NCERT Solutions Class 12 Chemistry Chapter – 2
NCERT Solutions Class 12 Chemistry Chapter 2 – Solutions
In everyday life, we rarely come across pure substances. The majority of them are a mix of two or more substances, and their composition determines their significance in life. For instance, the air we breathe is mostly composed of oxygen and nitrogen, and the water we drink has a trace amount of different salts dissolved in it. Our blood is also made up of a variety of components. Furthermore, gold, brass, bronze and stainless steel and other alloys are some of the best examples of mixes. Hence, the NCERT solutions class 12 Chemistry chapter 2 – Solutions, focuses on liquid solutions and their properties.
NCERT Solutions for Chemistry covers many different chapters making it a complicated subject for many students. It has solutions for all the chapters of class 12, which aims at enhancing the concept for students, in addition to suggesting a different approach to the topic covered in chapter 2 – solution. It helps students clarify their doubts on various topics and solve problems using a logical method. Chemistry class 12 chapter 2 NCERT Solutions helps students understand the chapter, study the question paper patterns and prepare for board exams.
The NCERT solutions class 12 Chemistry chapter 2 deals with types of solutions and their properties. Other major topics covered in this chapter are Raoult’s law and Henry’s law, the concentration of solutions, the vapour pressure of liquid solutions, abnormal molar masses, and colligative properties.
Key Topics Covered In NCERT Solutions Class 12 Chemistry Chapter 2
The key topics covered in NCERT solutions class 12 Chemistry chapter 2 – Solutions include
Exercise  Topic 
2.1  Introduction to Solutions 
2.2  Solute and Solvent 
2.3  Classification of solutions 
2.4  Strength of Solutions 
2.5  Some Important Relationships 
2.6  Henry’s Law 
2.7  Raoult’s Law 
2.8  Colligative properties 
Students can click on the above topics to refer to the study material, revision notes, and important questions that can help them during their examination. In addition, a brief of each topic is given below.
The key topics covered in NCERT solutions class 12 Chemistry chapter 2 in brief:
Exercise 2.1: Introduction to Solutions
Solution is a homogeneous mixture of two or more substances in the same or different physical phases. The substances forming the solution are called components of the solution. Based on the number of components, a solution of two components is called a binary solution.
Students may refer to NCERT solutions class 12 Chemistry chapter 2 solutions chapter for detailed information.
Exercise 2.2: Solute and Solvent
In a binary solution, the solvent is the component present in large quantities, while the other component is known as solute. For example, salt solution. The NCERT solutions class 12 Chemistry chapter 2 exercise 2.2 explains in detail solvent and solute. Students can refer to the study material in addition to important questions by clicking on the link provided above.
Exercise 2.3: Classification of solutions
Types of solutions are seen based on the physical state of solute and solvent. NCERT solutions class 12 Chemistry chapter 2 classification of a solution is as under.
Types of solution  Solute  Solvent  Examples 
Solid solutions  solid

Solid  Alloys 
Liquid  Solid  Mercury and Amalgam  
Gas  Solid  Dissolve gases in minerals  
Liquid solutions  Solid  Liquid  Glucose that has been dissolved in water 
Liquid  Liquid  Ethanol is a liquid that can be dissolved in water.  
Gas  Liquid  Water with oxygen dissolved in it  
Gas solutions  solid  Gas  Iodine vapour in the air 
Liquid  Gas  Water vapour in the air  
Gas  Gas  Air (combination of nitrogen and oxygen) 
If water is used as a solvent, the solution is called an aqueous solution, and if not, the solution is called a nonaqueous solution. Students can click here for more information on class 12 Chemistry chapter 2 exercise 2.3 provided by Extramarks. Extramarks gives a detailed explanation of all the topics in NCERT solutions class 12 Chemistry chapter 2.
Exercise 2.4: Strength of Solutions
The strength of solution refers to the amount of solute dissolved per unit of solution or solvent. There are several methods for determining the strength of a solution which are as follows.
 Mass Percentage (%w/w): “It denotes the mass of a component in 100 g of solution.
Mass % of a component = Mass of the component in the solution X 100
The total mass of solutions
 Volume Percentage (%v/v): “It represents the volume of a component in a solution of 100 mL.
Volume % of a component = Volume of component X 100
Total volume of solution
 Mass by Volume Percentage (% w / v): “It denotes the mass of the solute in grams in 100 mL of solutions.”
