Unit 1: The Solid State
Ques1:- Define the term 'amorphous'. Give a few examples of amorphous solids.
Ans:- If the Constituent structural units are not arranged in any regular pattern, then the solid formed is known as amorphous solid. These solids may have only short-range order. Since there is no regular pattern of arrangement of Constituent particles, these solids do not have definite geometry or shape. Amorphous solids can be prepared when the melts are rapidly cooled. Some well-known examples of amorphous solids are glass, plastics, rubber, fused silica, and polymers of high molecular mass such as starch, cellulose, etc. Though amorphous solids do not possess long-range of regularity, they may possess small regions of orderly arrangement. Such crystalline parts of amorphous solids are known as crystallites.
Ques2:- What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Ans:- Quartz is a three dimensional crystalline solid in which \(SiO_4 \) tetrahedra have only short-range order. Amorphous solids, either on heating or heating followed by slow cooling at a particular temperature change into a crystalline form. When quartz is melted and the liquid is cooled rapidly, an amorphous solid(glass) is formed.
Ques3:- Classify each of the following solids as ionic, metallic , molecular, network (covalent), or amorphous.
(i) Tetra phosphorus decoxide $\ce{(P4O10)}$
(ii) Ammonium phosphate $\ce{(NH4)3PO4}$
(iii) $\ce{SiC}$
(iv) $\ce{I2}$
(v)$\ce{P4}$
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) $\ce{Rb}$
(x) $\ce{LiBr}$
(xi) $\ce{SI}$
Ans:- According to the categories the above examples can be classified as:-
(i) Molecular solids : $\ce{(P4O10)}$, $\ce{I2}$, $\ce{P4}$
(ii) Ionic solids : $\ce{(NH4)3PO4,\, LiBr}$
(iii) Network (covalent) solids : Graphite, SiC and SI
(iv) Amorphous solid : Plastic
(v) Metallic solids : Brass and Rb
Ques4:- i) What is meant by the term 'coordination number'?
ii) What is the coordination number of atoms :
(a) in a cubic close-packed structure ?
(b) in a body-centred cubic structure ?
Ans:- i) Coordination number is defined as the number or atom surrounding a central atom or ion crystal. In other words, we can say that the Coordination number is the number closest neighbors of any Constituent particle.
ii)(a) The coordination number of atoms in a cubic close-packed structure is 12, because it is touching 6 atoms in its own plane and 3 just above it and 3 just below it.
(b) The coordination number of an atom in a body-centred cubic (bcc) structure is 8.
Ques5:- How can you determine the atomic mass of an unknown metal if you know its density and the dimensions of it's unit cell ? Explain.
Ans:- Suppose, the edge length of the unit cell = a
So, volume of the unit cell = a
Let the number of atoms in the unit cell is= z
Let d is the density of the unit cell and density of the unit cell and density of the unit cell= Mass of the unit cell/ volume of the unit cell
\( Mass \ of \ the \ unit \ cell= \) \(Number \ of \ atoms \ in \ the \ unit \ cell \times Mass \ of \ each \ atom \ of \ unit \ cell \)
=> \( Z \times m \)
Mass of each atom of the unit cell,
m = \( Atomic \ mass\over Avogadro's \ number \)
m = \( M \over N_A \)
So, Mass of the unit cell = \( Z \times M \)/ Na
Thus, density of the unit cell =\( {Z \times M} \over {Na \times a^3} \) gram per cm³
Hence we can say that atomic mass of unknown metal can be determined if the density and the other dimensions bof the unit cell are known.
Ques6:- 'Stability of a crystal is reflected in the magnitude of its melting point'. Comment.
Collect melting points of water, ethyl alcohol, diethyl ether and methane from a data book.
What can you say about the intramolecular forces between these molecules ?
Ans:- If the atoms or ions are packed in well ordered manner in the crystal, its melting point will be higher. If the packing is irregular then its melting point will be lower. So that, higher the melting point, more will be the stability of crystal.
