Notes On Gaseous State This page provides you all details about the gaseous state of BSc standard. Concerned problems will be solved at the end of each concept. Continue till the end to find the easiest explanation of every concept you need in this regard . Postulates of Kinetic Theory of Gases: 1. All gas consist of a very large number of minute particles, called molecules. 2. The gas molecules are extremely small in size and are separated by large distance. The actual volume of the gas molecules is thus negligible as compared to the total volume occupied by the gas. 3. The pressure exerted by the gas is due to the bombardment of the molecules on the walls of the vessel. 4. The gas molecules collide with one another and also with the walls of the vessels. These collisions are perfectly elastic and there is no loss of energy during these collisions. 5. The distance between the gas molecules are very large. Thus, there is no effective force of attraction or r
Solutions of Long and Short Questions on Solid States (CHSE and CBSE Class 12th)
Solid State
1.
Establish a relationship between edge lengths (a), nearest neighbouring distance (d) and radius of atom (r) for body centred cubic unit cell and calculate
its packing efficiency.
In Δ ABC, AC = √(AB2
+ BC2 ) = √(a2
+ a2) =
2 a2 = √2 a
Now in Δ ACD, AD =√(AC2
+ CD2) =
From the figure, AD = √3a
=> d = AD/2 = 2r = √3a/2
Packing
Efficiency: The
percent volume occupied by the atoms out of the total volume of the unit cell is
called the packing efficiency.
P.E = (Volume occupied by the atoms / Volume of the unit cell) X 100
The number of
atoms per BCC unit cell is equal to one.
P.E =
P.E =
2.
Discuss close packing in solids.
One
dimensional close packing: Here particles are packed closely in one direction (in only one axis say
X axis). Atoms are arranged in a row touching each other.
Two dimensional
close packing: When
particle are packed in two directions (in X and Y axis say) it is called two
dimensional close packing. This can be generated by stacking (placing) the rows
of closely packed spheres (one dimensional pack of atoms). This can occur in
two ways:
a.
b.
Two dimensional Hexagonal close packing: In this packing the atoms are packed
in such a manner that the atoms of the second row fit in the depressions of the
atoms of the first row.
Three
dimensional close packing: Atoms (spheres) are packed in all possible directions (X, Y and Z axis)
forming the three dimensional network of the crystal. This can be generated
when two dimensional layers of atoms are stacked one above another. Accordingly
we can have three types of packing, one from square close pack and two
from the hexagonal close pack in two dimensions.
A. The AAA
type three dimensional packing from two dimensional square close pack: In this case the spheres (atoms) of
the second layer are placed over the first layer in such a manner that the
spheres of the upper layer are exactly above those of the first layer.
Similarly we can place such layers one above another. Since all layers are
aligned similarly, each layer is denoted as A and we get AAA type three
dimensional close packing.
The unit cell obtained from this packing is simple cubic. The
coordination number is 6 and packing efficiency is 52.4%.
The rest two types of three dimensional packing from two
dimensional hexagonal close packing are due to the placing of the third layer
of spheres w.r.t. the first layer.
For both
of these packing the placing of the second layer over first layer is similar.
The second layer is placed over the first layer such that the spheres of
the second layer are placed in the depressions of the first layer. Since the
spheres of the two layers are aligned differently, the first layer is marked as
A and the second as B.
B. The
ABAB type close packing (Hexagonal close pack, HCP): When the third layer of atoms is
stacked over the second in such a manner that the tetrahedral voids of the
second layer is covered up without affecting the octahedral voids, the spheres
of the third layer are exactly aligned with those of the first layer, the ABAB
type packing is obtained. The coordination
number in this case is 12. The unit cell formed from such packing is called
hexagonal unit cell. The number of atoms per unit cell is 6. The packing
efficiency is 74%.
C. The
ABCABC type packing (cubic close pack, ccp): When the third layer of atoms is stacked over the
second in such a manner that its spheres cover the octahedral voids and thus
the spheres of the third layer are not aligned with those of either first or
second layer. A new layer C is formed and the packing is called ABCABC
type packing. The coordination number in this
case is 12. The unit cell formed from such packing is FCC unit cell. The number
of atoms per unit cell is 4. The packing efficiency is 74%.
3. What are defects (imperfections) in solids? Discuss Stoichiometric
defect.
