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![nBM
V
=
=
A
nBM
W
Unit -2
THEORY OF DILUTE SOLUTIONS
1) What
is
solution?
[1]
A:
It
is
a
homogenous
mixture
of
two
or
more
compounds.
2) What
is
dilute
solution?
[1]
A:
It
is
a
solution
in
which
solute
concentration
is
very
less.
3) Give
an
example
for
solid-‐solid
solution
[1]
A:
Copper
dissolved
in
gold.
4) Give
an
example
for
gas-‐gas
solution
[1]
A:
Mixture
of
oxygen
and
nitrogen
gases.
5) Give
an
example
for
gas-‐solid
solution
[1]
A:
Solution
of
hydrogen
in
palladium.
6) Give
an
example
for
liquid-‐solid
solution
[1]
A:
Amalgam
of
mercury
with
sodium.
7) Give
an
example
for
liquid-‐liquid
solution
[1]
A:
Ethanol
dissolved
in
water.
8) Give
an
example
for
solid-‐gas
solution
[1]
A:
Camphor
in
nitrogen
gas.
9) Define
mole
fraction
and
give
the
equation
to
calculate
it.
[2]
A:Mole
fraction
is
the
ratio
of
number
of
moles
of
one
component
to
the
total
number
of
moles
of
all
the
components
in
the
solution.
A
B
n
AX
n n
A
=
+
B
B
nBX
n n
A
=
+
10)Define
molarity
and
give
the
equation
to
calculate
it.
[2]
A:
Number
of
moles
of
the
solute
present
per
liter
solution
is
known
as
molarity.
11)Define
molality
and
give
the
equation
to
calculate
it.
[2]
A:
Number
of
moles
of
the
solute
present
perkgsolvent
is
known
as
molality.
12)Define
the
term
solubility
of
a
substance.
[1]
A:
Solubility
of
a
substance
is
its
maximum
amount
that
can
be
dissolved
in
a
specified
amount
of
solvent
at
a
specified
temperature
13)State
Henry’s
law.
[2]
A:
Henry’s
Law:
At
constant
temperature
solubility
of
a
gas
in
a
liquid
is
directly
proportional
to
the
partial
pressure
of
gas
present
above
the
solution.
OR
At
constant
temperature
the
partial
pressure
of
the
gas
in
vapor
phase
(p)
is
proportional
to
the
mole
fraction
of
the
gas
(x)
in
the
solution.
Mathematically p ∝ x ; p = KH x.
Where KH is Henry’s law constant. KH depends on the nature of the gas.
14)Write
the
plot
which
shows
relation
between
partial
pressure
of
a
gas
v/s
its
mole
fraction.
[2]
A:
15)Mention
the
factors
affecting
solubility
of
a
gas
in
liquid.
[2]
A:
1.
Temperature
2.
Pressure
16)Explain
how
temperatures
effect
the
solubility
of
a
gas
in
liquid.
[2]
A:
Solubility
of
gases
in
liquid
decreases
with
rise
in
temperature.
According
to
Le
Chatelier’s
Principle,as
dissolution
is
an
exothermic
process,
the
solubility
should
decrease
with
increase
of
temperature.
17)Explain
how
pressure
effects
the
solubility
of
a
gas
in
liquid.
[1]
A:
The
solubility
of
gases
increases
with
increases
of
pressure.
18)Mention
the
applications
of
Henry’s
law.
[3]
A:
(a)
To
increase
the
solubility
of
CO2
insoft
drink
and
soda
water,
the
bottle
is
sealed
under
high
pressure.
Mole
fraction.
Partial
pressure
of
a
gas](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-8-2048.jpg)
![(b)
To
avoid
bends,
as
well,
the
toxic
effects
of
high
concentration
of
nitrogen
in
the
blood,
the
tanks
used
by
scuba
divers
are
filled
with
air
dilute
with
helium.
(c)
At
high
altitudes
the
partial
pressure
of
oxygen
is
less
than
that
at
the
ground
level.
This
leads
to
low
concentrations
of
oxygen
in
the
blood
and
tissues
of
people
living
at
high
altitudes
or
climbers.
19)State
Raoult’slaw
of
liquid-‐liquid
dilute
solutions.
[2]
A:
The
partial
vapour
pressure
of
each
component
of
the
solution
is
directly
proportional
to
its
mole
fraction
present
in
solution.
Thus,
for
component
1
P1
⍺
x1
And
p1
=
p1
0x1
20)What
are
ideal
solutions?
[1]
A:
The
solution
which
obey
Raoul’s
law
over
the
entire
range
of
concentration
are
known
as
ideal
solution
21)Mention
the
characters
of
ideal
solutions.
[3]
A:
22)What
are
non-‐ideal
solutions?
[1]
A:
When
a
solution
does
not
obey
Raoult’s
law
over
the
entire
range
of
concentration,
then
it
is
called
non-‐ideal
solution.
23)Mention
the
types
of
non-‐ideal
solutions.
[1]
A:
There
are
two
types
(a)
Non-‐ideal
solution
with
positive
deviation
from
Raoult’s
law
(b)
Non-‐ideal
solution
with
negative
deviation
from
Raoult’s
law
24)Give
an
example
for
non-‐ideal
solution
with
positive
deviation
from
Raoult’s
law.
[1]
A:
Mixtures
of
ethanol
and
acetone
Ideal
I.
It
obeys
Raoults
law
is
obeyed
at
all
temperature
and
concentration
P
=
PA
+
PB
II. ∆
V
mix
=
O
i.e.,
there
is
no
change
in
volume
on
mixing
III. ∆Hmix
=
O
i.e.,
there
is
no
enthalpy
change
when
ideal
solution
formed
IV. It
doesn’t
form
azeotropic
mixture
V. Force
of
attraction
between
A―A,
B―B is similar
as A―B
25)Give
an
example
for
non-‐ideal
solution
with
negative
deviation
from
Raoult’s
law.
[1]
A:
An
example
of
this
type
is
a
mixture
of
phenol
and
aniline.
26)What
are
azeotropes?
Give
example.
[2]
A:
Azeotropes
are
binary
mixtures
having
the
same
composition
in
liquid
and
vapour
phase
and
boil
at
a
constant
temperature.
For
example:
ethanol-‐water
mixture
27)State
Raoult’s
law
of
relative
lowering
of
vapour
pressure.
[1]
A:
Relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute.
28)Define
colligative
property.
[1]
A:
The
properties
depend
on
the
number
of
solute
particles
irrespective
of
their
nature
relative
to
the
total
number
of
particles
present
in
the
solution.
Such
properties
are
called
colligative
properties
29)Mention
four
colligative
properties
of
dilute
solutions.
[2]
A:
Relative
lowering
of
vapour
pressure
I. Elevation
in
Boiling
point
II. Depression
in
Freezing
point
III. Osmotic
pressure
30)Define
the
term
relative
lowering
of
vapour
pressure.
[2]
A:
It
is
the
ratio
of
lowering
of
vapour
pressure
to
the
vapour
pressure
of
the
pure
solvent
o
o
P P
Relativeloweringof V.P
P
−
=
31)What
is
elevation
in
boiling
point?
[1]
A:
Elevation
in
boiling
point
is
the
difference
between
the
boiling
point
of
the
solution
containing
non-‐volatile
solute
and
the
boiling
point
of
the
pure
solvent
∆Tb
=
T
–
To
32)Give
the
relation
between
elevation
in
boiling
point
and
molecular
mass
of
solute.
[2]
A:
ΔTb = Kb
2
1 2
w 1000
w M
×
×
Where w2 is mass of solute, w1 is the mass of the solvent; M2 is molar mass of the solute](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-9-2048.jpg)
![33)Give
the
S.I.unit
of
ebullioscopic
constant
or
boiling
point
elevation
constant
or
molal
elevation
constant.
[1]
A:
The
unit
of
Kb
is
K
kg
mol-‐1
34)What
is
depression
infreezing
point?
[1]
A:
It
is
the
decrease
in
the
freezing
point
of
solution
when
non-‐volatile
solute
is
added
into
solvent.
35)Give
the
relation
between
depression
infreezing
point
and
molecular
mass
of
solute.[2]
A:
ΔTf = Kf
2
2
1
w
M
w
1000
∴ M2 = f 2
f 1
K 1000 w
T w
× ×
Δ ×
where M2 is molar mass of the solute.
Note: Values of Kf and Kb of the solvent depends on their molecular mass and ΔHfusion and ΔHvap
of the solvent respectively.
36)Give
the
S.I.unit
of
cryoscopic
constant.
[1]
A:
The
unit
of
Kf
is
K
kg
mol-‐1
37)Draw
the
plot
showing
elevation
in
boiling
point
in
a
solution.
[2]
A:
―∆Tb
―
Temperature/K
Vapour
pressure
Tb
Tb
0
Solution
Solvent
38)Draw
the
plot
showing
depression
in
freezing
point
in
a
solution.
[2]
―∆Tf―
Temperature/K
39)Define
osmosis.
[1]
A:
The
process
of
movement
of
solvent
particles
from
lower
concentration
to
higher
concentration
through
semi-‐permeable
membrane
to
attain
equilibrium
is
called
osmosis.
40)What
is
osmotic
pressure
and
give
its
relation
with
concentration
of
solution.
[2]
A:
The
amount
of
external
pressure
required
to
stop
the
osmosis.
=
CRT
Where:
=
osmotic
pressure,
R
=
gas
constant,
T
=
temperature,
C
=
concentration
of
solution.
41)
What
are
isotonic
solutions?
[1]
A:
Two
different
solutions
having
sameosmotic
pressure
are
called
isotonic
solutions
42)What
are
hypertonic
solutions?
[1]
A:
The
solution
having
more
osmotic
pressure
than
other
43)What
are
hypotonic
solutions?
[1]
A:
The
solution
having
less
osmotic
pressure
than
other
Tf
Tf
o
Vapour
pressure
Solution
Liquid
solvent
Frozen
solvent](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-10-2048.jpg)
![44)Explain
the
application
of
reverse
osmosis
in
desalination
of
water.
[2]
A:
When
pressure
more
than
osmotic
pressure
is
applied,
pure
water
is
squeezed
out
of
the
sea
water
through
the
membrane.
A
variety
of
polymer
membranes
are
available
for
this
purpose.
The
pressure
required
for
the
reverse
osmosis
is
quite
high.
A
workable
porous
membrane
is
a
film
of
cellulose
acetate
placed
over
a
suitable
support.
Cellulose
acetate
is
permeable
to
water
but
impermeable
to
impurities
and
ions
present
in
sea
water.
45)What
is
reverse
osmosis?
[1]
A:
Movement
of
solvent
particles
from
higher
concentration
to
lower
concentration
through
a
semi
permeable
membrane,
when
pressure
is
applied
greater
than
osmotic
pressure
46)What
is
abnormal
molar
mass?
[1]
A:
A
molar
mass
that
is
either
lower
or
higher
than
the
expected
or
normal
value
is
called
as
abnormal
molar
mass.
47)
Define
Vant
hoff
factor
Van’t Hoff factor ‘i’ to account for the extent of association or dissociation of a solute in a solvent
is
i =
Normal molar mass
Abnormal molar mass
or
i =
observed colligative property
calculated colligative property
or
i =
total number of moles of particles after association or dissociation
Number of moles of particles before association or dissociation
48)What
is
the
value
of
i
for
NaCl.
[1]
A:
2
49)What
is
the
value
of
i
for
K2SO4.
[1]
A:
3
50)What
is
the
value
of
i
for
sugar.
[1]
A:
1
51)What
is
the
value
of
i
for
glucose.
[1]
A:
1
52)On
what
factor
the
colligative
property
depends
on.
[1]
A:
It
depends
on
number
of
moles
of
solute
particles
but
not
on
the
nature
of
the
solute.
53)Write
the
mathematical
equation
of
Raoults
law
in
case
of
non-‐volatile
solute.
[1]
A:
If
one
of
the
components
(solute)
is
non-‐volatile
then
the
equation
of
Raoults
law
is.
PB=
O
P
=
PA
+
PB
P
=
PA
+
O
P
=
PA
54)Write
the
differentiate
between
non-‐ideal
solutions
with
positive
deviation
and
negative
deviation
from
Raoult’s
law
[2]
55)Define
lowering
of
vapour
pressure?
[1]
A:
It
is
defined
as
the
difference
between
the
vapor
pressure
of
the
solvent
in
pure
state
and
the
vapour
pressure
of
the
solution
∆P
=
Po
–
P
56)State
Roult’s
law
of
relative
lowering
of
vapour
pressure
[1]
A:
It
states
that
the
relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute
Positive
deviation
(a)In
this
solution
solvent
–
solute
interaction
is
weaker
than
solvent
–
solvent,
solute-‐solute
interactions
(b)
P
>
PA
+
PB
(c)
∆
V
>
O
(d)
∆H
=
positive
(e)
It
forms
azeotrope
with
minimum
boiling
point
Negative
deviation
(a)
In
this
solution
solvent
–
solute
interaction
is
stronger
than
solvent
–
solvent,
solute-‐solute
interactions
(b)
P
<
PA
+
PB
(c)
∆
V
<
O
(d)
∆H
=
negative
(e)It
forms
azeotrope
with
maximum
boiling
point
P
=
PA
O.
XA](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-11-2048.jpg)
![57)Why
sea
water
freezes
below
00C?
[1]
A:
Sea
water
freezes
below
00C
due
to
the
presence
of
the
non-‐volatile
solute
dissolved
in
the
water.
58)Derive
the
equation
to
calculate
molecular
mass
of
unknown
solute
using
Raoult’s
law
of
relative
lowering
of
V.P
[3]
A:
According
to
Raoult’s
law
relative
lowering
of
vapour
pressure
is
equal
to
the
mole
fraction
of
the
solute.
o
Bo
P P
X
P
−
=
o nP P B
o n nP BA
−
=
+
nB<<<nA
for
dilute
solution
So
we
can
neglect
nB
in
denominator
o nP P B
o nP A
−
=
B
B
A
A
W
o MP P
o WP
M
−
=
o MWP P B A
o W MP BA
−
=
B A
B
A
oW .M P
M oW P P
⎛ ⎞
= ⎜ ⎟
−⎝ ⎠
Numerical
problems
1. A
solution
containing
2.56
g
sulphur
in
100
g
CS2
gave
a
freezing
point
lowering
of
0.383
K.
Calculate
the
molar
mass
of
sulphur
molecules.
Given
Kf
of
CS2
=
3.83
K
kg
mol−
1
.
Ans.
ΔTf
=
0.383
K,
Kf
=
3.83
K
kg
mol−
1
ΔTf
=
Kf
×
m
;
ΔTf
=
Kf
×
2
2
1
W
M
W
1000
M2
(molar
mass
of
sulphur
molecules)
=
2.56 1000 3.83
100 0.383
× ×
×
=
256
g
mol−
1
2. 100
g
of
water
has
3g
of
urea
dissolved
in
it.
Calculate
the
freezing
point
of
the
solution.
Kf
for
water
=
1.86
K
kg
mol−
1
,
molar
mass
of
urea
=
60
g
mol−
1
,
freezing
point
of
water
=
273.15
K
(0°C)
Ans.
ΔTf
=
Kf
×
m
;
ΔTf
=
1.86
×
2
2
1
W
M
W
1000
ΔTf
=
1.86
×
3 1000
60 100
×
×
=
0.93
ΔTf
=
0
f fT T−
∴
Tf
=
273.15
−
0.93
=
272.22
or
−0.93°C
3. Human
blood
has
osmotic
pressure
of
7.2
atm
at
body
temperature
of
37°C.
Calculate
the
molar
concentration
of
solute
particles
in
blood.
Given
R
=
0.0821
L
atm
K−
1
.
Ans.
π
=
CRT
;
C
=
RT
π
T
=
273
+
37
=
310
K
C
(molar
concentration)
=
7.2
0.0821 310×
=
0.2828
M
4. Vapour
pressure
of
benzene
is
200
mm
of
Hg.
2g
of
a
non-‐volatile
solute
in
78
g
benzene
has
vapour
pressure
of
195
mm
of
Hg.
Calculate
the
molar
mass
of
the
solute.
Molar
mass
of
benzene
=
78
g
mol−
1
.
Ans.
2
1
nP P
P n
°−
=
°
;
2
2
1
1
W
MP P
WP
M
°−
=
°
200 195
200
−
=
2
2
M
78
78
;
Molar
mass
of
solute
(M2)
=
200 2
5
×
=
80
g
mol−
1](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-12-2048.jpg)
![Reaction
taking
place
⎯⎯→ 2+ -
( s)At anode Zn Zn + 2e
⎯⎯→+ -
2 4 3At cathode MnO + NH + e MnO( OH) + NH
NH3
produced
in
the
reaction
forms
a
complex
with
Zn2+
to
form
[Zn(NH3)4]2+
.
8. What
are
the
anode
and
cathode
of
lead
acid
battery?
What
is
the
electrolyte?
Write
the
reactions
taking
place
at
anode
and
cathode
and
the
overall
reaction
during
discharging
of
the
battery.
(3
M)
It
consists
of
lead
anode
and
a
grid
of
lead
packed
with
lead
dioxide
(PbO2)
as
cathode.
Electrolyte
is
38%
solution
of
sulphuric
acid.
The
reactions
taking
place
when
the
battery
is
in
use
are
2
( ) 4 ( ) 4( )
2
2 4 ( ) ( ) 4 ( ) 2 ( )
2
( ) 4 2 2
s aq s
aq aq s l
Anode Pb SO PbSO e
Cathode PbO s SO H e PbSO H O
− −
− + −
+ ⎯⎯→ +
+ + + ⎯⎯→ +
The
overall
reaction
is
( ) 2 ( ) 2 4 ( ) 4 ( ) 2 ( )2 2 2s s aq s lPb PbO H SO PbSO H O+ + ⎯⎯→ +
9. In
Hydrogen
oxygen
fuel
cell
(i)
Draw
the
schematic
diagram
mentioning
the
anode
and
cathode.
What
is
the
electrolyte?
Write
the
reaction
taking
place
at
each
electrodes
and
the
net
cell
reaction.
(4M)
In
this
hydrogen
and
oxygen
gases
are
bubbled
through
porous
carbon
electrodes
into
concentrated
aqueous
sodium
hydroxide
solution.
Catalyst
like
finely
divided
platinum
or
palladium
is
incorporated
into
the
electrodes
for
increasing
the
rate
of
electrode
reaction
Reaction
taking
place
are
⎯⎯→
⎯⎯→
- -
2 ( g) 2 ( l) ( aq)
- -
2( g) ( aq) 2 ( l)
Cathode O + 2H O + 4e 4OH
Anode 2H + 4OH 4H O + 4e
Overall
reaction
is
⎯⎯→2 ( g) 2 ( g) 2 ( l)2H +O 2H O
10. What
is
corrosion?
During
rusting
of
iron
write
the
anodic
and
cathodic
reactions.
Give
the
composition
of
rust.
(3M)
When
a
metal
is
exposed
to
the
atmosphere
it
is
slowly
attacked
by
the
constituents
of
the
environment
as
a
result
of
which
the
metal
is
slowly
lost
in
the
form
of
its
compound
.
This
is
called
corrosion.
Reaction
taking
place
are
⎯⎯→
⎯⎯→
2+ -
( s)
-
2 ( g) 2 ( l)
At Anode 2Fe 2Fe + 4e
At Cathode O + 4H +( aq)+ 4e 2H O
H+
are
produced
from
H2CO3
formed
due
to
dissolution
of
carbon
dioxide
from
air
into
water
The
Fe2+
ions
are
further
oxidised
by
atmospheric
oxygen
to
ferric
ion
which
are
ultimately
converted
to
hydrated
ferric
oxide
called
rust.
Composition
of
rust
is
(Fe2O3.xH2O).
11. A
conductivity
cell
when
filled
with
0.01M
KCl
has
a
resistance
of
747.5
ohm
at
25o
C.
When
the
same
cell
was
filled
with
an
aqueous
solution
of
0.05M
CaCl2
solution
the
resistance
was
876
ohm.
Calculate
(i)
Conductivity
of
the
solution
(ii
)Molar
conductivity
of
the
solution
(given
conductivity
of
0.01M
KCl
=
0.14114
sm-‐1
)
(3M)
Cell
constant
G*
=
Rk
=
747.5×0.14114
=0.105.5m-‐1
-1
-1cell constant 105.5m
Conductivity k = = = 0.1204Sm
R 876 ohm
2 -1
m
k 0.1204
Molar conductivity λ = = = 0.00241sm mol
1000C 1000 × 0.05
12. The
electrical
resistance
of
a
column
of
0.05M
NaOH
solution
of
diameter
1cm
and
length
50cm
is
5.55×103
ohm.
Calculate
its
(i)
resistivity
(ii)
conductivity
(iii)
molar
conductivity
(3M)
Cell
constant
o l
G =
a
l
=
50
cm
Diameter
=
1
cm
∴
radius
=
0.5
cm
Area
of
cross
section
A
=
πr2
=
3.14×(0.5)2
=
0.785
cm3
* -150
G = = 63.694 cm
0.785](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-18-2048.jpg)
![-2
1 1
Resistivity ρ = = = 87.135 Ω
k 1.148 ×10
m
-2
2 -1
1000k
Molarconductivityλ =
C
1000 ×1.148 ×10
=
0.05
= 229.6 S cm mol
13. Calculate
the
emf
of
the
cell
in
which
the
following
reaction
takes
place.
2
( ) ( )2 (0.002 ) (0.160 ) 2s sNi Ag M Ni M Ag+ +
+ ⎯⎯→ +
Given
that
Eo
cell
=
1.05V
2 2
( )
10 2
( )
[ ][ ]0.059
log
2 [ ][ ]
so
cell cell
s
Ni Ag
E E
Ni Ag
+
+
= −
But
[M]
for
any
element
is
taken
as
unity
⎡ ⎤⎣ ⎦
⎡ ⎤⎣ ⎦
2+
o
cell cell 10 2+
Ni0.059
E = E - log
2 Ag
( )
2
0.059 0.160
= 1.05 - log
2 0.002
=
0.914V
Unit-4
CHEMICAL KINETICS
Number of Hours of Teaching-9
Marks allotted-8
In part-D, 5mark question is split preferable in the form of 3+2
Definition :- The branch of chemistry which deals with study of reaction rate and their
mechanism is called chemical kinetics
Rate of a chemical reaction
Q. 1 What is rate of reaction? (1m)
Ans: Change in molar concentration of reactant or product in per unit time is called rate of
reaction.
Types of rate of reactions
For reaction R à P
Average rate = decrease in conc. R
Of reaction time taken
rav = - ∆[R]
∆t
Average rate = increase in conc. Of P
Reaction time taken
rav = + ∆[P]
∆t
8
Marks
Part –A
1x1=1
Part –B
1x2=2
Part-C
1x5=5](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-19-2048.jpg)
![Q2:- For the reaction RàP, the conc. of reactant changes from 0.03M to 0.02M in 25 min.
calculate average rate of the reaction using the unit of time in seconds.
rav= - ∆[R]= - (0.02-0.03 )
∆t 25x60
=-[-0.01]
1500
= 6.66x10-6
M/s
Q3: What is the SI Unit of rate of reaction ? (1m)
Ans: Mol /L /s
Factors influencing Rate of reaction
Q4 :- Mention any two factors which influence the rate of reaction . 2M
Ans 1) Pressure or conc. of reactants
2) temperature
3) catalyst.
Dependence of rate on concentration .
Q5.) What is rate law ? (1 m)
Ans: Representation of rate of reaction in terms of concentration of reactants is called rate law.
Rate expression and rate equation
Q.6) Define rate equation or rate expression (2m)
Ans: Expression in which reaction rate is given in terms of molar conc. of reactants with each
term raised to some power which may or may not be same as the stoichiometric coefficient of
the reacting species in a balanced chemical equation.
Q.7) Define rate constant of a reaction. (1m)
Ans: Rate constant is equal to rate of reaction when the product of the molar conc. of
reactants is unity.
Order of a Reaction
Q.8) Define order of a reaction. 1M
Ans: Sum of the powers of the concentration of the reactants in the rate equation is called order
of reaction.
Q.9) Calculate the overall order of a reaction which has the rate expression. 1M
Rate= K [A]1/2
[B]3/2
Ans: Order of reaction = 1/2 + 3/2
= 2
Q.10) What is elementary reaction ? (1m)
Ans: Reaction taking place in one step is called elementary reaction.
Q.11)What are complex reactions?(1m)
Ans: Reactions taking place in more than one step are called complex reaction.
Q.12) What is SI Unit of rate constant of nth order reaction ? (1m)
Ans: (mol)1-n
. Ln-1.s-1
Q.13) What is SI unit of rate constant of zero order reaction? (1m)
Ans: Mol/L/s
Q.14) What is the order of reaction whose unit of rate constant and rate of reaction are same ?
(1m)
Ans: Zero order.
Q.15) Identify the reaction order from the rate constant K=2.3x10-5
mol-1
.L.S-1
(1m)
Ans: Comparing the unit of rate constant with general unit
Mol-1
.L.S-1
with ( Mol)1-n
.Ln-1
.S-1
1-n= -1
n=2
Molecularity of a reaction
Q.16) Define molecularity of a reaction . (1 m)
Ans: The number of reacting species taking part in an elementary reaction which must colloid
simultaneously in order to bring about a chemical reaction is called molecularity of reaction.
Q.17) In a complex reaction which step controls the overall rate of reaction and
what is it called? (2m)
Ans: Slowest step, which is called rate determining step.
Q.18) The conversion of molecules X to Y follows second order kinetics .If conc. of X
Increased to three times,how will it affect the rate of formation of Y ? (1m)
Ans: Increased rate =( Increased conc.)n
=32
=9
Rate of formation of Y increases
by 9 times
Integrated rate equations
Q.19) Derive rate constant of zero order reaction (3m)
Ans: Consider a zero order reaction R--> P
Rate =-d[R] = K[R]o
dt
= - d[R] = K
dt
= d[R] = -kdt -------(1)](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-20-2048.jpg)
![Integrating equation (1) both sides
[R]= -kt+I ----------(2)
Where “I” is integration constant
At t=0 [R]=[R]o where
[R]o is initial concentration of reactant.
∴Eqn (2) becomes
I=[R]o
Substituting I in eqn-------- (2)
[R] = -Kt + [R]o
-Kt = [R]-[R]o
Kt = [R]o-[R]
K = [R]o-[R]
t
Q:20) Derive integrated rate equation for first order reaction? (4m)
Ans-Consider a first order reaction.
RàP
Rate = - d[R] = K[R]
dt
d[R]= - K[R]
dt
d[R] = - K. dt. -------(1)
[R]
Integrating eqn.(1)on both side
ln [R] = - Kt + I --------(2)
Where “I” is integration constant
At t=o [R]=[R]o which called initial Concentration reactant
Substituting the values in
ln[R]o= I
Equation (2) can be written as
ln[R] = -Kt + ln[R]o
Kt = ln[R]o – ln [R]
Kt = ln [R]o
[R]
Kt =2.303 log[R]o
[R]
K= 2.303 x log[R]o
t [R]
Log [R]o Slope = K
[R] 2.303.