Mass by volume % = Mass of solute in g X 100
The volume of solution in mL
Students may click here to access the NCERT solutions class 12 Chemistry chapter 2.
 Parts Per Million (ppm):
Parts per million = No. of parts of the component X 106
Total no. of all the components of sol.
Mass to mass, volume to volume, and mass to the volume are all ways to express concentration in parts per million. For more information, students may refer to NCERT solutions class 12 Chemistry chapter 2.
 Mole Fraction (x): “It denotes the number of moles of a solute in one mole of solution.”
Mole fraction = No. of moles of the component
Total no. of moles all the component
If the amount of moles A and B in the binary mixture is nA and nB, respectively, the mole fiction of A will be
xA =nA
_____________
nA + nB
 Molarity, M: It represents the number of moles of solute in 1 litre of solution.
Molarity, M = Moles of solute
Vol. of sol in L
Molarity is measured in milligrams per litre and is denoted by the letters ‘M ‘ or ‘Molar.’ “The density of a solution is its mass per unit volume.”
Density = Mass of solution
Vol. of solution
Students may refer to class 12 Chemistry solutions chapter 2 Exercise 2.4 provided by Extramarks for detailed notes on Molarity. In addition, Extramarks offers study notes, important questions, and sample question papers specific to NCERT solutions class 12 Chemistry chapter 2.
 Molality, m: It denotes the number of moles of solute present per kilogramme of solvent
Molality, m = Moles of solute
Mass of solvent in kg
Molality is measured in mol/kg, which can also be written as ‘m’ or ‘molal.’
 Normality, N: It denotes the number of solute equivalents in 1 litre of solution.
Normality, N = No. of Equivalents of solute
Vol. of sol. in L
No. of equivalents, eq = Weight_____
Equivalent weight(W/E)
E = M, where Z is the valency factor.
Z
Extramarks provides detailed study material as well as revision notes for NCERT solutions class 12 Chemistry chapter 2. Students may click here to access notes on Exercise 2.4
Exercise 2.5: Some Important Relationships
Some of the important relationships defined in class 12 Chemistry chapter 2 NCERT solutions are as under.
Dilution Law: When we dilute a solution with solvent, the amount of solute remains constant, and we can write:
M1 V1 =M2V2 & N1 V1=N2 V2
Molarity and Normality: Normality =Z X Molarity
Important Note: The temperature does not affect the mass percent, ppm, mole fraction, or molality; however, temperature influences molarity and normality. This is because the volume is affected by temperature, whereas mass is not.
Solubility: The maximum amount of solute that can be dissolved in a given amount of solvent given temperature is known as its solubility at that temperature.
The solubility of a solute in a liquid depends upon factors shown in below
 Nature of solute
 Nature of the solvent
 The temperature of the solution
 Pressure(Gases)
For a detailed explanation of solubility, students may refer to NCERT solutions class 12 Chemistry chapter 2 by clicking here.
Exercise 2.6: Henry’s Law
This law establishes a quantitative relation between pressure and solubility of a gas in a solvent. It applies only to gasliquid solutions. As per Henry’s law, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the mole fraction of the gas(x) in the solution.
The formula is p=KH x
where KH = Henry law constant
The value of Henry law constantly differs for different gases at the same temperature. The value of KH increases with the increase in temperature; therefore solubility of gases increases with decreasing temperature. Due to this reason, cold water is more sustainable for aquatic life than warm water.
Henry’s law applies to
 The manufacturing of aerated drinks
 During deep water diving
 Mountain climbers and those who live in high altitudes will benefit from this.
NCERT solutions class 12 Chemistry chapter 2 has detailed information on Henry’s law that students can access for free on Extramarks.
Exercise 2.7: Raoult’s Law
This law established a quantitative relation between the partial vapour pressure and the mole fraction of the solution. Raoult’s law follows only liquidliquid solutions. This law state is only for the solution of volatile liquids; the partial vapour pressure(p) of each component is directly proportional to its mole fractions(x).
So
px
Or p=p0x
where p0 is the vapour pressure component at the same temperature.
Exercise 2.8: Colligative properties
Colligative is a Latin word that means ‘together binds’. Colligative properties depend upon the number of solute particles in a solution, irrespective of their nature. More details of colligative properties are shown in NCERT class 12 chapter 2 solution.