Ans: The irregularities in the arrangement
of constituent particles in a crystalline substance are called the defects. The
defects can broadly be classified as point defects and line defects. The point
defects in crystal arise due to the absence of some ions (or atoms) from the
lattice sites or occupancy of vacant lattice sites or interstitial spaces. The
line defects are the irregularities from ideal arrangement in entire row of
lattice points.
Point defects can be classified into
three types:
(a)
Stoichiometric defects: These are the defects which do not disturb the stoichiometry
of the solids. Clearly the numbers of atoms or ions remain the same as in the
formula. These are also called intrinsic or thermodynamic defects. These
defects can be classified into the following categories:
(i) Vacancy defects: When some of the lattice sites are
vacant, the crystal is said to have vacancy defect. This results in the
decrease in density of the substance. This defect may arise due to the
heating of substance.
(ii) Interstitial defects: When some constituent particles (atoms or molecules)
occupy interstitial sites, the crystal is said to have interstitial defects.
This defect increases the density of the material.
Non ionic solids show vacancy and interstitial defects. The type of stoichiometric defects shown by ionic solids (keeping it in mind that they
maintain electrical neutrality) are Frenkel and Schottky defects.
(iii) Frenkel defects: in this defect the smaller ion (usually the cation) is
dislocated from its normal site to an interstitial site. Clearly it creates a
vacancy defect at its original site and an interstitial defect at the new
position. For this reason the defect is called the dislocation defect.
The density of substance is not affected by this defect. It is
shown by ionic solids in which there is a large difference in the size of ions.
For example, AgCl, AgBr, AgI and ZnS in which size of Zn2+ and Ag+
are small compared to corresponding anions.
(iv) Schottky defects: It is equivalent to vacancy defect in non-ionic solid
and decrease in density is observed. The number of missing cations and anions
remain the same to maintain the electrical neutrality. This is most common type
of defect in ionic solid.
Ionic solids in
which the difference in size of ions is very small, this type of defect is observed.
For example NaCl, KCl, CsCl.
4. What are defects (imperfections) in solids? Discuss Nonstoichiometric
defect.
Ans: See previous question for the answer
of the first part of this. Then write…
One of the point
defect is:
Non-Stoichiometric
defect: Non-Stoichiometric
compounds are those in which the numbers of positive and negative ions are not
in the same ratio as indicated by their chemical formulae. Examples of this
type includes Fe0.93O, Fe0.96O, Cu1.97S,
VOx (x ranging from 0.6 to 1.3) etc. Thus these compounds do not obey law
of constant proportions. This defect is of two types:
i.
Metal Excess defect:
·
Due to extra cations occupying the interstitial sites: Here an electron from a non metal
atom neutralizes the increased charge of the extra cation.
For Example: Zinc Oxide is white in colour at room temperature. On heating
it loses oxygen and turns yellow.
Now there is excess of zinc in
the crystal and the formula becomes Zn1+xO. The excess Zn2+
ions move to the interstitial sites and the electron to the neighbouring interstitial sites.
·
Due to anion vacancy: Alkali halides like NaCl and KCl show this type of defect. An
electron compensates the position of an anion. When crystal of sodium chloride
are heated in an atmosphere of sodium vapor sodium atoms are deposited on the
surface of the crystal, and the Cl- ions diffuse to the surface of
the crystal, combine with sodium to form sodium chloride. Thus sodium atoms
lose electrons to form Na+ ions and the released electrons diffuse
into the crystal and occupy anionic sites. This is the cause of metal excess
here (excess in Na+). The anionic sites occupied by electrons are
called F- centres (from Greek word Farbenzenter, the cause of coloured appearance of crystal). NaCl appears yellow. The n-type semiconductor
comes under this category where current flows due to extra electrons.
ii.
Metal deficiency defect:
·
Due to missing positive ion: In this case a cation is missing from its lattice sites and
the electrical neutrality is maintained by the adjacent metal ion with a higher
oxidation state.
For Example: FeO which is often found in a
composition of Fe0.96O is due to the missing of some Fe2+
ions and the loss positive charge is adjusted by required numbers of Fe3+
ions.
·
5.
What is impurity defect? Accordingly discuss semiconductors.
Ans: The irregularities in the arrangement
of constituent particles in a crystalline substance are called the defects. The
impurity defects are point defects which arise when foreign atoms are present
in place of host atoms. Addition of impurity to a crystalline substance is
called doping.