( Melting point s of water, ethyl alcohol, diethyl ether and methane are 273 K, 155.7 K, 156.8 K and 90.5 K respectively)
In general, higher the melting point, greater will be the forces of attraction which are holding the Constituent particles together and greater is the stability of the crystal.
i) Water has highest melting point because water molecules are associated with intermolecular hydrogen bonding.
ii) Methane has lower melting point because it is non polar and only forces present in them are weak Vander Waal's forces.
iii) Ethyl alcohol has higher melting point than methane and diethyl ether because there exist hydrogen bonding in ethyl alcohol molecules which is not present in diethyl ether and methane. However, ethyl alcohol has lower boiling point than water because there exist stronger hydrogen bonding in water than in alcohol.
iv) Diethyl ether is polar and it has higher melting point than methane because there exist greater intermolecular forces of attraction in diethyl ether which is not present in methane. The intermolecular forces present in diethyl ether are dipole dipole interaction.
Ques7:- How will 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?
Ans:- i) Hexagonal close packing and cubic close packing
In hexagonal close packing the different layers are packed in AB AB AB........pattern because every third layer is similar to the first layer, whereas, in cubic close packing the different layers are packed in ABC ABC.... pattern because every fourth layer is similar to the first layer.
In hexagonal close packing (hcp) structure, the third layer of atoms cover the tetrahedral voids of second layer whereas in ccp structure, the third layer of atoms covers the octahedral voids of the second layer.
ii) Crystal lattice and unit cell
Crystal lattice is a regular three dimensional arrangement of atoms, ions and molecules.
Unit cell is a portion of crystal lattice which on repetition leads to the formation of crystal structure or it is the smallest repeating unit in three-dimensional space from which the whole crystal lattice is made up.
iii) Tetrahedral void and octahedral void
Tetrahedral void is a void which is formed when the centre of four sphere are placed at the four corners of tetrahedron. And, octahedral void is the void surrounded by six spheres and the centre of each sphere is placed at the six corners of regular octahedron.
There are two Tetrahedral void and only one octahedral void for each sphere.
Ques8:- How many lattice points are there in one unit cell of each of the following lattices ?
i) face-centred cubic
ii) face - centred tetragonal
iii) body- centred
Ans:- Lattice point in fcc cubic or tetragonal is 14. However, particles per unit cell is 4 as explained below:
i) In face centred cubic, there are 8 lattice points on corners and 6 lattice point are on the faces of cube. However, particles per unit cell are 4 (i.e. \(8 \times {1 \over 8} + 6 \times {1 \over 2} \))
ii) In face centred tetragonal, there are 8 lattice points are on corners and 6 lattice point on six faces of cube. So, particle per unit cell is 4.
iii) In body centered cube, there are 8 lattice points on corners and 1 lattice point at body centre of cube. So, particle per unit cell is 2.
Ques9:- Explain :-
1) The basis of similarities and differences between Metallic and ionic crystal.
2) Ionic solids are hard and brittle.
Ans:- 1) Similarities:- i) Both ionic and Metallic crystal have electrostatic force of attraction.
ii) Both, ionic and Metallic crystals have high melting point.
Differences:- i) Both metallic and ionic solids conduct electricity. However, metallic solid conduct electricity in solid state whereas, ionic solid conduct electricity in molten state or in aqueous solution.
ii) There is decrease in conductance in metals with increase in temperature whereas, conductance in ionic solids increases in temperature.
2) Ionic solids are hard because there is strong electrostatic force of attraction between oppositely charged ions. However, they are brittle because bonds are non-directional and breakdown when force is applied.
Ques10:- Calculate the efficiency of packing in case of metal crystal for (i) simple cubic (ii) body-centred cubic (iii) face-centred cubic ( with the assumption that atoms are touching each other).
Ans:- i) for simple cubic
Ques11:- Silver crystallises in fcc lattice. If edge length of the cell is \(4.07 × 10^{–8} \) cm and density is 10.5 \( g cm^{–3 }\), calculate the atomic mass of silver.
Ans:- As we know that silver crystallises in fcc lattice, so, Z = 4
a ( edge length) = \(4.07 \times 10^{-8} \) cm
d = 10.5 g \(cm^3 \)
M = ?