NCERT solutions class 12 Chemistry chapter 2 are available on Extramarks website.Each solution has been explained in detail by the experts of Extramarks. In addition to chapter 2, students can access NCERT Solution for all other Chemistry chapters of class 12. Furthermore, students can click on the links provided below to access the study material of other classes.
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NCERT Solutions Class 12 Chemistry Chapter 2 Exercise & Answer Solutions
The NCERT Chemistry class 12 chapter 2 offers several exercises and answer solutions that are vital for students studying for an examination. The best way to study and practice these exercises mentioned in Chemistry class 12 starts is by going through the chapter entirely. Then it would help if students practised the numerical and reactions properly. Students must also be wellversed with the diagrams and formulas. The Extramarks answer solutions include diagrams and formulae wherever needed, that are aptly labeled. Students may refer to the solutions to help them better understand the concept and build a strong foundation for higher studies.
For Extramarks exercise questions, students may click on NCERT solutions class 12 Chemistry chapter 2 exercise questions.
Q.1 Define the term ‘amorphous’. Give a few examples of amorphous solids.
Ans.
Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharpedged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.
Q.2 What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Ans.
The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders. Quartz can be converted into glass by heating and then cooling it rapidly.
Q.3 Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide (P_{4}O_{10}) (vii) Graphite
(ii) Ammonium phosphate (NH_{4})_{3}PO_{4} (viii) Brass
(iii) SiC (ix) Rb
(iv) I_{2} (x) LiBr
(v) P_{4 }(xi) Si
(vi) Plastic
Ans.
Ionic→ (ii) Ammonium phosphate (NH_{4})_{3}PO_{4}, (x) LiBr
Metallic→ (viii) Brass, (ix) Rb
Molecular→ (i) Tetra phosphorus decoxide (P_{4}O_{10}), (iv) I_{2}, (v) P_{4}
Covalent (network) → (iii) SiC, (vii) Graphite, (ix) Si
Amorphous→ (vi) Plastic
Q.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic closepacked structure?
(b) in a bodycentred cubic structure?
Ans.
(i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.
(ii) The coordination number of atoms.
(a) in a cubic close packed structure is 12, and
(b) in a body centred cubic structure is 8
Q.5 How will you distinguish between the following pairs of terms:
(i) Hexagonal closepacking and cubic closepacking?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
Ans.
(i) A 2D hexagonal closepacking contains two types of triangular voids (a and b) as shown in figure 1. Let us call this 2D structure as layer A. Now, particles are kept in the voids present in layer A (it can be easily observed from figures 2 and 3 that only one of the voids will be occupied in the process, i.e., either a or b). Let us call the particles or spheres present in the voids of layer A as layer B. Now, two types of voids are present in layer B (c and d). Unlike the voids present in layer A, the two types of voids present in layer B are not similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is surrounded by 6 spheres and is called the octahedral void.
Fig1
Fig2
Fig3
Now, the next layer can be placed over layer B in 2 ways.
Case 1: When the third layer (layer C) is placed over the second one (layer B) in such a manner that the spheres of layer C occupy the tetrahedral voids c.
In this case we get hexagonal closepacking. This is shown in figure 4. In figure 4.1, layer B is present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is present over the voids b and layer C is present over the voids c. It can be observed from the figure that in this arrangement, the spheres present in layer C are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal closepacking are arranged in an ABAB….. pattern.
Fig4.1
Fig4.2
Case 2: When the third layer (layer C) is placed over layer B in such a manner that the spheres of layer C occupy the octahedral voids d.
In this case we get cubic closepacking. In figure 5.1, layer B is present over the voids a and layer C is present over the voids d. In figure 5.2, layer B is present over the voids b and layer C is present over the voids d. It can be observed from the figure that the arrangement of particles in layer C is completely different from that in layers A or B. When the fourth layer is kept over the third layer, the arrangement of particles in this layer is similar to that in layer A. Hence, we can say that the layers in cubic closepacking are arranged in an ABCABC….. pattern.
5.1
Fig5.2
The side views of hcp and ccp are given in figures 6.1 and 6.2 respectively.
Fig6.1
Fig6.2
(ii) The diagrammatic representation of the constituent particles (atoms, ions, or molecules) present in a crystal in a regular threedimensional arrangement is called crystal lattice.
A unit cell is the smallest threedimensional portion of a crystal lattice. When repeated again and again in different directions, it generates the entire crystal lattice.