Or
o time K= 2.303xSlope
Half life of a reaction
Q21)Define halfe life of a reaction . (1m)
Ans: The time in which the conc.of a reactant is reduced to one half of its initial conc. is called
half life of a reaction (t1/2)
Q:22) Show that half life of a zero order reaction is directly proportional to initial
concentration of reactant from integrated rate equation.
OR
Derive the relation between half life and rate constant of zero order reaction .(2m).
Ans:-Rate constant of zero order reaction is
K= [R]o – [R]
t
At half life t =t ½ & [R] = ½ [R]o
.: K= [R]o – ½ [R]o
t1/2
K=[R]o
2t½
t ½ =[R]o
2 K
OR
t ½ ∝ [R]o
Q.23) Show that half life of a first order reaction is independent of initial Conc. of reactant
from integrated rate equation (2m)
Or
Derive the relation between half life of a first order reaction and its rate constant . (2m)
Ans: Rate constant of first order reaction is
K= 2.303 x log [R]o
t [R]
At half life t=t ½ ,[R]=[R]o
2
.: K = 2.303 x log [R]o
t½ [R]o/2
K =2.303 x log 2
t ½
K= 2.303 x0.3010
t½
t½ = 0.693
K
K= 2.303 x0.3010
t½
t½ = 0.693
K](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-21-2048.jpg)
![Q.24)A first order reaction is found to have a rate constant 5.5x10-14
/s .Calculate the half life
of the reaction (2m)
Soln. : K= 5.5x10-14
/s t½ =?
t½ = 0.693
K
= 0.693
5.5x10-14
t½ = 1.26x1013
sec
Q:25) Show that the time required for 99/. Completion of a first order reaction is twice the
time required for the completion of 90% of reaction( 4m)
I set : [R]o= 100, [R]=[100-90]=10 t=t90%
IIset : [R]o= 100 [R]= [100-99]1 t=t99%
To be proved t99%= 2t90%
K= 2.303 x log [R]o
t [R]
Sub. I set values .
K= 2.303x log 100
t 90% 10
K = 2.303x log 10
t90%
K= 2.303 X 1 - (1)
t90%
Substituting II set values
K=2.303 x log 100
t99% 1
K= 2.303 x 2 --------------(2)
t99%
Comparing equations (1) & (2)
2.303x 1 = 2.303x2
t90% t99%
t99% = 2t90%
Pseudo first order reaction
Q:26 Define pseudo first order reaction . Give an example. (2m)
Ans: Chemical reactions which are not first order but behave as fist order reaction under
suitable conditions are called pseudo first order Reactions. Ex: Inversion of cane sugar.
C12 H22O11+H2O àC6H12O6 + C6H12 O6
Temperature dependence of the rate of a reaction
Q:27)How does rate of reaction vary with temperature? (1m).
Ans: Rate of reaction increases with increase of temperature.
Q:28) What happens to the rate constant of a reaction when temperature is increased by 10o
.?
Ans: Rate constant increases nearly by two times.
Q.29) Write Arrhenius equation which relates the rate constant , activation energy and
temperature . (1m)
Ans K= A e-Ea/RT
Energy of activation
Q.30) Define energy of activation (1m).
Ans: The minimum energy required for the reactants to form activated complex is called
Activation energy.
Q.31) How is activation energy related to rate of reaction? (1m)
Ans: Rate of reaction is inversely proportional to activation energy.
ie r ∝ 1
Ea
Q.32) How is activation energy affected by presence of positive catalyst? (1m)
Ans: Activation energy of a reaction decreases in presence of catalyst.
Q.33) On increasing 100
K temperature rate of reaction becomes double,
explain from the max well Boltzmann distribution curve. (2m)
On increasing 100
K temperature, substance Increases the fraction of molecules double,hence
rate of reaction doubles.
Q:34) How does positive catalyst increases the rate of reaction? (2m)
Ans positive catalyst decreases the activation energy by changing the
Path of the reaction,which increases the rate of reaction](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-22-2048.jpg)
![21. What is Brownian movement? How is it caused?
Zig– Zag movement of colloidal particles in a medium
is called Brownian movement.
Reason: Particles of the medium are very small and
are moving randomly in all directions. They collide
with the colloidal particles and transfer their kinetic
energy. Colloidal particles move slowly and randomly due to unequal bombardments
by the particles of the medium. This is seen as “Brownian Movement”. “This
property is a direct proof for the concept that liquid state of matter is made up of small
molecules, which are in random motion, does not allow the particles to settle and is
responsible for the stability of the sols”
22. Classify the following colloids into positively and negatively charged sols
Al (OH)3, CdS, AS2S3, Fe (OH)3, gum, clay, basic dyes, Sols of acidic dyes, sols of
starch, and metallic sulphides, sols of metals (Ag, Au), haemoglobin.
Positively charged sol negatively charged sol
Al (OH)3, Fe (OH)3, basic dyes,
haemoglobin,
CdS, AS2S3, Sols of acidic dyes, sols of
starch, gum, clay, and metallic sulphides
and sols of metals (Ag, Au)
23. How do colloidal particle become charged or acquire charge?
The charge on the colloidal particles may be due to (i) preferential adsorption of ions
from the medium or (ii) due to electron capture by sol particles during
electrodispersion of metals.(iii)formation of electrical double layer
24.What is Electrophoresis.
Movement of electrically charged colloidal particles
towards their oppositely charged electrodes when the
colloid is placed in an electric field is electrophoresis.
Positively charged particles move towards cathode and
negatively charged particles move towards anode.
25. Mention any two methods of Coagulation of lyophobic sol
i) Electrophoresis
ii) Mixing of two oppositely charged sols. E.g.: positively charged Fe(OH)3 sol with
negatively charged As2S3 sol
iii) Continuous dialysis
iv) Addition of electrolyte
v) By boiling
26. State and illustrate Hardy- Schulze rule.
Higher the valency of the flocculating ion added, greater is the coagulating power of
the ion.
Ex (1): In the coagulation of negatively charged sol (As2S3) the coagulation power of
the positively charged active ion is Na+
< Ba+2
< Al+3
.
Ex (2): In the coagulation of positively charged sol [Fe (OH)3] the coagulating power
of the negatively charged active ion is Cl−
< SO4
−
2
< PO4
−
3
< [Fe(CN)6]4−
.
Note: Higher the charge on the flocculating ion, lesser is the amount of the
electrolyte required to coagulate a sol.
27. Difine coagulating value or flocculating value
The minimum concentration of electrolyte in millimoles per litre required to cause
precipitation of a sol in 2 hours is called coagulating value. Smaller the coagulating
value, higher is the coagulating power of the ion.
28. What is protective action of a sol? Give an example.
The property of a lyophilic sol by which it protects the lyophobic sol from
precipitation, even upon adding an electrolyte to it, is called protective action of
lyophilic sol. Lyophilic sol particles form a coat or layer around the lyophobic sol and
hence protect them from the action of the electrolytes.
29. What are Emulsions? Give an example
A liquid in a liquid colloid is called an emulsion. If two immiscible liquids are shaken
well, a dispersion of one liquid in the other, an emulsion is obtained.
Eg: Milk, butter, vanishing cream
30.Write a note on formation of delta region.
River water flowing towards the sea picks up many colloidal particles (clay, mud,
humus, slit) with it. These particles are negatively charged. When the river water meets
the sea, the electrolytes (salts like NaCl, MgSO4 etc) in the sea causes the coagulation
of these colloidal particles. Thus clay, mud, humus gets precipitated and scattered at
these places to form delta region.
31.Write the application of colloids in purification of smoke using Cottrell
precipitator
Smoke (Colloidal dispersion of solid in gas) from industries contains carbon, dust, soot
and many others as colloidal particles. To remove these, electrostatic precipitator
called cottrell precipitator is used. The precipitator consists of metal plates attached to
a high potential. As the smoke enters the precipitator, the charged colloidal particles](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-28-2048.jpg)
![15. In the extraction of aluminium carbon anodes are replaced regularly. Why?
A: Because the carbon gets worn out as the oxygen liberated reacts with it to form
CO2.
II. TWO MARK QUESTIONS
1. What is the role of (i) lime stone in iron extraction and (ii) cryolite in aluminium
extraction.?
A: (i) Removes silica impurity as slag calcium silicate
(ii) Cryolite increases conductivity and reduces melting point of Al2O3.
2. Give the chemical reactions involved in
(a) Iron extraction 2 marks
A: C + O2 →CO2 ; CO2 + C →2CO,
Fe2O3 + CO →2FeO + CO2 ; FeO + CO →Fe + CO2
CaCO3 →CaO + CO2, ; CaO + SiO2 →CaSiO3.
(b) Aluminium extraction 2 marks
A: Cathode: Al3+ (melt) + 3e Al(l)
Anode: C(s) + O-2(melt) CO(g) + 2e
C(s) + 2O2-(melt) CO2 (g) + 4e
Overall reaction: 2Al2O3 + 3C 4Al + 3CO2
(c) Copper extraction 2 marks
A: 2FeS + 3O2 2FeO + 2SO2
FeO + SiO2 FeSiO3
2Cu2S + 3O2 2Cu2O + 2SO2
2Cu2O + Cu2S 6Cu + SO2
3. How is zinc obtained from ZnO?
A: ZnO is heated with coke at 1673K. Zinc and carbon monoxide are formed.
ZnO + C Zn + CO
4. Give equations for the extraction of gold using NaCN.
A: 8NaCN (aq)+ 4Au(s) + 2H2O(aq) + O2(g) 4Na[Au(CN)2] (aq) + 4NaOH(aq)
2Na[Au(CN)2] (aq) + Zn(s) 2Au(s) +Na2[Zn(CN)4](aq).
5. Give the principles involved in (i) zone refining (ii) liquation
A: (i) Zone refining: The impurities are more soluble in the melt than in the solid state
of the metal.
(ii) Liquation: melting point of metals is lower than the impurities.
6. What are the requirements for the compound to be purified by vapour phase refining?
A: (i) The metal should form a volatile compound with an available reagent,
(ii) The volatile compound should be easily decomposable.
7. How do you refine nickel by Mond’s process?
A: In this process, nickel is heated in a stream of carbon monoxide forming a volatile
complex, nickel tetracarbonyl:
The carbonyl is subjected to higher temperature so that it is decomposed giving the
pure metal:
8. How do you remove oxygen and nitrogen impurities from Zirconium.
A: By van Arkel method:
The crude metal is heated in an evacuated vessel with iodine. The metal iodide
volatilises
Zr + 2I2 ZrI4
The metal iodide is decomposed on a tungsten filament. The pure metal is thus
deposited on the filament.
ZrI4 Zr + 2I2
III. THREE MARK QUESTIONS:
1. Explain the concentration of bauxite ore.
A: Bauxite ore is concentrated by leaching. The steps involved are
i) Bauxite is concentrated by digesting the powdered ore in a concentrated solution of
sodium hydroxide at 473-573 K and 35 bar pressure. Al2O3 is leached as sodium
aluminate.
ii) Aluminate solution is neutralised by passing CO2. Hydrated Al2O3 is precipitated by
seeding.
iii) Hydrated Al2O3 is filtered, dried and heated to get pure Al2O3.
2. Write the equations involved in leaching of alumina.
A: Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq)
2Na[Al(OH)4](aq) + CO2(g) → Al2O3.xH2O(s) + 2NaHCO3 (aq)
Al2O3.xH2O(s) ∆ Al2O3 + xH2O
3. Two reactions are given below, which of these two happen at a temperature above
1073K. Justify.
FeO + C Fe + CO --------(A)
FeO + CO Fe + CO2--------(B)
A: FeO + C Fe + CO happens at a temperature above 1073K.
In the Ellingham diagram we can see that the C,CO line goes below while CO,CO2
goes above FeO line at temperature above 1073K. So, C is the reducing agent.
4. How do you extract aluminium from bauxite ore?
A: Diagram.](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-31-2048.jpg)
![Answer: E2O5 ( oxide with higher oxidation state is more acidic)
23.How does the acidic characters of 15th
group metal oxides varies down the group? 1M
Answer; The acidic character decreases on moving down a group.
24.Write the increasing order of acidic character of N2O5, P2O5, As2O5 and Sb2O5 1M
Answer: N2O5 >P2O5 >As2O5 > Sb2O5
25.How is dinitrogen prepared in the laboratory? 2M
Answer: In the laboratory, dinitrogen is prepared by treating an aqueous solution of
ammonium chloride with sodium nitrite.
NH4Cl (aq) + NaNO2 (aq) → N2 (g) + 2H2O (l) + NaCl (aq)
26.How is dinitrogen prepared from ammonium dichromate? 2M
Answer: thermal decomposition of ammonium dichromate gives dinitrogen.
(NH4)2Cr2O7 → N2 + 4H2O + Cr2O3
27.How does dinitrogen reacts with Mg? 2M
Answer: Dinitrogen reacts with Mg to form magnesium nitride.
N2 +3 Mg → Mg3N2
28.For the manufacture of ammonia by Haber’s process, write flow chart and balanced
equation along with conditions? 3M
Answer: On large scale, obtained by Haber’s process
Optimum condition:
Pressure = 200 × 105
Pa (about 200 atm)
Temperature ∼ 700 K
Catalysts used − Iron oxide with small amounts of K2O and Al2O3 to increase the rate
of attainment of
equilibrium.
29.How does ammonia react with zinc sulphate? 2M
Answer: Ammonia reacts with zinc sulphate to form white precipitate of zinc hydroxide.
ZnSO4(aq) + 2NH4OH(aq) → Zn(OH)2(s) + (NH4)2SO4(aq)
30.How does ammonia react with cupric ion? 2M
Answer: With Cu2+
ion Ammonia acts as lewis base and forms deep blue colored
cuprammonium complex.
Cu2+
(aq) + 4NH3(aq) → [Cu(NH3)4]2+
(aq)
(blue) (deep blue)
31.How is Nitric acid manufactured by Ostwald process? 3M
Answer: Nitric acid in Ostwald process manufactured by the oxidation of ammonia.
Nitric oxide thus formed combines with oxygen giving NO2.
2NO ( g ) + O2 ( g )→2NO2 ( g )
Nitrogen dioxide so formed, dissolves in water to give HNO3.
3NO2 ( g ) + H2O ( l ) → 2HNO3 ( aq ) + NO ( g )
Dilute nitric acid on distillation followed by dehydration using conc. sulphuric acid gives
98% nitric acid.
32.How is nitric acid prepared in laboratory? 2M
Answer: Nitric acid is prepared in the laboratory by heating KNO3 or NaNO3 with
concentrated H2SO4 in glass retort.
NaNO3 + H2SO4 → NaHSO4 + HNO3](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-34-2048.jpg)
![33.How does dilute nitric acid with copper? 2M
Answer: Dil. Nitric acid reacts with copper to form cupric nitrate with the liberation of
nitric oxide.
3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O
34.How does concentrated nitric acid with copper? 2M
Answer: Conc.Nitric acid reacts with copper to form cupric nitrate with the liberation of
nitrogen dioxide.
Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O
35.How does dilute nitric acid with zinc? 2M
Answer: Dil. Nitric acid reacts with zinc to form zinc nitrate with the liberation of nitrous
oxide.
4Zn + 10HNO3(dilute) → 4 Zn (NO3)2 + 5H2O + N2O
36.How does concentrated nitric acid with zinc? 2M
Answer: Conc.Nitric acid reacts with zinc to form zinc nitrate with the liberation of
nitrogen dioxide.
Zn + 4HNO3(conc.) → Zn (NO3)2 + 2H2O + 2NO2
37.How does concentrated nitric acid with iodine? 2M
Answer: Conc.Nitric acid oxidizes iodine to form iodic acid.
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
38.How does concentrated nitric acid with carbon? 2M
Answer: Conc.Nitric acid oxidizes carbon to carbon dioxide
C + 4HNO3 → CO2 + 2H2O + 4NO2
39.What is passivity? 1M
Answer: Some metals like aluminium and chromium do not dissolve in concentrated
nitric acid due to the formation of a protective layer of oxide on the surface of the
metal. This phenomena is called passivity of metals.
40.explain Brown ring test with equations . 3M
Answer: Dilute FeSO4 solution is added to an aqueous solution of nitrate ion.
concentrated H2SO4 is then added along the sides of the test tube. A brown ring is
observed at the interface between the solution and H2SO4 layers indicates the presence of
nitrate ion in the solution.
NO3
-
+ 3Fe2+
+ 4H+
→ NO + 3Fe3+
+ 2H2O
[Fe (H2O)6 ]2+
+ NO → [Fe (H2O)5 (NO)]2+
+ H2O
(brown)
41.Write the resonance structurs of a) NO b) NO2 c) N2O5 1M each
Answer:
a) Structure of NO:
b) Structure of NO2 :
c) Structure of N2O5 :
42.Distinguish between white and red phosphorus. 2M
Answer: (any two)
White phosphorus Red Phosphorus
It is a soft and waxy solid. It is a hard and crystalline solid.
It is poisonous. It is non-poisonous.
It is insoluble in water but soluble in
carbon disulphide.
It is insoluble in both water and carbon
disulphide.
Highly reactive It is relatively less reactive.
In both solid and vapour states, it exists as
a P4 molecule.
It exists as a chain of tetrahedral P4 units.
Less stable More stable
43.How is Phosphine prepared in the laboratory form white phosphorous? 2M
Answer: In the laboratory phosphene is prepared by heating white phosphorus with
concentrated NaOH solution in an inert atmosphere of CO2.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2 PO2
44.Give a reaction to support Basic nature of phosphine. 2M
Answer: Phosphine react with hydrogen bromide and forms phosphonium bromide.
PH3 + HBr → PH4 Br
45.How is Phosphorous trichloride is obtained from phosphorous and chlorine? 2M
Answer: Phosphorus tri chloride is obtained by passing dry chlorine over heated white
phosphorus.
P4 + 6Cl2 → 4PCl3
46.How is Phosphorous pentachloride is obtained from phosphorous and chlorine? 2M](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-35-2048.jpg)
![Unit 8
The d- and f- Block Elements
I. Answer the following questions. Each question carries one mark
1. Define transition elements.
Ans. Transition element is defined as the one which has incompletely filled d orbitals in its
ground state or in any one of its oxidation states.
2. What is the position of the d block elements in the periodic table?
Ans. The d block elements are in the middle of s and p blocks, comprising the groups 3 to 12.
They are the four rows of elements in the periods 4th (3d series), 5th (4d series), 6th ( 5d series)
and 7th ( 6d series).
3. Zinc, cadmium and mercury of group 12 are not regarded as transition metals, Why ?
Ans. Zinc, cadmium and mercury of group 12 have full d10 configuration ( d orbitals are
completely filled ) in their ground state as well as in their common oxidation states and hence,
are not regarded as transition metals
4. Why d- block elements are named as ‘transition elements ‘ ?
Ans. The d–block elements occupies the middle of the periodic table and their properties are
transitional between s– and p– block elements.
5.Write the general electronic configuration of d block elements.
Ans. [ Noble gas] (n-1)d1-10ns1-2
6. Write the general outer electronic configuration of d- block elements. .
Ans. The general outer electronic configuration of d- block elements is (n-1)d1–10 ns1–2
7. Write the general electronic configuration of f- block elements.
Ans. The general electronic configuration of f- block elements (Lanthanoids) is
[Xe] 4f1 – 14 5d0- 1 6s2
8.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation
state.
Ans. Cerium
9. The outer electronic configuration of Cr is 3d5 4s1 instead of 3d44s2, why?
Ans. Half filled (3d5) orbitals are relatively more stable, hence one electron of 4s orbital jumps
to 3d orbital.
10. The outer electronic configuration of Cu is 3d10 4s1 instead of 3d94s2 , why?
Ans. Completely filled (3d10 ) orbitals are relatively more stable, hence one electron of 4s
orbital jumps to 3d orbital.
11. Account for high melting point and boiling points of transition metals.
Ans. The melting and boiling points of transition metals are high because of the
involvement of greater number of electrons from (n-1)d orbitals in addition to the ns
electrons in the inter atomic metallic bonding.
12.What is the trend in melting points of transition metals in a series?
Ans.The melting points of the transition metals in a series rise to a maximum at the middle of
the series (i.e. Cr or Mo or W - element with d 5 configuration ) and fall regularly as the atomic
number increases.
13.Why do transition metals have higher enthalpies of atomization?
Ans. Involvement of a large number of unpaired electrons of d orbitals favour stronger inter
atomic interactions resulting in stronger bonds between the atoms of a metal and higher
enthalpies of atomization.
14.Name one 3d series elements, that do not show variable oxidation states.
Ans. Sc (+3)
15.Transition metals exhibit variable oxidation states in its compounds, why?
Ans. Transition metals exhibit variable oxidation states in its compounds due to the
availability of both ns & (n – 1 ) d electrons for bond formation.
16. Name 3d series metal which shows highest oxidation state.
Ans. The highest oxidation state shown by 3d series transiNa
tion metals is +7 by Mn
17. Name a metal in the 3d series of transition metals which exhibit +1 oxidation state
most frequently.
Ans. copper
18.What is the trend in oxidation state of transition metals ?
Ans. The oxidation state increases with increase in atomic number & reaches a maximum in
the middle and then decreases.
19. 3d series transition metals exhibit +2 as the most common oxidation state (except
Sc) why?
Ans. The +2 oxidation state, which commonly occurs for nearly all the transition metals is
due to the loss of their outer 4s electrons](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-41-2048.jpg)
![(i) Chromite ore is fused (FeCr2O4) with sodium or potassium carbonate in free access of
air to get sodium chromate .
4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2
(ii) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to
give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be
crystallised.
2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O
Potassium dichromate prepared by treating the solution of sodium dichromate with
potassium chloride.
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
Orange crystals of potassium dichromate crystallise out.
66. Describe the preparation of potassium permanganate from manganous dioxide.
Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and
an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates
in a neutral or acidic solution to give permanganate.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O
67.How does the acidified permanganate solution react with (a) iron(II) ions (b) oxalic acid and
(c) hydrogen sulphide ? Write the ionic equations for the reactions
In acid solutions:
(a) Fe2+ ion (green) is converted to Fe3+ (yellow):
5Fe2+ + MnO4– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+
(b) Oxalate ion or oxalic acid is oxidised at 333 K:
5C2O42− + 2MnO4- + 16H+ ——> 2Mn2+ + 8H2O + 10CO2
(c) Hydrogen sulphide is oxidised, sulphur being precipitated:
H2S —> 2H+ + S2−
5S2− + 2MnO4− + 16H+ ——> 2Mn2+ + 8H2O + 5S
UNIT 9 Topic: Coordination Compounds
1. State the postulates of Werner’s theory of coordination compounds. 3 Marks
Postulates:
1. Central metal ion in a complex shows two types of valences - primary valence and
secondary valence.
2. The primary valence is ionisable and satisfied by negative ions.
3. The secondary valence is non ionisable. It is equal to the coordination number of the
central metal ion or atom. It is fixed for a metal. Secondary valences are satisfied by
negative ions or neural molecules (ligands).
4. The primary valence is non directional. The secondary valence is directional. Ions or
molecules attached to satisfy secondary valences have characteristic spatial
arrangements. Secondary valence decides geometry of the complex compound.
2. What are the limitations of Werner’s theory of coordination compounds?
3 Marks
This theory fails to explain why,
a) a few elements have the property to form coordination compounds
b) the bonds in coordination compounds have directional properties
c) coordination compounds have characteristic magnetic and optical properties.
3. Write one difference between double salts and complex salts with respect to
their ionisation. Give one example for each type of salt. 2 Marks
Double salt Complex salt
Double salt is stable only in solid state, but
dissociate into simple ions completely in
solution state.
Complex salt is stable both in solid and
solution state and does not dissociate
completely in solution state.
E.g: KCl.MgCl2.6H2O E.g: K4[Fe(CN)6]
4. Define Coordination entity of coordination compounds. 2 Marks
A coordinate entity constitutes a central metal atom or ion bonded to a fixed number of
ions or molecules (ligands).
e.g: [Fe(CN)6]4-.](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-45-2048.jpg)
![5. What is central metal ion in a coordination compound? Give an example.
2 Marks
The metal atom or ion in a coordination entity to which, a fixed number of ions or
molecules (ligands) are bound in a definite geometrical arrangement around it is called
central metal ion. It is a Lewis acid.
e.g: In K4[Fe(CN)6] , central metal ion is Fe2+.
6. What are ligands? Give an example. 2 Marks
The ions or molecules bonded to the central metal atom or ion in a coordination entity
are called ligands. Ligands are Lewis bases.
e.g: In [Ni(CN)4]2-, ligand is CN- ion.
7. Define the term coordination number of a central metal atom or ion in a
complex compound. 2 Marks
The coordination number of central metal atom or ion in a complex is the number of
ligand donor atoms to which the metal is directly bonded.
In K4[Fe(CN)6] , coordination number of Fe2+ is 6.
8. Define coordination sphere of coordination compounds. 2 Marks
The central metal atom or ion and the ligands of the complex compound are written
within square bracket. This is called coordination sphere of coordination compounds.