·
·
Impurity defect in covalent solid: In case of covalent solids such as
silicon or germanium (group 14 elements), which have four valence electrons,
the elements which are added as impurities may have more than four electrons
(group 15 elements) or less than four electrons (group 13 elements) in their
valence shell. The defects thus introduced in the crystals are called electronic
defects. This electronic defect may also appear in pure substance.
·
The electronic defect is the cause of electrical conductivities in Semiconductors:
Semiconductors are
materials in which the energy gap between V.B and C.B is very small of
about 0.7ev to 1.1ev. In this case the electrons from V.B can jump to the C.B
and accordingly show less electrical conductivity as compared to the metals.
Ex: Si. Depending on whether electronic defect is in pure substance or in
impure substance semi conductors are classified into two broad types:
·
Intrinsic semiconductor: The electronic defect in pure covalent crystals is referred
to as intrinsic semiconductors. These are insulators at room
temperature and become semiconductors at high temperature. When temperature
increases, some covalent bonds are broken and free electrons are produced. When
electrons move from bond positions, holes (electron vacancies) are produced,
then the electrons from neighbouring atoms flow into the hole sites creating new
holes at their sites. This process continues which is the cause of flow of
electricity. Ex: Si or Ge
· Extrinsic semiconductors: The conductivity of pure substance like silicon or germanium can be increased by introduction of required quantity of impurity into them. The introduction of impurity is called doping. Extrinsic semiconductors can further be classified into two types depending on whether higher valent or lower valent elements are doped:
·
· p-type semiconductors: When group 14 elements like Si or Ge is doped with group 13 elements like boron, aluminium or gallium (lower valent elements.) p-type semiconductor is formed. The three valence electrons of boron are utilized in the formation of three covalent bonds with silicon. Thus holes are created at the sites where the fourth electron is missing. This is called electron hole or electron vacancy. Thus electrons from neighbouring atoms flow into that sites (holes) and the process continues and show conductivity. Since positive holes are the cause of conductivity, these doped materials are called as p-type semiconductors.
6.
Discuss band theory of metals and accordingly classify conductors,
semiconductors and insulators.
Ans: Metals conduct electricity with the
help of valence electrons in them. The atomic orbitals of the metals having
equivalent energy combine to form molecular orbitals (bonding and antibonding
M.Os). The antibonding M.Os are higher in energy than bonding M.Os. The energy
bands formed during the formation of M.Os decide whether the valence electrons
can jump to the conduction band, and accordingly conductors, semiconductors and
insulators are classified.
Let us consider a piece of lithium
crystal which consists of large number of lithium atoms. Li has an E.C 1s2
2s1. The half filled 2s orbital take part in bonding with other Li
atoms as follows.
The bonding molecular orbital (B.M.O)
contains two electrons while the antibonding molecular orbital (A.B.M.O)
remains vacant.
Similarly we can have M.O diagram for
three Li atoms, then for four and finally for n numbers of Li atoms. In Li
crystal containing n atoms (Lin), the n number of valence atomic
orbital (2s) on combination produces number of bonding and antibonding M.Os.
i.e., n number of closely spaced energy levels called energy bands.
The bands formed from atomic orbitals
containing valence electrons are called valence band. The empty bands
formed from empty atomic orbitals in valence shell are called the conduction
band. If the two bands do not overlap then the energy difference between
them is called the forbidden band or energy gap. This is the basis of
classification of conductors, semiconductors and insulator.
The relationship
between these bands in various types of solids is as follows:
Conductors: In conductor the valence band and the
conduction band are partially overlapped with each other, so the electron can
easily pass (and accordingly flow) from the V.B to C.B when a little electrical
potential is applied.
Semiconductors: The energy gap between V.B and C.B is
very small of about 0.7ev to 1.1ev. In this case the electrons from V.B can
jump to the C.B and accordingly show less electrical conductivity as compared
to the metals. Ex: Si
Insulators: Due to large energy gap between V.B
and C.B electrons in these materials can’t pass into the C.B and no current
flow is observed.
7.
Discuss magnetic properties in solids.
Ans: The magnetic behaviour of materials
arises due to the orbital and spinning motions of the electrons. As electrons
are charged particles, their orbital and spinning motions produces a resultant
magnetic field and magnetic moment (act like tiny bar magnet). The magnetic
moment generated due to the orbital motion of electrons is called orbital
magnetic moment and that due to the spin motion is called the spin magnetic
moment.
