not easily fit into the interstitial sites.
1744859_1830817_ans_764cc4a6ad96462ea286e865c8c80c2e.png

1744859_1830817_ans_764cc4a6ad96462ea286e865c8c80c2e.png

What are crystals; How can we make growing crystals?

Prepare a saturated solution of a solute like copper sulfate. When we allow it to evaporated slowly, the crystals are appeared. Select a suitable crystal and hang it in the solution with a twain. After a few days, the crystal can be seen grown as well..

The crystalline parts of the amorphous solids are called ....... .

The crystallites are small (can be microscopic) crystal which are formed during the crystallization process. Due to short range strong attraction force possessed by amorphous solids, small parts of amorphous solids are crystallized, they are known as crystallites.

An element X (atomic weight $$= 24$$ g/mol) forms a face centered cubic lattice. If the edge length of the lattice is $$4\times10^{-8}$$ cm and the observed density is $$2.40 \times10^{3}$$ g cm$$^{-3}$$, then the percentage occupancy of lattice points by element X is : (Use $$N_A=6\times10^{23}$$)

Given that,

Atomic weight of element $$X=24$$ $${g/mol}$$

And element $$X$$ is in FCC.

We know that,

Density of unit cell $$=\cfrac{Z\cdot M}{a^3\times N_A}$$

where, $$Z=$$ Number of atoms of elements in unit cell.

$$M=$$ Molecular weight

$$a^3=$$ Volume of unit cell

$$N_A=$$ Avogadro's Number

$$a=$$ Edge length

In FCC $$H-$$ atom is present.

So, Density $$=\cfrac{4\times 24}{64\times 6\times 10^{23}\times 10^{-24}}$$ $$g/cm^3=0.25\times 10^1$$ $${g/cm^3}$$

Calculated Density $$=2.5$$ $${g/cm^3}$$

$$Percentage\quad occupancy=\cfrac{Observed\quad density}{Calculated\quad density}=\cfrac{2.4}{2.5}\times 100=96\%$$

In face centred cubic (fcc) crystal lattice, edge length is $$400$$ pm. Find the diameter (in pm) of a greatest sphere which can be fit into the interstitial void without distortion of a lattice.
(Give your answer to the nearest integer)

In fcc lattice, the largest void present is octahedral void.

If the radius of void sphere is $$R$$ and of lattice sphere is $$r$$, then
$$\displaystyle \mathrm{r}=\frac{\sqrt{2}\times 400}{4}=141.42$$ pm

Applying condition for octahedral void

$$2(r+R)=$$ edge length, $$a$$

$$\therefore 2R=a-2r$$

$$2R=400-(2\times 141.42)=117.16$$

Hence, the diameter of greatest sphere is $$117.16$$ pm.

Match the crystal system/unit cells mentioned in Column I with their characteristic features mentioned in Column II.
Indicate your answer by darkening the appropriate bubbles of the 4 X 4 matrix given in the ORS.

(A) Simple cubic and face-centred cubic unit cells have cell parameters as $$a=b=c$$ and $$\alpha =\beta=\gamma$$.
They belong to same crystal system.
(B) Cubic and rhombohedral unit cells have have cell parameters $$a=b=c$$ and $$\alpha =\beta=\gamma$$.
They are two crystal systems.
(C) Cubic and tetragonal unit cells are two crystal systems.
(D) Hexagonal and monoclinic unit cells are two crystal systems.
They have have only two crystallography angles of $$90^{\mathrm{o}}$$.

This section contains questions each with two columns-I and II. Match the items given in column I with that in column II (unoccupied volume).

hcp and ccp unit cells represent the closest packed structure.

The volume occupied (as a fraction of total volume) is 0.78
The volume unoccupied (as a fraction of total volume) is $$1-0.78 = 0.22$$.

For bcc unit cell, the volume occupied (as a fraction of total volume) is 0.62.
The volume unoccupied (as a fraction of total volume) is $$1-0.62 = 0.38$$.

For sc unit cell, the volume occupied (as a fraction of total volume) is 0.52.
The volume unoccupied (as a fraction of total volume) is $$1 -0.52 = 0.48$$.

For dc unit cell, the volume occupied (as a fraction of total volume) is 0.34 .
The volume unoccupied (as a fraction of total volume) is $$1-0.34 = 0.66$$.