(iii) A void surrounded by 4 spheres is called a tetrahedral void and a void surrounded by 6 spheres is called an octahedral void. Figure 1 represents a tetrahedral void and figure 2 represents an octahedral void.
Q.6 How many lattice points are there in one unit cell of each of the following lattice?
(i) Facecentred cubic
(ii) Facecentred tetragonal
(iii) Bodycentred
Ans.
(i) There are 14 (8 from the corners + 6 from the faces) lattice points in facecentred cubic.
Lattice points per unit cell = 8 x 1/8 + 6 x ½
= 4
(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in facecentred tetragonal.
Lattice points per unit cell = 8 x 1/8 + 6 x ½
= 4
(iii) There are 9 (1 from the centre + 8 from the corners)
Lattice points per unit cell = 8 x 1/8 + 1
= 2
Q.7 Explain
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Ans.
(i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositelycharged ions. Hence, both have high melting points.
The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.
(ii) The constituent particles of ionic crystals are ions. These ions are held together in threedimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle.
Q.8 A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q?
Ans.
It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell
=\mathrm{8*}\frac{1}{8}=1
It is also given that the atoms of P are present at the bodycentre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:Q =1:1
Hence, the formula of the compound is PQ.
The coordination number of both P and Q is 8.
Q.9 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Ans.
Semiconductors are substances having conductance in the intermediate range of 10^{6} to 10^{4} ohm^{−1}m^{−1}.
The two main types of semiconductors are:
(i) ntype semiconductor
(ii) ptype semiconductor
ntype semiconductor: The semiconductor whose increased conductivity is a result of negativelycharged electrons is called an ntype semiconductor. When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an ntype semiconductor is generated.
Si and Ge have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, P and As contain five valence electrons each. When Si or Ge is doped with P or As, the latter occupies some of the lattice sites in the crystal. Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Si or Ge atoms. The remaining fifth electron becomes delocalized and increases the conductivity of the doped Si or Ge.
ptype semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a ptype semiconductor. When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a ptype of semiconductor is generated.
Perfect crystal of Si n type semiconductor p type semiconductor
When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positivelycharged plate through electron holes.However, it will appear as if the electron holes are positivelycharged and are moving toward the negativelycharged plate.
Q.10 Classify each of the following as being either a ptype or an ntype semiconductor:
(i) Ge doped with In
(ii) B doped with Si.
Ans.
(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a ptype semiconductor.
(ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an ntype semiconductor.
Q.11 How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Ans.
By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.
Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z’ be the number of atoms in the unit cell.
Now, density of the unit cell = mass of the unit cell / volume of the unit cell
\begin{array}{l}\Rightarrow d=\frac{Zm}{{a}^{3}}\left(i\right)\\ \left[\text{Since mass of the unit cell = Number of atoms in the unit cell \xd7Atomic mass}\right]\end{array}
Q.12 ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
Ans.
Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.
The melting points of the given substances are:
Solid water → 273 K
Ethyl alcohol → 158.8 K
Diethyl ether → 156.85 K
Methane → 89.34 K
Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest.
Q.13 Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) bodycentred cubic
(iii) facecentred cubic (with the assumptions that atoms are touching each other).
Ans.
(i) Simple cubic
In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be ‘a’ and the radius of each particle be r.
So, we can write:
a = 2r
Now, volume of the cubic unit cell =a^{3}
=(2r)^{3}
=8r^{3}
We know that the number of particles per unit cell is 1.
$\begin{array}{l}\text{=}\frac{\text{4}}{\text{3}}{\text{\pi r}}^{\text{3}}\\ \text{Therefore, packing efficiency}\\ \text{=}\frac{\text{Valume of one particle}}{\text{Volume of cubic unit cell}}\text{\xd7100\%}\\ \text{Hence, packing efficiency}\\ \\ \text{=}\frac{\text{1}}{\text{6}}\text{\pi \xd7100\%}\\ \text{=}\frac{\text{1}}{\text{6}}\text{\xd7}\frac{\text{22}}{\text{7}}\text{\xd7100\%}\\ \text{=52.4\%}\end{array}$
(ii) Bodycentred cubic
It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.