M = d × Na × a³/ Z
\(M = {10.5 \times 6.022 \times 10^{23} \times {(4.07 \times 10^{-8})}^3 \over 4} \)
M = 107.8 u
So atomic mass of silver = 107.8 u
Ques12:- A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the corners of the cube and P at the body centre. What is the formula of the compound ? What are the formula of the compound? What are the Coordination numbers of P and Q?
Ans:- Since atoms Q are at the corners of the cube and each atom at the corners is shared by 8 unit cells, therefore, the contribution of atoms at corners = 8× 1/8 = 1
Atoms P are at the body centre and atoms at the body centre is not shared by any unit cell, therefore, the contribution due to atoms at body centre = 1
So the formula of the compound is PQ and the coordination numbers of P and Q are 8 each.
Ques13:- Niobium crystallises in body-centred cubic structure. If density is \( 8.55 g \ cm^{–3} \), calculate atomic radius of niobium using its atomic mass 93 u.
Ans:- M, atomic mass of niobium = 93 u
And value of Z = 2
d = 8.55 g \(cm^3\)
We know that, \(d = {Z \times M \over d \times Na}\)
\(a^3= {2 \times 93 \over 8.55 \times 6.022 \times 10^{23}} \)
a = 330.4 pm
For bcc crystal structure, \(r = {√3a \over 4}\)
So, \(r = {1.732 \times 330.4 \over 4} \)
r = 143.1 pm
Ques14:- If the radius of the octahedral void is r and radius of atoms in close packing is R, derive relation between r and R.
Ans:- The radius of octahedral void = r
and the radius of sphere = R
In an isoceles triangle ABC, as you can see in figure,
\({AC \over AB} = {√2\over1}\) .......(i)
But, AB = 2R
and, AC = R + 2r + R
AC = 2R + 2r
putting these values in eqn (i), we get
\({2R + 2r \over 2R} = {√2\over1} \)
\( {r \over R} = √2 -1 \)
r = 0.414 R
So, we can say that the radius of an octahedral void is 0.414 times the radius of sphere.
Ques15:- 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:- As, copper crystallises into fcc lattice, so, Z= 4
\(a = 3.61 \times 10^{-8}\) cm
We know that, d = Z×M/a × Na
\(d = {4 \times 63.5 \over 6.022 \times 10^{23} \times {(3.61 × 10^{-8})}^3} \)
\(d = {2540 \over 283.35}\)
d = 8.97 g \(cm^3\)
So we can say that the value of calculated density is nearly same as measured value.
Ques16:- Analysis show that nickel oxide has formula \(Ni_{0.98} O_{1.00}\) . What fraction of the nickel exist as \(Ni^{+2}\) and \(Ni^{+3}\) ions?
Ans:- Nickel oxide is \(Ni_{0.98} O_{1.00}\)
Let\(Ni^{+2}\) in nickel oxide is x
So, \(Ni^{+3}\) ion will be 0.98-x
Since, the total charge on nickel oxide is zero. So,
\(2 Ni^{+2} + 3 Ni^{+3} - 1 \times O^{-2} = 0 \)
2× x + 3(0.98 -x) - 2 = 0
2x + 2.94 - 3x - 2 = 0
- x = -0.94
x= 0.94
So, percent fraction of \(Ni^{+2}\) is,
= \({0.94 \over 0.98} \times 100 \)
= 96%
And, percent fraction of \(Ni^{+3} \) is,
= 100 - 96
= 4%
Ques17:- What is a semiconductor? Describe the main two types of semiconductors and contrast their conduction mechanism.
Ans:- Semiconductor are those solids which have conductivity in between those of conductors and insulators.
Depending on the process of doping there are two types of semiconductors:-
i) p-type semiconductor
ii) n-type semiconductor
• When silicon crystal is doped with a group 13 element such as Boron, Aluminium, Ga and In, then, the semiconductor formed is known as p-type semiconductor.
• When silicon crystal is doped with group 15 elements, such as, P, As, Sb and Bi then the semiconductor formed is known as the n-type semiconductor.