The magnetic moment (denoted as µB)
is measured in unit of Bohr Magneton (BM), given by the formula µB =
eh/4πmc B.M where e = charge of electron, h = Plank’s constant, m = mass of
electron, c = velocity of light. Thus µB = 9.274 x 10 -24
J T -1. Spin magnetic moment of each electron is indicated as
± µB (due to two spinning possibilities) and the orbital
magnetic moment is indicated as ml µB (ml
is the magnetic quantum number of the electron).
Based on the response to the external magnetic
field, the substances are classified into different categories as follows:
·
Paramagnetic substances: These are materials weakly attracted by the magnet (by
external magnetic field). These substances have permanent dipole moment (µ)
due to unpaired electrons, represented by the formula µ =
Examples:
O2 , Cu2+,
Fe3+, Na, K etc.
·
Diamagnetic substance: These substances are weakly repealed
by magnet. The electrons are paired up in orbitals (magnetic moments are
cancelled by opposite spins) in these materials and hence have no permanent
magnetic moment. Examples include Zn, Cd, Hg, H2O, NaCl etc
(containing even number of electrons).
Domain:
It is the region
within the magnetic material in which the magnetization is in a uniform
direction i.e., the individual magnetic moments of atoms grouped together and
aligned in the same direction. The magnetic domains are indicated by arrows. On
the basis of arrangement in the domains the magnetic substances are classified
as follows:
·
Ferromagnetic: These
substances are strongly attracted by magnet (external magnetic field) and don’t
lose their magnetism even after the removal of external magnetic fields. The
domains in these materials are arranged in parallel direction. This can be
considered as an extreme case of paramagnetism. These substances show a strong
tendency to get magnetization. Examples include Fe, Co, Ni, CrO2 .
Curie
temperature: The temperature above which ferromagnetic
substances behave like paramagnetic (due to randomization of domains) is called
the Curie temperature.
·
Antiferromagnetic: Substances which are expected to show paramagnetism or ferromagnetism but
are nonmagnetic are called antiferromagnetic compounds. The magnetic moments
are mutually cancelled out as the domains are arranged in parallel and
antiparallel directions in equal numbers. Examples include MnO, MnO2.
NiO etc.
·
Solid State: Questions
of 2 & 3 marks
1.
What are the seven crystal system? What are the unit
cell parameters (a,b,c,α,β,γ) of cubic and hexagonal unit cells?
Ans: The seven crystal system are:
cubic, tetragonal, orthorhombic (or rhombic), monoclinic, triclinic,
rhombohedral (or trigonal) and the hexagonal.
The unit cell parameters for cubic unit
c ells are, a=b=c and α=β=γ =900
and the parameters for hexagonal unit cell are a=b≠c and α=β =900
but γ =1200
2.
What is the difference between crystalline and
amorphous substance? Refer book
3.
Define crystal lattice (space lattice), lattice points
and unit cell. Refer book
4.
How can you classify crystalline solids based on the
nature of constituent particles and the binding forces between them.
Ans: based on the nature of constituent particles and the binding force between
them, crystalline substance can be classified
into the following categories:
a)
Ionic solid b) molecular solids c) covalent or network
solids d) metallic solids
a)
Ionic Solids: The constituent particles are ions
and the binding force is ionic bond. The characteristics of these types of
solids are: high melting points, conduct electricity in aq. or in fused state,
soluble in polar solvents, hard and brittle.
b)
Molecular Solids: The constituent particles are
molecules and these are further classified into three types based on the nature
of molecules: a) Nonpolar molecular solids: The constituent
particles are either like atoms of noble gas or non-polar molecules like H2,
Cl2, I2, CH4 etc. The binding force is weak
van der Waals forces. The characteristics of these solids are: soft, non-conductors of electricity. b) Polar molecular solids: The constituent
particles are polar molecules like HCl, SO2 etc. The binding force
is the dipole induced dipole interaction which is some what stronger than that
between non-polar molecular solids. The characteristics are almost similar to
non-polar molecular solids. C) hydrogen bonded molecular solids: The constituent
particles are molecules (containing H atoms) which are bound together by strong
hydrogen bonds. The characteristics of these compounds are soft solids, non-conductor of electricity, m.p and b.p higher than those of molecular solids.
c)
Covalent or Network solids: The constituent
particles are non-metallic atoms linked together by covalent bonds. Example are
diamond, graphite, silicon carbide (SiC) etc. The characteristics of these
compounds are strong, hard, brittle, extremely high melting point and
insulators (graphite is exception to it).
d)
Metallic Solids: The constituent
particles are positively charged metal ions and the free electrons. The binding
force is due to the attractive force between the metal ions and free electrons.