From ∆FED, we have:
$\begin{array}{l}{\mathrm{b}}^{2}={\mathrm{a}}^{2}+{\mathrm{a}}^{2}\\ \Rightarrow {\mathrm{b}}^{2}=2{\mathrm{a}}^{2}\\ \Rightarrow \mathrm{b}=\sqrt{2a}\\ \mathrm{Again},\text{}\mathrm{from}\text{}\mathrm{\Delta AFD},\text{}\mathrm{we}\text{}\mathrm{have}:\\ {\mathrm{c}}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\\ \Rightarrow {\mathrm{c}}^{2}={\mathrm{a}}^{2}+2{\mathrm{a}}^{2}(\mathrm{Since}{\mathrm{b}}^{2}=2{\mathrm{a}}^{2})\\ \Rightarrow {\mathrm{c}}^{2}=3{\mathrm{a}}^{2}\\ \Rightarrow \mathrm{c}=\sqrt{3\mathrm{}}\\ \mathrm{Let}\text{}\mathrm{the}\text{}\mathrm{radius}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{atom}\text{\hspace{0.17em}}\mathrm{be}\text{\hspace{0.17em}}\mathrm{r}.\\ \mathrm{Length}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{body}\text{\hspace{0.17em}}\mathrm{diagonal},\text{}\mathrm{c}\text{}=\text{}4\mathrm{r}\\ \Rightarrow \sqrt{3a}=4\mathrm{r}\\ \Rightarrow \mathrm{a}=\frac{4\mathrm{r}}{\sqrt{3}}\\ \mathrm{orr}=\frac{\sqrt{3\mathrm{a}}}{4}\\ \mathrm{Volume}\text{}\mathrm{of}\text{}\mathrm{cube},{\mathrm{a}}^{3}={\left(\frac{4\mathrm{r}}{\sqrt{3}}\right)}^{3}\\ \mathrm{A}\text{}\mathrm{body}\mathrm{centred}\text{}\mathrm{cubic}\text{}\mathrm{lattice}\text{}\mathrm{contains}\text{}2\text{}\mathrm{atoms}.\\ \mathrm{So},\text{}\mathrm{volume}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{occupied}\text{}\mathrm{cubic}\text{}\mathrm{lattice}=2\mathrm{\pi}\frac{4}{3}{\mathrm{r}}^{3}\\ =\frac{8}{3}{\mathrm{\pi r}}^{3}\\ \therefore \mathrm{Packingefficiency}=\frac{\mathrm{volumeoccupiedbytwospheresintheunitcell}}{\mathrm{Totalvolumeoftheunitcell}}\times 100\mathrm{\%}\\ =\frac{\frac{8}{3}{\mathrm{\pi r}}^{3}}{{\left(\frac{4}{\sqrt{3}}\mathrm{r}\right)}^{3}}\times 100\mathrm{\%}\\ =\frac{\frac{8}{3}{\mathrm{\pi r}}^{3}}{\frac{64}{\sqrt[3]{3}}{\mathrm{r}}^{3}}\times 100\mathrm{\%}\\ =68\mathrm{\%}\end{array}$
(iii) Facecentred cubic
Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.
$\begin{array}{l}\mathrm{From}\text{}\mathrm{\Delta ABC},\text{}\mathrm{we}\text{}\mathrm{have}:\\ {\mathrm{AC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}\\ \Rightarrow {\mathrm{b}}^{2}={\mathrm{a}}^{2}+{\mathrm{a}}^{2}\\ \Rightarrow {\mathrm{b}}^{2}=2{\mathrm{a}}^{2}\\ \Rightarrow \mathrm{b}=\sqrt{2\mathrm{a}}\\ \mathrm{Let}\text{}\mathrm{r}\text{}\mathrm{be}\text{}\mathrm{the}\text{}\mathrm{radius}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{atom}.\\ \mathrm{Now},\text{}\mathrm{from}\text{}\mathrm{the}\text{}\mathrm{figure},\text{}\mathrm{it}\text{}\mathrm{can}\text{}\mathrm{be}\text{}\mathrm{observed}\text{}\mathrm{that}:\\ \mathrm{b}=4\mathrm{r}\\ \Rightarrow \sqrt{2\mathrm{a}}=4\mathrm{r}\\ \Rightarrow \mathrm{a}=\sqrt[2]{2\mathrm{r}}\\ \mathrm{Now},\text{}\mathrm{volume}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{cube},{\mathrm{a}}^{3}={\left(\sqrt[2]{2\mathrm{r}}\right)}^{3}\\ \mathrm{We}\text{}\mathrm{know}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\mathrm{number}\text{}\mathrm{of}\text{}\mathrm{atoms}\text{}\mathrm{per}\text{}\mathrm{unit}\text{}\mathrm{cell}\text{}\mathrm{is}\text{}4.\\ \mathrm{So},\text{}\mathrm{volume}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{occupied}\text{}\mathrm{unit}\text{}\mathrm{cell}=4\mathrm{\pi}\frac{4}{3}{\mathrm{r}}^{3}\\ \therefore \mathrm{Packingefficiency}=\frac{\mathrm{Volumeoccupiedbysheresintheunitcell}}{\mathrm{Totalvolumeoftheunitcell}}\times 100\mathrm{\%}\\ =\frac{4\mathrm{\pi}\frac{4}{3}{\mathrm{r}}^{3}}{{\left(\sqrt[2]{2\mathrm{r}}\right)}^{3}}\times 100\mathrm{\%}\\ =\frac{\frac{16}{3}{\mathrm{\pi r}}^{3}}{\sqrt[16]{2{\mathrm{r}}^{3}}}\times 100\mathrm{\%}\\ =74\mathrm{\%}\end{array}$
Q.14 Silver crystallises in fcc lattice. If edge length of the cell is 4.07×10^{−8} cm and density is 10.5g cm^{−3}, calculate the atomic mass of silver.