9. What are homoleptic complexes? Give an example. 2 Marks
Homoleptic complexes are the complexes in which central metal ion or atom is bound to
only one type of donor groups.
e.g: K4[Fe(CN)6]
10. What are heteroleptic complexes? Give an example. 2 Marks
Homoleptic complexes are the complexes in which central metal ion or atom is bound to
more than one type of donor groups.
e.g: [Co(NH3)5Cl]SO4
11. Classify the following ligands into unidentate, didentate and polydentate
ligands. NH3, EDTA, oxalate. 3 Marks
NH3 Unidentate
EDTA Polydentate
Oxalate Didentate
12. Give the IUPAC name for the following compounds.
a) K4[Fe(CN)6] potassium hexacyanidoferrate(II)
b) [Cu(NH3)4] SO4 tetramminecopper(II) sulphate
c) [Co(NH3)5Cl]SO4 pentamminechloridocobalt(III) sulphate
d) K3[Fe(C2O4)3] potassium trioxalatoferrate(III)
e) [CoCl2(en)2]+ dichloridobis(ethane-1,2-diamine)cobalt(III)
f) [Co(NH3)5(NO2)]Cl2 pentamminenitrito-N-cobalt(III) chloride
g) [Co(NH3)5(ONO)]Cl2 pentamminenitrito-O-cobalt(III) chloride
h) [Ni(CO)4] tetracarbonylnickel(0)
13. What are ambidentate ligands? Give one example. 2 Marks
Ligands which have two donor atoms, but can bond to central metal atom or ion through
only one donor atom are called ambidentate ligands.
e.g: NO2
-
, SCN
-
etc.,
14. What is geometrical isomerism in complexes? Give an example. 2 Marks
It is a phenomenon in which two complex compounds have the same molecular formula
and same chemical bonds, but different geometrical arrangement of the ligands. The two
forms are called cis and trans forms.
e.g: [Pt(NH3)2Cl2]
Cl NH3
Pt
Cl NH3
Cis isomer
Cl NH3
Pt
NH3 Cl
Trans isomer](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-46-2048.jpg)
![15. Explain optical isomerism in complexes with an example. 2 Marks
It is a phenomenon in which two complex compounds have the same molecular formula
and same chemical bonds but cannot be superposed on one another. They differ in the
rotation of plane of plane polarised light. These isomers are called enantiomers. The two
forms are called dextro (d) and laevo (l).
e.g:
16. What is linkage isomerism? Give an example. 2 Marks
Linkage isomerism is type of isomerism in which two complex compounds differ in the
donor atoms for ligands (different ligating atoms). It is shown by complex compounds
containing ambidentate ligands.
e.g: [Co(NH3)5NO2]Cl2 and [Co(NH3)5(ONO)]Cl2
17. Indicate the type of isomerism in the following set of complex compounds.
a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl
b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 2 Marks
a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl Ionisation isomerism
b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 Linkage isomerism
18. Explain coordination isomerism in complexes. Give one example.
Coordination isomerism is type of isomerism due to interchange of ligands between
cationic and anionic entities of different metal ions present in a complex.
e.g: [Co(NH3)6] [Cr(CN)6] and[Cr(NH3)6] [Co(CN)6]
19. Explain ionization isomerism in complexes. Give an example.
Ionisation isomerism is type of isomerism in which two complex compounds produce
different ions in solution form.
e.g: [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl
20. What is solvate isomerism? Explain with an example.
Solvate isomerism is a type of isomerism in which complex compounds differ in the
number of water molecules acting as ligands and water of hydration.
e.g: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2. H2O
21. Explain the formation of [CoF6]3- . Is this complex paramagnetic? 3 Marks
Co, Z=27 [Ar]3d7 4s2 4p0 4d0
Co3+ [Ar] 3d6 4s0 4p0 4d0
It undergoes sp3d2 hybridisation.
When F - ligand attacks the central metal ion, Co3+
This complex uses outer d orbital (4d) for hybridisation. It is an outer orbital
complex. It has unpaired electrons. [CoF6]3- is paramagnetic. This complex is called high
spin or spin free complex.](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-47-2048.jpg)
![22. Give the geometry, hybridization and magnetic property of [Co(NH3)6]3+ based
on VBT. 3 Marks
Co, Z=27 [Ar]3d7 4s2 4p0
Co3+ [Ar] 3d6 4s0 4p0
When NH3 ligand attacks the central metal ion Co3+ , pairing of electrons in 3d orbital
occurs against Hund’s rule. Co3+ can undergo d2sp3 hybridization.
This complex uses inner d orbital (3d) for hybridisation. It is an inner orbital complex. It
has no unpaired electrons. [Co(NH3)6]3+ is diamagnetic. This complex is called low spin or
spin paired complex.
It has octahedral geometry.
23. Using VBT, explain the type of hybridization, geometry and magnetic property
of [NiCl4]2-. 3 Marks
Ni, Z =28 [Ar] 3d8 4s2 4p0
Ni2+, [Ar] 3d8 4s0 4p0
It undergoes sp3 hybridisation.
Four pair of eelctrons from 4 Cl- ions
This complex has unpaired electrons. It is paramagnetic.
[NiCl4]2- has tetrahedral structure.
24. Explain the hybridization, geometry and magnetic property in the complex
compound [Ni(CN)4]2-. 3 Marks
Ni, Z =28 [Ar] 3d8 4s2 4p0
Ni2+, [Ar] 3d8 4s0 4p0
When CN- ligand attacks the central metal ion Ni2+, unpaired electrons in 3d orbital are
paired up against Hund’s rule. It undergoes dsp2 hybridisation.
Four pair of electrons from 4 CN- ligands.
This complex has no unpaired electrons. It is diamagnetic.
[Ni(CN)4]2- has square planar structure.
25. What are the limitations of Valence Bond Theory? 2 Marks
This theory,
a) has a number of assumptions
b) does not give quantitative interpretation of magnetic data
c) does not explain colour shown by complexes.
d) could not predict tetrahedral or square planar shape for the coordination number 4
e) could not distinguish between strong and weak ligands.
26. Why [CoF6]3- is called an outer orbital complex? 1 Mark
In this complex, Co3+ uses outer d orbital (4d) for hybridization. Therefore it is called an
outer orbital complex.](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-48-2048.jpg)
![27. What are inner orbital complexes? Give an example. 2 Marks
Inner orbital complex is one where the central metal ion uses inner d orbital {(n-1)d
orbital} for hybridsation.
e.g: [Co(NH3)6]3+
28. Explain the salient features of crystal field theory. 2 Marks
This theory considers ligands as point charges in case of anionic ligands and dipoles in
case of neutral molecules. The bond formed between central metal ion and the ligands is
purely ionic.
29. What is crystal field splitting? Explain crystal field splitting in octahedral
entities using energy level diagram. 3 Marks
In an isolated gaseous central metal atom or ion, all the five d orbitals are having same
energy. i.e they are degenerated. In the presence of attacking ligands, it becomes
asymmetric and the d orbitals lose degeneracy, resulting in splitting of d orbitals. This is
called crystal field splitting.
In an octahedral complex, six ligands surround the central metal ion. dx2 – y2 and dz2
orbitals (called eg set) are directed along the direction of ligands and experience more
repulsion. They have more energy. dxy, dyz and dzx orbitals (called t2g set) are directed
between the axes of attacking ligands and experience lesser repulsion by the ligands.
They have lesser energy. The energy separation between two split sets is denoted as ∆o .
The energy of eg orbitals increase by 3/5 ∆o and that of t2g set decrease by 2/5 ∆o .
30. What is spectrochemical series? Arrange the following ligands in the increasing
order of their field strength.
Br-, I-, H2O, CO, F- 2 Marks
Spectrochemiccal series is the arrangement of ligands in the order of increasing field
strength.
Correct order for the given set is,
I- < Br- < F- < H2O < CO
31. Draw a figure to show the splitting of d orbitals in a tetrahedral crystal
field. 2 Marks](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-49-2048.jpg)
![UNIT.10
HALOALKANES AND HALOARENES
ONE MARKS QUESTIONS
1. What are haloalkanes? [1]
A: Haloalkane is a derivative obtained by replacing hydrogen atom of alkane by halogen atom.
2. What is the hybridization of the carbon attached with vinylic halides (or) aryl halides?
[1]
A: sp2hybridisation.
3. Among phosphorus trihalides which halides are generated insitu ? [1]
A: PBr3, PI3
4. Free radical halogenation of hydrocarbons is not a best method to prepare haloalkanes.
Why? [1]
A: Because this method gives mixture of isomeric mono and poly haloalkanes, which is difficult
to separate as pure compounds.
5. Why fluoro compounds cannot be prepared from electrophilic substitution reaction?
[1]
A: Due to high reactivity of fluorine.
6. What happens to the boiling point of isomeric haloalkanes with increase in
branching?[1]
A: decreases.
7. How density and atomic mass of halogen atoms in haloalkanes are related? [1]
A: Directly
8. Why tertiary alkyl halide undergoes SN1 reaction very fast? [1]
A: Because of the high stability of tertiary carbocation
9. What is the order of reactivity of alkyl halides towards SN1 and SN2 reaction [1]
A: Towards SN1 reaction order of reactivity is 30> 20 > 10
Towards SN2 reaction order of reactivity is 10> 20> 30.
10.Allylic and benzylic halides are highly reactive towards SN1 reaction. Why? [1]
A: Because the carbocation formed from allylic and benzylic halides is more stable due to
resonance.
11.Between SN 1and SN2 reaction which one proceeds with complete stereo chemical
inversion? [1]
A: SN2 reaction.
12.Between SN1 and SN2 reactions which one proceeds with racemisation?
[1]
A: SN1 reaction.
13.What are optically active compounds?
[1]
A: Certain compounds rotate the plane polarized light when it is passed through their solutions
are called as optically active compounds.
14.What are dextro rotatory compounds?
[1]
A: The compound which rotates plane polarized light in clockwise direction is called as dextro
rotatory compound.
15.What is a laevo rotatory compound? [1]
A: The compound which rotates plane polarized light in anticlockwise direction is called as laevo
rotatory compound.
16.What are optical isomers?
[1]
A: The dextro and laevo rotatory isomers of a compound are called as optical isomers.
17.What is asymmetric carbon (or) stereocentre [1]
A: A carbon atom attached with four different substituent groups is called as asymmetric carbon
(or) stereocentre.
18.What are chirals? [1]
A: The objects which are non-superimposable on their mirror image are said to be chirals.
19.What are achiral molecules? [1]
A: The molecules which are, superimposable on their mirror images are called achiral
molecules.
20.Between propan – 2- ol and butan – 2 – ol, identify the chiral molecule?
[1]
A: Butan – 2- ol.
21.What are enantiomers? [1]
A: The stereo isomers related to each other as non-superimposable mirror images are called
enantiomers.
22.What are racemic mixtures? [1]
A: A mixture containing two enantiomers in equal proportions will have zero optical rotation, are
called as racemic mixture (or) racemic modification.](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-50-2048.jpg)
![23.What is racemization? [1]
A: The process of conversion of enantiomer into a racemic mixture is known as racemization.
24.Between α and β- hydrogen which one is removed during dehydrohalogenation of alkyl
halides? [1]
A: β - Hydrogen.
25.What are organo-metallic compounds? [1]
A: Most organic chlorides, bromides and iodides react with certain metals to give compounds
containing carbon-metal bonds are known as organo – metallic compounds.
26.Write the general formula of Grignard reagent? [1]
A: RMgX.
27.Name the product formed when Grignard reagent treated with water?
[1]
A: Alkanes
28.What is the hybridization of carbon atom in C-X bond of alkyl halides? [1]
A: sp3hybridisation
29.Mention the hybridization of carbon atom in C-X bond of aryl halides? [1]
A: sp2hybridisation
30.Why SN1mechanism is ruled out in haloarenes? [1]
A: In case of haloarenes, the phenyl cation formed as a result of self-ionization will not be
stabilized by resonance and therefore, SN1mechanism is ruled out.
31.What happens to the reactivity of haloarenes towards nucleophilic substitution when
electron withdrawing group present at ortho or para position? [1]
A: Increases.
32.What is wurtz-fitting reaction? [1]
A: A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in
dry ether and is called as wurtz-fitting reaction.
33.Expand DDT [1]
A: Dichlorodiphenyltrichloro ethane
34.Among chloral, chloroform, DDT, carbontetrachlorides , which has more number of
chlorides ?
A;DDT
[1]
Cl
Cl
H
Cl
Cl
Cl
O2N
OH
NO2
NO2
Br
CH3
H3CH2C
H
CH3
H CH2CH3
+ Br
Θ
2-bromobutane
secondary haloalkane C
R
H
R X
primary haloalkane C
H
H
R X
35. Identify the product.
A:
36.Write the structure of DDT [1]
A:
TWO MARKS QUESTIONS
37.Explain the classification of monohaloalkanes containing Sp3C – X bond
[2]
A: These are classified into three types](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-51-2048.jpg)
![X
X X
H3C
X
Tertiary haloalkane C
R
R
R X
X CH2 X
CH2 X
38.What is the difference between allylic acid and benzylic halides
[2]
A:
39.What are vinylic halides? Give example?
[2]
A: These are the compounds in which the halogen atom is bonded to a Sp2-hybridised carbon
atom of a carbon-carbon double bond.
40.What are aryl halides? Give example?
[2]
A: These are the compounds in which the halogen atom is bonded to the Sp2hybridised carbon
atom of an aromatic ring.
Allylichalide
These are the compounds where the
halogen atom is bonded to a Sp3-hybridised
carbon atom next to carbon-carbon double
bond.
Benzylic halides
These are the compounds in which the
halogen atom is bonded to an Sp3-hybridised
carbon atomnext to an aromatic ring.
Br
Br
Br
C
H3C
CH3
H3C CH2 Cl
CH2Cl
C
+
X
-
CH3 CH
Cl
Cl
CH2 CH2
ClCl
41.Write the IUPAC name of the following compounds [2]
a) b) c)
d) CH2=CH-Cl
A: a) 1, 3, 5 – tribro,obenzece
b) 1, chloro 2, 2 – dimethyl propane
c) Chloro phenyl methane
d) Chloroethene
42.Explain nature of C-X bond in haloalkanes? [2]
A: Haloalkanes are polar in nature because halogen atoms are more electronegative than
carbon as a result the carbon atom bears a partial positive charge whereas the halogen atom
bears a partial negative charge
43.What are geminal halides? Give example? [2]
A: These are the dihalo compounds where two same halogen atoms are present on the same
carbon atom
Ex:
44.What are vicinal halides? Given example? [2]
A: These are the dihalo compounds where two same halogen atoms are present on the adjacent
carbon atoms
Ex:](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-52-2048.jpg)
![CH3
+ Cl2
Fe
dark
CH3
Cl
Toluene
O-chloro toluene
+
Cl
CH3
P-chloro toluene
N2
+
X
-
X
Cu2X2 + N2
Bebzene diazonium halide haloarene X = Cl, Br
45.R-OH + A R – Cl + H2O
Identify A and B in the given reaction [2]
A: R – OH + HCl R-Cl + H2O
A = HCl,
B = ZnCl2
46.How do you prepare chloroalkane using reaction between alcohol and phosphorus
pentachloride?
[2]
A: R – O – H + PCl5 R – Cl + HCl + POCl3
47.How do you prepare chloroalkane using alcohol and thionylchloride? Mention
advantage of this reaction?
[2]
A: R – OH +SOCl2 R – Cl + SO2↑ + HCl↑
48.Explain the reaction between toluene and chlorine?
[2]
A:
49.Why iodination of arenes by electrophilic substitution requires an oxidizing agent?
[2]
A: Because iodination of arenes is a reversible reaction due to formation of biproduct HI and
presence of an oxidizing agent oxidises the HI formed, there by prevents reversible reaction.
50.Explain sandmeyer’s reaction to prepare haloarenes (or) how do you convert benzene
diazonium salt into haloarenes?
[2]
A:
ZnCl2
B
N2
+
Cl
-
I
KI
+ N2
+ CH3CH3 CH3
I
In this reaction iodination of benzene diazonium halide does not requires cuprous halide
51.Identify major product in the given reaction and give reason?
[2]
CH3 – CH = CH2 + H – I CH3 – CH2– CH2– I
Iodopropane 2- Iodo propane
A: 2- Iodo propane
Because 2- Iodo propane involves stable 2°
- carbocation.
52.What happens when ethane reacts with bromine in presence of CCl4. Write the reaction
[2]
A: This reaction gives 1.2- dibromoethane product.
CH2 = CH2 + Br2 BrCH2 – CH2Br
53.What is finkelstein reaction? Give its general reaction?
[2]
A: The reaction in which an alkyl chloride (or) bromides reacts with sodium iodide in dry
acetone gives alkyl iodides is called finkelstein reaction.
R – X + NaI R – I + NaX
X= Cl, Br
54.How do you get fluoro methane from chloro (or) bromo methane and name the
reaction? [2]
A: CH3 – Br + AgF CH3 – F + AgBr
Bromomethane. Silver fluoride fluoro methane
This reaction is called as swart’s reaction.
55.Why the boiling point of halides are higher than hudrocarbons of comparable molecular
mass [2]
A: Because of greater polarity as well as higher molecular mass as compared to the parent
hydrocarbon, the inter molecular forces of attraction are stronger in the halogen derivatives.
CCl4](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-53-2048.jpg)
![Nu
C
X
C Nu + X
OH +
Cl OH Cl
H
HO
H H HH H H
H
+ Cl
H
(Slow step)r [OH ] [CH 3 - Cl]
order = 1 + 1 = 2
56.Arrange the halides R-Br, R-Cl, R-F, R-I containing same alkyl group in the increasing
order of their boiling point [1]
A: R-F < R – Cl< R-Br < R – I .
57.In isomeric dihalobenzenes, the para - isomers has high melting point than ortho and
meta – isomers. Why? [2]
A: Because para – isomers are symmetric in nature and fits in crystal lattice better as compared
to ortho and meta – isomers.
58.Haloalkanes are less soluble in water even though they are polar in nature. Why?
[2]
A: For a haloalkane to dissolve in water, energy is required to overcome and break the hydrogen
bond between water molecules. Less energy is released when new attractions are set up
between the haloalkane and water molecules as these are not as strong as the original
hydrogen bonds in water.
59.Whyhaloalkanes undergoes nucleophilic substitution reaction?
[2]
A: In haloalkanes due to difference in electronegativity of carbon and halogen, carbon bears
partial positive charge and attracts nucleophile and halogen atom departs as halide ion
60.Write SN2 mechanism [2]
61.In SN2 reaction transition state cannot be isolated. Why? [2]
CH3 - CH2 - CH = CH - CH3 CH3 - CH2 - CH2 - CH - CH2 HC3 - CH2 - CH2 - CH = CH2
OH OH
Θ ΘBr
HPent - 2 - ene (81%) 2 - Bromo pentane Pent - 1 - ene (19%)
CH3
H CH2
CH2
OH
CH3
+ HCl
heat
CH3
H CH2
CH2
Cl
CH3
+ H - OH
(-) - 2 methylbutan -1-of (+) -1-chloro-2-methylbutane
A: In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and
the outgoing leaving group; as a result carbon atom in transition state is simultaneously bonded
to five atoms and therefore is unstable.
62.Why tertiary haloalkanes are less reactive towards SN2 reaction?
[2]
A: Because SN2 reaction requires the approach of the nucleophile to the carbon bearing the
leaving group, the presence of bulky substituent on (or) near the carbon atom like in 30 –
haloalkane have a dramatic inhibiting effect.
63.What is retention of configuration? Give example [2]
A: Retention of configuration is the preservation of integrity of the spatial arrangement of bonds
to an asymmetric centre during a chemical reaction.
64.Write the reaction between 2-bromopentane with alcoholic solution of potassium
hydroxide and mention the major product in thereaction?
[2]
A:
Pent-2-ene is the major product.
65.During dehydrohalogenation of 2-bromopentane, Pent-2-ene is the major product.
Why? [2]
A: According to saytzeff rule alkene which has the greater number of alkyl groups attached to
the doubly bonded carbon atoms are stable.
66.How do you prepare ethyl magnesium bromide from ethyl bromide? [2]
A: CH3CH2Br + Mg CH3CH2Mg Br
Dry Ether](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-54-2048.jpg)
![:X: x
+
X
+
X
+..
Θ
Θ
Θ
Cl Cl
NO2
Cl
NO2
+
Conc. HNO3
CONC. H2SO4
Chlorobenzene 1 - chloro - 2 - nitro
- benzene (minor) 1 - chloro - 4 - nitrobenzene
(major)
Cl Cl
CH3
Cl
CH3
+
anhydrous
Chlorobenzene 1 - chloro - 2 - methyl
- benzene (minor) 1 - chloro - 4 - methylbenzene
AlCl3+ CH3 - Cl
Chloromethane
Cl
(i) NaoH, 623 K, 300 atm
(ii) H
+
OH
:Cl: Cl
+
Cl
+
Cl
Θ
Θ
Θ
..
67.Aryl halides are extreamely less reactive towards nucleophilic substitution reaction
than alkyl halides. Why? [2]
A: In haloarenes, the electron pairs on halogen atom are in conjugation with π- electrons of the
ring and C-Cl bond aquires partial double bond character due to resonance. As a result, the
bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive
towardsnucleophilic substitution reaction.
68.How do you convert chlorobenzene to phenol? [2]
A:
69.Write the resonance structure to show the halogenatom present on the haloarenes
areortho and para directing group? [2]
A:
70.Write the reaction involved in the nitration of chlorobenzene? [2]
A:
71.Explain methylation of chlobenzene and name the reaction? [2]
X
+ 2Na + 2NaXdry ether2
Diphenyl
X
+ 2Na + R - X
R
+ 2NaX
dry ether
haloarene haloalkane alkylarene
(CH3)3 CBr
CH3
CH3 CH3
+ + Br (slow)
Θ
CH3
CH3 CH3
+ + OH
Θ
(CH3)3 COH (fast)
72.Write general equation of wurtz-fitting reaction? [2]
A:
73.Explain fitting reaction with general equation? [2]
A: Two arylhalides reacts with sodium in presence of dry ether gives diphenyl. This reaction is
called as fitting reaction.
74.What are freons? Give an example [2]
A: The fluro, chloro compounds of methane or ethane collectively called as freons
Ex: CCl2 F2
THREE MARKS QUESTIONS
75.Explain SN1machanism [3]
A: (CH3)3CBr + ŌH (CH3)3 C OH + BrΘ
2-bromo-2-methylpropane 2-methylpropan – 2 -ol
Step (1): formation of carbocation
Step (2): attack of nucleophile on carbocation](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-55-2048.jpg)
![H Br
CH3
C6H13
+ OH OH
CH3
H
C6H13
+ Br
Θ Θ
Br
CH3
H3CH2C
H
CH3
H CH2CH3
+ Br
Θ
2-bromobutane
CH3
CH2CH3CH3
Θ
OHHO H
CH2CH3
CH3
OH
Θ
OH
CH3
H
H3CH2C
(+) - Butan - 2 - ol (-) Butan - 2 - ol
Rate of the reaction depends on slow step of the reaction
r α [(CH3)3C Br]1
Order = 1
76.With example explain SN2reactions of optically active halides areaccompanied by
inversion of configuration.
[3]
A: SN2 mechanism of optically active halides has the inverted configuration because nucleophile
attaches itself on the side opposite to the one where the halogen atom is present. For example
when (-) – 2 – bromooctane is allowed to react with sodium hydroxide, (+) – octan – 2 – ol is
formed with the – OH group occupying the position opposite to what bromide had occupied.
77.SN1 reaction of optically active halides accompanied by racemisation. Explain with
example?[3]
A: SN1 reaction of optically active halides accompanied by racemization (but not 100%
racemization) because carbocation formed in the slow step being sp2hybridised is planar. As a
result the attack of the nucleophile may be accomplished from either side resulting in a mixture
of products, one having the same configuration and the other having opposite configuration.
Ex:
(Please make correction of CH3 as H in the transition state)
1
Unit 11. Alcohols, Phenols and Ethers
One mark questions
1. Name the alcohol which is used for polishing wooden furniture.
Ans: Ethanol
2. What are alcohols?
Ans: Hydroxyl derivatives of aliphatic compounds are called alcohols.
3. What is the IUPAC name of ?
Ans: Ethane-1, 2-diol
4. Write the structure of 2-methyl cyclopentanol.
Ans:
5. Name the simplest hydroxyl derivative of benzene.
Ans: Phenol
6. What is the IUPAC name of Resorcinol?
Ans: Benzene-1, 3-diol
7. What is the common name of CH3OC2H5?
Ans: Ethylmethyl ether
8. Write the formula of anisole.
Ans: C6H5OCH3 or
9. What is the IUPAC name of anisole?
Ans: Methoxybenzene.
10. Write the IUPAC name of CH2 = CH CH2OH
Ans: prop-2-en-1-ol](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-56-2048.jpg)
![18. What are thermosetting polymers? Give Examples
Thermosetting polymers are cross linked and heavily branched molecules. On heating they
undergo extensive cross linking and become hard and infusible. These cannot be reused.
Ex: Bakelite, urea formaldehyde resin, etc.
19. Name the monomer and write the partial structure of polythene?
Monomer of polythene - Ethene or Ethylene
Partial structure - (CH2CH2)n
20. Name the monomer and write the partial structure of Nylon-6 ?
Monomer of Nylon-6 - caprolactum
Partial structure - [-CO-(CH2)5-NH-]n
21. Name the monomers and write the partial structure of Nylon- 6,6
Monomer of Nylon-6,6 - Hexamethylene diamine and Adipic-acid
Partial structure - [-OC-(CH2)4-CONH-(CH2)6-NH-]n
22. Name the monomers and write the partial structure of terylene (Dacron)
Monomer of terylene - Ethylene glycol and terephthatic-acid
Partial structure -
23. Name the monomer and write the partial structure of Bakalite ?
Monomer of Bakalite - Phenol and formaldehyde
Partial structure -
RUBBER:
24. Name the monomer present in natural rubber.Write the partial structure
Natural rubber is a polymer of cis-2-methyl-1,3-butadiene (isoprene). ( cis-poly-isoprene). Its
partial structure is
25. Define Synthetic rubber? Give one example
Synthetic rubber is defined as any valcanisable rubber like polymer capable of getting
stretched to twice its length and returns to its original length, size and shape when the stretching
force is withdrawn
Ex: Neoprene, Buna-S, Buna-N
26. What is Valcanisation ?
The process of heating natural rubber with sulphur or sulphur containing compounds at
about 415k for a few hours in order to give strength and elasticity to natural rubber is called
valcanisation.
27. Explain the preparation of Neoprene? Write the equation.
When chloroprene (2-chloro-1,3-butadiene) is heated in the presence of peroxide catalyst,
polychloroprene or neoprene is formed](https://image.slidesharecdn.com/chemistrychapterwiseimpquestions-170213020000/75/Chemistry-chapter-wise-important-questions-74-2048.jpg)
Uploaded bySrikanth KS
Chemistry chapter wise important questions
The document contains a series of questions and answers related to the properties and types of solids, focusing on crystalline and amorphous solids, defects, bonding, and unit cells in crystallography. It addresses various concepts such as melting points, types of defects in solids, magnetic properties, and calculations related to density and atomic structure. Additionally, it includes comparisons between different types of solids and their characteristics.