Give the total score of the correct statements of the following.
Statements Score
a. In an antifluorite structure, cations are present in all tetrahedral voids. $$1$$
b. If the radius of cation is $$0.35$$ pm and that of anion is $$0.95$$ pm, then the co-ordination number of the crystal is $$4$$. $$2$$
c. An atom or ion is transfered from a lattice site to an interstitial position in Frenkel defect. $$3$$
d. The density of a crystal always decreases in point defects. $$4$$

Statements (a), (b) and (c) are correct. So, the total score $$= 1+ 2 + 3=6$$
Statement (a, b) are factual.
In an antifluorite structure, cations are present in all tetrahedral voids.
If the radius of cation is $$0.35$$ pm and that of anion is $$0.95$$ pm, then the coordination number of the crystal is $$4$$.
$$\dfrac {r_\oplus }{r_\circleddash }=\dfrac {0.35}{0.95}=0.368$$
The radius ratio lies in the range of $$0.225-0.414$$, which corresponds to tetrahedral void and the coordination number of tetrahedral void $$= 4$$
An atom or ion is transferred from a lattice site to an interstitial position in Frenkel defect.
The density of a crystal may or may not decrease in point defects.

Which stoichiometrie defect increases the density of a solid?

Schottky defect : The absence of a cation or an anion from the position which it is expected to occupy in the periodic arrangement of ions since the solid with a Schottky defect contains lesser ions as compared to perfect crystal. The density of crystal exhibiting Schottky defect will be less as compared to that of perfect crystal.

If $$NaCl$$ is doped with $$10^{-3}$$ $$mol$$ percent of $$SrCl_{2}$$, what is the concentration of cation vacancies?

$$NaCl$$ is doped with $$10^{−3}$$ mol % of $$SrCl_2$$.

$$100$$ moles of $$NaCl$$ are doped with $$0.001$$ moles of $$SrCl_2$$.

$$1$$ mole of $$NaCl$$ is doped with $$\displaystyle \frac {0.001}{100} = 10^{-5}$$ moles of $$SrCl_2$$ or $$10^5$$ moles of $$Sr^{2+}$$ cations.
One $$Sr^{2+}$$ ion creates one cation vacancy.

$$10^{-5}$$ moles of $$Sr^{2+}$$ will create $$10^{-5}$$ moles of cation vacancies which correspond to $$\displaystyle10^{-5} \times 6.023 \times 10^{23} = 6.023 \times 10^{18} $$ cation vacancies per mole of $$NaCl$$.

When heated above $$916^{\circ}C$$, iron changes its crystal structure from body-centered cubic to cubic closed packed structure. Assuming that the metallic radius of the atom does not change, calculate the ratio of the density of the bcc crystal to that of the ccp crystal.

In body-centred packing, the efficiency of packing is $$67.92$$%. In the cubic closed packing, the packing efficiency is $$74.02$$%.
Let $$\rho_{1}$$ be the density when packing efficiency is $$74.02$$% and $$\rho_{2}$$ is the density when packing efficiency is $$67.92$$%.
$$\dfrac {\rho_{2}}{\rho_{1}} = \dfrac {67.92}{74.02} = 0.918$$.

You are given marbles of diameter $$10\ mm$$. They are to be placed such that their centers are lying in a square bond by four lines each of length $$40\ mm$$. What will be the arrangement of marbles in a plane, so that, the maximum number of marbles can be placed inside the area? Sketch the diagram and derive the expression for the number of marbles per unit area.

In order to accomodate maximum number of spheres, there should be hcp (hexagonal closed packing).
Area of square having spherical marbles $$= 16\ cm^{2}$$
Maximum number of spheres $$= 14 (full) + 8 (half)$$
Number of spheres $$= 18$$
Per unit area (per $$cm^{2}) = \dfrac {18}{16} = 1.125$$
Length $$PQ$$ of square $$= 4\ cm$$
Length $$PR = 5 + 4\times 5\sqrt {3} = 40\ mm = 4\ cm$$
$$CD = 10\sin 60^{\circ} = \dfrac {10\sqrt {3}}{2} = 5\sqrt {3}$$.

Distinguish between crystalline solid and amorphous solid.

$$\textbf{CRYSTALLINE SOLIDS}$$	$$\textbf{AMORPHOUS SOLIDS}$$
Long-range order	Short-range order
Sharp melting point	Do not have sharp melting point i.e. melt over a range of temperature
Undergo regular breakages i.e. can be cleared along with definite plans	Undergo irregular breakage
Anisotropic in nature i.e. properties like thermal, expansion, conductivity will be different in a different direction.	Isotropic in nature i.e. properties like thermal, expansion, conductivity are independent of direction
True solids	Pseudo solids or super cooled liquids

$$BaTiO_3$$ crystallizes in the perovskite structure. This structure may be described as a cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying the face centers and titanium ions occupying the centers of the unit cells.
(a) If titanium is described as occupying holes in the $$BaO$$ lattice, what type of hole does it occupy?
(b) What fraction of the holes of this type does it occupy?
(c) Can you suggest a reason why it has certain holes of this type but not the other of the same type?