Ans.
It is given that the edge length, a = 4.077 × 10^{−8} cm
Density, d = 10.5g cm^{−3}
As the lattice is fcc type, the number of atoms per unit cell, z=4
We also know that, N_{A} = 6.022 × 10^{23} mol^{−1}
Using the relation:
$\begin{array}{l}\mathrm{d}=\frac{\mathrm{zM}}{{\mathrm{a}}^{3}{\mathrm{N}}_{\mathrm{A}}}\\ \Rightarrow \mathrm{M}=\frac{{\mathrm{da}}^{3}{\mathrm{N}}_{\mathrm{A}}}{\mathrm{z}}\\ =\frac{10.5{\mathrm{gcm}}^{3}\times {(4.077\times {10}^{8}\mathrm{cm})}^{3}\times 6.022\times {10}^{23}{\mathrm{mol}}^{1}}{4}\\ =107.13{\mathrm{gmol}}^{1}\\ \text{Therefore, atomic mass of silver =107.13u}\end{array}$
Q.15 Niobium crystallizes in bodycentred cubic structure. If density is 8.55g cm^{−3}, calculate atomic radius of niobium using its atomic mass 93 u.
Ans.
It is given that the density of niobium, d = 8.55g cm^{−3}
Atomic mass, M = 93g mol^{−1}
As the lattice is bcc type, the number of atoms per unit cell, z = 2
We also know that, N_{A} = 6.022 × 10^{23} mol^{−1}
Applying the relation:
$\begin{array}{l}\mathrm{d}=\frac{\mathrm{zM}}{{\mathrm{a}}^{3}{\mathrm{N}}_{\mathrm{A}}}\\ \Rightarrow {\mathrm{a}}^{3}=\frac{\mathrm{zM}}{{\mathrm{dN}}_{\mathrm{A}}}\\ =\frac{2\times 93{\mathrm{gmol}}^{1}}{8.55{\mathrm{gcm}}^{3}\times 6.022\times {10}^{23}{\mathrm{mol}}^{1}}\\ =3.612\times {10}^{23}{\mathrm{cm}}^{3}\\ \mathrm{so},\mathrm{}\mathrm{a}=3.306\times {10}^{8}\mathrm{cm}\\ \mathrm{For}\text{bodycentred cubic unit cell:}\\ \mathrm{r}=\frac{\sqrt{3}}{4}\mathrm{a}\\ =\frac{\sqrt{3}}{4}\times 3.306\times {10}^{8}\mathrm{cm}\\ =\text{1}.\text{432}\mathrm{}\times \mathrm{}\text{1}{0}^{\text{9}}\mathrm{}\text{cm}\\ =\text{}.\text{1432}\mathrm{}\text{nm}\end{array}$
Q.16 If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
Ans.