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Chemistry chapter wise important questions
- 1. UNIT-1 SOLIDS QUESTIONSCARRYING ONE MARK: 1. Which type of solid is anisotropic in nature? Ans: Crystalline solids are anisotropic in nature 2. Which type of solids is called as super cooled liquids or pseudo solids? Ans:Amorphous solids are called super cooled liquids 3. A solid has a sharp melting point, and then to which type of solids does it belong? Ans: Crystalline solids 4. Which type of solids has long range orderly arrangement of constituent particles? Ans: Crystalline solids 5. Sodium chloride and quartz belong to which type of solid? Ans: Crystalline solids 6. A solid shows different values for refractive index when measured in different directions. - Identify the type of solid Ans: Crystalline solids 7. When a solid is cut with a sharp edged tool, they cut into two pieces and the newly generated surfaces are plain and smooth. – Identify the type of solid. Ans: Crystalline solids 8. Which type of force of attractions is present between the molecules in polar molecular solids? Ans: Dipole –dipole interactions 9. Which type of force of attractions is present between the molecules in non-polarmolecular solids? Ans: London forces or Dispersion forces 10. Which type of force of attractions is present between the particlesin ionic solids? Ans: Electrostatic force of attraction or coulombic force of attraction 11.Solid SO2 and solid NH3belong to which type of molecular solids? Ans: Polar molecular solids 12. What is crystal lattice? Ans: The regular three dimensional array of lattice points in space is called crystal lattice 13. What is a unit cell? Ans: It is the smallest repeating unit which when arranged in three dimension gives the crystal lattice. 14. How many types of primitive unit cells are present? Ans: Three types 15. What is a primitive cubic unit cell? Ans: The cubic unit cell in which the particles/atoms are present only at the eightcorner of the cube is called primitive cubic unit cell. 16. Define the co-ordination number of a particle in solids. Ans:It is the total number of nearest neighboring particles to a given particle. 17. What is the number of octahedral voids generated, if the number of close packed spheres is N? Ans: N 18. What is the number of tetrahedral voids generated, if the number of close packed spheres is N? Ans: 2N 19. What is the co-ordination number of a particle in a tetrahedral void? Ans:Four 20. Among Schottky and Frenkel defect, which type of defect decreases the density of the crystal? Ans: Schottky defect
- 2. 21. Whatare point defects? Ans: Deviations from the ideal arrangement around a particular point or an atomin a crystalline solid 22. What are F-centers? Ans: The anionic sites occupied by the unpaired electrons are called F- centre. 23. To which colour potassium chloride crystal turns, when excess potassium ionis present? Ans: Violet 24. Name the type of non-Stoichiometric defect observed when white ZnO turns yellow on heating. Ans: Metal excess defect 25. Name the non-Stoichiometric defect responsible for the composition of ferrous oxide to be Fe0.95O1. Ans:Metal deficiency defect 26. Which type of point defect is observed when NaCl containing little SrCl2 is crystallized? Ans:Impurity defect 27. Which defect is also called as dislocation defect? Ans:Frenkel defect 28. What is doping? Ans: The process of increasing the conductivity of an intrinsic semiconductors by adding asuitable impurity is called doping 29. What type of semiconductors are obtained when silicon doped with boron impurity? Ans: p-type semiconductor 30. Name the unit used to measure magnetic moment. Ans: Am2( 1Bohr magneton= 9.27x10-24 Am2) 31. What are diamagnetic substances? Ans: These are the substances which are repelled by the magnetic field 32. What are ferromagnetic substances? Ans: These are the substances which are strongly attracted by the magnet 33. How body diagonal and radius of a sphere(r) are related in bcc unit cell? Ans:4r =√2a 34. Give an example for Ferromagnetic substance. Ans: Fe Co Ni Gd CrO2 35. Give an example for Diamagnetic substance. Ans: H2O, NaCl, and C6H6 QUESTIONS CARRYING TWO MARKS: 1. How crystalline solids differ from amorphous solids in their melting point? Ans:Crystalline solids have sharp melting point whereas amorphous solid do not have a sharp melting point 2. Write any two differences between crystalline solids and amorphous solids? Crystalline solid Amorphous solid 3-D long range orderly arrangement of particles No orderly arrangement of constituent particles Sharp Melting point Do not have sharp M P ( Softening temperature) True solids having definite shape Pseudo solids having irregular shapes They have a well-defined cleavage planes Do not have cleavage planes Anisotropic in nature Isotropic in nature 3. What is meant by anisotropy? What type of solids show this nature? Ans: The physical properties like refractive index, coefficient of thermal expansion, when measured in different directions gives different value for a crystalline solid hence it is anisotropic in nature. Crystalline solids 4. What are the nature of particles and the force of attractions between the particles in non-polar molecular solids? Ans: In a non-polar molecular solids, the constituent particles are non-polar moleculeslike H2, Cl2, I2 and even atoms like Ar, Ne, Xe etc. The nature of force of attraction is weak dispersion force or London force.
- 3. 5. Whatare the nature of particles and the force of attractions between the particles in polar molecular solids? Ans: In a polar molecular solids, the constituent particles are formed by polar covalent bond like HCl, SO2. The nature of force of attraction is dipole-dipole attractions 6. What are the nature of particles and the force of attractions between the particles in hydrogen bonded molecular solids? Ans: In a hydrogen bonded molecular solids, the constituent particles are polar molecules capable of forming hydrogen bond like water. The nature of force of attraction is hydrogen bonding 7. What are point defects? Mention the types Ans: Point defects are the irregularities in the arrangement of constituent particlesaround a point or a lattice site in a crystalline substance. These are of three types. 1. Stoichiometric defects. 2. Non-stoichiometric defect 3. Impurity defect. 8. What are the differences between Schottky and Frenkel defect? Schottky defect a. Shown by ionic solidscontaining similar-sized cationsand anions (having high coordination number) b. An equal number of cations and anions are missing to maintain electrical neutrality c. Decreases the density of the substance d. Example, NaCl, KCl , CsCl, and AgBr Frenkel defect a. Shown by ionic solids containing largedifferences in the sizes of ions, (having less coordination number) b. Created when the smaller ion (usually cation) is dislocated from its normal site to an interstitial site c. No change in density of the crystal. creates a vacancy defect as well as an interstitial defect .Also known as dislocation defect d. Example: AgCl, AgBr, AgI and ZnS 9. What are the nature of particles and the force of attractions between the particles in ionic solids? Ans: The nature of the particles is ions (both cation and anion). The nature of the force of attraction is electrostatic force or coulombic force 10. What are the nature of particles and type of bonding in network solids? Ans: The nature of the particles is atoms. The bonding is covalent bond. 11. Classify the following into polar and non-polar molecular solids: Ar, HCl, I2 and SO2 Ans: Non-polar molecular solids: Ar, I2 Polar molecular solids:HCl, SO2 12. Calculate the number of particles present per unit cell in an FCC unit cell. Ans:Contribution of corner particle = 8 x 1/8 = 01 Contribution of a particle at the centreof face = 6 x ½ = 03 Total number particle /unit cell = 04 13. Calculate the number of particles present per unit cell in a BCC unit cell. Ans:Contribution of corner particle = 8 x 1/8 = 01 Contribution of a particleat the centre = 1 x 1 = 01 Total number particle /unit cell = 02 14. Calculate the number of particles present per unit cell in a simple cubic unit cell. Ans:Contribution of corner particle = 8 x 1/8 =01 Total number particle /unit cell = 01 15. Mention the two characteristics of a unit cell. Ans: Two characteristics of unit cells are a. Edge length b. Axial angles 16. What is the relation between edge length (a) and radius of the sphere (r) infcc unit cell? What is itspacking efficiency? Ans: The relationship between edge length and radius of the sphere are a=2 2 r Packing efficiency is 74% 17. What is the relation between edge length (a) and radius of the sphere (r) in bcc unit cell? What is its packing efficiency? Ans: The relationship between edge length and radius of the sphere are a= !! ! Packing efficiency is 68 % 18. How many tetrahedral and octahedral voids is present, if the number of sphere is N? Ans: The number of tetrahedral void is 2N The number of octahedral void is N 19. Explain Schottky defect. Give an example. Ans: The defect which arises due to missing of equal number of cations and anions from the crystal lattice is called Schottky defect. Ex. NaCl, KCl ,CsCl, AgBr
- 4. 20. ExplainFrenkel defect. Give an example. Ans: The defect in which an ion (generally cation) leaves the original site and occupies the interstitialsite is called Frenkel defect. E. AgCl, AgBr, AgI 21. How Schottky defect and Frenkel defect affect the density of the crystal? Ans: In Schottky defect density of the crystal decreases. In Frenkel defect the density of the crystal remains same. 22. Mention the two types of Non-stoichiometric defects in solids? Ans: Metal excess defect and metal deficiency defect. 23. What is F- center? What colour is imparted to the NaCl crystal, due to the presence of excess sodium? Ans: The anionic sites occupied by the unpaired electrons are called F- Centre The colour of NaCl crystal is Yellow 24. Write the formula to calculate the density of the unit cell and explain the terms. Ans: z = number of particles present per unit cell d = 𝒛𝑴 𝒂 𝟑 𝑵 𝑨 M = Molecular mass , d = density NA = Avogadro’s number a = Edge length. 25. What are n-type and p-type semiconductors? Ans: n-type semiconductor is obtained by doping of the crystal of a group 14 element such as Si or Ge, with a group 15 element such as P or As(pentavalent). Conductivity increases due to negatively charged electrons. p-type semiconductor is obtained by doping of the crystal of a group 14 element such as Si or Ge, with a group 13 element such as B, Al or Ga( trivalent). Conductivity increases as a result of electron hole 26. An ionic compound is formed by two elements A and B. The cat ions A are in ccp arrangement and those of anions B occupy all the tetrahedral voids. What is the simplest formula of the compound? Ans: Since cations are in ccp arrangement, the total number cat ions A = 4 The number of tetrahedral voids is double the number of particles = 8 All the tetrahedral voids are occupied by anions B. The number of elements of B = 8 Hence the formula of the ionic compound is A4B8 or AB2 27. A compound is formed by two elements X and Y. The element X forms ccp and atoms of Y occupy 1/3 rd of tetrahedral voids. What is the formula of the compound? Ans: Since element X are in ccp arrangement, the number of X per unit cell = 4 The number of tetrahedral void = 8 But only 1/3 rd is occupied by Y, therefore 8 x1/3 = 8/3 Hence the formula of the compound is X4Y8/3 = X12Y8 or X3Y2 28. Gold(atomic radius=0.144nm)crystallizesin a face centered unit cell. What is the length of the side of the cell? Ans: For FCC the edge length and radius of sphere arerelated by the equation, r = 0.144nm a=2 2 r a = ? = 2 2 x 0.144 nm = 2x1.414 x 0.144 = 0.40723nm. 29. Silver forms ccp lattice and X- ray studies of its crystals show that the edge lengthof its unit cell is 408.6pm. Calculate the density of silver (atomic mass = 107.9 u) Ans: d = !" !!!! d= 4 x 107.9/(4.08)3 x10-24 x 6.022 x1023 d = 431.6/40.899 d = 10.5528g/cm3 30. X- ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.6 x10-8cm. In a separate experiment, copperis determined to have a density of 8.92g/cm3,calculate the atomic mass of copper. Ans: d = !" !!!! M = d a3 NA/Z = 8.92 x(3.6)3x10-24 x 6,022 x1023/4 = 250.61/4 M = 62.6525 u 31. The edge of fcc unit cell of platinum is 392 pm and density is 21.5 g/cm3, calculate the Avogadro number. Ans: d = !" !!!! NA = Z x M/ d a3 = 4 x 195.08/21.5 x (3.92)3x 10—24 = 780.32/1295.08 x10—24 NA= 6.025 x1023
- 5. 32. Aunit cell of sodium chloride has four formula units. The edge length of the unit cell is 0.564 nm. What is the density of sodium chloride? Ans: d = !" !!!! d = 4 x 58.5/(5.64)3 x10-24 x 6.022 x1023 d = 234/108.038 d = 2.165 g/cm3 33. A body centered cubic element having density 10.3 g/cm3, has a edge length of 314pm. Calculate the atomic mass of the element (Avogadro’s number= 6.023x1023/mol) Ans: d = !" !!!! M = d x a3 xNA/Z = 10.3 x (3.14)3x 10-24 x 6.022 x1023/2 M = 96.01u 34. Calcium metal crystallizes in a face centered cubic lattice with edge length of 0.556nm. Calculate the density of the metal. (Atomic mass of calcium = 40g/mol and Avogadro number= 6.022 x1023mol-1) Ans: d = !" !!!! d = 4 x 40/(5.56)3 x10-24 x 6.022 x1023 d = 160/103.50 d = 1.54 g/cm3 35. Copper crystallizes into afcc lattice with edge length 3.61 x10-8cm. Calculate the density of the of the crystal (Atomic mass of copper =63.5g/mol and Avogadro number= 6.022 x1023mol-1 ) Ans: d = !" !!!! d = 4 x 63.5/(3.61)3 x10-24 x 6.022 x1023 d = 254/28.33 d = 8.9 g/cm3 36. Silver crystallizes in a face centered cubic structure. If the edge length is 4.077 x10-8cm and density is 10.5 g/cm3, calculate the atomic mass of silver. Ans: d = !" !!!! M = d a3 NA/Z = 10.5 x (4.077)3x10-24 x 6,022 x1023/4 = 103.57/4 The atomic mass of silver M = 107.09 u 37. The density of Li atoms is 0.53g/cm3.The edge length of Li is 3.5 A0. Find out the number of Li atoms in a unit cell (N0= 6.022 x1023/mol& M= 6.94) Ans:d = !" !!!! Z = d x 𝑎! 𝑁!/𝑀 = 0.53 x (3.5)3 x10–24x 6.022 x1023/6.94 = 2 The number of lithium atoms in unit cell is 2 Questions carrying THREE marks 1. Calculate the packing efficiency in simple cubic unit cell Edge length of the cube = a = 2r Volume of the cubic unit cell= a3 = (2r)3= 8r3 volume of one particle(sphere) = ! ! 𝝅r3 The number of particles per unit cell =1 Total volume occupied by one sphere = ! ! 𝜋𝑟! Packing efficiency= !"#$% !"#$%& !""#$%&' !" !"# !"!!"! !"#$%& !" !"#$! !"#$ !"## × 100 = ! ! !!!×! ! !! ×100 = 52.4%
- 6. 2. Calculatethe packing efficiency in face centered cubic unit cell edge length of the cube be ‘a’ In ABC AC2 = BC2 + AB2 b2 = a2 + a2 b2 = 2a2 b = 2 a Let the radius of the atom = r Length of the diagonal of ABC, b= 4r 2 a = 4r a = 2 2 r Edge length of the cube =a=2 2 r Volume of the cubic unit cell= a3 = 2 2 r ! volume of one particle(sphere) = ! ! 𝝅r3 The number of particles per unit cell =4 Total volume occupied by four spheres = 4 × ! ! 𝜋𝑟! packing efficiency= !"#$% !"#$%& !""#$%&' !" !"#$ !"!!"!# !"#$%& !" !"#$! !"#$ !"## × 100 = ! ! !!!×! ! !! ! × 100 = !" ! !!! !" !!! × 100 = 74 % 3. Calculate the packing efficiency in body centered cubic unit cell In ABG, b2 = a2 + a2 ⇒ b2 = 2 a2 In, AGD, C2 = a2 + b2 C2 = a2 + 2a2 C2 = 3a2⇒ C = 3𝑎 Radius of the atom = r. Length of the body diagonal, C=4r 3𝑎 = 4r a = !! ! Edge length of the cube =a= !! ! Volume of the cubic unit cell= a3 = !! ! ! volume of one particle(sphere) = ! ! 𝝅r3 The number of particles per unit cell =2 Total volume occupied by two spheres = 2 × ! ! 𝜋𝑟! packing efficiency= !"#$% !"#$%& !""#$%&' !" !"# !"!!"!# !"!"#$ !" !"#$! !"#$ !"## × 100 = ! ! !!!×! ! ! ! ! × 100 = ! ! !!! !" ! ! !! × 100 = 68% 4.Based on band theory explain conduction in metals, insulators and semiconductors Conduction of electricity in metals: In metals, the valence shell is partially filled, so this valence band overlaps with a higher energy unoccupied conduction band so that electrons can flow easily under an applied electric field. Conduction of electricity in insulators: In insulators, the valence shell is empty, so the gap between the valence band and conduction band is very large. so that electrons cannot flow under an applied electric field. Conduction of Electricity in Semiconductors In semiconductors, the gap between the valence band and conduction band is so small that some electrons may jump to the conduction band. Electrical conductivity of semiconductors increases with increase in temperature. Substances like Si, Ge show this type of behaviour, and are called intrinsic semiconductors. 5. How are solids classified on the basis of the force of attraction? Ans: a. Molecular solids: Particles are held by a. London forces (in non-polar solids) ex : Benzene, Argon, P4O10, I2, P4 b. Dipole - dipole interaction ( in polar solids) ex: Urea, Ammonia c. Hydrogen bonding (in hydrogen bonded solids) ex: ice b. Ionic solids a. Particles are held by ionic bond b. Conduct electric current in aqueous solution or molten state c. Examples: NaCl, MgO, ZnS d. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity. c. Covalent or network solids: a. Particles are held by covalent bonding. Examples SiO2 (quartz), diamond, d. Metallic solids: a. Particles are held by metallic bond. b. These are electrical conductors, malleable, and ductile. Examples: Fe, Cu, 6. What are point defects? Explain the types. Ans: Point defects are the irregularities in the arrangement of constituent particles around a point or an atom in a crystalline substance. These are of three types. 1. Stoichiometric defects: Do not disturb stoichiometry of the solid. These are also called intrinsic or thermodynamic defects Ex : Frenkel defect, Schottky defect 2. Non-stoichiometric defects: This defect alters the stoichiometric ratio of the constituent elements i) Metal excess defect a. Metal excess defect due to anionic vacancies: b. Metal excess defect due to the presence of extra cations at interstitial sites: ii) Metal deficiency defect a. By cation vacancy 3. Impurity defect.
- 7. 7. Whatare diamagnetic, paramagnetic and ferromagnetic substances? 1. Paramagnetic substance: The substance which are attracted by the magnet. The magnetic character is temporary and is present as long as the external magnetic field is present. Ex; O2, Cu2+, Fe3+, Cr3+ NO. 2. Diamagnetic substance: The substance which are weakly repelled by the magnetic field TiO2, H2O,NaCl.This property is shown by those substance which contain fully –filled orbitals (no unpaired electrons) 3. Ferro magnetic substance: The substance which are strongly attracted by the magnet. They show permanent magnetism even in the absence of magnetic field. Ex : Fe Co Ni Gd& CrO2 8. An element with molar mass 2.7 x 10-2 kg/mol forms a cubic unit cell with edge length 405pm. If its density is 2.7 x 103 kg/m3, what is the nature of the cubic unit cell Ans: d = !" !!!! Z = d x 𝑎! 𝑁!/𝑀 = 2.7 x103 x(405)3 x 10—27 x 6.022 x1023/2.7 x 10-2 = 4 Since there are 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face centered or cubic close packed structure (ccp) 9. Niobium crystallises in body-centered cubic structure. If density is 8.55g/cm3, calculate atomic radius of niobium, given that its atomic mass is 93 u. Ans: d = !" !!!! a3 = !" ! !! = 2 x 93/8.55 x6.022 x1023 = 36.1 x106 a = (36.1)1/3 x102 =330 pm For BCC r = ! ! a r = ! ! x 330 r = 143pm 10. An element has a body-centered cubic (bcc) structure with cell edge of 288pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208g of the element? Ans: d = !" !!!! M = d a3 NA/Z = 7.2 x (2.88)3x10-24 x 6,022 x1023/2 = 103.57/2 M = 51.78 u 51.78 g (1mole) contains 6.022 x1023 atoms Therefore 208g contains 4.01 x 6.022 x1023 = 24.187 x1023 atoms.
- 8. nBM V = = A nBM W Unit -2 THEORY OFDILUTE SOLUTIONS 1) What is solution? [1] A: It is a homogenous mixture of two or more compounds. 2) What is dilute solution? [1] A: It is a solution in which solute concentration is very less. 3) Give an example for solid-‐solid solution [1] A: Copper dissolved in gold. 4) Give an example for gas-‐gas solution [1] A: Mixture of oxygen and nitrogen gases. 5) Give an example for gas-‐solid solution [1] A: Solution of hydrogen in palladium. 6) Give an example for liquid-‐solid solution [1] A: Amalgam of mercury with sodium. 7) Give an example for liquid-‐liquid solution [1] A: Ethanol dissolved in water. 8) Give an example for solid-‐gas solution [1] A: Camphor in nitrogen gas. 9) Define mole fraction and give the equation to calculate it. [2] A:Mole fraction is the ratio of number of moles of one component to the total number of moles of all the components in the solution. A B n AX n n A = + B B nBX n n A = + 10)Define molarity and give the equation to calculate it. [2] A: Number of moles of the solute present per liter solution is known as molarity. 11)Define molality and give the equation to calculate it. [2] A: Number of moles of the solute present perkgsolvent is known as molality. 12)Define the term solubility of a substance. [1] A: Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature 13)State Henry’s law. [2] A: Henry’s Law: At constant temperature solubility of a gas in a liquid is directly proportional to the partial pressure of gas present above the solution. OR At constant temperature the partial pressure of the gas in vapor phase (p) is proportional to the mole fraction of the gas (x) in the solution. Mathematically p ∝ x ; p = KH x. Where KH is Henry’s law constant. KH depends on the nature of the gas. 14)Write the plot which shows relation between partial pressure of a gas v/s its mole fraction. [2] A: 15)Mention the factors affecting solubility of a gas in liquid. [2] A: 1. Temperature 2. Pressure 16)Explain how temperatures effect the solubility of a gas in liquid. [2] A: Solubility of gases in liquid decreases with rise in temperature. According to Le Chatelier’s Principle,as dissolution is an exothermic process, the solubility should decrease with increase of temperature. 17)Explain how pressure effects the solubility of a gas in liquid. [1] A: The solubility of gases increases with increases of pressure. 18)Mention the applications of Henry’s law. [3] A: (a) To increase the solubility of CO2 insoft drink and soda water, the bottle is sealed under high pressure. Mole fraction. Partial pressure of a gas
- 9. (b) To avoid bends, as well, the toxic effects of high concentration of nitrogen in the blood, the tanks used by scuba divers are filled with air dilute with helium. (c) At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. 19)State Raoult’slaw of liquid-‐liquid dilute solutions. [2] A: The partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Thus, for component 1 P1 ⍺ x1 And p1 = p1 0x1 20)What are ideal solutions? [1] A: The solution which obey Raoul’s law over the entire range of concentration are known as ideal solution 21)Mention the characters of ideal solutions. [3] A: 22)What are non-‐ideal solutions? [1] A: When a solution does not obey Raoult’s law over the entire range of concentration, then it is called non-‐ideal solution. 23)Mention the types of non-‐ideal solutions. [1] A: There are two types (a) Non-‐ideal solution with positive deviation from Raoult’s law (b) Non-‐ideal solution with negative deviation from Raoult’s law 24)Give an example for non-‐ideal solution with positive deviation from Raoult’s law. [1] A: Mixtures of ethanol and acetone Ideal I. It obeys Raoults law is obeyed at all temperature and concentration P = PA + PB II. ∆ V mix = O i.e., there is no change in volume on mixing III. ∆Hmix = O i.e., there is no enthalpy change when ideal solution formed IV. It doesn’t form azeotropic mixture V. Force of attraction between A―A, B―B is similar as A―B 25)Give an example for non-‐ideal solution with negative deviation from Raoult’s law. [1] A: An example of this type is a mixture of phenol and aniline. 26)What are azeotropes? Give example. [2] A: Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. For example: ethanol-‐water mixture 27)State Raoult’s law of relative lowering of vapour pressure. [1] A: Relative lowering of vapour pressure is equal to the mole fraction of the solute. 28)Define colligative property. [1] A: The properties depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. Such properties are called colligative properties 29)Mention four colligative properties of dilute solutions. [2] A: Relative lowering of vapour pressure I. Elevation in Boiling point II. Depression in Freezing point III. Osmotic pressure 30)Define the term relative lowering of vapour pressure. [2] A: It is the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent o o P P Relativeloweringof V.P P − = 31)What is elevation in boiling point? [1] A: Elevation in boiling point is the difference between the boiling point of the solution containing non-‐volatile solute and the boiling point of the pure solvent ∆Tb = T – To 32)Give the relation between elevation in boiling point and molecular mass of solute. [2] A: ΔTb = Kb 2 1 2 w 1000 w M × × Where w2 is mass of solute, w1 is the mass of the solvent; M2 is molar mass of the solute
- 10. 33)Give the S.I.unit of ebullioscopic constant or boiling point elevation constant or molal elevation constant. [1] A: The unit of Kb is K kg mol-‐1 34)What is depression infreezing point? [1] A: It is the decrease in the freezing point of solution when non-‐volatile solute is added into solvent. 35)Give the relation between depression infreezing point and molecular mass of solute.[2] A: ΔTf = Kf 2 2 1 w M w 1000 ∴ M2 = f 2 f 1 K 1000 w T w × × Δ × where M2 is molar mass of the solute. Note: Values of Kf and Kb of the solvent depends on their molecular mass and ΔHfusion and ΔHvap of the solvent respectively. 36)Give the S.I.unit of cryoscopic constant. [1] A: The unit of Kf is K kg mol-‐1 37)Draw the plot showing elevation in boiling point in a solution. [2] A: ―∆Tb ― Temperature/K Vapour pressure Tb Tb 0 Solution Solvent 38)Draw the plot showing depression in freezing point in a solution. [2] ―∆Tf― Temperature/K 39)Define osmosis. [1] A: The process of movement of solvent particles from lower concentration to higher concentration through semi-‐permeable membrane to attain equilibrium is called osmosis. 40)What is osmotic pressure and give its relation with concentration of solution. [2] A: The amount of external pressure required to stop the osmosis. = CRT Where: = osmotic pressure, R = gas constant, T = temperature, C = concentration of solution. 41) What are isotonic solutions? [1] A: Two different solutions having sameosmotic pressure are called isotonic solutions 42)What are hypertonic solutions? [1] A: The solution having more osmotic pressure than other 43)What are hypotonic solutions? [1] A: The solution having less osmotic pressure than other Tf Tf o Vapour pressure Solution Liquid solvent Frozen solvent
- 11. 44)Explain the application of reverse osmosis in desalination of water. [2] A: When pressure more than osmotic pressure is applied, pure water is squeezed out of the sea water through the membrane. A variety of polymer membranes are available for this purpose. The pressure required for the reverse osmosis is quite high. A workable porous membrane is a film of cellulose acetate placed over a suitable support. Cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. 45)What is reverse osmosis? [1] A: Movement of solvent particles from higher concentration to lower concentration through a semi permeable membrane, when pressure is applied greater than osmotic pressure 46)What is abnormal molar mass? [1] A: A molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass. 47) Define Vant hoff factor Van’t Hoff factor ‘i’ to account for the extent of association or dissociation of a solute in a solvent is i = Normal molar mass Abnormal molar mass or i = observed colligative property calculated colligative property or i = total number of moles of particles after association or dissociation Number of moles of particles before association or dissociation 48)What is the value of i for NaCl. [1] A: 2 49)What is the value of i for K2SO4. [1] A: 3 50)What is the value of i for sugar. [1] A: 1 51)What is the value of i for glucose. [1] A: 1 52)On what factor the colligative property depends on. [1] A: It depends on number of moles of solute particles but not on the nature of the solute. 53)Write the mathematical equation of Raoults law in case of non-‐volatile solute. [1] A: If one of the components (solute) is non-‐volatile then the equation of Raoults law is. PB= O P = PA + PB P = PA + O P = PA 54)Write the differentiate between non-‐ideal solutions with positive deviation and negative deviation from Raoult’s law [2] 55)Define lowering of vapour pressure? [1] A: It is defined as the difference between the vapor pressure of the solvent in pure state and the vapour pressure of the solution ∆P = Po – P 56)State Roult’s law of relative lowering of vapour pressure [1] A: It states that the relative lowering of vapour pressure is equal to the mole fraction of the solute Positive deviation (a)In this solution solvent – solute interaction is weaker than solvent – solvent, solute-‐solute interactions (b) P > PA + PB (c) ∆ V > O (d) ∆H = positive (e) It forms azeotrope with minimum boiling point Negative deviation (a) In this solution solvent – solute interaction is stronger than solvent – solvent, solute-‐solute interactions (b) P < PA + PB (c) ∆ V < O (d) ∆H = negative (e)It forms azeotrope with maximum boiling point P = PA O. XA
- 12. 57)Why sea water freezes below 00C? [1] A: Sea water freezes below 00C due to the presence of the non-‐volatile solute dissolved in the water. 58)Derive the equation to calculate molecular mass of unknown solute using Raoult’s law of relative lowering of V.P [3] A: According to Raoult’s law relative lowering of vapour pressure is equal to the mole fraction of the solute. o Bo P P X P − = o nP P B o n nP BA − = + nB<<<nA for dilute solution So we can neglect nB in denominator o nP P B o nP A − = B B A A W o MP P o WP M − = o MWP P B A o W MP BA − = B A B A oW .M P M oW P P ⎛ ⎞ = ⎜ ⎟ −⎝ ⎠ Numerical problems 1. A solution containing 2.56 g sulphur in 100 g CS2 gave a freezing point lowering of 0.383 K. Calculate the molar mass of sulphur molecules. Given Kf of CS2 = 3.83 K kg mol− 1 . Ans. ΔTf = 0.383 K, Kf = 3.83 K kg mol− 1 ΔTf = Kf × m ; ΔTf = Kf × 2 2 1 W M W 1000 M2 (molar mass of sulphur molecules) = 2.56 1000 3.83 100 0.383 × × × = 256 g mol− 1 2. 100 g of water has 3g of urea dissolved in it. Calculate the freezing point of the solution. Kf for water = 1.86 K kg mol− 1 , molar mass of urea = 60 g mol− 1 , freezing point of water = 273.15 K (0°C) Ans. ΔTf = Kf × m ; ΔTf = 1.86 × 2 2 1 W M W 1000 ΔTf = 1.86 × 3 1000 60 100 × × = 0.93 ΔTf = 0 f fT T− ∴ Tf = 273.15 − 0.93 = 272.22 or −0.93°C 3. Human blood has osmotic pressure of 7.2 atm at body temperature of 37°C. Calculate the molar concentration of solute particles in blood. Given R = 0.0821 L atm K− 1 . Ans. π = CRT ; C = RT π T = 273 + 37 = 310 K C (molar concentration) = 7.2 0.0821 310× = 0.2828 M 4. Vapour pressure of benzene is 200 mm of Hg. 2g of a non-‐volatile solute in 78 g benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. Molar mass of benzene = 78 g mol− 1 . Ans. 2 1 nP P P n °− = ° ; 2 2 1 1 W MP P WP M °− = ° 200 195 200 − = 2 2 M 78 78 ; Molar mass of solute (M2) = 200 2 5 × = 80 g mol− 1
- 13. 5. 500 g of water containing 27 g of a non-‐volatile solute will boil at 100.156°C. Calculate the molar mass of the solute. Given boiling point of water = 100°C, Kb = 0.52 K kg mol− 1 . Ans. ΔTb = Kb × m ; ΔTb = Kb × 2 2 1 W M W 1000 Molar mass of solute (M2) = 0.52 27 1000 500 0.156 × × × = 180 g mol− 1 . Unit 3 Electrochemistry One mark questions 1. What is an electrolyte? An electrolyte is a compound which conducts electricity either in its aqueous solution or in its molten state. e.g Acids HCl, CH3COOH, HNO3 Bases NaOH, NH4OH Salts CuSO4, NaCl etc 2. Define conductivity of an electrolytic solution. Conductivity of a solution of an electrolyte is the conductance of a solution placed between two electrodes each of one square meter area kept at a distance of 1 meter apart. 3. Write the S.I unit for conductivity. SI unit for conductivity is Sm-‐1 . 4. Give the S.I unit for molar conductivity. Sm2 mol-‐1 5. State Kohlrausch Law. The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. 6. Define electrode potential. The potential difference developed between the electrode (metal) and the electrolyte (solution containing its own ions) when both the metal and the solution are in equilibrium is called electrode potential. 7. Define standard electrode potential. Standard electrode potential is the electrode potential when the concentrations of all the species involved is unity (1M) and if a gas is involved its pressure should be 1 bar. 8. Write Nernst Equation. ⎡ ⎤⎣ ⎦ n+ n+ o 10 n+( M / M) ( M / M) 0.059 1 E = E - log n M 9. State Faradays second law of electrolysis. The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.