Ques18:- Non-stichiometric cuprous oxide, \(Cu_2O\) 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 p-type semiconductor?
Ans:- Since the ratio of Cu : O in \(Cu_2O\) is slightly less than 2:1, it means \(Cu_2O\) is a non- stichiometric crystal. Since positive charge is less than negative charge, therefore, positive ion will move and this will result in formation of p-type semiconductor. In this oxide, some of \(Cu^+\) ion have been replaced by \(Cu^{+2}\) ion and to maintain neutrality, one \(Cu^{+2}\) ion will replace two \(Cu^{+}\) ions. Due to this replacement, positive holes will be created and movement of these holes will cause conduction in Non-stichiometric cuprous oxide and it will behave as a p-type semiconductor.
Ques19:- Ferric oxide crystallises in hexagonal close- packed array of oxide ion with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Ans:- We know that in a closed packed array of anions, then, number of octahedral voids/holes per anion(oxide ion) = 1
Only 2/3 of these holes are occupied by $\ce{Fe^{+3}}$ ion.
So, number of $\ce{Fe^{+3}}$ ion = \({2 \over 3} \times1 = {2\over 3}\)
\(O^{-2}\) form cubic closed packed lattice, so, number of \(O^{-2}\) ions = \(8 \times {1\over8} = 1\)
The formula of ferric oxide is \(Fe^{2/3} O\).
The simplest whole number ratio is 2:3
So, the formula of ferric oxide is \(Fe_2 O_3\) .
Ques20:- Classify each of the following as being either a p-type or n-type semiconductor :
i) Ge doped with In.
ii) B doped with SI.
Ans:- i) Ge doped with In is a p-type semiconductor because, Ge is an element of group 14 whereas In is an element of group 13. All the 3 valence electrons present in In will get bonded with 3 of the 4 valence electrons of Ge. Since, the fourth bond of Ge contains only one electron deficient bond or a hole and as a result of this, a p-type semiconductor is formed by doping of Ge with In.
ii) Boron, B doped with SI is a n-type semiconductor because, B is an element of group 13 whereas Si is an element of group 14. All the three valence electrons in B will get bonded with three out of four electrons of silicon. The fourth electron of silicon will be responsible for the conductivity of the semiconductor is formed by doping of B with silicon.
Ques21:- Gold ( atomic radius= 0.414nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
Ans:- Given that, radius, r = 0.414 nm, Z= 4, a =?
In face centred unit cell, diagonal of a face = 4r = a√2
a = \(4r\over√2\) = 2√2r
a = \(2\times 1.414 \times 0.414 \) nm
a = 0.4072 nm
Thus, the length of side of a unit cell, a = 0.4072 nm.
Ques22:- 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) In conductors, valence band and conduction band either overlap or there is very small energy gap between them and thus, electrons can easily go into conduction band from the valence band. The movement of electrons makes the substance conductor.
However, in an insulator, there is large energy gap between valence band and conduction band and thus, electrons can not move from valence band into conduction band.
ii) In semiconductors, there is small energy gap between valence band and conduction band than the energy gap between valence band and conduction band in insulator but larger than that of conductors. At room temperature, they are not good conductor of electricity. However, when the temperature is increased, some of the electrons get sufficient energy to move from valence band to conduction band and this result in increase of conductivity. In conductors the energy gap between valence band and conduction band is very small and electron can easily flow into the conduction band and thus, they are very good conductor of electricity.
Ques23:- Explain the following terms with suitable example :
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials, and,
(iv) F-centres .
Ans:- i) Schottky defect:- In Schottky defect, equal number of cations and anions are missing from the crystal lattice. Electrical conductivity is maintained and decrease in density takes place here.
ii) Frenkel defect:- In Frenkel defect, cations are located in Interstitial sites in the crystal lattice. There is no decrease in density takes place and electrical conductivity is maintained.
iii) Interstitials:- When an atom with small enough radius sits/fits in Interstitial hole in a metal lattice, then, the compound formed is known as Interstitial compound. These compounds are industrially very important.
iv) F-centres:- In metal excess defect, due to anionic vacancies anion removed from lattice site leaving behind holes, and, electron entrapped in these holes to maintain electrical neutrality. These electrons are known as F-centres. These F-centres are responsible for the colour in the crystal.