These free electrons are the cause of flow of electricity. The characteristics
of these compounds are high thermal and electrical conductivity, metallic lustre, malleable, ductile and high melting points.
5.
Prepare a table to compare various crystalline substances w.r.t. their type, constituent particles, binding force, example,
physical nature such as hard, soft, electrical conductivity, m.p etc. refer book and the
above question
6. Calculate the number
of atoms per various cubic unit cells. Refer class practice note.
7. A compound formed by
elements X and Y crystallises in the cubic structure where y atoms are at the
corners and the X at the centres of faces. Determine the formula of the
compound. Refer class practice notes
8. Sodium crystallises in a BCC unit cell.
Calculate the approximate number of unit cells in 9.2g of sodium. Refer class practice
notes
9. What are tetrahedral and octahedral voids?
The number of tetrahedral voids = 2n
The number of octahedral voids = n (where n = numbers of atoms per unit cell)
10. How are the atoms
packed in three dimension or how many ways atoms can be packed in three
dimension?
Ans: We can have four types of three dimensional close packing of atoms. They
are:
a.
The AAA type three dimensional close packing from two
dimensional square close pack: In this case the spheres (atoms) of the second layer are
placed over the first layer in such a manner that the spheres of the upper
layer are exactly above those of the first layer and so on. The packing efficiency
is 52.4% as the unit cell obtained in this packing is simple
cubic.
b.
The ABAB type three dimensional close packing from two
dimensional hexagonal close pack: The second layer is placed over the first layer such that the
spheres of the second layer are placed in the depressions of the first layer.
When the third layer of atoms is stacked over the second in such a manner that
the tetrahedral voids of the second layer is covered up without affecting the
octahedral voids, the spheres of the third layer are exactly aligned with those
of the first layer, the ABAB type packing is obtained. The coordination number in this case is 12. The unit
cell formed from such packing is called hexagonal unit cell. The number of
atoms per unit cell is 6. The packing efficiency is 74%.
c.
The ABCABC type packing (cubic close pack, ccp) from two dimensional hexagonal close pack: When the third layer of atoms is
stacked over the second in such a manner that its spheres cover the octahedral
voids and thus the spheres of the third layer are not aligned with those of
either first or second layer. A new layer C is formed and the packing is
called ABCABC type packing. The coordination
number in this case is 12. The unit cell formed from such packing is FCC unit
cell. The number of atoms per unit cell is 4. The packing efficiency is 74%.
d.
The BCC (body centred cubic) type packing: Atoms are so place
that the unit cell formed is BCC. The coordination number in this case is 8.
The packing efficiency is 68%.
11. In a FCC crystal
edge length is 580 pm. Calculate d and r. Ans: For answering all
related questions prepare a table showing type of unit cell and relation
between a, d and r.
12.
Ans:
consider
the figure, atoms of radius (R) creating octahedral void of radius (r). The
fifth and sixth atoms which are present above and below the plane of the four
atoms are not shown.
ABC is
aright angled triangle, => BC2 = AB2 + AC2 => (2R)2 = (R+r)2 +
(R+r)2 = 2(R+r)2 =>
2R2 = (R+r)2
=>
(√2R)2 = (R+r)2
=> √2R = R+r => r = R
(√2-1) => r = 0.414R
13.
Ans:
A tetrahedral void is represented by the figure where three atoms A, E, F form
the triangular base and the fourth atom (B) lies at the top. The tetrahedral
void comes within these four spheres.
In
right angled triangle ABC, AB = √ (AC2 + BC2)= √(a2
+ a2)= √2a
The
spheres A and B are actually touching each other, => AB =2R = √2a
=>
R = a/√2 -----eqn 1
In Δ
ABD, AD = √ AB2 + BD2
= √{(√2a)2 + a2}= √3a
In the
body diagonal octahedral void is present which touches atoms A and D. Thus AD
= R+r+r+R
=>
AD = 2 (R+r) = √3a => R+r = √3a/2
----- eqn 2
Dividing eqn 2
by 1, (R+r)/R = (√3a/2 ) / (a/√2) =
√3/√2
=>1 + (r/R) =
√3/√2 => r/R = (√3/√2) -1 => r = 0.225R
14. A solid crystallises in NaCl type structure. If the radius of the cation A is 100pm then calculate
the radius of anion B.