(a) Since $$Ti^{+4}, $$ is very small in size it will occupy octahedral hole present in the body center of FCC lattice.

(b) It occupies $$1$$ out of $$4$$ octahedral hole i.e.$$\cfrac {1}{4}$$.

(c) Since a lattice has definite symmetry and composition. These both have been changed if $$Ti^{+4}$$ have occupied octahedral hole present at edge centers. Also since $$Ti^{+4}$$ is very small, it fits perfectly into body centered octahedral hole.

An element 'X' (At. mass $$=40$$g $$mol^{-1}$$) having f.c.c structure, has unit cell edge length of $$400$$ pm. Calculate the density of 'X' and the number of unit cells in $$4$$g of 'X'. $$(N_A=6.022\times 10^{23}mol^{-1})$$.

$$ \text { edge length } = \text { 400 pm} = \text { 400 }\times 10^{-10} \text { cm}= \text { 4 }\times 10^{-8} \text { cm}$$

$$ \text { Density } = \dfrac { \text { Atomic mass} \times \text { Number of atoms in 1 unit cell } }{ \text { Avogadro's number} \times ( \text { edge length} )^3 } $$

$$ \text { Density } = \dfrac {40g mol^{-1} \times \text { 4 } }{ 6.022\times 10^{23}mol^{-1} \times ( \text { 4 }\times 10^{-8} \text { cm} )^3 } $$

$$ \text { Density } =\text {4.15 g/cm}^3 $$

Volume of 4 g of 'X' $$=\dfrac { \text { 4 g }}{\text {4.15 g/cm}^3}=\text {0.964 cm}^3$$

Volume of one unit cell $$=( \text { edge length} )^3= ( \text { 4 }\times 10^{-8} \text { cm} )^3=6.4 \times 10^{-23}\text { cm}^3$$

The number of unit cells in 4g of 'X' $$= \dfrac {\text {0.964 cm}^3}{6.4 \times 10^{-23}\text { cm}^3}=1.5 \times 10^{22}$$

Using the data given below, find the type of cubic lattice to which the crystal belongs.
$$Fe$$ $$V$$ $$Pd$$
$$a$$ in pm $$286$$ $$301$$ $$388$$
$$\rho$$ in gm $$cm^{-3}$$ $$7.86$$ $$5.96$$ $$12.16$$

The theoretical density of the crystal is given by

$$\rho = \cfrac {nM}{N_AV}= \cfrac {nM}{N_A a^3}$$

$$n=$$ number of molecules / atoms / ions in unit cell

$$M=$$ Molar mass of substance

$$V=$$ Volume of unit cell

$$a=$$ edge length of unit cell

For $$Fe$$:-

$$n= \rho \cfrac {N_AV}M$$

Here, $$\rho = 7.86 \ g/cm^3 \ ; \ N_A= 6.022 \times 10^{23} \ ; a= 286 \times 10^{-10} \ cm \ ; \ M=56 \ g$$

$$\Rightarrow n = \cfrac {7.86 \times 6.022 \times 10^{23} \times (286)^3 \times (10^{-10})^3}{56}= 2$$

So it has body centered cubic lattice (BCC)

For $$V$$:-

$$\rho = 5.96 \ g/cm^3 \ ; \ N_A= 6.022 \times 10^{23} \ ; a= 301 \times 10^{-10} \ cm \ ; \ M=51 \ g$$

$$\Rightarrow n = \cfrac {5.96 \times 6.022 \times 10^{23} \times (301)^3 \times (10^{-10})^3}{51}= 2$$

So unit cell of $$V$$ has body centered cubic lattice (BCC)

For $$Pd$$:-

$$\rho = 12.16 \ g/cm^3 \ ; \ N_A= 6.022 \times 10^{23} \ ; a= 388 \times 10^{-10} \ cm \ ; \ M=106 \ g$$

$$\Rightarrow n = \cfrac {12.16 \times 6.022 \times 10^{23} \times (388)^3 \times (10^{-10})^3}{106}= 4$$

So unit cell of $$Pd$$ is face centered cubic lattic (FCC)

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 $$1.00\ cm^{3}$$ of aluminium?