A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ∆POQ is rightangled
∠POQ = 90^{0}
Now, applying Pythagoras theorem, we can write:
\begin{array}{l}{\text{PQ}}^{\text{2}}={\text{PO}}^{\text{2}}+{\text{OQ}}^{\text{2}}\\ \Rightarrow {\left(\text{2R}\right)}^{\text{2}}={\left(\text{R}+\text{r}\right)}^{\text{2}}+{\left(\text{R}+\text{r}\right)}^{\text{2}}\\ \Rightarrow {\left(\text{2R}\right)}^{\text{2}}=\text{2}{\left(\text{R}+\text{r}\right)}^{\text{2}}\\ \Rightarrow {\text{2R}}^{\text{2}}={\left(\text{R}+\text{r}\right)}^{\text{2}}\\ \Rightarrow \sqrt{\text{2}}\text{R}=\text{R}+\text{r}\\ \Rightarrow r=\sqrt{2}RR\\ \\ \Rightarrow \text{r}=0.\text{414 R}\end{array}
Q.17 Copper crystallises into a fcc lattice with edge length 3.61×10^{−8} cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^{−3.}
Ans.
Edge length, a = 3.61 × 10^{−8} cm
As the lattice is fcc type, the number of atoms per unit cell, z=4
Atomic mass, M = 63.5 g mol^{−1}
We also know that, N_{A} = 6.022 × 10^{23} mol^{−1}
Applying the relation:
$\begin{array}{l}\mathrm{d}=\frac{\mathrm{zM}}{{\mathrm{a}}^{3}{\mathrm{N}}_{\mathrm{A}}}\\ =\frac{4\times 63.5\mathrm{g}\mathrm{}{\mathrm{mol}}^{1}}{{(3.61\times {10}^{8})}^{3}\times 6.022\times {10}^{23}{\mathrm{mol}}^{1}}\\ =8.97\mathrm{g}\mathrm{}{\mathrm{cm}}^{3}\end{array}$
The measured value of density is given as 8.92g cm^{−3}. Hence, the calculated density 8.97g cm^{−3} is in agreement with its measured value.
Q.18 Analysis shows that nickel oxide has the formula Ni_{0.98}O_{1.00. }What fractions of nickel exist as Ni^{2+} and Ni^{3+} ions?
Ans.
The formula of nickel oxide is Ni_{0.98}O_{1.00.}
Therefore, the ratio of the number of Ni atoms to the number of O atoms,
Ni : O = 0.98 : 1.00 = 98 : 100
Now, total charge on 100 O^{2−}ions = 100 × (−2)
= −200
Let the number of Ni^{2+} ions be x.
So, the number of Ni^{3+} ions is 98 − x.
Now, total charge on Ni^{2+} ions = x(+2)
= +2x
And, total charge on Ni^{3+} ions = (98 − x)(+3)
= 294 − 3x
Since, the compound is neutral, we can write:
2x + (294 − 3x) + (−200) = 0
⇒ −x + 94 = 0
⇒ x = 94
Therefore, number of Ni^{2+} ions = 94
And, number of Ni^{3+} ions = 98 – 94 = 4
Hence, fraction of nickel that exists as Ni^{2+} = 94/98
= 0.959
And, fraction of nickel that exists as Ni^{3+} = 4/98
= 0.041
Alternatively, fraction of nickel that exists as Ni^{3+} = 1 − 0.959
= 0.041
Q.19 Nonstoichiometric cuprous oxide, Cu_{2}O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a ptype semiconductor?
Ans.
In the cuprous oxide (Cu_{2}O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu^{+} ions is slightly less than twice the number of O^{2−} ions. This is because some Cu^{+} ions have been replaced by Cu^{2+} ions.
Every Cu^{2+} ion replaces two Cu^{+} ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a ptype semiconductor.
Q.20 Ferric oxide crystallises in a hexagonal closepacked array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Ans.
Let the number of oxide (O^{2−}) ions be x.
So, number of octahedral voids = x
It is given that two out of every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+) ions=2/3 x
Therefore, ratio of the number of Fe3+ ions to the number of O^{2−} ions,
Fe^{3+} : O^{2 } = 2/3 x : x
= 2/3 : 1
= 2 : 3
Hence, the formula of the ferric oxide is Fe_{2}O_{3}.
Q.21 Gold (atomic radius = 0.144 nm) crystallizes in a facecentred unit cell. What is the length of a side of the cell?
Ans.
For a facecentred unit cell:
a= √2 r
It is given that the atomic radius, r = 0.144 nm
So, a=√2 × 0.144
=0.407nm
Hence, length of a side of the cell =0.407 nm
Q.22 In terms of band theory, what is the difference
(i) Between a conductor and an insulator
(ii) Between a conductor and a semiconductor
Ans.