- 14. 10. Define cell potential. Cell potential is the potential difference between the two electrodes of the galvanic cell. 11. Define EMF of the cell. It is the difference between the electrode potential of the cathode and anode when no current is drawn through the cell. 12. What is Fuel cell? Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane etc directly into electrical energy are called fuel cells. 13. Give a method to prevent rusting. Rusting may be prevented by barrier protection like painting, metal plating etc. 14. Write the relationship between cell potential and Gibb’s energy o o r cellG nFEΔ = − 15. Write the relationship between equilibrium constant and Eo cell 0.059 logo cell c V E K n = 2 mark questions 1. What are redox reactions? Give an example. Reactions in which both oxidation and reduction taken place simultaneously are called redox reactions. e.g ⎯⎯→2+ 2+ Zn+Cu Zn +Cu In this Zn is oxidised to Zn2+ Cu2+ is reduced to Cu 2. Mention any two factors on which the conductivity of an electronic conductor depends. The electronic conductance depends on (i) The nature and structure of the metal (ii) The number of valence electrons per atom. (iii) Temperature (it decreases with increase in the temperature) (any two) 3. Mention any two factors on which the conductivity of an electrolytic conductor depends. The conductivity of electrolytic solution depends upon (i) The nature of the electrolyte (ii) Size of the ions produced and their solvation. (iv) The nature of the solvent and its viscosity. (iv) Concentration of the electrolyte and (v) Temperature (increases with increase in temperature (any two) 4. Give two difference between the conductivity of an electronic conductor and electrolytic conductor. 1. On passing direct current composition of electronic conductor does not change but that of electrolytic conductor changes. 2. On increasing the temperature in case of electronic conductor conductivity decreases in case of electrolytic conductor conductivity increases. 5. What is a strong electrolyte? Give an example. A strong electrolyte is an electrolyte that dissociates completely into ions at moderate concentrations of its aqueous solution Ex: acids HCl, H2SO4, HNO3 Base NaOH, KOH Salts NaCl, CuSO4 (any salt) 6. What is a weak electrolyte? Give an example. A weak electrolyte is an electrolyte that dissociates partially into ions in its aqueous solution. Ex: CH3COOH, NH4OH 7. Define molar conductivity. How is it related to conductivity? Molar conductivity of a solution at a given concentration is the conductance of the volume V of a solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. It is represented by λm λm = kv where k is conductivity and v is volume of the solution containing 1 mole of the electrolyte or If λm is in Sm2 mol-‐1 and k in Sm-‐1 m k λ = 1000C where C is conc. in mol L-‐1 or When λm is in S cm2 mol-‐1 and k is in Scm-‐1 m 1000k λ = C 8. How does conductivity of a solution change with change in concentration of the solution? Give reason. Conductivity of a solution decreases with decrease in concentration of the solution due to decrease in the number of ions per unit volume of the solution.
- 15. 9. Define limiting molar conductivity. Write the relationship between molar conductivity and limiting molar conductivity. Limiting molar conductivity is the molar conductivity of a solution when concentration approaches zero or molar conductivity at infinite dilution. 1 o 2 m mλ = λ - AC where λm is molar conductivity and λo m is limiting molar conductivity, C is concentration in mole/L and A is constant which depends on nature of the electrolyte, solvent and temperature. 10. Draw a graph of molar conductivity verses square root of the molar concentration for KCl and CH3COOH mentioning clearly each. 11. How is limiting molar conductivity for a strong electrolyte found out by extrapolation method? Prepare four solutions of given strong electrolyte of different concentrations. Measure the conductivities of each solutions using conductivity cell and calculate the molar conductivities of each solution. Plot a graph of molar conductivity verses square root of the molar concentration for these solutions. A straight line is obtained which is to be extrapolated back so as to touch the vertical axes .This point of intersection on the vertical axes gives the limiting molar conductivity. 12. State and illustrate Faradays first law of electrolysis. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte either through its aqueous solution or molten state. If w is the mass of the substance deposited and Q is the current passed in coulombs w ∝ Q But Q = I t where I is the current strength in ampere and t is time in seconds. 13. Conductivity of 0.01 M NaCl solution is 0.12 Sm-‐1 . Calculate its molar conductivity. -2 2 m k 0.12 λ = = =1.2 ×10 Sm / mol 1000C 1000 × 0.01 14. The molar conductivity of 0.1M nitric acid is 630 S cm2 /mol. Calculate its conductivity. m -1 1000k λ = C 1000k 630 = 0.1 630 × 0.1 ∴ k = = 0.063 Scm 1000 15. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of nickel is deposited at the cathode? (Mol mass of Ni = 58.7) Q = I t = 5×20×60 = 6000C ⎯⎯⎯⎯⎯→2+ - Ni + 2e Ni 2 × 96500C 193000C 58.7g For 193000C of electricity mass of nickel obtained = 58.7g For 6000C of electricity 6000 × 58.7 =1.812g 193000 16. How long it will take for the deposition of 0.2g of silver when silver nitrate solution is electrolysed using 0.5 ampere of current (Mol mass of Ag = 108) ⎯⎯→+ - Ag + e Ag 96500C 108g For 108g of silver to be deposited current required is 96500C. For 0.2g of Ag But Q = I t 0.2 × 96500 =178.7C = Q 108 Q 178.7 t = = = 357.4 se I 0.5 17. The cell in which the following reaction occurs 3 2 ( ) ( ) ( ) 2( )2 2 2aq aq aq sFe I Fe I+ − + + ⎯⎯→ + Has Eo cell = 0.236V at 298K. Calculate the standard Gibb’s energy and the equilibrium constant for the cell reaction. n = 2 Δ.Go = -‐nFEo = -‐ 2×96500×0.236 = -‐ 45548 J 0.059 logcellE K n =
- 16. 0.059 0.236 log 2 K= 2 0.236 log 8 0.059 K × = = Taking the antilog K = 108 18. Write the reaction taking place at cathode and anode when aqueous solution of copper sulphate is electrolysed using copper electrodes. 2 ( ) ( ) 2 ( ) ( ) t 2 t 2 anode cathode oxdn s aq redn aq s A Cu Cu e A Cu e Cu + − + − ⎯⎯⎯→ + + ⎯⎯⎯→ Thus copper from anode dissolves and an equivalent amount of pure copper is deposited on cathode. This technique is used in electrolytic refining of crude copper. 19. Write the reaction taking place at anode and cathode when molten NaCl is electrolysed. When molten sodium chloride is electrolysed using inert electrodes 22 2t anode oxdn redn A Cl Cl e At cathode Na e Na − − + − ⎯⎯⎯→ + + ⎯⎯⎯→ Thus chlorine gas is liberated at anode and Sodium metal is formed at cathode. 20. Write the reaction taking place when aqueous solution of NaCl is electrolysed. When aqueous solution of NaCl is electrolysed, 2 NaCl Na Cl H O H OH + − + − ⎯⎯→ + +à àÜá àà The reaction taking place at cathode is 2( ) 1 2 aq gH e H+ − + ⎯⎯→ The reaction taking place at anode is ⎯⎯→- - aq 2 aq 1 Cl Cl + e 2 21. What is a primary battery/cell ?Give an example. Primary battery is one in which reaction occurs only once and cannot be recharged. Eg Dry cell or Leclanche cell and Mercury cell 22. What is a secondary battery/cell ? Give an example. Secondary battery is one which can be recharged by passing current through it in opposite direction,so that it can be Reused. Eg: Lead storage battery and Nickel cadmium cell. 23. Eo Cu = +0.34V and Eo Zn =-‐0.76V. Daniel cell is obtained by coupling these two electrodes. (i) represent the cell symbolically (ii) calculate the EMF of the cell (i) Daniel cell can be represented as Zn/ Zn2+ (aq) || Cu2+ (aq) /Cu (ii) EMF of Daniel cell Eo cell = Eo R -‐ Eo L = Eo Cu -‐ Eo Zn = 0.34-‐(-‐0.76) = 1.10V 24. Calculate the molar conductivity of a solution of MgCl2 at infinite dilution given that the molar ionic conductivities of 2+ - o 2 -1 o 2 -1 ( Mg ) ( Cl ) λ =106.1 Scm mol and λ = 76.3 Scm mol 2+ - 2 o o o MgCl Mg Cl 2 -1 λ = λ + 2λ =106.1+ 2( 76.3) = 258.7 Scm mol 25. The resistance of a conductivity cell containing 0.001 M KCl solution at 298K is 1500Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298K is 0.146×10-‐3 Scm-‐1 ? Cell constant G*= Rk =resistance × conductivity =0.146×10-‐3 Scm-‐1 ×1500S-‐1 = 0.219 cm-‐1 Question carrying 3 or 4 marks 1. Explain the construction of Daniel cell. Write the reaction taking place at anode and cathode and the net cell reaction. (3 mark) To prepare Daniel cell get a zinc electrode by dipping zinc rod in 1M ZnSO4 solution. Get a copper electrode by dipping a copper plate in 1 M CuSO4 solution. Couple these two electrodes using a salt bridge to get Daniel cell. Reactions taking place 2 2 2 2 ( ) ( ) ( ) ( ) t anode 2 cathode 2 oxdn redn s aq aq s A Zn Zn e At Cu e Cu Net cell reaction Zn Cu Zn Cu + − + − + + ⎯⎯⎯→ + + ⎯⎯⎯→ + ⎯⎯→ +
- 17. 2. With a labeled digram explain standar hydrogen electrode. Represent it symbolically. Write the reduction reaction at the anode. What is its electrode potential? (4 marks) It consists of a platinum electrode coated with platinum black. The electrode is dipped in 1M HCl. Pure hydrogen gas is bubbled through it under a pressure of 1 bar. S.H.E is represented as Pt(s) |H2 (g)(1bar) |H+ (aq)(1M) The reduction reaction taking place is 2 1 ( ) ( ) 2 H aq e H g+ − + ⎯⎯→ S.H.E is assigned an electrode potential of 0.0 V at all temperatures. 3. Explain the use of standard hydrogen electrode in measuring the standard electrode potentials of copper and zinc electrode (4 mark) Construct a standard electrode of the given metal by dipping the pure metal in 1M solution of its own ion at 25o C Couple this standard electrode with SHE using a salt bridge to get galvanic cell. Measure the emf of the cell using suitable instrument like potentiometer. Eo = Eo R – Eo L One of the electrodes of the cell is SHE and its electrode potential is 0.0V. So the electrode potential of the given electrode will be the emf of the cell in magnitude. If reduction takes place at the given electrode its Eo will be +ve but if oxidation takes place at the given electrode is Eo will be –ve. e.g if SHE is coupled with standard copper electrode reduction takes place at copper electrode cell can be represented as Pt (s) |H2(g. 1bar)|H+ (aq1M)||Cu2+ (aq.1M)|Cu 2+ + 2 o o o cell Cu / Cu H / H E = E - E 2+ 2+ o o Cu / Cu Cu / Cu 0.34 = E - 0 ∴E = 0.34V If SHE coupled with standard zinc electrode oxidation takes place at zinc electrode. Cell can be represented as 2 ( .1 ) ( ) ( ) 2( .1 ) ( .1 )/aq M s s g bar aq MZn Zn Pt H H+ + ⏐⏐ ⏐ ⏐ 2 2/ / o o o H H Zn Zn E cell E E+ += − 2 2 // 0.76 0 0.76o ZnZn Zn Zn E E V+ += − ∴ = − 4. How is Kohlrausch law helpful in finding out the limiting molar conductivity of a weak electrolyte? (3 m) Let us try to calculate λo m for a weak electrolyte CH3COOH. Select three strong electrolytes whose λo m can be found by extrapolation method in such a way that if we subtract λo m for one electrolyte from the sum of λo ms of the remaining two electrolyte λo m for CH3COOH can be obtained. The three electrolytes to be selected are CH3COONa, HCl & NaCl 3 3 o o o o CH COOH CH COONa HCl NaClλ = λ + λ - λ 5. The values of limiting molar conductivities (λo m) for NH4Cl, NaOH and NaCl are respectively 149.74; 248.1 and 126.4 Scm2 mol-‐1 . Calculate the limiting molar conductivity of NH4OH (3M) 4 4 o o o o NH OH NH Cl NaOH NaClλ = λ + λ - λ = 149.74+248.1-‐126.4 = 271.44 Scm2 mol-‐1 6. Calculate the equilibrium constant for the reaction at 298K 2 ( ) ( ) ( )2 ( ) 2s aq sCu Ag aq Cu Ag+ + + ⎯⎯→ + Given that Eo Ag+ /Ag = 0.80V and Eo (Cu2+ /Cu) = 0.34V 0.059 logo cell cE K n = log 0.059 o c nE cell K∴ = 2 ( / ) ( / ) o o o cell Ag Ag Cu Cu E E E+ += − =0.80-‐0.34=0.46V 2 0.46 log 15.59 0.059 cK × = = Taking the antilog Kc =3.92×1015 7. In Leclanche cell (dry cell) what are anode and cathode? What is the electrolyte used? Write the reactions at each electrode. What is the role of zinc chloride? It consists of a zinc container as an anode. A graphite rod surrounded by a mixture of manganese dioxide and carbon powder is cathode. The space between the electrodes is filled with electrolyte a moist paste of ammonium chloride and zinc chloride
- 18. Reaction taking place ⎯⎯→ 2+ - ( s)At anode Zn Zn + 2e ⎯⎯→+ - 2 4 3At cathode MnO + NH + e MnO( OH) + NH NH3 produced in the reaction forms a complex with Zn2+ to form [Zn(NH3)4]2+ . 8. What are the anode and cathode of lead acid battery? What is the electrolyte? Write the reactions taking place at anode and cathode and the overall reaction during discharging of the battery. (3 M) It consists of lead anode and a grid of lead packed with lead dioxide (PbO2) as cathode. Electrolyte is 38% solution of sulphuric acid. The reactions taking place when the battery is in use are 2 ( ) 4 ( ) 4( ) 2 2 4 ( ) ( ) 4 ( ) 2 ( ) 2 ( ) 4 2 2 s aq s aq aq s l Anode Pb SO PbSO e Cathode PbO s SO H e PbSO H O − − − + − + ⎯⎯→ + + + + ⎯⎯→ + The overall reaction is ( ) 2 ( ) 2 4 ( ) 4 ( ) 2 ( )2 2 2s s aq s lPb PbO H SO PbSO H O+ + ⎯⎯→ + 9. In Hydrogen oxygen fuel cell (i) Draw the schematic diagram mentioning the anode and cathode. What is the electrolyte? Write the reaction taking place at each electrodes and the net cell reaction. (4M) In this hydrogen and oxygen gases are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution. Catalyst like finely divided platinum or palladium is incorporated into the electrodes for increasing the rate of electrode reaction Reaction taking place are ⎯⎯→ ⎯⎯→ - - 2 ( g) 2 ( l) ( aq) - - 2( g) ( aq) 2 ( l) Cathode O + 2H O + 4e 4OH Anode 2H + 4OH 4H O + 4e Overall reaction is ⎯⎯→2 ( g) 2 ( g) 2 ( l)2H +O 2H O 10. What is corrosion? During rusting of iron write the anodic and cathodic reactions. Give the composition of rust. (3M) When a metal is exposed to the atmosphere it is slowly attacked by the constituents of the environment as a result of which the metal is slowly lost in the form of its compound . This is called corrosion. Reaction taking place are ⎯⎯→ ⎯⎯→ 2+ - ( s) - 2 ( g) 2 ( l) At Anode 2Fe 2Fe + 4e At Cathode O + 4H +( aq)+ 4e 2H O H+ are produced from H2CO3 formed due to dissolution of carbon dioxide from air into water The Fe2+ ions are further oxidised by atmospheric oxygen to ferric ion which are ultimately converted to hydrated ferric oxide called rust. Composition of rust is (Fe2O3.xH2O). 11. A conductivity cell when filled with 0.01M KCl has a resistance of 747.5 ohm at 25o C. When the same cell was filled with an aqueous solution of 0.05M CaCl2 solution the resistance was 876 ohm. Calculate (i) Conductivity of the solution (ii )Molar conductivity of the solution (given conductivity of 0.01M KCl = 0.14114 sm-‐1 ) (3M) Cell constant G* = Rk = 747.5×0.14114 =0.105.5m-‐1 -1 -1cell constant 105.5m Conductivity k = = = 0.1204Sm R 876 ohm 2 -1 m k 0.1204 Molar conductivity λ = = = 0.00241sm mol 1000C 1000 × 0.05 12. The electrical resistance of a column of 0.05M NaOH solution of diameter 1cm and length 50cm is 5.55×103 ohm. Calculate its (i) resistivity (ii) conductivity (iii) molar conductivity (3M) Cell constant o l G = a l = 50 cm Diameter = 1 cm ∴ radius = 0.5 cm Area of cross section A = πr2 = 3.14×(0.5)2 = 0.785 cm3 * -150 G = = 63.694 cm 0.785
- 19. -2 1 1 Resistivity ρ= = = 87.135 Ω k 1.148 ×10 m -2 2 -1 1000k Molarconductivityλ = C 1000 ×1.148 ×10 = 0.05 = 229.6 S cm mol 13. Calculate the emf of the cell in which the following reaction takes place. 2 ( ) ( )2 (0.002 ) (0.160 ) 2s sNi Ag M Ni M Ag+ + + ⎯⎯→ + Given that Eo cell = 1.05V 2 2 ( ) 10 2 ( ) [ ][ ]0.059 log 2 [ ][ ] so cell cell s Ni Ag E E Ni Ag + + = − But [M] for any element is taken as unity ⎡ ⎤⎣ ⎦ ⎡ ⎤⎣ ⎦ 2+ o cell cell 10 2+ Ni0.059 E = E - log 2 Ag ( ) 2 0.059 0.160 = 1.05 - log 2 0.002 = 0.914V Unit-4 CHEMICAL KINETICS Number of Hours of Teaching-9 Marks allotted-8 In part-D, 5mark question is split preferable in the form of 3+2 Definition :- The branch of chemistry which deals with study of reaction rate and their mechanism is called chemical kinetics Rate of a chemical reaction Q. 1 What is rate of reaction? (1m) Ans: Change in molar concentration of reactant or product in per unit time is called rate of reaction. Types of rate of reactions For reaction R à P Average rate = decrease in conc. R Of reaction time taken rav = - ∆[R] ∆t Average rate = increase in conc. Of P Reaction time taken rav = + ∆[P] ∆t 8 Marks Part –A 1x1=1 Part –B 1x2=2 Part-C 1x5=5
- 20. Q2:- For thereaction RàP, the conc. of reactant changes from 0.03M to 0.02M in 25 min. calculate average rate of the reaction using the unit of time in seconds. rav= - ∆[R]= - (0.02-0.03 ) ∆t 25x60 =-[-0.01] 1500 = 6.66x10-6 M/s Q3: What is the SI Unit of rate of reaction ? (1m) Ans: Mol /L /s Factors influencing Rate of reaction Q4 :- Mention any two factors which influence the rate of reaction . 2M Ans 1) Pressure or conc. of reactants 2) temperature 3) catalyst. Dependence of rate on concentration . Q5.) What is rate law ? (1 m) Ans: Representation of rate of reaction in terms of concentration of reactants is called rate law. Rate expression and rate equation Q.6) Define rate equation or rate expression (2m) Ans: Expression in which reaction rate is given in terms of molar conc. of reactants with each term raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. Q.7) Define rate constant of a reaction. (1m) Ans: Rate constant is equal to rate of reaction when the product of the molar conc. of reactants is unity. Order of a Reaction Q.8) Define order of a reaction. 1M Ans: Sum of the powers of the concentration of the reactants in the rate equation is called order of reaction. Q.9) Calculate the overall order of a reaction which has the rate expression. 1M Rate= K [A]1/2 [B]3/2 Ans: Order of reaction = 1/2 + 3/2 = 2 Q.10) What is elementary reaction ? (1m) Ans: Reaction taking place in one step is called elementary reaction. Q.11)What are complex reactions?(1m) Ans: Reactions taking place in more than one step are called complex reaction. Q.12) What is SI Unit of rate constant of nth order reaction ? (1m) Ans: (mol)1-n . Ln-1.s-1 Q.13) What is SI unit of rate constant of zero order reaction? (1m) Ans: Mol/L/s Q.14) What is the order of reaction whose unit of rate constant and rate of reaction are same ? (1m) Ans: Zero order. Q.15) Identify the reaction order from the rate constant K=2.3x10-5 mol-1 .L.S-1 (1m) Ans: Comparing the unit of rate constant with general unit Mol-1 .L.S-1 with ( Mol)1-n .Ln-1 .S-1 1-n= -1 n=2 Molecularity of a reaction Q.16) Define molecularity of a reaction . (1 m) Ans: The number of reacting species taking part in an elementary reaction which must colloid simultaneously in order to bring about a chemical reaction is called molecularity of reaction. Q.17) In a complex reaction which step controls the overall rate of reaction and what is it called? (2m) Ans: Slowest step, which is called rate determining step. Q.18) The conversion of molecules X to Y follows second order kinetics .If conc. of X Increased to three times,how will it affect the rate of formation of Y ? (1m) Ans: Increased rate =( Increased conc.)n =32 =9 Rate of formation of Y increases by 9 times Integrated rate equations Q.19) Derive rate constant of zero order reaction (3m) Ans: Consider a zero order reaction R--> P Rate =-d[R] = K[R]o dt = - d[R] = K dt = d[R] = -kdt -------(1)
- 21. Integrating equation (1)both sides [R]= -kt+I ----------(2) Where “I” is integration constant At t=0 [R]=[R]o where [R]o is initial concentration of reactant. ∴Eqn (2) becomes I=[R]o Substituting I in eqn-------- (2) [R] = -Kt + [R]o -Kt = [R]-[R]o Kt = [R]o-[R] K = [R]o-[R] t Q:20) Derive integrated rate equation for first order reaction? (4m) Ans-Consider a first order reaction. RàP Rate = - d[R] = K[R] dt d[R]= - K[R] dt d[R] = - K. dt. -------(1) [R] Integrating eqn.(1)on both side ln [R] = - Kt + I --------(2) Where “I” is integration constant At t=o [R]=[R]o which called initial Concentration reactant Substituting the values in ln[R]o= I Equation (2) can be written as ln[R] = -Kt + ln[R]o Kt = ln[R]o – ln [R] Kt = ln [R]o [R] Kt =2.303 log[R]o [R] K= 2.303 x log[R]o t [R] Log [R]o Slope = K [R] 2.303. Or o time K= 2.303xSlope Half life of a reaction Q21)Define halfe life of a reaction . (1m) Ans: The time in which the conc.of a reactant is reduced to one half of its initial conc. is called half life of a reaction (t1/2) Q:22) Show that half life of a zero order reaction is directly proportional to initial concentration of reactant from integrated rate equation. OR Derive the relation between half life and rate constant of zero order reaction .(2m). Ans:-Rate constant of zero order reaction is K= [R]o – [R] t At half life t =t ½ & [R] = ½ [R]o .: K= [R]o – ½ [R]o t1/2 K=[R]o 2t½ t ½ =[R]o 2 K OR t ½ ∝ [R]o Q.23) Show that half life of a first order reaction is independent of initial Conc. of reactant from integrated rate equation (2m) Or Derive the relation between half life of a first order reaction and its rate constant . (2m) Ans: Rate constant of first order reaction is K= 2.303 x log [R]o t [R] At half life t=t ½ ,[R]=[R]o 2 .: K = 2.303 x log [R]o t½ [R]o/2 K =2.303 x log 2 t ½ K= 2.303 x0.3010 t½ t½ = 0.693 K K= 2.303 x0.3010 t½ t½ = 0.693 K
- 22. Q.24)A first orderreaction is found to have a rate constant 5.5x10-14 /s .Calculate the half life of the reaction (2m) Soln. : K= 5.5x10-14 /s t½ =? t½ = 0.693 K = 0.693 5.5x10-14 t½ = 1.26x1013 sec Q:25) Show that the time required for 99/. Completion of a first order reaction is twice the time required for the completion of 90% of reaction( 4m) I set : [R]o= 100, [R]=[100-90]=10 t=t90% IIset : [R]o= 100 [R]= [100-99]1 t=t99% To be proved t99%= 2t90% K= 2.303 x log [R]o t [R] Sub. I set values . K= 2.303x log 100 t 90% 10 K = 2.303x log 10 t90% K= 2.303 X 1 - (1) t90% Substituting II set values K=2.303 x log 100 t99% 1 K= 2.303 x 2 --------------(2) t99% Comparing equations (1) & (2) 2.303x 1 = 2.303x2 t90% t99% t99% = 2t90% Pseudo first order reaction Q:26 Define pseudo first order reaction . Give an example. (2m) Ans: Chemical reactions which are not first order but behave as fist order reaction under suitable conditions are called pseudo first order Reactions. Ex: Inversion of cane sugar. C12 H22O11+H2O àC6H12O6 + C6H12 O6 Temperature dependence of the rate of a reaction Q:27)How does rate of reaction vary with temperature? (1m). Ans: Rate of reaction increases with increase of temperature. Q:28) What happens to the rate constant of a reaction when temperature is increased by 10o .? Ans: Rate constant increases nearly by two times. Q.29) Write Arrhenius equation which relates the rate constant , activation energy and temperature . (1m) Ans K= A e-Ea/RT Energy of activation Q.30) Define energy of activation (1m). Ans: The minimum energy required for the reactants to form activated complex is called Activation energy. Q.31) How is activation energy related to rate of reaction? (1m) Ans: Rate of reaction is inversely proportional to activation energy. ie r ∝ 1 Ea Q.32) How is activation energy affected by presence of positive catalyst? (1m) Ans: Activation energy of a reaction decreases in presence of catalyst. Q.33) On increasing 100 K temperature rate of reaction becomes double, explain from the max well Boltzmann distribution curve. (2m) On increasing 100 K temperature, substance Increases the fraction of molecules double,hence rate of reaction doubles. Q:34) How does positive catalyst increases the rate of reaction? (2m) Ans positive catalyst decreases the activation energy by changing the Path of the reaction,which increases the rate of reaction
- 23. Collision theory ofchemical reactions. Q:35) What is effective collision? How is it related to rate of reaction? (2m). Ans. Collision in which molecules colloid with sufficient kinetic energy and proper orientation so as to form products is called effective collision. It is directly proportional to the rate of reaction. Q:36)How is activation energy calculated by plotting graph ln K against 1/T ? (2m) Q:37)Write Arrhenius equation at different Temperature and rate constants. Ans: log K2/K1 = Ea X T2-T1 2.303RT T1 T2 Q:38)The rate constants of a reaction at 500K. and 700K are 0.02s-1 and 0.07s-1 respeetively calculate the activation energy. (3m) Ans: log K2/K1 = Ea X T2-T1 2.303RT T1 T2 log 0.07 = Ea x 700- 500 0.02 2.303x8.314 500x 700 0.544 = Ea x 5.714 x10-4 19.15 Ea= 0.544x19.15 5.714 x10-4 Ea= 18230.8 J = Ea= 18. 2308 KJ. UNIT -5 SURFACE CHEMISTRY A. Short answer questions carrying 1 mark 1. What is adsorption A surface phenomenon wherein there is accumulation of molecules on the surface (than in the bulk) of a solid or a liquid. 2. Why solids in finely divided state are good adsorbent? Solids in finely divided state have large surface area, as surface area increases adsorbing power increases. 3. What is desorption? The process of removing an adsorbed substance from a surface on which it is adsorbed is called desorption. 4. Name the substance used to decolour the solution of raw sugar. Animal charcoal. 5. Name of the phenomenon in which both the adsorption and desorption takes place simultaneously. Sorption 6. Why is adsorption always exothermic? During adsorption there is always decrease in residual forces on the surface, hence adsorption is always exothermic. Or There is decrease in surface energy which appears as heat, hence adsorption is always exothermic. 7. Name catalyst used in the conversion of alcohols into gasoline (petrol) Zeolite ZSM-5 (Zeolite Sieve of molecular porosity-5) 8. Name the colloidal system in which dispersed phase is solid and dispersion medium is liquid Sol 9. Name the dispersed phase in gel Liquid 10. Give an example for oil in water emulsion Milk, Vanishing cream
- 24. 11. What typeof colloidal emulsion is present in butter Water in oil (W/O) 12. What is the dispersion medium in gel? Solid 13. Between Na2SO4 and Na3PO4 which has greater power to coagulate a positively charged colloid? Na3PO4 14. Alum is added to muddy drinking water. Why? Alum is added to muddy drinking water to coagulate 15. What is the dispersed phase in milk? Oil or liquid 16. A liquid is dispersed in a gas. Name the type of colloid obtained. Liquid aerosal 17. Name the instrument designed by Zigmondy. Ultramicroscope 18. Movement of the dispersion medium in an electric field by preventing the movement of colloidal particles by suitable method.Name the phenomenon Electroosmosis 19. The process by which colloidal particles aggregate, become bigger and settle down. Name the phenomenon Coagulation 20.What happens when an electrolyte is added to lyophobic sol? Coagulation or precipitation 21. Name the phenomenon, when an electrolyte having a common ion is added to freshly prepared precipitate? Peptization B. Answer questions carrying 2 marks ADSORPTION 1. What are adsorbate and adsorbent? Give an example. Molecules (substances) that accumulates on the surface is called adsorbate. The material on the surface of which adsorption takes place is called adsorbent. Example: Ni adsorbs H2. Ni is the adsorbate, H2 is the adsorbent 2. Give two examples for adsorption. i) When animal charcoal is added to methylene blue, charcoal adsorbs the dye. ii) Air becomes dry in the presence of silica gel because silica gel adsorbs water molecules on the surface iii) A small pillow of silica gel in a box adsorbs moisture in the box keeps the air dry. (Any two) 3. Give differences between adsorption and absorption. Adsorption Absorption 1. A substance gets concentrated on the surface of a solid or liquid. 2. It increases with increase in surface area. Example: adsorption of water by silica gel. A substance gets uniformly distributed through the bulk of solid or liquid. It remains unaffected by increase in surface area. Example: Absorption of water by anhydrous CaCl2. 4. Of SO2 (critical temperature 630K) and CH4 (critical temperature 190K) which gas will be adsorbed readily on the surface of 1 gram of activated charcoal. Justify the answer. SO2 gas Easily liquefiable gases with higher critical temperature are readily adsorbed as the theVander Waal’s forces are stronger near critical temperature. 5. What is the effect of temperature on physical and chemical adsorption? Physical adsorption decreases with increase in temperature. Chemical adsorption increases with increase intemperature. 6. Mention any two applications of adsorption. i) In the production of high vaccum ii) In gas mask, to adsorb poisonous gases iii) In the separation of noble gases using activated charcoal iv) Removal of colouring matter from solutions v) In adsorption chromatography to analyse a given