Ans: In NaCl the anions
(in this case B of radius r) occupy the lattice sites and the cation (in this
case A of radius r) occupy the octahedral voids. We know r = 0.414R
=> 100pm = 0.414R
=>
R = 100/0.414 = 241pm
15. A compound is formed by two elements X and Y.
Atoms Y occupy the lattice points (or in the ccp arrangement) and the X atoms
occupy all octahedral voids. What is the formulas of the compound? Refer
class practice copy
16. Two elements A and
B form a compound. Atoms of element B form hcp lattice and the atoms of A
occupy 2/3rd of tetrahedral voids. What is the formula of the
compound? Refer class practice copy
17. In the mineral,
spinel (MgAl2O4), oxide ions form ccp structure, Mg2+
occupy the tetrahedral voids and Al3+ ions occupy the
octahedral voids. Calculate what percentage of tetrahedral and octahedral voids
are filled in by Mg2+ and Al3+?
Ans: Since ccp (FCC) arrangement, oxide ions (at
lattice points) should be 4 ( =n ) per unit cell. Accordingly the number of Mg2+
ions in tetrahedral voids should be = 2n
2x4 = 8. But in the formula only one Mg2+ ion present per four oxide
ion. The fraction is 1/8 and the percentage of tetrahedral void filled in
=(1/8)x100 = 12.5%.
Similarly number of Al3+ (in the octahedral void) is 2 per four
oxide ion whereas it shoud be = n = 4. Thus the fraction is 2/4 and the
percentage is (2/4)x100 = 50%
18. In corundum, oxides ions are arranged in
hexagonal close packing and aluminium ions occupy 2/3rd of the
octahedral voids. What is the formula of the compound? Try Yourself
19. Calculate the
approximate number of unit cells in 1g of ideal NaCl crystal.
Hints:
one
unit cell of NaCl crystal contains four NaCl formula unit. Mass of one formula
unit = 58.5g
20. What is radius ratio?
How is it linked to coordination number and structure of ionic crystal?
Ans:
The
ratio of the radius of the cation (r+) to radius of the anion (r-)
is called the radius ratio.
Mathematically,
radius ratio = r+/r-
The
co-ordination number of an ion is defined as the number of oppositely charged
ions surrounding it.
The formula
indicates that the larger the size of the cation, greater is the radius ratio
and ultimately a higher coordination number. The coordination number can be used
to determine the structure of the crystal. The following table lists all these
things:
|
Radius Ratio |
C.N |
Structure |
Structure type |
Example |
|
1-0.732 |
8 |
Body centred cubic |
Caesium chloride |
CsI, CsBr, TlBr,
TlCl |
|
0.732-0.414 |
6 |
Octahedral |
Sodium chloride |
NaBr, KBr, MgO,
MnO, CaO, CaS |
|
0.414-0.225 |
4 |
Tetrahedral |
Sphalerite, ZnS |
CuCl, CuBr, CuI,
BaS, HgS |
|
0.225-0.155 |
3 |
Planar triangular |
|
B2O3 |
21. A compound AB has
NaCl type structure. If the cationic radius 75pm what would be the maximum and
minimum sizes of the anions filling the voids?
Ans:
NaCl
has an octahedral type structure. Radius ratio = r+/r- =
0.732-0.414. Given that r+ = 75 pm
Minimum
value of r- = r+/0.732 = 75/0.732 =102.5pm
Maximum
value of r- = r+/0.414 = 75/0.414 = 181.2pm
22. A solid A+B-
has a NaCl type structure. If the anion has a radius of 250pm what should
be the
radius
of the cation? Can a cation of radius 180pm be slipped into the tetrahedral
site of the said crystal? Hints: Calculate the radius ratio. Then find
whether the value falls in the range of 0.414-0.225.
23. Prepare a table showing relationship between
a (edge length), (d)nearest neighbouring distance and radius of atom (r) in
various cubic unit cells.
Ans:
d =2r for all cubic unit cells
|
Simple cubic |
Body centred cubic |
Face centred cubic |
|
d=a |
d = √3a/2 = 0.866a |
d = a/√2 = 0.707a |
|
r = a/2 |
r = √3a/4 = 0.433a |
r = a/2√2 =
0.3535a |
24. Establish a relationship
between a, d and r in bcc unit cell. Refer class practice copy.
25. Establish a relationship
between a, d and r in fcc unit cell. Refer class practice copy.