Ques24:- Aluminum crystallises in a cubic close-packed structure. Its metallic radius is 125pm.
i) What is the length of the side of the unit cell ?
ii) How many unit cells are there in 1.00 cm³ of aluminum ?
Ans:- i) For a cubic close close-packed structure, the radius r of an atom is related to the edge length of the unit cell by the relation :
4r = a√2
So, a = 4r /√2
a = 2√2r
a = 2 × 1.414 × 125 pm
a = 353.5 pm
ii) Volume of one unit cell,
\(a^3 = 353.5 \times 10^{-10} \) cm³
\(a^3 = 4.42 \times 10^{-23} \ cm^3 \).
Number of unit cells in 1 cm³ of aluminum,
=> \(1 \over 4.42 \times 10^{-23}\)
=> 2.262 × 10²² .
Ques25:- If NaCl is doped with \(10^{–3}\) mol % of $\ce{SrCl2}$, what is the concentration of cation vacancies?
Ans:- The charge on \(Sr^{+2}\) ion is +2 and on \(Na^{+}\) ion is +1. Thus, when one \(Sr^{+2}\) cause one Na plus to leave the crystal lattice of NaCl, one cation vacancy is created because one of the cation vacancy created by the removal of \(2Na^{+}\) ion is occupied by \(Sr^{+2}\) ion.
Doping of NaCl with \(10^{–3}\) mol % of $\ce{SrCl2}$ indicates that 100 moles of NaCl is doped with $\ce{10^{-3} \ moles \ of \ SrCl2}$.
So, 100 moles of NaCl are doped with $\ce{SrCl2 = 10^{-3}}$ moles.
and, 0.1 mole of NaCl are doped with $\ce{SrCl2}$ = \({10^{-3} \over 100} = 10^5\) moles.
Since, each \(Sr^{+2}\) ion cause one cation vacancy, so, 1 mole of \(Sr^{+2}\) ions cause cation vacancies = \(10^{-5} \times 6.022 \times 10^{23} \) per mol
= \(6.022 × 10^{18} \) per mol
So, we can say that the concentration of cations vacancies = \(6.022 × 10^{18} \) per mol.
Ques26:- Explain the following with suitable examples :
i) Ferromagnetism
ii) Paramagnetism
iii) Ferrimagnetism
iv) Antiferromagnetism
v) 12-16 and 13-15 compounds.
Ans:- i) Ferrimagnetism:- The substance which remains magnetised even in absence of magnetic field is known as Ferromagnetic substance and the phenomenon is known as Ferromagnetism. For example Fe, Co, Ni etc., are Ferromagnetic substances.
ii) Paramagnetism:- Those substances which are attracted by magnetic field are known as the Paramagnetic substances and the phenomenon is known as the Paramagnetism.\(Cu^{+2}, Fe^{+2} \) etc. are examples of Paramagnetic substances or we can say that the substances which have unpaired electron shows Paramagnetism.
iii) Ferrimagnetism:- Those substances in which magnetic dipoles are parallel and antiparallel but unequal in numbers, are known as Ferrimagnetic substances and the property is called Ferrimagnetism. For example, $\ce{Fe3O4}$, ferrites of formula $\ce{MFe2O4}$ where M = Mg, Cu, Zn etc.
iv) Antiferromagnetism:- Those substances which expected to possess Paramagnetism or Ferromagnetism on the basis of unpaired electrons but they possesses zero magnetic moment are called antiferromagnetic substances and process is known as Antiferromagnetism. It arises due to the presence of equal number of magnetic moments in opposite direction which cancel each other.
v) 12-16 and 13-15 compounds:-
a) 12-16 compounds:- Compounds formed between the elements of group 12 and elements of group 16 are called 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe are 12-16 compounds.
b) 13-15 compounds:- Compounds formed between elements of group 13 and elements of group 15 are called 13-15 compounds.