i) In CCP,

$$4r=a\sqrt{2}$$

$$4\times 125=500=a\sqrt2$$

$$a=353.6$$ pm $$=3.536\mathring A$$

ii) Let the number of atoms be $$n$$

$$V=n\times V_m$$

$$10^{-6}=n\times (3.536)^{3}\times 10^{-30}$$

$$n=\cfrac {10^{24}}{(3.536)^3}=\cfrac {10^{24}}{44.21}=2.262\times 10^{22}$$ atoms

Drop a crystal of copper sulphate or potassium permanganate into a glass of hot water and another containing cold water. Do not stir the solution. Allow the crystals to settle at the bottom.
What do you observe just above the solid crystal in the glass?
What happens as time passes?
What does this suggest about the particles of solid and liquid?
Does the rate of mixing change with temperature? Why and how?

When crystals are added in a glass of cold and hot water, the following observations are observed:

i) In a glass of cold water, the crystals settle down to the bottom and in case of hot water, it slowly spreads. The crystals can be observed dissolving in the water just above and blue/ purple color is observed.

ii) After some time, the hot water becomes blue/purple and in cold water, no appreciable color change is observed because of less dissolution of crystals.

iii) From the above observation, we can conclude that particles of matter move continuously and possess kinetic energy. The particles of matter also have diffusion property.

iv) As the kinetic energy of particles depends on temperatures and increases with the increase of temperature. So, the rate of mixing also increases as observed in the case of hot and cold water.

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2g/c$$m^3$$. How many atoms present in 208g of the element.

Volume of unit cell $$=(288\ pm)^3=(288\times10^{-10}\ cm)^3=2.389\times10^{-23}\ cm^3$$

Volume of 208 g of the element $$=\dfrac{Mass}{Density}=\dfrac{208}{7.2}=28.89\ cm^3$$

Number of unit cells $$=\dfrac{Total\ Volume}{Volume\ of\ a\ unit\ cell}=\dfrac{28.89}{2.389\times10^{-23}}=12.09\times10^{23}$$

For a BCC structure, number of atoms per unit cell $$=2$$

$$\therefore$$ Number of atoms present in 208 g $$=$$ No. of atoms per unit cell $$\times$$ No. of unit cells

$$=2\times12.09\times10^{23}$$

$$=24.18\times10^{23}$$

$$=2.418\times10^{24}$$

Why is glass considered a super cooled liquid?

Glass is basically an amorphous solid. It does not form a crystalline structure. So, the constituent particles of the glass can move. In regular solids, there is no movement of constituent particles under normal conditions.Due to this fluidity property,$$\text{ glass is called as supercooled liquid}$$. Glass can be considered as a liquid of extremely high viscosity.The evidence of the fact can be seen in the windows getting thicker at bottom over a period of time.

Find the void space in fluorite structure per unit volume of unit cell.

In fluorite structure $$(CaF_2)$$, the cations $$(Ca^{2+})$$ ion forms ccp arrangement and anions $$(F^{2-}) ion occupy all tetrahedral voids.

So, along face diagonal, cations will be touching each other. If $$r_f$$ is radius of cations and $$a$$ is the edge length of the cube, then,

$$\sqrt 2 a = 4r_+ \Rightarrow a = 2 \sqrt 2 r_+$$

So, $$a^3 = 16 \sqrt 2 r_+^3$$

Now, if $$r_- $$ is the radius of the anion. So, for tetrahedral voids, we have

Radius of cation $$(r_+)= \cfrac 1{0.225} $$ (For this crystal type)

Radius of anion $$(r_-)

So, packing fraction $$= \cfrac {4 \times \cfrac 43 \times (r_+)^3 + 8 \times \cfrac 43 \pi (r_-)^3}{16 \sqrt 2 r_+^3}$$

$$= \cfrac {\pi}{3 \sqrt 2} + \cfrac {2 \pi}{3 \sqrt 2} \left (\cfrac {r_-}{r_+}\right)^3$$

$$= \cfrac {\pi}{3 \sqrt 2} [1+ 2 \times (0.225)^3] = 0.749$$

So, void fraction $$= 1-0.749 = 0.251$$

Void % $$= 25.1 $$%

Why do hexagonal and trigonal lattice do not have bcc and fcc variations?

For trigonal system crystal parameters are given as $$a=b=c$$ and $$\alpha = \beta = \gamma \neq 90^o$$. So, the length of the sides and angles of unit cell are equal but the shape is harder to visualize because $$\alpha, \ \beta, \ \gamma$$ can be anything.