(i) The valence band of a conductor is partiallyfilled or it overlaps with a higher energy, unoccupied conduction band.
On the other hand, in the case of an insulator, the valence band is fullyfilled and there is a large gap between the valence band and the conduction band.
(ii) In the case of a conductor, the valence band is partiallyfilled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field.
On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.
Q.23 Explain the following terms with suitable examples:
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) Fcentres
Ans.
(i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 10^{6} Schottky pairs per cm^{3 }at room temperature. Ionic substances containing similarsized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.
(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.
(iii) Interstitials: Interstitial defect is shown by nonionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.
(iv) Fcentres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called Fcentres. These unpaired electrons impart colour to the crystals. For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating Fcentres.
Q.24 Aluminium crystallizes in a cubic closepacked 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?
Ans.
$\begin{array}{l}\left(\mathrm{i}\right)\text{For cubic closepacked structure:}\\ \text{a=2}\sqrt{2\mathrm{r}}\\ =2\sqrt{2}\times 125\mathrm{pm}\\ =353.55\mathrm{pm}\left(\text{approximately}\right)\\ \\ \left(\mathrm{ii}\right)\text{Volume of one unit cell=}{\left(354\mathrm{pm}\right)}^{3}\\ =4.4\times {10}^{7}{\mathrm{pm}}^{3}\\ =4.4\times {10}^{7}\times {10}^{30}{\mathrm{cm}}^{3}\\ =4.4\times {10}^{23}{\mathrm{cm}}^{3}\\ \mathrm{There}{\text{fore, number of unit cells in 1.00 cm}}^{3}=\frac{1.00{\mathrm{cm}}^{3}}{4.4\times {10}^{23}{\mathrm{cm}}^{3}}=2.27\times {10}^{22}\end{array}$
Q.25 If NaCl is doped with 10^{−3} mol % of SrCl_{2}, what is the concentration of cation vacancies?
Ans.
It is given that NaCl is doped with 10^{−3} mol % of SrCl_{2}.
This means that 100 mol of NaCl is doped with 10^{−3} mol of SrCl_{2}.
$\begin{array}{l}\text{Therefore,1 mol of NaCl is doped with}\frac{{10}^{3}}{100}{\text{mol of SrCl}}_{2}\\ ={10}^{5}{\text{mol of SrCl}}_{2}\\ {\text{Cation vacancies produced by one Sr}}^{2}+\mathrm{ion}=1\\ \therefore \mathrm{concentration}{\text{of the cation vacancies produced by 10}}^{5}{\text{mol of Sr}}^{2+}\text{ions}\\ {\text{=10}}^{5}\times 6.022\times {10}^{23}\\ =6.022\times {10}^{18}{\mathrm{mol}}^{1}\\ {\text{Hence, the concentration of cation vacancies created by SrCl}}_{2}{\text{is 6.022\xd710}}^{18}\mathrm{per}\text{mol of NaCl.}\\ \end{array}$
Q.26 Explain the following with suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Antiferromagnetism
(v) 1216 and 1315 group compounds.
Ans.
(i) Ferromagnetism: The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetized even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO_{2}.
In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains and each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance, the domains are randomlyoriented and so, their magnetic moments get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.
Schematic alignment of magnetic moments in ferromagnetic substances
(ii) Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances are O_{2}, Cu^{2t}, Fe^{3t}, and Cr^{3t}.
Paramagnetic substances get magnetized in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed. To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.
(iii) Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and antiparallel directions, in unequal numbers, are said to have ferrimagnetism. Examples include Fe_{3}O_{4} (magnetite), ferrites such as MgFe_{2}O_{4} and ZnFe_{2}O_{4}.
Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances. On heating, these substances become paramagnetic.
Schematic alignment of magnetic moments in ferromagnetic substances
(iv) Antiferromagnetism: Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositelyoriented. The oppositelyoriented domains cancel out each other’s magnetic moments.
Schematic alignment of magnetic moments in antiferromagnetic substances
(v) 1216 and 1315 group compounds: The 1216 group compounds are prepared by combining group 12 and group 16 elements and the 1315 group compounds are prepared by combining group 13 and group 15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (InSb), aluminium phosphide (AlP), and gallium arsenide (GaAs) are typical compounds of groups 1315. GaAs semiconductors have a very fast response time and have revolutionized the designing of semiconductor devices. Examples of group 1216 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.
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