26. Certain element
crystallises in the face centred cubic lattice having edge length of the unit
cell 620pm. What is the nearest neighbouring distance and the radius of the atom?
Try yourself
27. An ionic compound AB has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance. Hints: interionic distance = d = √3a/2 = 0.866a for bcc
28. Sodium crystallises
in bcc lattice of edge 4.29A0. Calculate the radius of the atom and
the body diagonal.(JEE main 2015). Hints: body diagonal= 4r
29. In a face centred
cubic lattice, edge length is 400pm. Find the diameter of the greatest sphere which can be
fitted into the interstitial void? Hints: the octahedral hole is larger
than the tetrahedral hole. For fcc, R (radius of atom at lattice points) = a/2√2 = 0.3535a. For octahedral hole the range is
0.732-0.414. Thus the largest radius of octahedral hole = 0.414R
30. If the unit cell of
a mineral has ccp array of oxygen atoms with m fraction of octahedral voids
occupied by aluminium ions and n fraction of tetrahedral voids occupied by
magnesium ions, then m and n respectively are? (JEE advanced 2016)
Ans:
for ccp, z=4
No. of
oct. voids = 4 => no. of Al3+ ions = 4m
No. of
tet. Voids = 8 => no. of Mg2+ ions = 8n
Formula
of the compound = Al4mMg8nO => 4m (+3) + 8n(+2) +
4(-2) = 0 => 4(3m+4n-2) = 0
3m+4n=2
=> m =1/2 and n = 1/8
31. An alloy of copper
silver and gold form a cubic lattice, cu atoms make the ccp. If Ag atoms are located at the edge centres and
Au atoms at the body centre then what will be the formula? Try
32. The ionic radius of
A+ and B‑ ions are 0.98x 10-10
m and 1.81x 10-10 m. the coordination number of each ion are? (NEET
2016) Try
33. For calculating
density related numerical such as Calculation of edege length, interionic
distance, radius, Avogadro’s number, atomic mass and nature of unit cell from
density, memorise density = zM/a3NA x
10-30 Unit
= g/cm3.
34. Copper crystallises
in fcc lattice. Atomic radius of copper is 128pm. Calculate the density of
copper. (Atomic mass of copper =63.5u)
Ans:
unit
cell = fcc, => z = 4, r = 128pm, M
= 63.5, a = 2√2r =2x1.414x128pm = 361.98pm
=>
a3 = (361.98)3x10-30 cm3
Density
= zM/a3NA x 10-30 =
(4x63.5) / [(361.98)3x10-30 cm3 x 6.02
x 1023] = 8.9 g cm-3
35. CsCl has cubic
structure having density 3.99 g
cm-3. Calculate the interionic distance. (Cs = 133u)
Ans:
Unit
cell = bcc, Since the unit cell contains one formula unit, => z = 1, M
= 133+35.5 = 168.5 u
Density
= zM/a3NA x 10-30
=> a3
= zM/NA x density = 1 x 168.5 / (6.02 x 1023 x
3.99) = 70.15 x10-24 cm3
=>
a = (70.15)1/3 x 10-8 cm = (70.15)1/3 x 102 pm
Now
let x = (70.15)1/3 => log
x = 1/3 log 70.15 = 1/3 x 1.8640 = 0.6153
=>
x = antilog 0.6153 = 4.124
=>
a = 4.124 x 102 pm = 412.4 pm
=>
interionic distance = d = √3a/2 = 0.866a =0.866 x 412.4 = 357pm
36. Density of sodium
chloride is 2.165 g cm-3. The interionic distance is 281 pm.
Calculate the value of Avogadro’s number using above data.
Ans:
unit
cell = fcc and one unit cell contains 4 NaCl units => z = 4
M =
58.5u, density =2.165 g cm-3
Edge
is the distance between two Cl- ions within which one Na+ ion is present at the centre of the edge.
=> interionic distance = d = 2 x 281 = 562pm
=> Density
= zM/a3NA x 10-30 =>
N0 = 6.09 x1023
37. An element has a
body centred cubic structure with cell edge
of 288pm. The density of the element is 7.2 g cm-3. How many
atoms are present in 518 g of the element?
Ans
: calculate
the atomic mass (M) and then apply mole concept.
38. the density of KBr
is 2.75 g cm-3 and the edge length is 654 pm. Predict the nature of
the unit cell. (K = 39u, Br =80u). calculate z get the nature
39. Thallium chloride
(TlCl) crystallises in a cubic lattice of edge length 385pm and density 7gcm-3.