For hexagonal system, the crystal parameters are given as $$a=b \neq c; \ \alpha = \beta =90^o$$ and $$\gamma=120^o$$. So, we have got one restriction to keep $$\gamma =120^o$$

As, in case of the cubic unit cell, we have greater symmetry so we can have bcc and fcc variations but due to lack of such symmetry in trigonal and hexagonal systems we cannot have such variations.

What is the packing efficiency for end centred unit cell?

In end centered unit cell the atoms are present on the corners and on the centers of any two opposite faces. So, effective number of atoms in unit cell $$= \cfrac 18 \times 8 + \cfrac 12 \times 2= 1+1=2$$

Now, at one of the faces toms along face diagonals are touching each other. If $$a$$ is the edge length of cube and $$r$$ is the radius of the atom.

$$\Rightarrow \sqrt 2 a=4r$$

So, volume occupied $$= 2 \times \cfrac 43 \times \pi \times r^3$$

Total volume $$= a^3= \left (\cfrac {4r}{\sqrt 2}\right )^3= \cfrac {64}{2 \sqrt 2} r^3 = \cfrac {32}{\sqrt 2} r^3$$

So, $$ packing \ efficiency = \cfrac {volume \ occupied}{total \ volume} \times 100$$%

$$= \cfrac {\cfrac 43 \times \pi \times r^3 \times 2 }{\cfrac 32 \sqrt 2 \times r^3} \times 100$$%

$$=37.02$$%

In terms of bond theory, explain the difference between a conductor (metal) , semiconductor and insulator. Give examples.

According to band theory a partially filled valance band will show conduction means will behave as conductor fully filled band show insulation means will behave as insulator for example lithium behave as conductor.

In case of semiconductors,the gap between the valence band and conduction band is small,therefore,some electrons may jump to the conduction band and shows some conductivity.

For example silicon germanium.

Examine the defective crystal lattice given below and answer the following questions:
(a) Name of defect present in ionic solid.
(b) Out of AgCl and NaCl, which is most likely to show this type of defect and why?
(c) Why the defect is also as dislocation defect?

This is Frankel defect

AgCl shows Frenkel defect while NaCl does not. ... Frenkel defect is shown by those in which there is large difference in the size of cations and anions. Hence,Frenkel defect is shown by AgCl due to small size of Ag+ ion but not by NaClbecause alkali metal ions can not fit into interstital sites.

this defect is call dislocation defect because in this a cation genarrly is dislocated from its site.

Metallic gold crystallises in a face-centredcubic lattice. The length of the cubicunit cell is a = 4.0 mass of gold = 197 u)
(a) What is the closest distance between gold atoms?
(b) How many "nearest neighbours" does each gold atom have at a distance calculated in (a)?
(c) What is the density of gold?

(A)

(B)Each sphere is surrounded by 12 nearest neighbour.

Thus in case of metallic gold crystal each gold atom has 12 nearest neighbour.

(C)

If the radius of the octahedral void is $$r$$ and the radius of the atoms in the close packing is $$R,$$ derive relationship between $$r$$ and $$R$$.

1943356_1163171_ans_d00fd0f803d444a8acc3cfc3cbc0cbb4.png

Ag forms $$CCP$$ lattice with edge length $$408.6$$pm calculate density of silver. $$(Atomic mass =107.9\mu)$$

$$a=408.6,pm=408.6 \times 10^{-12}m=408.6\times 10^{-10}cm$$
$$m=107.9$$
$$Z=4$$ for CCP
mass of Ag=$$\dfrac{107.9}{6.023 \times 10^{23}}=17.93\times 10^{23}g$$
$$d=\dfrac{ZM}{a^3NA}$$
$$d=\dfrac{4\times 17.93\times 10^{-23}}{(408.6\times 10^{-10})^36.023 \times 10^{23}}$$
$$d=1.74\times 10^{-23}$$ $$g/cm^3 $$

What type of stichiometric defect is shown by $$KCl$$ and Why?

1331830_1316825_ans_851c0d228f8a4e0aa9b7e1f8b15c5597.jpg

What is the relation between edge length and angles between the edges in triclinic crystal system.

$$a \neq b\neq c$$, $$\alpha \neq \beta \neq \gamma$$
2041492_1360811_ans_a8bc6415e7284593b4c1a32dcadf02e6.jpg

Give reasons:
In stoichiometric defects, $$NaCl$$ exhibits Schottky defect and not Frenkel defect.

1657237_1385301_ans_0546a710520648d8ba93527e9b9c9674.jpg

There are three solids made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have highest inertia? Give reason for your answer.