- 25. CATALYSIS 1. What iscatalysis? Give an example. A substance that accelerates the rate of a reaction without itself remaining unchanged chemically and quantitatively is a catalyst. The phenomenon is catalysis. E.g.: 2KClO3 2MnO ⎯⎯⎯→2KCl + 3O2 MnO2 is a catalyst. 2. What are promoters and poisons with respect to a catalytic process? Promoters are substance that increases the activity of a catalyst. E.g.: In Haber’s process molybdenum acts as a promoter for iron used as a catalyst. A catalytic poison is one that decreases the efficiency or activity of a catalyst. E.g.: In Haber’s process CO if present in the mixture of H2 and N2, poisons the iron catalyst. 3. What is homogeneous catalysis? Give an example. When reactants and catalyst are in the same phase the process is homogeneous catalysis. E.g.: a) 2SO2(g) + O2(g) ( )gNO ⎯⎯⎯→ 2SO3(g) Here the reactants (SO2 and O2) and catalyst (NO) are all gases. b) Acid hydrolysis of cane sugar is also an example for homogeneous catalysis. Here the reactants sugar solution, water and the catalyst dil. HCl are in the same phase (aqueous solution) C12H22O11(aq) + H2O(l) H+ ⎯⎯→C6H12O6 + C6H12O6 (both are in aq solution) Sucrose glucose fructose 4. What is heterogenous catalysis? Give an example. A catalytic process in which reactants and catalyst are in different phases are known as heterogenous catalysis. E.g.: 1. N2(g) + 3H2(g) (s)Fe ⎯⎯⎯→2NH3 Here the reactants are gases, catalyst iron is a solid 2. Vegetable oil (l) + H2(g) ( )sNi ⎯⎯⎯→Vanaspathi ghee Here reactants and catalyst are in different phases. 5. Write a note on a) activity b) selectivity of solid catalysts. a) Activity: The activity (efficiency) of a solid catalyst depends on how strongly the reactants are chemisorbed on it. It is found that elements (metals) in group 7-9 of the periodic table show greater catalytic activity for hydrogenation reactions. E.g.: 2H2(g) + O2(g) Pt ⎯⎯→2H2O (l) b) Selectivity: For a given set of reactants, different catalyst may yield different products. This is selectivity of a catalyst. E.g.: CO(g) + 3H2(g) Ni ⎯⎯→CH4(g) + H2O(g) CO(g) + H2(g) Cu ⎯⎯→H−CHO Ni is selective to convert water gas to CH4 whereas Cu converts water gas into formaldehyde. In otherwords Ni catalyses the conversion of water gas to CH4 but cannot catalyse to convert water gas to formaldehyde. Catalyst is highly selective in nature i.e a given substance can act as a catalyst only in a particular reaction and not for all the reactions. 6. What is shape selective catalysis? Give an example. A catalytic reaction that depends on pore structure of the catalyst and size of the reactant and product molecules is called shape selective catalysis. E.g.: zeolites. 7. Write a note on zeolites as shape selective catalysts. Zeolites are aluminosilicates with 3D nework of Al-O-Si frame with honey comb like structure. This structure makes them to act as shape selective catalyst depending on pore size in them and on the size of reactant and products. Many zeolites are synthesized for selective catalytic activity. E.g.: 1) Zeolite ZSM-5 (Zeolite Sieve of molecular porosity- 5) converts alcohols into gasoline (petrol) by dehydrating alcohols. 2) Many zeolites are used in petroleum industry in cracking of hydrocarbons and in isomerisation. 8. What are enzyme catalysis or biochemical catalysis? Give an example for enzyme catalysis. Enzymes are proteins, which catalyse large number of reactions that maintain life processes in both plants and animals. Hence they are biochemical catalysts and the phenomenon is called as biochemical catalysis. Inversion of cane sugar in the presence of enzyme invertase into glucose and fructose 9. Give two examples for enzyme-catalysed reaction. a) Conversion of starch into maltose 2(C6H10O5)n (aq) + nH2O (l) diastase enzyme ⎯⎯⎯⎯⎯→nC12H22O11 (aq) Starch maltose b) Urea into ammonia and carbon dioxide NH2CONH2(aq) + H2O (l) urease ⎯⎯⎯→2NH3(g) + CO2(g) c) In human beings enzyme pepsin converts proteins into peptides and pancreatic trypsin enzyme converts proteins into amino acids.
- 26. d) Milk isconverted into curds by lactobacilli enzyme present in the curd which is added to milk. 10. Mention characteristics of enzyme catalysis. a) Their efficiency as catalyst is very high. b) They are highly specific in their action. c) They are highly active at optimum temperature and optimum pH. d) Their activity increases in presence of activatiors and coenzyme. e) Their activity decreases in presence of inhibitors and poisons. 11. Write the two steps involved in mechanism of enzyme catalysis. a) An enzyme binds to the substrate to form an activated complex: E + S ⎯⎯→ES* b) Decomposition of activated complex to form the product: ES* ⎯⎯→E + P. COLLOIDS 1. What is a colloid? Colloid (Colloidal system or solution) is a heterogeneous system in which one substance is dispersed as very fine particle in another substance called dispersion medium. The size of the particle is larger than the true solution but smaller than the suspended particle i.e their diameter ranges between 1nm to 1000nm. 2. Name the 2 phases of colloidal system Dispersion medium and dispersed phase (colloidal particles) 3. What is Dispersion Medium and dispersed phase for a colloid? Give an example. The continuous medium in which the colloidal particles are dispersed is called Dispersion Medium. The discontinuous phase which the substance is dispersed as colloidal particles is called dispersed phase. Eg: Milk is a colloid in which fat globules form the dispersed phase, water is the dispersion medium. 4.Classify the colloids based on the physical state Based on the physical state of dispersed phase and dispersion medium colloids are classified into 8 types Dispersed Phase Dispersion Medium Name of the Type Example solid solid Solid sol Ruby glass, gems solid liquid sol Ink, Paint, Gold Sol solid gas Solid aerosol Dust, Smoke, Soot in Air, liquid solid gel curds, jam, silica gel, butter liquid liquid emulsion Milk, Cream, Cod Liver Oil liquid gas Liquid aerosol Fog, Mist, Cloud gas solid Solid foam Foam rubber, Pumice stone gas liquid foam Shaving cream, soap lather 5. What is a Sol? Give an example. It is a colloid wherein the dispersed phase is a solid and dispersion medium is a liquid. Eg: Sulphur dispersed in Water. Sulphur (solid) is the dispersed phase, water is the dispersion medium. If the dispersion medium is water, alcohol and benzene, sol is called aqua sol (hydrosol), alcosol and benzosol. 6. How are colloids classified based on the affinity of the dispersed phase towards dispersion medium Based on the affinity of dispersed phase towards dispersion medium, sols are classified as lyophilic and lyophobic sols (colloids). 7. What is lyophylic sol? Give an example Lyophilic sol (colloid): These are sols in which the (colloidal particle) dispersed phase has affinity towards dispersion medium. (Intrinsic colloids). If the dispersion medium is separated from the dispersed phase, these sols can be formed by remixing them. Hence these are called reversible sols. Eg: Starch dispersed in water Albumin dispersed in water Gum or gelatin in suitable solvent. 8. What is lyophobic sol? Give an example Lyophobic sol (colloid): These are sols in which the dispersed phase has no affinity towards the dispersion medium (extrinsic colloids). Once precipitated or if the dispersion medium is separated from the dispersed phase, these sols cannot be formed by remixing them. Hence these are called irreversible sols. Eg: Sulphur dispersed in water, gold sol. 9.Distinguish between lyophillic to lyophobic sols (Any two) Property Lyophilic Lyophobic 1 Affinity towards dispersion medium High affinity Low affinity 2 Method of preparation Easily formed on mixing or heating the dispersed phase with dispersion medium Special methods are used
- 27. 3 Stability stableHighly unstable 4 Reversibility Reversible Irreversible 5 Solvation of colloidal particles. Highly Solvated Not solvated 6 Addition of electrolyte Does not easily coagulate Gets easily coagulated 10.Classify the colloids based on type of particles of the dispersed phase Multimolecular colloid, Macromolecular colloid, associated colloid (micelles) 11.Write a note on Multimolecular colloids with an example If large number of atoms or smaller molecules of a substance aggregate together to form particles having size in the colloidal range then the colloidal system is known as multimolecular colloid. E.g.: A gold sol contains colloidal gold particles each made up of large number of gold atoms. Sulphur sol contains an aggregate of thousands of S8 sulphur molecules. 12.Write a note on Macromolecular colloids with an example Macromolecule (polymers) in a suitable solvent form solutions in which the size of the macromolecules may be in the colloidal range and the system is known as macromolecular colloids. E.g.: starch, cellulose, enzymes, proteins, nylon, polystyrene in a suitable solvent. 13. Write a note on Associated colloids /micelles with an example Some substances at low concentration behave as strong electrolytes (true solution), but at higher concentrations aggregate to form colloidal particles. Such substances form associated colloid. Aggregate of molecules thus formed is called a micelle. Formation of micelle takes place if i) the temperature is above Kraft temperature (TK) ii) concentration is greater than critical micelle concentration (CMC). If an associated colloid (micelle) is diluted, it behaves as a strong electrolyte. E.g.: Surface active agents like soaps and detergents form associated colloids. These have both lyophilic and lyophobic groups. For soaps, critical micelle concentration is 10− 4 to 10− 3 mol 14. Write equations for the preparation of Sulphur sol and Ferric hydroxide sol I. SO2 + 2H2S oxidation ⎯⎯⎯⎯→ 3S (sol) + 2H2O 2. FeCl3 + 3H2O hydrolysis ⎯⎯⎯⎯→ Fe(OH)3(sol) + 3HCl 15. How is a metal sol prepared by Bredig’s arc process? This process involves both dispersion and condensation. Sol of metals like gold, platinum and silver can be prepared by this method. Two metal electrodes of a metal are dipped in water and an electric arc is struck between them. Intense heat of the arc causes the metal to vapourise.The vapours condenses to form metal particles of colloidal size. Thus metal sol is obtained. 16. What is peptization? Give an example The process of converting a freshly prepared precipitate (suspension) into a colloid by adding a electrolyte having a common ion is called peptization. The electrolyte added is called a peptizing agent. During peptisation, the precipitate adsorbs one of the ions (positive or negative) of the electrolyte. This causes the precipitates to break into smaller particles of colloidal size. 17. What is electro dialysis? The process of increasing the rate of dialysis, under the influence of an electric field is called electro dialysis. The process can be used iff the impurity is an electrolyte. In presence of the electric field, the ions diffuse faster (through parchment paper) towards the oppositely charged electrodes. 18. Write a note on ultrafiltration It is a process of separating collidal particles from the solvent (dispersion medium) and all other soluble solutes present in collidal solution using specially prepared ultrafilters. An ultra filter paper (made by soaking filter paper in 4% nitro cellulose in alcohol and ether and later hardening it by using formaldehyde) allows all other particles except the colloidal particles to filter (pass) through it. To speed it up pressure or suction can be applied. The colloidal particles left on the ultrafilter paper are then stirred into fresh dispersion medium to get the pure colloidal solution. 19. Describe Tyndall effect Scattering of light by colloidal particles in the medium is called Tyndall effect. The path of light in the colloidal medium becomes visible when observed at right angles. The illuminated path within the medium is called Tyndall cone. 20. Write two conditions in which tyndal effect can be clearly observed. Tyndall effect is clearly observed when a) size of colloidal particles matches with the wave length of light used b) there is large difference in refractive index between dispersed phase and medium
- 28. 21. What isBrownian movement? How is it caused? Zig– Zag movement of colloidal particles in a medium is called Brownian movement. Reason: Particles of the medium are very small and are moving randomly in all directions. They collide with the colloidal particles and transfer their kinetic energy. Colloidal particles move slowly and randomly due to unequal bombardments by the particles of the medium. This is seen as “Brownian Movement”. “This property is a direct proof for the concept that liquid state of matter is made up of small molecules, which are in random motion, does not allow the particles to settle and is responsible for the stability of the sols” 22. Classify the following colloids into positively and negatively charged sols Al (OH)3, CdS, AS2S3, Fe (OH)3, gum, clay, basic dyes, Sols of acidic dyes, sols of starch, and metallic sulphides, sols of metals (Ag, Au), haemoglobin. Positively charged sol negatively charged sol Al (OH)3, Fe (OH)3, basic dyes, haemoglobin, CdS, AS2S3, Sols of acidic dyes, sols of starch, gum, clay, and metallic sulphides and sols of metals (Ag, Au) 23. How do colloidal particle become charged or acquire charge? The charge on the colloidal particles may be due to (i) preferential adsorption of ions from the medium or (ii) due to electron capture by sol particles during electrodispersion of metals.(iii)formation of electrical double layer 24.What is Electrophoresis. Movement of electrically charged colloidal particles towards their oppositely charged electrodes when the colloid is placed in an electric field is electrophoresis. Positively charged particles move towards cathode and negatively charged particles move towards anode. 25. Mention any two methods of Coagulation of lyophobic sol i) Electrophoresis ii) Mixing of two oppositely charged sols. E.g.: positively charged Fe(OH)3 sol with negatively charged As2S3 sol iii) Continuous dialysis iv) Addition of electrolyte v) By boiling 26. State and illustrate Hardy- Schulze rule. Higher the valency of the flocculating ion added, greater is the coagulating power of the ion. Ex (1): In the coagulation of negatively charged sol (As2S3) the coagulation power of the positively charged active ion is Na+ < Ba+2 < Al+3 . Ex (2): In the coagulation of positively charged sol [Fe (OH)3] the coagulating power of the negatively charged active ion is Cl− < SO4 − 2 < PO4 − 3 < [Fe(CN)6]4− . Note: Higher the charge on the flocculating ion, lesser is the amount of the electrolyte required to coagulate a sol. 27. Difine coagulating value or flocculating value The minimum concentration of electrolyte in millimoles per litre required to cause precipitation of a sol in 2 hours is called coagulating value. Smaller the coagulating value, higher is the coagulating power of the ion. 28. What is protective action of a sol? Give an example. The property of a lyophilic sol by which it protects the lyophobic sol from precipitation, even upon adding an electrolyte to it, is called protective action of lyophilic sol. Lyophilic sol particles form a coat or layer around the lyophobic sol and hence protect them from the action of the electrolytes. 29. What are Emulsions? Give an example A liquid in a liquid colloid is called an emulsion. If two immiscible liquids are shaken well, a dispersion of one liquid in the other, an emulsion is obtained. Eg: Milk, butter, vanishing cream 30.Write a note on formation of delta region. River water flowing towards the sea picks up many colloidal particles (clay, mud, humus, slit) with it. These particles are negatively charged. When the river water meets the sea, the electrolytes (salts like NaCl, MgSO4 etc) in the sea causes the coagulation of these colloidal particles. Thus clay, mud, humus gets precipitated and scattered at these places to form delta region. 31.Write the application of colloids in purification of smoke using Cottrell precipitator Smoke (Colloidal dispersion of solid in gas) from industries contains carbon, dust, soot and many others as colloidal particles. To remove these, electrostatic precipitator called cottrell precipitator is used. The precipitator consists of metal plates attached to a high potential. As the smoke enters the precipitator, the charged colloidal particles
- 29. gets neutralized andprecipitated on the metal plates. Gases free from colloidal impurities are led into chimney. 32. Write the application of colloids in the Purification of drinking water Drinking water if muddy contains negatively charged clay, sand, mud as colloidal particles dispersed in it. When alum is added to this, Al+3 ions of the alum causes the coagulation of the negatively charged muddy colloidal particles which settle down as a precipitate. The upper layers of clear clean water are decanted. Thus water gets purified. C. Questions carrying 3 marks 1. What happens to ΔH, ΔS and ΔG during the process of adsorption? i) Adsorption is always an exothermic process, because there is decrease in surface energy. ∴ ΔH is negative (enthalpy decreases). ii) When a gas is adsorbed on a liquid or solid, freedom of movement of gas molecules decreases. ∴ entropy decreases. ΔS is negative. iii) Adsorption is a spontaneous process hence ΔG must be negative. ΔG = ΔH − TΔS. For adsorption ΔH = negative, ΔS = negative. Therefore ΔH must be more negative than TΔS being positive so that ΔG becomes negative. 2.Write any three differences between two types of adsorption of gases on solids. Physiosorption (physical adsorption) Chemisorption (chemical adsorption) 1. Accumulation of gas on a solid due to weak van der Waal’s forces. 2. This is not specific, as force between adsorbate and adsorbent is van der Waals forces which is universal. 3. The process is reversible. 4. Gases that can be easily liquefied (high critical temperature) are readily absorbed. 5. Enthalpy of adsorption is low, as the forces involved are weak (ΔH is negative but low) 6. Adsorption decreases with increase in temperature. Low temperature favours better adsorption. 7. Under high pressure, it leads to multimolecular layers of adsorption. 1. Accumulation of gas on a solid due to chemical bond (covalent or ionic) 2. It is highly specific as there is chemical bonding between adsorbate and adsorbent. 3. Process is irreversible. 4. Gases that can form chemical compounds with adsorbent are specifically adsorbed. 5. Enthalpy of adsorption is high, as the forces involved are strong (ΔH is negative, very high) 6. Adsorption process involves high energy of activation, therefore increases with increase in temperature. 7. It leads to unimolecular layer of adsorption even at high pressure. (Any 3 of the above) 3. Classify the following colloids to their respective type of colloids a. Smoke b. Cod liver oil c. gems. a. Smoke-Solid aerosal b. Cod liver oil-Emulsion c. gems-solid sol 4.Write the mechanism of micelle formation considering soap as an example Soap is sodium or potassium salt of higher fatty acid RCOO− Na+ . In water RCOO− Na+ dissociates into RCOO− and Na+ . RCOO− has two parts. R is long hydrocarbon chain and is a non- polar tail (hydrophobic). COO− is polar-ionic head (hydrophilic). At low concentration COO− group will be dissolved in water and R chains away from water and remain at the surface. At critical micelle concentration, the anions are pulled into the water. They aggregate to form spherical shape in which hydrocarbon chains point to the interior and COO− projects outwards of the sphere. Such an aggregate is called a micelle. 5. Write a note on Cleansing action of soap a. It is due to formation of micelle by soap. b. Soap molecules form a micelle around oil droplet (dirt) in such a way that hydrophobic R is in the oil and hydrophilic −COO− projects out into water. c. The oil droplet thus gets pulled into water and gets detached from dirty cloth (material to be washed). d. Soap thus helps in emulsification of oil and fat in the dirt, which is then washed away with water. a) Grease on cloth b) Stearate ions arranging around the grease droplet and c) grease droplet surrounded by stearate ions (micelle formed)
- 30. 6. Write anote on Dialysis. A process of purifying a lyophobic sol by removing particles of true solution (ions or molecules) by their preferential diffusion through parchment paper or animal membrane is called Dialysis. The membrane is called a dialyser. Particles of true solution pass through the membrane but not the colloidal particles. Process: The sol to be purified is taken in a parchment bag. The bag is suspended in a tank, in which water is circulated. Particles of true solution diffuse out from the bag. Water flowing in the tank carries away these particles. The sol gets purified and stabilized. 7. Describe how colloidal particles acquire charge by preferential adsorption of ions The colloidal particle in a lyophobic sol tends to adsorb cations or anions from the medium and hence become positively or negatively charged sols. They show a preference to adsorb a common ion from the medium. Ex: (a) when potassium iodide solution is slowly added to silver nitrate solution, the silver iodide sol formed adsorbs Ag+1 (present in plenty) and becomes positively charged. (AgI/ Ag+1 ) Ex: (b) When silver nitrate solution is slowly added to potassium iodide solution, silver iodide sol formed adsorbs I-1 (present in plenty)ions from the medium and becomes negatively charged sol (AgI / I-1 ). 8. Mention two types of emulsion. Give example for each i) Oil in water or ii) water in oil emulsion. For oil in water emulsion, water is the dispersion medium, oil the dispersed phase. E.g.: milk, vanishing cream. In milk, liquid fat is dispersed in water. For water in oil emulsion, water is the dispersed phase, oil is the dispersion medium. E.g.: butter, cream. UNIT- 6 PRINCIPLES AND PROCESSES OF EXTRACTION OF METALS. I. ONE MARK QUESTIONS: 1. Name an important ore of Aluminium . Ans: Bauxite 2. Give the composition of copper pyrites. A: CuFeS2 3. What is meant by concentration of ores? A: The process of removal of earthy impurities from the ore. 4. Name the electrolyte used in the extraction of aluminium. A: Molten Al2O3 + Cryolite + CaF2 5. Sulphide ores are roasted before reduction. Why? A: To convert sulphides to oxides so that reduction is easy. 6. What are the products formed when calcium carbonate is calcined? A: CaO + CO2 7. Give the composition of copper matte. A: Cu2S + FeS 8. How is FeO removed during the extraction of copper? A: It is removed as ironsilicate FeSiO3 using SiO2. 9. What do you mean by blister copper? A: The solidified copper obtained has blistered appearance due to the evolution of SO2 and so it is called blister copper. 10. During froth floatation process, name the component that comes along with the froth. A: Ore 11. Why do we add collectors during froth floatation? A: To enhance non-wettability of ore particles by water. 12. Haematite ore particles are heavier than gangue. Suggest a suitable method for its concentration. A: Gravity separation 13. What is the importance of roasting and calcination. A: This is done to get the metal in its oxide form so that reduction can be done easily. 14. Give an example of a metal that can be extracted by electrolytic method. A: Aluminium ( Or Sodium, magnesium)
- 31. 15. In theextraction of aluminium carbon anodes are replaced regularly. Why? A: Because the carbon gets worn out as the oxygen liberated reacts with it to form CO2. II. TWO MARK QUESTIONS 1. What is the role of (i) lime stone in iron extraction and (ii) cryolite in aluminium extraction.? A: (i) Removes silica impurity as slag calcium silicate (ii) Cryolite increases conductivity and reduces melting point of Al2O3. 2. Give the chemical reactions involved in (a) Iron extraction 2 marks A: C + O2 →CO2 ; CO2 + C →2CO, Fe2O3 + CO →2FeO + CO2 ; FeO + CO →Fe + CO2 CaCO3 →CaO + CO2, ; CaO + SiO2 →CaSiO3. (b) Aluminium extraction 2 marks A: Cathode: Al3+ (melt) + 3e Al(l) Anode: C(s) + O-2(melt) CO(g) + 2e C(s) + 2O2-(melt) CO2 (g) + 4e Overall reaction: 2Al2O3 + 3C 4Al + 3CO2 (c) Copper extraction 2 marks A: 2FeS + 3O2 2FeO + 2SO2 FeO + SiO2 FeSiO3 2Cu2S + 3O2 2Cu2O + 2SO2 2Cu2O + Cu2S 6Cu + SO2 3. How is zinc obtained from ZnO? A: ZnO is heated with coke at 1673K. Zinc and carbon monoxide are formed. ZnO + C Zn + CO 4. Give equations for the extraction of gold using NaCN. A: 8NaCN (aq)+ 4Au(s) + 2H2O(aq) + O2(g) 4Na[Au(CN)2] (aq) + 4NaOH(aq) 2Na[Au(CN)2] (aq) + Zn(s) 2Au(s) +Na2[Zn(CN)4](aq). 5. Give the principles involved in (i) zone refining (ii) liquation A: (i) Zone refining: The impurities are more soluble in the melt than in the solid state of the metal. (ii) Liquation: melting point of metals is lower than the impurities. 6. What are the requirements for the compound to be purified by vapour phase refining? A: (i) The metal should form a volatile compound with an available reagent, (ii) The volatile compound should be easily decomposable. 7. How do you refine nickel by Mond’s process? A: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl: The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal: 8. How do you remove oxygen and nitrogen impurities from Zirconium. A: By van Arkel method: The crude metal is heated in an evacuated vessel with iodine. The metal iodide volatilises Zr + 2I2 ZrI4 The metal iodide is decomposed on a tungsten filament. The pure metal is thus deposited on the filament. ZrI4 Zr + 2I2 III. THREE MARK QUESTIONS: 1. Explain the concentration of bauxite ore. A: Bauxite ore is concentrated by leaching. The steps involved are i) Bauxite is concentrated by digesting the powdered ore in a concentrated solution of sodium hydroxide at 473-573 K and 35 bar pressure. Al2O3 is leached as sodium aluminate. ii) Aluminate solution is neutralised by passing CO2. Hydrated Al2O3 is precipitated by seeding. iii) Hydrated Al2O3 is filtered, dried and heated to get pure Al2O3. 2. Write the equations involved in leaching of alumina. A: Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq) 2Na[Al(OH)4](aq) + CO2(g) → Al2O3.xH2O(s) + 2NaHCO3 (aq) Al2O3.xH2O(s) ∆ Al2O3 + xH2O 3. Two reactions are given below, which of these two happen at a temperature above 1073K. Justify. FeO + C Fe + CO --------(A) FeO + CO Fe + CO2--------(B) A: FeO + C Fe + CO happens at a temperature above 1073K. In the Ellingham diagram we can see that the C,CO line goes below while CO,CO2 goes above FeO line at temperature above 1073K. So, C is the reducing agent. 4. How do you extract aluminium from bauxite ore? A: Diagram.
