Chemistry

### Exercise with Detailed Solutions

Exercise with Detailed Solutions
Q 1885178967

Define the term 'amorphous'. Give a few
examples of amorphous solids.
Class 12 Exercise 1 Q.No. 1
Solution:

Amorphous solids are those substances, in
which there is no regular arrangement of its
constituent particles, (i.e., ions, atoms or
molecules). The arrangement of the constituting
particles has only short range order, i.e., a regular
and periodically repeating pattern is observed
over. short distances only, e.g., glass, rubber and
plastics.
Q 1815180060

What makes a glass different from a solid such
as quartz? Under what conditions could quartz
be converted into glass?
Class 12 Exercise 1 Q.No. 2
Solution:

Glass is made up of SiO_4 tetrahedral units. These
constituent particles have short range order only.
Quartz is also made up of SiO_4 tretrahedral units.
On heating it softens and melts over a wide range
of temperature. It is a crystalline solid having
long range ordered structure. It has a sharp
melting point.
Quartz can be converted into glass by first
melting and then rapidly cooling it.
Q 1815280160

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:
(i) Tetra phosphorusdecoxide(P_4O_ 10)
(ii) Ammonium phosphate, (NH_4)_3PO_4
(iii) SiC
(iv) I_2
(v) p_4
(vi) Plastics
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si
Class 12 Exercise 1 Q.No. 3
Solution:

Q 1845791663

Niobium crystallises in a body centred cubic
structure. If density is 8.55  g cm^-3, calculate
atomic radius of niobium, using its atomic mass
93u.
Class 12 Exercise 1 Q.No. 3
Solution:

 a^3 = (M *Z)/(d* N_a xx 10 ^-30) = (93 xx2)/(8.55 xx 6.02 xx 10^23 xx 10^-30)

= 3.61 xx 10^7

:. a = (3.61 xx 10^7)^(1/3) = (36.1 xx 10^6)^(1/3)

= 3.304 xx 10^2 p m = 330.4 p m
Q 1845280163

(i) What is meant by the term 'coordination
number'?
(ii) What is the coordination number of atom
(a) in a cubic close packed structure?
(b) in a body centred cubic structure?
Class 12 Exercise 1 Q.No. 4
Solution:

(i) The number of nearest neighboures of a
particle are called its coordination number.

(ii) (a) 12 (b) 8
Q 1815480360

How can you determine the atomic mass of an
unknown metal if you know its density and the
dimensions of its unit cell? Explain.
Class 12 Exercise 1 Q.No. 5
Solution:

Let the edge length of a unit cell = a
Density= d
Molar mass= M
Volume of the unit cell = a^3
Mass of the unit cell
=No. of atoms in unit cell x Mass of each atom
=Zxxm

Mass of an atom present in the unit cell

=m = M/N_a

:. d = text(Mass of unit cell)/text(Volume of unit cell) = (Z * m )/a^3 = (Z *M)/(a^3N_a)

:.  Atomic mass,  M = (d * a^3 * N_a)/Z
Q 1835480362

'Stability of a crystal is reflected in the
magnitude of its melting points'. 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?
Class 12 Exercise 1 Q.No. 6
Solution:

Higher the melting point, greater are the forces
holding the constituent particles together and
thus greater is the stability of a crystal. Melting
points of given substances are following.
Water = 273 K, Ethyl alcohol = 155.7 K,
Diethylether = 156.8 K, Methane= 90.5 K.

The intermolecular forces present in case of water
and ethyl alcohol are mainly due to the hydrogen
bonding which is responsible for their high
melting points. Hydrogen bonding is stronger in
case of water than ethyl alcohol and hence water
has higher melting point then ethyl alcohol.
Dipole-dipole interactions are present in case of
diethylether. The only forces present in case of
methane is the weak van der Waal's forces (or
London dispersion forces).
Q 1875480366

How will you distinguish between the following
pairs of terms:
(i) Cubic close packing and hexagonal close
packing?
(ii) Crystal lattice and unit cell?
(iii) Tetrahedral void and octahedral void?
Class 12 Exercise 1 Q.No. 7
Solution:

(i) Cubic close packing;: When the third layer is
placed over the second layer in such a way
that the spheres cover the octahedral voids,
a layer different from first (A) and second (B)
is produced. If we continue packing in this
manner, then a packing is obtained where the
spheres in every fourth layer will vertically
aligned. This pattern of packing spheres is
called ABCABC .. ... pattern or cubic close
packing.
Hexagonal close packing: When a third layer
is placed over the second layer in such a
manner that the spheres cover the tetrahedral
void, a three dimensional close packing is
obtained where the spheres in every third or
alternate layers are vertically aligned. If we
continue packing in this manner, then the
packing obtained would be called ABAB .......
pattern or hexagonal close packing.