Inertia mainly depends on mass .The inertia is simply the mass of the object. Since steel has the highest density the steel solid has the highest mass and therefore the most inertia.

How 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?

Hexagonal Close Packing

1. When the tetrahedral voids of the second layer are covered by the spheres of the third layer. So, that the spheres of the third layer are exactly aligned with those of the first layer, we get a pattern of the sphere which is repeated in alternate layers. This pattern can be written in the form of an ABAB..... pattern. This structure is called hexagonal closed packing (HCP).

2. Ex :- Mg ( Magnesium ) , Ti ( Titanium ) , etc.

Cubic Close Packing

1. When the third layer is placed above the second layer in a manner such that its sphere covers the octahedral voids and the sphere of the third layer is not aligned with the first or the second layer, the arrangement is called ′C′ type. This pattern is written as ABCABC........ This structure is called cubic closed packing(CCP).

2. Ex:- Au ( Gold ), Ag ( Silver ), etc.

Crystal Lattice

1. A lattice is a regular repeated three-dimensional arrangement of atoms, ions, or molecules in a metal or other crystalline solid.

2. Lattice is a very large, complex structure.

3. A lattice has large numbers of constituents (atoms, molecules, or ions).

Unit Cell

1. A unit cell is a simple arrangement of spheres (atoms, molecules, or ions) that resemble the repeating pattern of a lattice.

2. The unit cell is the simplest and smallest repeating unit of a lattice.

3. A unit cell has a small number of constituents.

Tetrahedral Void

1. Tetrahedral void is a simple triangular void in a crystal and is surrounded by four spheres arranged tetrahedrally around it.

2. Tetrahedral Voids can be found in substances having a tetrahedral arrangement in their crystal system.

3. Tetrahedral Voids can be observed in the edges of the unit cell.

4. The coordination number of the tetrahedral void is four.

5. There are two tetrahedral voids per sphere in the crystal lattice.

Octahedral Void

1. Octahedral void is a double triangular void with one triangle vertex upwards and the other triangle vertex downwards and is surrounded by six spheres.

2. Octahedral Voids can be found in substances having an octahedral arrangement in their crystal system.

3. Octahedral voids can be observed in the center of the unit cell.

4. The coordination number of the octahedral void is six.

5. There is one octahedral void per sphere in the crystal lattice.
Draw structure of tetrahedral and octahedral void. - Sarthaks eConnect | Largest Online Education Community

Draw structure of tetrahedral and octahedral void. - Sarthaks eConnect | Largest Online Education Community

Why are solids incompressible?

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Which of the following is true about the value of refractive index of quartz glass?
(a) Same in all directions
(b) Different in different directions
(c) Cannot be measured
(d) Always zero

Hint:

Isotropic properties:- Isotropic properties refers to the properties of a material which is independent of the direction which means that properties are same in all the directions.

Quartz glass is amorphous solid i.e., isotropic in nature. Therefore, value of any physical property (e.g. refractive index) would be same along any direction.

Therefore, The correct answer is option (A).

Draw a diagram of 'two dimension square closed packing'.

In two-dimensional close packing, row of closed packed spheres are stacked to obtain atwo-dimensional pattern. This stacking is done intwo ways: Square close packing: The second row can be placed exactly below the first row in a close packing.

1204683_1513333_ans_cc72e3d2f6404dafaf357dc11c67faa1.jpg

1204683_1513333_ans_cc72e3d2f6404dafaf357dc11c67faa1.jpg

Out of NaCl and AgCl, which one shows Frenkel defect and why?

AgCl show Frenkel defects and NaCl is Schottky defects.

Frenkel defect is shown by those in which there is large difference in the size of cations and anions. Hence, Frenkel defect is shown by AgCl due to small size of Ag$$^+$$ ion but not by NaCl because alkali metal ions cannot fit into interstital sites.

Calculate the packing fraction for the $$K$$ unit cell. $$K$$ crystallizes in a body-centred cubic unit cell.

In BCC no. of effective atoms $$=2$$ and $$\sqrt {3}a = 4r \Rightarrow a = \dfrac {4\pi}{\sqrt {3}}$$.

$$P.F. =\dfrac{2\times \text{Volume of K atom}}{\text{Volume of the unit cell}} =\dfrac {2\times \dfrac {4}{3} \pi r^{3}}{\left (\dfrac {4}{\sqrt {3}}r\right )^{3}} = 0.68$$

The density of solid argon is $$1.65g/{ cm }^{ 3 }$$ at $$-{ 233 }^{ o }C$$. if the argon atoms are assumed to be spheres of radius $$1.54\times { 10 }^{ -8 }cm$$, the approximate percentage of empty space in solid argon is $$(Ar=40)$$

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Calculate the percentage of vacant space in a Si unit cubic cell. The unit-cell content for Si is $$8$$ and $$r = \dfrac {\sqrt {3}a}{8}$$.