- 32. Aluminium is extractedfrom bauxite ore by Hall-Heroult process. the electrolyte is purified Al2O3 +Na3AlF6 + CaF2 . Na3AlF6 + CaF2 lowers the melting point of the mix and brings conductivity. The fused matrix is electrolysed. Steel cathode and graphite anode are used. The overall reaction may be taken as: 2Al2O3 + 3C 4Al + 3CO2 5. How do you extract iron from roasted haematite ore. Explain with diagram. A: Diagram Iron extraction is carried out in blast furnace and different reactions takes place at different temperatures. Ore, limestone and coke are fed into the furnace. Hot air is blown from the bottom. Coke is burnt to give high temperature. At the top of the furnace at lower temperature iron oxide is reduced to iron by carbon monoxide and at high temperatue at the botton iron oxide is reduced by carbon. Calcium carbonate forms CaO which removes silica as calcium silicate. 6. Name the methods used in the refining of (a) tin (b) copper (c) germanium (d) Titanium A: (a) tin- liquation (b) copper- electrolytic refining (c) germanium-zone refining (d) Titanium-van Arkel refining Or vapour phase refining 7. Explain van Arkel method of refining of zirconium. A: The crude metal is heated in an evacuated vessel with iodine. The metal iodide volatilises Zr + 2I2 ZrI4 The metal iodide is decomposed on a tungsten filament. The pure metal is thus deposited on the filament. ZrI4 Zr + 2I2 8. Explain magnetic separation method of concentration of ore. A: Priniciple: This is based on differences in magnetic properties of the ore and the gangue. The powdered ore is carried on a conveyer belt which passes over a magnetic roller. Magnetic substances stick to the roller while non magnetic substances fall and form a heap. Once the roller moves the magnetic substances come out of the influence of the magnetic roller and fall off forming a separate heap. 9. Give the principle of froth floatation process. How can we separate ZnS and PbS present in an ore using froth floatation process? A: Principle: This is based on the differences in the wetting properties of ore and gangue. ZnS and PbS present in the ore can be separated by using depressants like NaCN. It selectively prevents ZnS from coming to the froth but allow PbS to come with the froth.
- 33. Unit 7 p- BLOCKELEMENTS: 1. Classify the following 1 5 th group p-block elements in to nonmetals/metalloids /metal. 1)Nitrogen 2) Phosphorus 3) Arsenic 4) Antimony 5) Bismuth 1M each Answer: Nonmetals: Nitrogen and Phosphorus Metalloids: Arsenic and Antimony Metal: Bismuth 2. Write the formula of 1M each 1) chile salt petre 2) Indian salt petre 3)apatite mineral 4)chlorapetite 5) Fluorapetite Answer: 1) NaNO3 2) K NO3 3) Ca9(PO4)6 CaX2 4) Ca9(PO4)6 CaCl2 5) Ca9(PO4)6 CaF2 3. Write the valence shell electronic configuration of 15th group elements. 1M Answer; ns2 np3 4. There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed. Give reason. 1M Answer: This is due to the presence of completely filled d and/or f orbital in heavier members. 5. Ionization enthalpy decreases down the group 15. Give reason. 1M Answer: Due to gradual increase in atomic size. 6. The ionization enthalpy of the group 15 elements is much greater than that of group 14 and group 16 elements in the corresponding periods. Give reason. 1M Answer: Because of the extra stable half-filled p orbital electronic configuration and smaller size. 7. How does electronegativity of 15th group elements varies down the group? 1M Answer: decreases 8. Mention the common Oxidation states of p block elements. 1M Answer: common ox.state of these elements are-3,+3 and +5 9. How is stability of oxidation states of 15th group elements varies? 1M Answer: -3 in their covalent compounds. In addition to the −3 state, N and P also show −1 and −2 All the elements present in this group show +3 and +5 oxidation states. Stability of +5 ox. State decreases and that of +3 ox. State increases due to inert pair effect. 10.Nitrogen atom has five valence electrons but it does not form NCl5. 1M Answer: Because of absence of d-orbitals it can’t expand its covalency from 3 to 5. 11.Nitrogen does not form pentahalides.Why? 1M Answer: Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalide. 12.Why is Nitrogen an inert gas? 1M Answer: Nitrogen exists as triply bonded diatomic non polar molecule. Due to short internuclear distance between two nitrogen atoms the N ≡ N bond strength is very high. It is, therefore, very difficult to break the bond. 13.Why nitrogen exhibits anomalous behavior? 2M Answer: Due to smaller size, high electronegativity, high ionisation enthalpy and non- availability of d-orbitals, nitrogen shows anomalous behavior. 14.Mention any three anomalous properties of nitrogen. 3M Answer; (any three of the following) 1)Nitrogen forms pπ – pπ bonds where as other members not. 2) nitrogen exists as diatomic molecule with a triple bond 3)The single N−N bond is weaker than P−P bond due to small bond length. 4) lower catenation tendency. 5)cannot form dπ – pπ bonds like phosphorus. 15.Why R3P=O exist but R3N=O does not? 1M Answer: Due to the absence of d orbitals in valence shell of nitrogen, nitrogen cannot form d π–p π bond. Hence R3N=O does not exist. 16.Catenation property of nitrogen is less than phosphorus. Why? 1M Answer: Due to strong pπ–pπ overlap in Nitrogen and weaker N-N bond than the single P- P bond. 17.Write the formula of hydrides formed by 15th group elements? 1M Answer: EH3 18.How does the stability of 15th group metal hydride varies down the group? 1M Answer:The stability of hydrides decreases on moving down from NH3 to BiH3. 19.Why is NH3 basic while BiH3 is only feebly basic. 1M Answer: NH3 is basic due to smaller size & high electro negativity of Nitrogen. 20.Ammonia has higher boiling point than Phosphine. Explain. 1M Answer: Ammonia (NH3) form hydrogen bond but Phosphine (PH3) does not. Hence boiling point of ammonia is higher than that of phosphene. 21.Write the formula of two types of oxides formed by 15th group elements? 1M Answer: E2O3 and E2O5 22.Out of E2O3 and E2O5 which is acidic? 1M
- 34. Answer: E2O5 (oxide with higher oxidation state is more acidic) 23.How does the acidic characters of 15th group metal oxides varies down the group? 1M Answer; The acidic character decreases on moving down a group. 24.Write the increasing order of acidic character of N2O5, P2O5, As2O5 and Sb2O5 1M Answer: N2O5 >P2O5 >As2O5 > Sb2O5 25.How is dinitrogen prepared in the laboratory? 2M Answer: In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite. NH4Cl (aq) + NaNO2 (aq) → N2 (g) + 2H2O (l) + NaCl (aq) 26.How is dinitrogen prepared from ammonium dichromate? 2M Answer: thermal decomposition of ammonium dichromate gives dinitrogen. (NH4)2Cr2O7 → N2 + 4H2O + Cr2O3 27.How does dinitrogen reacts with Mg? 2M Answer: Dinitrogen reacts with Mg to form magnesium nitride. N2 +3 Mg → Mg3N2 28.For the manufacture of ammonia by Haber’s process, write flow chart and balanced equation along with conditions? 3M Answer: On large scale, obtained by Haber’s process Optimum condition: Pressure = 200 × 105 Pa (about 200 atm) Temperature ∼ 700 K Catalysts used − Iron oxide with small amounts of K2O and Al2O3 to increase the rate of attainment of equilibrium. 29.How does ammonia react with zinc sulphate? 2M Answer: Ammonia reacts with zinc sulphate to form white precipitate of zinc hydroxide. ZnSO4(aq) + 2NH4OH(aq) → Zn(OH)2(s) + (NH4)2SO4(aq) 30.How does ammonia react with cupric ion? 2M Answer: With Cu2+ ion Ammonia acts as lewis base and forms deep blue colored cuprammonium complex. Cu2+ (aq) + 4NH3(aq) → [Cu(NH3)4]2+ (aq) (blue) (deep blue) 31.How is Nitric acid manufactured by Ostwald process? 3M Answer: Nitric acid in Ostwald process manufactured by the oxidation of ammonia. Nitric oxide thus formed combines with oxygen giving NO2. 2NO ( g ) + O2 ( g )→2NO2 ( g ) Nitrogen dioxide so formed, dissolves in water to give HNO3. 3NO2 ( g ) + H2O ( l ) → 2HNO3 ( aq ) + NO ( g ) Dilute nitric acid on distillation followed by dehydration using conc. sulphuric acid gives 98% nitric acid. 32.How is nitric acid prepared in laboratory? 2M Answer: Nitric acid is prepared in the laboratory by heating KNO3 or NaNO3 with concentrated H2SO4 in glass retort. NaNO3 + H2SO4 → NaHSO4 + HNO3
- 35. 33.How does dilutenitric acid with copper? 2M Answer: Dil. Nitric acid reacts with copper to form cupric nitrate with the liberation of nitric oxide. 3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O 34.How does concentrated nitric acid with copper? 2M Answer: Conc.Nitric acid reacts with copper to form cupric nitrate with the liberation of nitrogen dioxide. Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O 35.How does dilute nitric acid with zinc? 2M Answer: Dil. Nitric acid reacts with zinc to form zinc nitrate with the liberation of nitrous oxide. 4Zn + 10HNO3(dilute) → 4 Zn (NO3)2 + 5H2O + N2O 36.How does concentrated nitric acid with zinc? 2M Answer: Conc.Nitric acid reacts with zinc to form zinc nitrate with the liberation of nitrogen dioxide. Zn + 4HNO3(conc.) → Zn (NO3)2 + 2H2O + 2NO2 37.How does concentrated nitric acid with iodine? 2M Answer: Conc.Nitric acid oxidizes iodine to form iodic acid. I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O 38.How does concentrated nitric acid with carbon? 2M Answer: Conc.Nitric acid oxidizes carbon to carbon dioxide C + 4HNO3 → CO2 + 2H2O + 4NO2 39.What is passivity? 1M Answer: Some metals like aluminium and chromium do not dissolve in concentrated nitric acid due to the formation of a protective layer of oxide on the surface of the metal. This phenomena is called passivity of metals. 40.explain Brown ring test with equations . 3M Answer: Dilute FeSO4 solution is added to an aqueous solution of nitrate ion. concentrated H2SO4 is then added along the sides of the test tube. A brown ring is observed at the interface between the solution and H2SO4 layers indicates the presence of nitrate ion in the solution. NO3 - + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O [Fe (H2O)6 ]2+ + NO → [Fe (H2O)5 (NO)]2+ + H2O (brown) 41.Write the resonance structurs of a) NO b) NO2 c) N2O5 1M each Answer: a) Structure of NO: b) Structure of NO2 : c) Structure of N2O5 : 42.Distinguish between white and red phosphorus. 2M Answer: (any two) White phosphorus Red Phosphorus It is a soft and waxy solid. It is a hard and crystalline solid. It is poisonous. It is non-poisonous. It is insoluble in water but soluble in carbon disulphide. It is insoluble in both water and carbon disulphide. Highly reactive It is relatively less reactive. In both solid and vapour states, it exists as a P4 molecule. It exists as a chain of tetrahedral P4 units. Less stable More stable 43.How is Phosphine prepared in the laboratory form white phosphorous? 2M Answer: In the laboratory phosphene is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2. P4 + 3NaOH + 3H2O → PH3 + 3NaH2 PO2 44.Give a reaction to support Basic nature of phosphine. 2M Answer: Phosphine react with hydrogen bromide and forms phosphonium bromide. PH3 + HBr → PH4 Br 45.How is Phosphorous trichloride is obtained from phosphorous and chlorine? 2M Answer: Phosphorus tri chloride is obtained by passing dry chlorine over heated white phosphorus. P4 + 6Cl2 → 4PCl3 46.How is Phosphorous pentachloride is obtained from phosphorous and chlorine? 2M
- 36. Answer: Phosphorus pentachlorideis prepared by the reaction of white phosphorus with excess of dry chlorine. P4 + 10Cl2 → 4PCl5 47.How does Phosphorous trichloride react with water? 2M Answer: PCl3 hydrolyses in the presence of moisture to give phosphorus acid. PCl3 + 3H2O →H3PO3 + 3HCl 48.How does Phosphorous pentachloride react with water? 2M Answer: PCl5 in presence of waterhydrolyses to POCl3 and finally gets converted to phosphoric acid. PCl5 + H2O → POCl3 + 2HCl POCl3 + 3H2O → H3PO4 + 3HCl 49.Write the formula, structure ,reducing property and basicity of Hypophosphorus acid. 3M Answer: Formula- H3PO2 Reducing property: Reducing agent as it contains two P – H linkage structure: Basicity: one( as it contains only one P-OH linkage) 50.Write the formula, structure ,reducing property and basicity of Orthophosphorus acid. 3M Answer: Formula- H3PO3 Reducing property: Reducing agent as it contains one P – H linkage structure: Basicity: Two( as it contains two P-OH linkage) 51.Write the formula, structure ,reducing property and basicity of Orthophosphoric acid. 3M Answer: Formula- H3PO4 Reducing property: Not a Reducing agent as it does not have P – H linkage structure: Basicity: Three( as it contains three P-OH linkage) 52.How do you account for the reducing behavior of H3PO2 on the basis of its structure? 1M Answer: In H3PO2, two H atoms are bonded directly to P atom which imparts reducing character to the acid. 53.Classify the following 1 6 th group p-block elements in to nonmetals/metalloids / metal. 1)Oxygen 2) Sulphur 3) selenium 4) Tellurium 5) Polonium 1M each Answer: Nonmetals: Oxygen and Sulphur Metalloids: selenium and Tellurium Metal: Polonium 54.Name the 16th group p-block element which is radioactive in nature. 1M Answer: Polonium 55.Write the valence shell electronic configuration of 16th group elements. 1M Answer; ns2 np4 56.Mention the Oxidation state of oxygen. 1M Answer; Oxygen exhibits the oxidation state of−2 in metal oxides, −1 (H2O2), zero (O2 and O3) and +2 (OF2). 57.Write a note on Anomalous Behavior of Oxygen. 3M It is due to its Small size, High electronegativity and absence of d-orbitals oxygen shows anomalous properties. 1. Strong hydrogen bonding is present in H2O, which is not found in H2S. 2. Absence of d-orbitals in oxygen limits its covalence to four and in practice rarely exceeds two. On the other hand, other elements of the group can expand their covalence beyond four.
- 37. 58.How is Oxygenobtained from potassium chlorate? 2M Oxygen can be obtained by heating potassium chlorate in presence of MnO2. 59.Write the chemical equation for the reaction of Oxygen with aluminum /carbon /methane. 1M each Answer: 60.What are acidic oxides? What type of oxides are acidic in nature? Give example. 3M Answer: Acidic oxides are those oxides which combine with water to give an acid. Non-metal oxides and Oxides of some metals in higher oxidation state are acidic in nature. Example for non metal acidic oxides− SO2, Cl2O7, CO2, N2O5 Examples for metal oxides which are acidic − Mn2O7, CrO3, V2O5 61.What are basic oxides? What type of oxides are basic in nature? Give example. 3M Answer: Basic oxides are those oxides which combine with water to give bases. Metal oxides are basic in nature. Examples for metal oxides which are basic- Na2O, CaO, BaO 62.What are amphoteric oxides? Give example. 2M Answer: Amphoteric oxides are those oxides which show the characteristics of both acidic as well as basic oxides . Example − Al2O3 63.Illustrate amphoteric nature of Al2O3 with suitable reactions. 2M Answer: 64.What are neutral oxides? Give example. 2M Answer: Neutral oxides arethose oxides which are neither acidic nor basic. Examples − CO, NO, N2O 65.How is Ozone prepared from oxygen? 2M Answer: A slow dry stream of oxygen is passed through a silent electrical discharge. Oxygen partially gets converted into ozone. 66.Why is high concentrations of ozone can be explosive? 2M Answer: High concentrations of ozone can be explosive because the decomposition of O3 to O2 results in the liberation of heat (ΔH = − ve) and an increase in entropy (ΔS = + ve), leading to large negative value of ΔG. 67.How does ozone react with PbS? write equation. 2M Answer: Ozone oxidizes lead sulphide to lead sulphate. PbS + 4O3 → PbSO4 + 4 O2 68.How does ozone react with NO? write equation. 2M Answer: Nitric oxides reacts with ozone to give nitrogen dioxide and oxygen O3 + NO → NO2 + O2 69.In the preparation of H2SO4 by Contact Process, why is SO3 not absorbed directly in water to form H2SO4? 1M Answer: SO3 is not dissolved in water directly as the process is highly exothermic & the H2SO4 obtained is in the form of a mist which cannot be condensed easily. 70.Which form of the sulphur is stable at room temperature? 1M Answer:Rhombic sulphur. ( α sulphur) 71.Which form of the sulphur is stable above 369K? 1M Answer:Monoclinic sulphur( β sulphur) 72.Explain the laboratory method of preparation of SO2 from 2 3SO . 2M Answer: sulphites are treated with dil H2SO4 to get SO2 i) SO3 - (aq) +2H+ (aq) H2O +SO2 73., What happens when Sulphrur dioxide is treated with (2 Marks) i)NaOH ii) Cl2 Answer:i) 2NaOH +SO2 Na2SO3 +H2O Na2SO3 + H2O 2NaH SO3 ii) SO2+ Cl2 SO2Cl2( Salphuryl chloride) 74.Give any two reactions to show that SO2 is a reducing agent. 2M Answer: 2Fe3+ + SO2+ 2 H2O2Fe2+ + SO4 2- +4H+ 5SO2+2MnO4 - + 2H2O 5SO4 2- +4H+ +2Mn2+ 75.How is the presence of SO2 detected? Answer: SO2 discharges pink colour of KMnO4 due to the reaction 5SO2+2MnO4 - +2H2O 5SO4 2-+ 4H+ +2Mn2+ 76.Draw the structure of i) Sulphurus acid ii) Sulphuric acid (iii)peroxo sulphuric acidiv) pyrosulphuric acid( oleum). 1M each
- 38. Answer:(i) (ii) (iii)(iv) 77.Name the catalyst used in the manufacture of sulphuric acid by contact process 1M Answer: V2O5 78.Write chemical equations in the manufacture of sulphuric acid by contact process with the conditions required. (3 Marks) Answer: 2SO2 +O2 2SO3 At 720K, temperature & 2 bar pressure. SO3 + H2SO4 H2S2O7 79.Explain the manufacture of H2SO4 by contact process from purified SO2. 3M Answer: Purified SO2 is passed through catalytic converter containing V2O5 at 720K,and 2 bar pressure. SO3 is obtained. 2SO2 +O2 2SO3 SO3 obtained is dissolved in to get oleum in absorption tower. SO3 + H2SO4 H2S2O7 Oleum is carefully diluted with water to get sulphuric acid. 80.Draw the flow chart for manufacture of H2SO4 by contact process 2M 81.What happens when Concentrated H2SO4 is added to 2M Each i)CaF2. ii) Sugar. Answer: i) CaF2+ H2SO4 CaSO4 + 2HF ii) C12H22O11 12C + 11 H2O( Charring of sugar- Dehydrating property) 82.Give an example to show that Conc H2SO4 is a strong oxidizing agent. 1M Each Answer: Cu + 2 H2SO4( Hot ,Conc) CuSO4 + SO2 + 2H2O 3S + 2 H2SO4( Hot ,Conc) 3 SO2 + 2H2O C + 2 H2SO4( Hot ,Conc) CO2+ 2SO2 + 2H2O 83.Name the halogens 1M Answer:Flourine, Chlorine,bromine,iodine,asyatine. 84.Which is the radioactive halogen? 1M Answer: Astatine 85.Name the halogen present in sea weeds. 1M Answ: Iodine. 86.Write the outermost electronic configuration of halogens. 1M Answer: ns2 np5 . 87.Give reason (1 M each) i) Halogens have very high ionization enthalpy in the corresponding period. ii) Halogens have Maximum negative electron gain enthalpy in the corresponding period iii) Negative electron gain enthalpy of fluorine is less than that of chlorine. iv) Enthalpy of dissociation of F2 is less than Cl2. v) Fluorine is stronger oxidizing agent than chlorine. vi) Fluorine exhibits only -1 oxidation state.