Predict the nature of the unit cell. Z comes out to be 1. It does not mean simple
cubic. Note that its an ionic crystal. So Z = 1 means one formula unit which id
similar to CsCl (bcc unit cell).
40.
41. The edge length of
the unit cell (NaCl type structure) is 5.14A0. Assuming anion anion
contact calculate the ionic radius
of Cl‑ ion.
Ans:
Since
NaCl type structure, the interionic distance in LiCl is = 5.14/2 = 2.57 A0
AC =
√AB2 + BC2 = √ (2.57)2 + (2.57)2 =
3.63 A0
=>
radius of Cl-ion = 3.63/2 = 1.81 A0
42. Aluminium metal
forms a cubic close pack structure. Atomic radius is 125 x 10-12m.
Calculate edge length of the unit cell. How many unit cells are there in 1m3
of aluminium?
43. Draw a diagram showing various types of defects.
44. What are the
differences between Schottky defect and Frenkel defect?
45. What are the
differences between n-type semiconductor and p-type semiconductor?
46. Compare conductor,
semiconductor and insulator.
47. If NaCl is dopped with 10-3 mol% of SrCl2,
what will be the concentration of cation vacancies?
Ans: From given data, 100 moles of NaCl = 10-3 mol SrCl2
=>1 mole of NaCl = 10-3/100 = 10-5
mol/mol of NaCl
Each Sr2+ can replace two Na+ and
occupy one of the position of Na+ ion therefore creating one vacancy
=>The moles of cation vacancy = moles of Sr2+
dopped = 10-5 mol/mol of NaCl
=>The concentration of cation vacancies = 10-5
x 6.02 x 1023 numbers per mole of NaCl
48. If Al3+
ions replace Na+ ions at the edge centres of NaCl lattice then
calculate the vacancies in 1 mole NaCl.
Ans: 1 mole NaCl contains =6.02 x 1023 Na+
ions
Total numbers Na+
ions contributed from edge centres = ¼ x 12 = 3
Total numbers of Na+ ions in one unit cell = 3 + 1(at the body
centre) = 4
For every 4 Na+ ions, the ions present at edge
centres = 3
Al3+ ions replace Na+ ions at the edge
centres only
This means that Na+ ions which have been replaced
in one mole of NaCl = 3/4th x 6.02 x 1023
= 4.157 x 1023
Each Al3+ ion can replace 3 Na+ ions
and occupy one of the position of Na+ ion, therefore creating two
vacancies => 2/3rd of the
positions of Na+ ions (which are replaced by Al3+ ions )
remains vacant.
Hence the number of vacancies in one mole of NaCl = 2/3rd
x 4.157 x 1023 = 3.01 x 1023
49. The composition of a
sample of nickel oxide is Ni0.93O1.0 . What fraction of
nickel exist as Ni2+ and Ni3+ ?
Ans: The formula NiO suggests there is one Ni2+
for each Ni3+. When some of the Ni2+ ions are replaced
(two Ni3+ ions can replace three Ni2+ions) by Ni3+
ions, in a way that out of three Ni2+
ions, two positions occupied by Ni3+ ions and one position remain
vacant. This vacancy decrease the number of Ni2+ ions per oxygen
atom and formula becomes Ni0.93O1.0.
Now, out of total 93 Ni atoms let number of Ni3+ ions = x and Ni2+
= 93-x
Applying electrical neutrality of ionic crystal,
3x + 2 (93-x) = 2 x 100 => x = 14 = Ni3+ and Ni2+ = 93-14 = 79
=>fraction of Ni2+ = 79/93
=> percentage of Ni2+ = (79/93) x 100 = 85%
50. Why is coordination number of 12 not found in ionic crystal?
51. Why do NaCl and CsCl
have different structures?
52. What type of defect
can arise when a solid is heated? Which physical properties is affected by
it and in what way? Ans: Vacancy defect, explain next.
53. What type of stoichiometric defect is shown by (i) ZnS (ii) AgBr
Ans: Due to large difference in size between Zn2+
and S2-, shows Frenkel defect.
AgBr shows both Schottky and Frenkel defect.
54. Which type of defect
produces colour in ionic crystal? Explain with suitable example.
Ans: Metal excess defect and due to f centres. Refer book
55. How can you classify
magnetism in material according to their domain?
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