No. of atom unit cell $$= 8$$

Radius $$= \dfrac {\sqrt{3}a}{8}$$

Packing efficiency $$= \dfrac {\text{Volume of atom}}{\text{Volume of unit cell}}$$

$$p.e = \dfrac {8\times \dfrac {4}{3}nr^{3}}{a^{3}}$$

$$r = \dfrac {\sqrt{3}a}{8}$$

$$=\dfrac {8\times \dfrac {4}{3}\pi \left(\dfrac {\sqrt {3}a}{8}\right)^{3}}{a^{3}}$$

$$p.e = \dfrac {8\times \dfrac {4}{3}\pi \left(\dfrac {\sqrt {3}}{8}\right)^{3}a^{3}}{a^{3}}$$

$$p.e = 34\%$$

Percentage of vaccant space $$= 100 - 34 = 66\%$$.

Match Column I with Column II and Column III
Column I Column II Column III
(A) $$NaCl$$ (P) FCC, anion in all tetrahedral voids (W) Cation (6), anion (6)
(B) $$Ca{F}_{2}$$ (Q) FCC, cation in all octahedral voids (X) Cation (8), anion (4)
(C) $$ZnS$$ (Zinc blende) (R) FCC, cation in alternate tetrahedral voids (Y) Cation (6), anion (6)
(D) $${Na}_{2}O$$ (S)FCC, cation in all tetrahedral voids (Z) Cation (6), anion (6)

$$A\rightarrow Q,W$$
$$B\rightarrow P, X$$
$$C\rightarrow R,Z$$
$$D\rightarrow S,Y$$

You are given marble of diameter $$10mm$$. They are to be placed such that their centres are lying in a square bond by four lines each of length $$40mm$$. The maximum number of marbles which may be placed inside this area is

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A simple cubic lattice consists of eight identical spheres of radius $$R$$ in contact, placed at the corners of a cube. What is the volume of the cubical box that will just enclose these eight spheres and what fraction of this volume is actually occupied by the spheres?

Consider one face as shown in fig.

To enclose all these spheres we have to increase the side length by $$2R$$

$$\therefore$$ New side length $$= 2R + 2R = 4R$$

$$\therefore$$ Volume of such cube $$= (4R)^{3} = 64R^{3}$$

Volume of sphere $$= 8\times \dfrac {4}{3} \pi R^{3}$$

$$\%$$ volume occupied by atom $$= \dfrac {8\times (\dfrac{4}{3}) \times \pi R^{3}}{64R^{3}} \times 100 = \dfrac {\pi}{6} \times 100 = 52.33\%$$.

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Rutile, a mineral that contains only titanium and oxygen atoms, has a structure that can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral voids.
$$(Ti=48)$$
$$ab=$$ the mass percentage of titanium in rutile
$$c=$$ magnitude of oxidation state of titanium in rutile
$$d=1$$ if the oxidation state of $$Ti$$ is positive and $$2$$ if the oxidation state of $$Ti$$ is negative
The value $$abcd$$ is

No. of Oxygen atoms in a unit cell of Rutile = Z (CCP) = 4

No. of Titanium atoms in a unit cell of Rutile = $$\mathrm{\cfrac{Z(Octahedral \ of \ CCP)}{2} = \cfrac{4}{2} }$$ = 2

So, the simplest whole no. ratio of titanium and oxygen is 1:2

Therefore, Rutile is $$\mathrm{TiO_2}$$

Mass % of Ti in Rutile = $$\mathrm{\cfrac{Mass \ of \ Ti}{Mass \ of \ Rutile} = \cfrac{48}{80} = 60 \ \%}$$

$$\mathrm{\implies ab = 60}$$

Oxidation state of Ti Rutile = +4

$$\implies\mathrm{c = 4}$$

Since oxidation state is positive

$$\implies \mathrm{d=1}$$

Therefore, the value of $$\mathrm{abcd = 6041}$$

Define void .

The empty spaces present between the atoms or the ions when they are packed within the crystal are called voids .

What are types of lattice imperfections found in crystals ?

(a) Stoichiometric defects , viz ., Schottky defect and Frenkel defect (b) Non-stoichiometric defects , viz ., metal excess , metal deficiency , and (c) Impurity defects .