(ii) Crystal lattic: It is a regular arrangement of
the constituent particles (i.e., ions, atoms or
molecules) of a crystal in three dimensional
space.
Unit cell: The smallest three dimensional
portion of a complete space lattice which
when repeated over and over again in
different directions produces the complete
crystal lattice is called the unit cell.

(iii) Tetrahedral void: A simple triangular void is
a crystal is surrounded by four spheres and
is called a tetrahedral void.
Octahedral void: A double triangular void is
surrounded by six spheres and is called a
octahedral void.
Q 1815580460

How many lattice points are there is one unit
cell of each ofthe following lattices?
(i) Face centred cubic
(it) Face centred tetragonal
(iir) Body centred cubic
Class 12 Exercise 1 Q.No. 8
Solution:

Lattice points in face centred cubic
(i) and face centred tetragonal
(ii) = 8 (at comers) +6 (at face centres)= 14

Particle per unit cell = 8 xx 1/8 + 6 + 6 xx 1/2 =4

Lattice points in body centred cubic
(iii)= 8 (at corners)+ 1 (at body centre)= 9
Particles per unit cell = 8 xx1/2 + 1/8 xx 1 =2
Q 1835580462

Explain:
(i) The basis of similarities and differences
between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Class 12 Exercise 1 Q.No. 9
Solution:

(i) Metallic and ionic crystals
Similarities:

(a) There is electrostatic force of attraction in
both metallic and ionic crystals.
(b) Both have high melting points.
(c) Bonds are non-directional in both the cases.

Differences:

(a) Ionic crystals are bad conductors of
electricity in solids state as ions are not free
to move. They can conduct electricity only
in the molten state or in aqueous solution.
Metallic crystals are good conductors of
electricity in solid state as electrons are free
to move.
(b) Ionic bond is strong due to strong
electrostatic forces of attraction.
Metallic bond may be strong or weak
depending upon the number of valence
electrons and the size of the kernels.

(ii) Ionic solids are hard and brittle.Ionic solids
are hard due to the presence of strong
electrostatic forces of attraction. The
brittleness in ionic crystals is due to the non-directional
bonds in them.
Q 1815880760

Calculate the efficiency of packing in case of a
metal crystal for
(i) simple cubic,
(ii) body centred cubic, and
(iii) face centred cubic

(with the assumptions that atoms are touching each other).
Class 12 Exercise 1 Q.No. 10
Solution:

Packing efficiency: It is the percentage of total
space filled by the particles.
(i) In simple cubic lattice:
Here a=2r
No. of spheres per unit cell = 1
Volume of spheres = 4/3 pi r^3
Volume of cube = a^3 = (2 r)^3 = 8 r^3

:. Packing efficiency
= (4/3 pi r^3)/(8r^3) = 0.524, i.e. 52.4 %

In bcc: AD= 4r in fig (i)

From right angled triangle ABC,

AC = sqrt(AB^2 +BC^2) = sqrt(a^2 +a^2) = sqrt2a

= sqrt(AC^2 +CD^2) = sqrt(2a^2 +a^2) = sqr (3a^2) = sqrt3 a

:. sqrt3 = 4r, a= (4r)/sqrt3

:. Volume of unit cell = a^3 =((4r)/sqrt3)^3 = (64r^3)/(3sqrt3)

No. of spheres per unit cell = 2

:. Volume of two spheres= 2 xx 4/3 pir^3 = 8/3pi r^2

:. Packing efficiency

= (8/3 pi r^3)/(64 r^3// sqrt3) = 0.68, i.e., 68 %

(iii) In fcc: Let the edge length of unit cell= a in fig (iii)