Calculate the packing factor for spheres occupying (a) a body-centred cubic structure, and (b) a simple cubic structure, where closest neighbours in both cases are in contact.

$$P.F. = \dfrac {\text{No. of atoms in each unit cell}\times \text {volume of each atom}}{\text {volume of unit cell}}$$

$$P.F_{BCC} = \dfrac {2\times \dfrac {4}{3} \pi r^{3}}{\left (\dfrac {4r}{\sqrt {3}}\right )^{3}} = 0.68$$ (In $$BCC,\ 4r = \sqrt {3}a$$ and number of atoms in unit cell $$=2$$)

$$P.F_{SCC} = \dfrac {1 \times \dfrac {4}{3}\pi r^{3}}{(2r)^{3}} = 0.524$$ ( In $$SCC,\ 2r = a$$ and number of atoms in unit cell $$=1$$).

What are primitive unit cells and what are non- primitive unit cells ?

Primitive unit cells have one atom per unit cell . On the other hand , non-primitive unit cell have more than one atom per unit cell .

Explain the following term with suitable examples:
Interstitials.

Interstitials:

Atoms or ions which occupy the normally vacant interstitial positions in a crystal are called interstitials.

How will you distinguish between the following pairs of terms :
Crystal lattice and unit cell

The regular three dimensional arrangement of identical points in the space which represent how the constituent particles (atoms , ions , molecules ) are arranged in a crystal is called a crystal lattice .
A unit cell is the smallest portion of a crystal lattice , which when represented over and again in different directions produces the complete crystal lattice .

What type of stoichiometric defects is shown by KCl and why?

KCl shows Schottky defect as the cation $$ K^+ $$ and anion $$ Cl^- $$ are of almost similar sizes.

What are stoichiometric defects or intrin sic defects in ionic crystals ?

Stoichiometric defects are those defects in which the ratio of cations to anions remains the same are represented by the molecular formula.

Answer the following questions .
What happens when $$ CdCl_2 $$ vis doped with $$ AgCl $$ ?

It results impurity defect . Each $$ Cd^{2+} $$ replaces two $$ Ag^+ $$ ions . It occupies the site of one ion and other site remains vacant .

Calculate the efficiency of packing in the case of a metal crystal for body-centred cubic.
(with the assumption that atoms are touching each other).

Let us suppose the edge length $$= a$$ & radius of each sphere $$= r$$ then there are $$8$$ spheres at the corners & $$1$$ in the body of the unit cell

Therefore number of spheres per unit cell $$=(8\times \dfrac18)+1=2$$

Also, $$\sqrt3a=4r,$$ where a is the edge length, and r is the radius of an atom.

The total volume of unit cell $$=a^3.$$

$$\text{Packing efficiency} =\dfrac{\text{Volume of two spheres}}{\text{Total volume of unit cell}}\times 100$$

$$= \dfrac{\dfrac83\pi r^3}{\dfrac{64}{3\sqrt{3}}\times r^3} \times100 = 68\%$$

$$\text{Therefore volume occupied in bcc arrangement }= 68\%$$

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What are interstitials in a crystal ?

Atoms or ions that fill the normal vacant interstitial voids in a crystal are called interstitials .

Examine the given defective crystals :
$$ X^+ $$ $$ Y^- $$ $$ X^+ $$ $$ Y^- $$ $$ X^+ $$
$$ Y^- $$ $$ Z^{2+} $$ $$ Y ^- $$ $$ X^+ $$ $$ Y^- $$
$$ X^+ $$ $$ Y^- $$ $$ Y^- $$ $$ X^+ $$
$$ Y^- $$ $$ X^+ $$ $$ Y- $$ $$ X^+ $$ $$ Y^- $$

Write the term used for this type of defect .

Impurity defect.

Examine the given crystal :

$$ X^+ $$ $$ Y^- $$ $$ X^+$$ $$ Y^-$$ $$X^+$$
$$ Y^- $$ $$ X^+$$ $$Y^- $$ $$X^+ $$ $$Y^- $$
$$ X^+ $$ $$ Y^- $$ $$ X^+$$ $$ e^-$$ $$X^+$$
$$ Y^- $$ $$ X^+$$ $$Y^- $$ $$X^+ $$ $$Y^- $$

Answer the following questions :
Give an example of the compound which shows this type of defect .

When crystals of NaCl are heated in an atmosphere of Na vapours of when crystals of KCl are heated in an atmosphere of K vapours .

In seven possible crystal system how many crystal system have more than one Bravais lattice ?