Let the radius of each sphere= r

:. AC =4r

From right angled triangle ABC,

AC = sqrt (AB^2 + BC^2) = sqrt(a^2 + a^2) = sqrt(2a^2) = sqrt2 a

:. sqrt 2a = 4 r

:. a = (4r)/sqrt2

:. Volume of the unit cell

= a^3= (4/sqrt2 r)^3 = (64r^3)/(2sqrt2) = (32r^3)/sqrt2

No. of unit cell in fcc= 4

:. Volume of four spheres = 4 xx 4/3 pi r^3 = 16 /3 pi ^3

:. Packing efficiency

(16 pir^3//3)/(64 r^3// 3 sqrt3 ) =0.74, i.e, 74 %
Q 1885880767

Silver crystallises in fee lattice. If edge length of
the cell is 4.07 xx 10^-8 cm and density is 10.5 g
cm^(-3) calculate the atomic mass of silver.
Class 12 Exercise 1 Q.No. 11
Solution:

M = (d * a^3 * N_a )/Z = (10.5 xx ( 4.07 xx 10^-8)^3 xx 6.023 xx 10^23)/4

= 107 . 09 g mol^-1
Q 1815880769

Silver crystallises in fcc lattice. If edge length of
the cell is 4.07 xx 10^-8 cm and density is 10.5 g
cm^(-3). Calculate the atomic mass of silver.
Class 12 Exercise 1 Q.No. 11
Solution:

M = (d * a^3 * N_a )/Z = (10.5 xx ( 4.07 xx 10^-8)^3 xx 6.023 xx 10^23)/4

= 107 . 09 g mol^-1
Q 1865591465

A cubic solid is made up of two elements P and
Q. Atoms of Q are at the comers of the cube and
P at the body centre. What is the formula of the
compound? What are the coordination numbers
of P and Q?
Class 12 Exercise 1 Q.No. 12
Solution:

As atoms Q are present at the eight corners of
the cube, therefore, the contribution of atoms of

Q in the unit cell =1/8 xx 8 =1

As atom P is present at the body centre, therefore,
the contribution of atoms of P in the unit cell= 1.
:.  Ratio of atoms of P: Q = 1 : 1
Hence, the formula of the compound= PQ
The coordination number of each P and Q = 8.
Q 1825191961

of the atoms in close-packing is R, derive relation
between rand R.
Class 12 Exercise 1 Q.No. 14
Solution:

A sphere is fitted into the octahedral void as
shown in the diagram.
DeltaABC is a right angled triangle.

:. BC^2 = AB^2 + AC^2
(2R)^2= (R + r)^2 + (R+ r)^2

(2R)^2= 2 (R + r)^2

=> (2R)^2/2 = (R +)^2

(sqrt2 R)^2 = (R +r)^2

=> sqrt2 R = R +r

r = sqrt 2 R -R

r = R(sqrt2 -1)

r = R (1.414 -1)

r = 0.414 R
Q 1816301279

Copper crystallises into a fcc lattice with edge
length 3.61 xx 10^-8 cm. Show that the calculated
density is in agreement with its measured value
of 8.92 gcm^-3
Class 12 Exercise 1 Q.No. 15
Solution:

d = (Z xx M)/(a^3 xx N_a)

= (4 xx 63.5)/((3.61 xx 10^-8) ^3 xx 6.023 xx 10^23) where { M=63.5 for Cu}

= 8.96  g cm^-3.
This calculated value of density is closely in
agreement with its measured value of 8.92  g cm^-3
Q 1816701679

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?
Class 12 Exercise 1 Q.No. 16
Solution:

The formula Ni_(0.98) O_(1.00) implies that 98 Ni atoms
are associated with 1 00 O atoms.
Let out of 98 Ni atoms present, x atoms are present
as Ni^(2+) ions and (98 -x) are present as Ni^(3+) ions.

:. Total charge on x Ni^(2+) ions and (100 -x) Ni^(3+)
ions should balance the charge on 100 O^(2-) ions.
Hence, x xx 2 + (98 - x) xx 3 = 1 00 x 2
2x+294-3x =200 x=94.

:. Fraction of Ni atoms present as

Ni^(2+) = 94 /98 xx 100 = 95 .91 %

and fraction of Ni atoms present as

Ni^(3+) = (98 -94)/98 xx 100 = 4/98 xx 100 = 4.081 %
Q 1806801778

What is a semiconductor? Describe the two main
types of semiconductors and contrast their
conduction mechanism.
Class 12 Exercise 1 Q.No. 17
Solution:

Semiconductors are those substances whose
conductance lies between that of conductors and
insulators. Semiconductor are of two types: (a)
n-type (b) p-type.
(a) n-type: When group 14 elements like silicon
and germanium (having four electrons in
the valence shell) are doped with group 15
element like phosphrous or arsenic (having
five electrons in the valence shell), few
lattice sites in Si or Ge is substituted by
atoms of P or As. As group 15 atoms have
five electrons in the valence shell, after
forming four covalent bonds with
neighbouring four Si atoms, fifth extra
electron is free and gets delocalized. These
delocalized electrons, increases the
conductivity of Si or Ge. As the increase in
conductivity is due to the negatively
charged electrons, the silicon or germanium
crystals doped with electron-rich impurities
are called n-type semiconductor.

(b) When group 14 elements like silicon or
germanium, having four electrons in the
valence shell, is doped with group 13
elements like boron, aluminium or gallium,
few sites of Si or Ge atoms are substituted
by atoms of B, AI or Ga. As group 13 atoms
have three el~l::trons in the valence shell,
they form three covalent bonds with the
neighbouring three Si or Ge atoms. Thus, a
hole is created at the site where fourth
electron is missing (called electron hole or
electron vacancy). As electron from the
electron hole than a hole is created at the
site from where electron has jumped. As
this continues, the electron holes will move
in a direction opposite to that of the flow
of electrons. When electricity in applied,
the electrons move towards the postively
charged plate and electron holes move
towards the negatively charged plate as if
they carry positive charge. Hence, electron
deficit doped Si or Ge are called p-type
semiconductors.
Q 1816001870

Non-stoichiometric cuprous oxide, Cu_2 O can be
prepared in the 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?
Class 12 Exercise 1 Q.No. 18
Solution:

The stoichiometric ratio slightly less than 2 : 1 in
Cu_2O shows that some cuprous (Cu^+) ions have
been replaced by cupric (Cu^(2+)) ions. In order to
maintain electroneutrality, every two (Cu^+) ions
will be replaced by one Cu^(2+) ion thereby creating
a hole. As the conduction will be due to the
presence of these positive holes, hence it is a
p-type semiconductor.
Q 1836101972

Ferric oxide crystallises in a hexagonal close
packed array of oxide ions with two out of every
three octahedral holes occupied by ferric ions.
Derive the formula of the ferric oxide.
Class 12 Exercise 1 Q.No. 19
Solution:

Let the number of oxide ions (O^(2-)) in the packing = x

:. Number of octahedral voids = x

As out of every three octahedral holes, two are occupied by ferric ions, therefore, the number of ferric ions present = 2/3 xx x = (2x)/3

:. Ratio of Fe^(+3): O^(2-) = (2x)/3 : x , 2x : 3x = 2 :3

Hence, the formula of ferric oxide is Fe_2O_3
Q 1816101979

Classify each of the following as being either a
p-type or n-type semiconductor:
(i) Ge doped with ln (ii) B doped with Si
Class 12 Exercise 1 Q.No. 20
Solution:

(i) In belongs to group 13, hence an electron
deficient hole is created and therefore, it is a
p-type semiconductor.

(ii) B belongs to group 13 and has 3 electrons in
valence shell. When doped with Si, having
4 electrons in valence shell an electron deficit
hole is created and therefore in this case also
a p-type semiconductor is formed
Q 1866112075

Gold (atomic radius= 0.144 nm) crystallises in
a face centred unit cell. What is the length of a
side of the cell?
Class 12 Exercise 1 Q.No. 21
Solution:

In case of fcc, a= 2-sqrt2r = 2 xx 1.414 xx 0.144 nm

=0.407 nm.
Q 1836212172

In terms of band theory, what is the difference
(i) between a conductor and an insulator.
(ii) between a conductor and a semiconductor?
Class 12 Exercise 1 Q.No. 22
Solution:

(i) In case of conductor, the energy gap between
the valence band and the conduction band
is very small or there is overlapping between
valence and conduction band whereas in
case of insulator the energy gap between the
valence band and the conduction band is
very large.

(ii) In case of conductor, there is very small
energy gap or there is overlapping between
valence and conduction band but in a
semiconductor there is always a small energy
gap between valence and conduction band.
Q 1816312270

Explain the following terms with suitable
examples:
(i) Schottky defect
(ii) Frenkel defect
(iii) InterstitiaI and
(iv) F-centres
Class 12 Exercise 1 Q.No. 23
Solution:

(i) Schottky defect occurs when a pair of ions
of opposite charges, i.e., cations and anions,
are missing from the ideal lattice. In NaCI,
there are approximately 10^6 schottky pairs
per cm^3 at room temperature. There is one
schottky defect for 10^16 ions. The presence
of a large number of schottky defects in a
crystal lowers its density. Schottky defect is
shown by ionic substances in which cation
and anion are of almost similar sizes. For
example, NaCI, KCl, CsCI and AgBr.

(ii) Frenkel defects is a combination of two basic
types of point defects : Schottky and
interstitial. It occurs when an ion leaves its
position in the lattice and occupies an
interstitial site leaving a gap in the crystal. It
is shown by ionic substances having large
difference in the size of oppositly charged
ions, For example, ZnS, AgCI, AgBr and Agl
due to small size of Zn^(2+)  and Ag^+ ions. It may
be noted that AgBr shows both Frenkel as
well as schottky defects.

(iii) Interstitials are the atoms or ions which
occupy the normally vacant interstitial sites
in a crystal. Non ionic solids are examples of
this defect

(iv) When there is an excess of metal ions in non-stoichiometric
compounds, the crystal lattice
has vacant anion sites. These sites are
occupied by electrons. The anion sites
occupied by electrons are called F-centres.
Alkali halides like NaCI, KCl and LiCl show
this type of defect.
Q 1816512470

Aluminium crystallises in a cubic close-packed
structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit
cell?
{ii) How many unit cells are there in 1.00 cm^3 of
aluminium?
Class 12 Exercise 1 Q.No. 24
Solution:

Aluminium crystallises in ccp structure which is
same as fcc structure.

(i) For fcc, a= 2sqrt2 r

=2xx 1.414xx 125 p m=354 p m

(ii) Volume of one unit cell = a^3 = (354 xx 10^-12)^3

= (354 xx 10^-10 cm)^3

= 4.44 xx 10^-23  cm^3

Number of unit cells in 1  cm^3 = 1/(4.44 xx 10 ^-23)

= 2.25 xx 10^22 .
Q 1836712672

lf NaCI is doped with 10^-3 mol % SrCl_2, what is
the concentration of cation vacancies?
Class 12 Exercise 1 Q.No. 25
Solution:

Doping of NaCl with 10^-3 mol% SrCl_2 means that
100 moles of NaCI are doped with 10^-3 mol of
SrCl_2.

:. 1 mole of NaCI is doped with SrCl_2

10^-3/100 mole = 10^-5 mole

Each Sr^(2+) introduces one cation vacancy,
therefore, concentration of cation vacancies
= 10^-5 xx N_A

= 10^-5 mol xx 6.023 xx 10^23 mol^-1 = 6.023 xx 10^18
Q 1806812778

Explain the following with suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Anti-ferromagnetism
(v) 12-16and 13-15 group compounds
Class 12 Exercise 1 Q.No. 26
Solution:

(i) Ferromagnetism is the phenomenon shown
by substances which are strongly attracted
by magnetic field. These substances show
magnetism even in the absence of a magnetic
field. Examples are Fe, Co, Ni and CrO_2

(il) Paramagnetism is the phenomenon shown
by substances which are attracted by a
magnetic field but they lose their magnetism
in the absence of a magnetic field.
Examples are Cu^(2+), Fe^(3+), O_2, NO and CuO.

(iii) Ferrimagnetism is the phenomenon shown
by substances in which the number of
magnetic moments aligned .in parallel and
antiparallel directions are unequal in
numbers. Examples are Fe_3O_4 and ferrites of
fommla MFe_2O_4, where M = Mg, Cu and Zn.

(iv) Anti-ferromagnetism is the phenomenon
shown by substances in which equal number
of magnetic moments are aligned in opposite
directions so as to give net zero moment.
Examples are MnO, MnO_2 and Mn_2O_3.

(v) 12-16 compo.unds : Compounds formed
between elements of group 12 and group 16
are called 12-16 compounds e.g., ZnS, HgTe,
CdSe, etc.
13-15 compounds : Compounds formed
between elements of group 13 and group 15
are called 13-15 compounds e.g., GaAs, AlP,
etc.