- 39. Answer:i) Due tothe ns2 np5 configuration, they have little tendency to loose electrons. ii) They have only one electron less than the stable noble gas configuration. iii) Due the very small size of fluorine atom. iv) Due the very small size of fluorine. v) Due to the high electro negativity of fluorine atom it readily accepts an electron. vi) Due to non availability of d- orbital. 88.Write the chemical equation 1M Each i) When F2 is treated with Cl- ,Br- & I- ii) When Cl2 is treated with Br- & I- iii) When Br2 is treated with I- iv) When F2 is treated with H2O v) When Cl2 is treated with H2O Answer: i) F2 +2X- 2F- +X2 ( X= Cl, Br, or I) ii) Cl2 +2X- 2Cl- +X2 ( X= Br, or I) iii) Br2 +2I- 2Br- +I2 ( X= Cl, Br, or I) iv) 2F2 + 2H2O 4H+ (aq) +4F- (aq) +O2. v) 2Cl2 + 2H2O 4HCl(aq) +HOCl(aq) 89.Mention the three reasons for the anomalous behavior of fluorine. 3M Answer: Due to its small size, highest electro negativity, low F—F bond dissociation enthalpy & non- availability of d- orbitals in the valence shell of fluorine. 90.Give any three examples to show anomalous behavior of fluorine. 3M. Answer: i) ionisation enthalpy, electronegativity, electrode potential are higher for F ii) Ionic & covalent radii, m.pt, b.pt, bond dissociation enthalpy,electron gain enthalpy lower than expected. iii) F forms only one halo acid iv) HF is liquid, other hydrogen halides are gases. 91.How is chlorine prepared from KMnO4. Write the chemical equations involved. 2M Answer: By the action of HCl on KMnO4, 2KMnO4 + 16 HCl 2KCl +2MnCl2 + 8H2O +5Cl2 92. What happens when Concentrated chlorine is treated with i)Alluminium ii) sulphur S8 iii) H2S iv) excess of NH3 v) cold & dilute NaOH vi) hot & concNaOH vii) Dry slaked lime. 1M each Answer: (i)with Al: 2Al + 3 Cl2 2AlCl3 (ii) with S: S8 +4 Cl2 2S2Cl2 (iii) With H2S: H2S +Cl2 2HCl+S (iV) With NH3: 8NH3 +3Cl2 6NH3Cl + N2 ( excess) NH3 + 3Cl2 6NCl3 + 3HCl ( excess) (v) With NaOH: 2NaOH+Cl2 NaCl + NaOCl+H2O ( cold & dil) (hypochlorite) (vi) 6NaOH+3Cl2 5NaCl + NaOCl3+3H2O ( hot & conc) ( chlorate) (vii) With Ca(OH)2 : 2Ca(OH)2+2Cl2Ca(OCl)2+CaCl2 +2H2O (dry slaked lime) ( bleaching powder) 93.Give any one example for oxidizing property of chlorine with FeSO4, Na2SO3. (2M each) Answer:2 FeSO4+H2SO4 + Cl2 Fe2(SO4)3+ 2HCl ( Ferrous) ( Ferric) Na2SO3 + Cl2+H2O Na2SO4 + 2HCl ( Sulphite) ( Slphate) 94.Give the reason for the bleaching action of chlorine. 1M Answer: Due to the oxidation Cl2+H2O 2HCl + O Coloured sub + O Colourles substance. 95.Give the composition of bleaching powder. 1M Answer: Ca(OCl)2.CaCl2. Ca(OH)2 .2H2O. 96.How is HCl is prepared in the laboratory? 2M Answer: NaCl + H2SO4 NaHSO4+ HCl at 420K NaHSO4+ NaCl Na2SO4 + HCl at 823K HCl is dried using Conc H2SO4 97.Give the composition of aqua regia? Write the ionic equation when it is treated with gold/ platinum. 3M Answer: Aqua regia: 3:1 part conc HCl & conc HNO3
- 40. Dissolves noble metals Au+4H+ +NO3 - +4Cl- AuCl4 - +NO + 2 H2O 3Pt+16H+ +4NO3 - +18Cl- 3PtCl6 - +4NO + 8 H2O 98.What happens when hydrochloric acid is treated with NH3 1M Answer: NH3 +HCl NH4Cl ( White fumes) 99.Write the structure of i) Hypochlorus acid ii) Chlorus acid iii) Chloric acid iv) Perchloric acid. 1Meach Answer: Answer: 100. What are interhalogen compounds? Give an example. Why they are more reactive than individual halogen. 3M Answer:When two different halogen atoms react inter halogen compounds are formed. Eg: ClF3, ICl, BrF5 ,lF7 Reactivity is more compared with halogens because X- X’ bond is weaker than X-X bond in pure halogens. 101. How is following interhalogen compound prepared? i) ClF3 ii) ICl3 iii) BrF5 1Meach Answer: 473K i) Cl2+F2 2ClF3 ii) I2+Cl2 2ICl iii) Br2+5F2 2BrF5 ( excess) 102. Name i) the radioactive noble gas ii) most abundant noble gas. 1M Answer: i) Radon ii) Argon 103. Why noble gases are chemically inert? 1M Answer: Stable completely filled orbitals are there. 104. Why noble gases have maximum ionization enthalpy in the corresponding period. 1M Answer: Stable completely filled orbitals are there 105. Why noble gases have positive electron gain enthalpy . 1M Answer:Stable completely filled orbitals are there 106. Which is the first noble gas compound synthesized? 1M Answer: Xe+ PtF6 - 107. Who prepared first noble gas compound? 1M Answer: Neil Bartlett 108. Write the chemical equations to prepare following compounds with the conditions required. i) XeF6 ii) XeO3 iii) XeO2F2. 1M each 573K, 60-70 bar Answer: i) Xe(g) + 3F2(g) 3XeF6(s) ii) XeF6 + 3H2O XeO3 + 6HF iii) XeF6 + 2H2O XeO2F2 + 4HF 109. Write/ Name the structure of i) Xe F2 ii) XeF4 iii) XeF6 iv) XeOF4 v) XeO3. 1M each Answer: i) linear ii) sqare panar iii) Distorted octa hedral iV) Square pyramidal) trigonal pyramidal 110. Noble gases have very low boiling point .Why? 1M Answer:They are mono atomic due to weak dispersion forces, hence have low boiling points.
- 41. Unit 8 The d-and f- Block Elements I. Answer the following questions. Each question carries one mark 1. Define transition elements. Ans. Transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of its oxidation states. 2. What is the position of the d block elements in the periodic table? Ans. The d block elements are in the middle of s and p blocks, comprising the groups 3 to 12. They are the four rows of elements in the periods 4th (3d series), 5th (4d series), 6th ( 5d series) and 7th ( 6d series). 3. Zinc, cadmium and mercury of group 12 are not regarded as transition metals, Why ? Ans. Zinc, cadmium and mercury of group 12 have full d10 configuration ( d orbitals are completely filled ) in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals 4. Why d- block elements are named as ‘transition elements ‘ ? Ans. The d–block elements occupies the middle of the periodic table and their properties are transitional between s– and p– block elements. 5.Write the general electronic configuration of d block elements. Ans. [ Noble gas] (n-1)d1-10ns1-2 6. Write the general outer electronic configuration of d- block elements. . Ans. The general outer electronic configuration of d- block elements is (n-1)d1–10 ns1–2 7. Write the general electronic configuration of f- block elements. Ans. The general electronic configuration of f- block elements (Lanthanoids) is [Xe] 4f1 – 14 5d0- 1 6s2 8.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. Ans. Cerium 9. The outer electronic configuration of Cr is 3d5 4s1 instead of 3d44s2, why? Ans. Half filled (3d5) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital. 10. The outer electronic configuration of Cu is 3d10 4s1 instead of 3d94s2 , why? Ans. Completely filled (3d10 ) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital. 11. Account for high melting point and boiling points of transition metals. Ans. The melting and boiling points of transition metals are high because of the involvement of greater number of electrons from (n-1)d orbitals in addition to the ns electrons in the inter atomic metallic bonding. 12.What is the trend in melting points of transition metals in a series? Ans.The melting points of the transition metals in a series rise to a maximum at the middle of the series (i.e. Cr or Mo or W - element with d 5 configuration ) and fall regularly as the atomic number increases. 13.Why do transition metals have higher enthalpies of atomization? Ans. Involvement of a large number of unpaired electrons of d orbitals favour stronger inter atomic interactions resulting in stronger bonds between the atoms of a metal and higher enthalpies of atomization. 14.Name one 3d series elements, that do not show variable oxidation states. Ans. Sc (+3) 15.Transition metals exhibit variable oxidation states in its compounds, why? Ans. Transition metals exhibit variable oxidation states in its compounds due to the availability of both ns & (n – 1 ) d electrons for bond formation. 16. Name 3d series metal which shows highest oxidation state. Ans. The highest oxidation state shown by 3d series transiNa tion metals is +7 by Mn 17. Name a metal in the 3d series of transition metals which exhibit +1 oxidation state most frequently. Ans. copper 18.What is the trend in oxidation state of transition metals ? Ans. The oxidation state increases with increase in atomic number & reaches a maximum in the middle and then decreases. 19. 3d series transition metals exhibit +2 as the most common oxidation state (except Sc) why? Ans. The +2 oxidation state, which commonly occurs for nearly all the transition metals is due to the loss of their outer 4s electrons
- 42. 20. Why transitionmetals and their compounds shows paramagnetic behavior ? Ans. The transition metal ions are generally containing one or more unpaired electrons in them & hence their compounds are generally paramagnetic. 21. Name an of alloys of transition metals with non transition metals. Ans. Brass ( Cu & Zn) or Bronze ( Cu & Sn) 22.What is the action of neutral or faintly alkaline permanganate solution on iodide ? Ans. Alkaline permanganate solution oxidize iodide to iodate. 23. What happens when potassium permanganate is heated to 513 K ? Ans. Potassium permanganate decomposes at 513K to potassium manganate, manganese dioxide and oxygen. 24. What is the principal oxidation state exhibited by the lanthanoids? Ans. The principal oxidation state of lanthanoids is +3. 25.Write the spin-only formula used to calculate the magnetic moment of metal ions. Ans. The magnetic moment is determined by using the spin only formula, where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). 26. Why is Sc3+ (or Zn2+ ) diamagnetic? Ans. Sc3+(Z=21) 3d0 no unpaired electron, n=0, μ=0. (or Zn2+(Z=30) 3d10 no unpaired electron, n=0, μ=0) 27. What is the most common oxidation state of lanthanoids and actionoids? Ans. The most common oxidation state of lanthanoids and actionoids is +3. 28.What is Actinoid contraction? Ans. There is a gradual decrease in the size of atoms or M3+ ions across the series. This is known as the actinoid contraction. 29.Actinoid contraction is more than lanthanoid contraction. Give reason. Ans. The actinoid contraction is, more than lanthanoid contraction due to poor shielding by 5f electrons from nuclear charge. 30. Actionoids show larger number of oxidation states than lanthanoids. Why? Ans. In actinoids 5f, 6d and 7s levels are of comparable energies ,hence electrons from these orbitals are available to lose or share. 31. Give one use of Mischmetall . Ans. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. 32. Why transition metals forms alloys readily? Ans. Transition metals readily form alloys with other transition metals because of their similar radii 33. Give one use of transition metal alloy. Ans.Ferrous alloys containing chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels. II. Answer the following questions. Each question carries TWO marks. 34. Name two characteristic properties exhibited by d – block elements due to their partly filled d orbitals. Ans. The characteristic properties exhibited by d – block elements due to their partly filled d orbitals are variable; (i) Oxidation states (ii) Formation of coloured ions. 35. Name two typical metallic properties displayed by transition elements. Ans.High tensile strength, ductility malleability, high thermal and electrical conductivity and metallic luster etc. 36. What are interstitial compounds? Give example. Ans. Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of transition metals. Example; TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. 37. Give any two physical characteristics of interstitial compounds. Ans.Two physical characteristics of interstitial compounds are: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard and they retain metallic conductivity. 38.Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27). Ans. M (z= 27 , 3d7 4s2) M+2 (3d7 4s0) hence it has 3 unpaired electrons n= 3 = 3(3+2) = 3.87 BM 39. The second ionisation enthalpy is high for Cr and Cu , why? Ans: The second ionisation enthalpy is unusually high values for Cr and Cu because when M+ ion ionize to M+2 ion , the d5 and d10 configurations of the M+ ions (i.e Cr+ or Cu+) are disrupted, with considerable loss of exchange energy .
- 43. 40. Why firstionisation enthalpy of Cr is lower than that of Zn ? IE1 of Cr is lower, because removal of an electron from Cr does not change the d (3d5 4s1 to 3d5 4s0 ) configuration . Cr (z= 24 , 3d5 4s1) Cr+ (3d5 4s0) ------ IE1 IE1 value for Zn is higher, because removal of electron from 4s level needs more energy. Zn (z= 30 , 3d10 4s2) Zn+ (3d10 4s1) ------ IE1 IE1 (Zn) > IE1 (Cr) 41.Give two characteristics of transition metal alloys. . The alloys are hard and have high melting points. 42.What is the action of heat on potassium permanganate ? Give equation. Ans. It decomposes at 513K to potassium manganate, manganese dioxide and oxygen. 2KMnO4 K2MnO4 + MnO2 + O2 43. What is the action of neutral or faintly alkaline permanganate solution on iodide ?Give equation. Ans. Alkaline permanganate solution oxidize iodide to iodate. I- + 2MnO4- + H2O IO3- + 2MnO2 + 2OH- 44. Explain the oxidising action of acidified potassium dichromate on (iron(II) salts) Fe +2 ions and write the ionic equations for the reaction. Ans. Acidified potassium dichromate oxidises iron(II) salts to iron(III). Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O 45. The transition metals generally form coloured compounds, why? Ans. The compounds of transition elements shows colour due to presence of unpaired electron & ability to undergo d-d transition. When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. 46. Give reason “ transition metals and their many compounds acts as good catalysts”. Ans. Transition metals and their many compounds acts as good catalysts,it is due to (i) partially filled (n-1) d orbital (ii)variable oxidation state and provide a suitable surface for the reaction to take place. 47.Explain giving reason “transition metals form a large number of complex compounds”. Ans.Transition metals form a large number of complex compounds due to (i) Small size & high charge density of the ions of transition metals. (ii) presence of vacant d orbitals of suitable for bond formation. 48. What is the effect of increasing pH on a solution of potassium dichromate? Ans.On increasing the pH of the solution Potassium dichromate (orange) becomes potassium chromate (yellow) Cr2O72- + 2OH- → 2CrO42- + H2O 49.What happens when H2S is passed into potassium dichromate in acidic medium? Give the equation. Ans. H2S gets oxidized to sulphur Cr2O7 −2 + 14H+ + 6e− → 2Cr+3 + 7H2O 3H2S → 6H+ + 3S + 6e− Cr2O7−2 + 3H2S + 8H+ → 2Cr+3 + 7H2O + 3S 50. What is ‘disproportionation’ of an oxidation state ? Give one example of disproportionation reaction in aqueous solution. Ans. A particular oxidation state , which is relatively less stable compared to other oxidation states , under goes disproportion. Manganese (VI) which is relatively less stable changes over to manganese (VII) and manganese (IV) in acid solution. 3 MnO4-2 + 4H+ MnO2 +2MnO4-+ 2 H2O 51. What is lanthanoid contraction? Write any one consequence of lanthanoid contraction. Ans. Steady decrease in the size of lanthanides with increase in atomic number is known as lanthanoid contraction. Due to lanthanoid contraction radii of members of 3rd transition series are very much similar to corresponding members of 2nd series. 52.Write any two consequences of lanthanoid contraction. Ans. Two consequences of lanthanoid contrations are (i) The radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. Ex. The almost identical radii of Zr (160 pm) and Hf (159 pm) & Nb (146pm) & Ta (146pm) (ii) Difficulty in separation of lanthanoids due to similarity in chemical properties. 53. Name the two series of f-block. Ans. The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium.
- 44. 54. The chemistryof actionoids is more complicated than lanthanoids. Why? Ans. The actinoids are radioactive elements having half lifes varying. Some members can be prepared only in nanogram quantities. These facts render their study more difficult. 55.Write two comparisons of variability in oxidation states of transition metals and non transition elements (p- block elements) ? Ans. 1. In transition elements , variable oxidation state differ from each other by unity, whereas in case of non transition elements , oxidation state differ by units of two.(For example Fe exhibits o.s of +2 and +3 . similarly copper exhibits two o.s of +1 and +2 . on the other hand, Sn, Pb exhibit o.s of +2 and +4.) 2. In transition elements, higher o.s are more favoured in elements of higher atomic mass, whereas in p-block elements lower o.s are favoured by heavier members ( due to inert pair effect, For example Mo(VI) and W(VI) are more stable than Cr(VI). On the other hand Pb(II) is more stable than Sn(II)) 56. What happens when (a) A lanthonoid reacts with dilute acids ? (b) A lanthonoid reacts with water? Ans.(a) When lanthonoid reacts with dilute acids , it liberates hydrogen gas. (b)When lanthonoid reacts with water , it forms lanthanoid hydroxide and liberate hydrogen gas. 57. What is the gas liberated when i) crystals of potassium permanganate is heated to 513K ? ii) acidified potassium permanganate is treated with oxalate ion at 333K? Ans. i) When crystals of potassium permanganate is heated to 513K Oxygen (O2) gas is liberated . ii) Acidified potassium permanganate when treated with oxalate ion at 333K liberates Carbon dioxide (CO2) gas. 58. What is the composition of mischmetall? Give its one use. Ans. The composition of mischmetall is lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint 59. Show the interconversion of chromate and dichromate The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. At pH less than 7: Chromate (yellow) on adding acid becomes dichromate (orange) 2CrO42- + 2H+ → Cr2O72- + H2O At pH more than 7 : Dichromate (orange) on adding base becomes Chromate (yellow) Cr2O72- + 2OH- → 2CrO42- + H2O 60.How does the neutral or faintly alkalline potassium permanganate solution react with (a) Iodide (b) thiosulphite? Write the ionic equations for the reactions In neutral or faintly alkaline solutions: (a) The oxidation of iodide to iodate: 2MnO4− + H2O + I− ——> 2MnO2 + 2OH− + IO3− (b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4– + 3S2O32– + H2O ——> 8MnO2 + 6SO42– + 2OH– III. Answer the following questions. Each question carries THREE marks 61.Name the metal of the 1st row transition series that i) has highest value for magnetic moment ii) has zero spin only magnetic moment in its +2 oxidation state. iii) exhibit maximum number of oxidation states. Ans. i) Chromium ii) Zinc iii) Manganese 62.Transition metals form a large number of complex compounds.Give reason. Ans. Transition metals for complex compounds due to, i) small sizes of metal cations ii) their ionic charges and iii) availability of d orbitals for bond formation. 63.Explain the trend in atomic size of 3d series of transition elements with reason. Ans. With increase in atomic number in 3d series - atomic size decreases (Sc to Cr) , then remain almost constant (Cr to Cu) and increase slightly at the end (Cu to Zn). Reason: In the beginning of the series the screening (or shielding effect) effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases , hence atomic size radius decreases. In the middle of the series, increase in nuclear charge and increase in screening effect balance each other. So atomic radii become almost constant. Increase in atomic radii towards the end is due to the electro – electron repulsions causes the expansion of electron cloud. 64. Explain trend in Ionisation Enthalpies of 3d series of transition elements . Ans.Ionisation enthalpy increase along each series of the transition elements from left to right. However many small variations, IE of Chromium is lower because removal of an electron from Chromium does not change in the d (3d5 4s1 to 3d5 4s0 ) configuration. I.E value for Zn (3d10 4s2) is higher because an electron is removed from 4s level which needs more energy. 65. How is potassium dichromate prepared from iron chromite ore? Ans.Potassium dichromate is manufactured from chromite ore (FeCr2O4).
- 45. (i) Chromite oreis fused (FeCr2O4) with sodium or potassium carbonate in free access of air to get sodium chromate . 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 (ii) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised. 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O Potassium dichromate prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Orange crystals of potassium dichromate crystallise out. 66. Describe the preparation of potassium permanganate from manganous dioxide. Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 3MnO42– + 4H+ → 2MnO4– + MnO2 + 2H2O 67.How does the acidified permanganate solution react with (a) iron(II) ions (b) oxalic acid and (c) hydrogen sulphide ? Write the ionic equations for the reactions In acid solutions: (a) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe2+ + MnO4– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+ (b) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O42− + 2MnO4- + 16H+ ——> 2Mn2+ + 8H2O + 10CO2 (c) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H+ + S2− 5S2− + 2MnO4− + 16H+ ——> 2Mn2+ + 8H2O + 5S UNIT 9 Topic: Coordination Compounds 1. State the postulates of Werner’s theory of coordination compounds. 3 Marks Postulates: 1. Central metal ion in a complex shows two types of valences - primary valence and secondary valence. 2. The primary valence is ionisable and satisfied by negative ions. 3. The secondary valence is non ionisable. It is equal to the coordination number of the central metal ion or atom. It is fixed for a metal. Secondary valences are satisfied by negative ions or neural molecules (ligands). 4. The primary valence is non directional. The secondary valence is directional. Ions or molecules attached to satisfy secondary valences have characteristic spatial arrangements. Secondary valence decides geometry of the complex compound. 2. What are the limitations of Werner’s theory of coordination compounds? 3 Marks This theory fails to explain why, a) a few elements have the property to form coordination compounds b) the bonds in coordination compounds have directional properties c) coordination compounds have characteristic magnetic and optical properties. 3. Write one difference between double salts and complex salts with respect to their ionisation. Give one example for each type of salt. 2 Marks Double salt Complex salt Double salt is stable only in solid state, but dissociate into simple ions completely in solution state. Complex salt is stable both in solid and solution state and does not dissociate completely in solution state. E.g: KCl.MgCl2.6H2O E.g: K4[Fe(CN)6] 4. Define Coordination entity of coordination compounds. 2 Marks A coordinate entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules (ligands). e.g: [Fe(CN)6]4-.
- 46. 5. What iscentral metal ion in a coordination compound? Give an example. 2 Marks The metal atom or ion in a coordination entity to which, a fixed number of ions or molecules (ligands) are bound in a definite geometrical arrangement around it is called central metal ion. It is a Lewis acid. e.g: In K4[Fe(CN)6] , central metal ion is Fe2+. 6. What are ligands? Give an example. 2 Marks The ions or molecules bonded to the central metal atom or ion in a coordination entity are called ligands. Ligands are Lewis bases. e.g: In [Ni(CN)4]2-, ligand is CN- ion. 7. Define the term coordination number of a central metal atom or ion in a complex compound. 2 Marks The coordination number of central metal atom or ion in a complex is the number of ligand donor atoms to which the metal is directly bonded. In K4[Fe(CN)6] , coordination number of Fe2+ is 6. 8. Define coordination sphere of coordination compounds. 2 Marks The central metal atom or ion and the ligands of the complex compound are written within square bracket. This is called coordination sphere of coordination compounds. 9. What are homoleptic complexes? Give an example. 2 Marks Homoleptic complexes are the complexes in which central metal ion or atom is bound to only one type of donor groups. e.g: K4[Fe(CN)6] 10. What are heteroleptic complexes? Give an example. 2 Marks Homoleptic complexes are the complexes in which central metal ion or atom is bound to more than one type of donor groups. e.g: [Co(NH3)5Cl]SO4 11. Classify the following ligands into unidentate, didentate and polydentate ligands. NH3, EDTA, oxalate. 3 Marks NH3 Unidentate EDTA Polydentate Oxalate Didentate 12. Give the IUPAC name for the following compounds. a) K4[Fe(CN)6] potassium hexacyanidoferrate(II) b) [Cu(NH3)4] SO4 tetramminecopper(II) sulphate c) [Co(NH3)5Cl]SO4 pentamminechloridocobalt(III) sulphate d) K3[Fe(C2O4)3] potassium trioxalatoferrate(III) e) [CoCl2(en)2]+ dichloridobis(ethane-1,2-diamine)cobalt(III) f) [Co(NH3)5(NO2)]Cl2 pentamminenitrito-N-cobalt(III) chloride g) [Co(NH3)5(ONO)]Cl2 pentamminenitrito-O-cobalt(III) chloride h) [Ni(CO)4] tetracarbonylnickel(0) 13. What are ambidentate ligands? Give one example. 2 Marks Ligands which have two donor atoms, but can bond to central metal atom or ion through only one donor atom are called ambidentate ligands. e.g: NO2 - , SCN - etc., 14. What is geometrical isomerism in complexes? Give an example. 2 Marks It is a phenomenon in which two complex compounds have the same molecular formula and same chemical bonds, but different geometrical arrangement of the ligands. The two forms are called cis and trans forms. e.g: [Pt(NH3)2Cl2] Cl NH3 Pt Cl NH3 Cis isomer Cl NH3 Pt NH3 Cl Trans isomer
- 47. 15. Explain opticalisomerism in complexes with an example. 2 Marks It is a phenomenon in which two complex compounds have the same molecular formula and same chemical bonds but cannot be superposed on one another. They differ in the rotation of plane of plane polarised light. These isomers are called enantiomers. The two forms are called dextro (d) and laevo (l). e.g: 16. What is linkage isomerism? Give an example. 2 Marks Linkage isomerism is type of isomerism in which two complex compounds differ in the donor atoms for ligands (different ligating atoms). It is shown by complex compounds containing ambidentate ligands. e.g: [Co(NH3)5NO2]Cl2 and [Co(NH3)5(ONO)]Cl2 17. Indicate the type of isomerism in the following set of complex compounds. a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 2 Marks a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl Ionisation isomerism b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 Linkage isomerism 18. Explain coordination isomerism in complexes. Give one example. Coordination isomerism is type of isomerism due to interchange of ligands between cationic and anionic entities of different metal ions present in a complex. e.g: [Co(NH3)6] [Cr(CN)6] and[Cr(NH3)6] [Co(CN)6] 19. Explain ionization isomerism in complexes. Give an example. Ionisation isomerism is type of isomerism in which two complex compounds produce different ions in solution form. e.g: [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl 20. What is solvate isomerism? Explain with an example. Solvate isomerism is a type of isomerism in which complex compounds differ in the number of water molecules acting as ligands and water of hydration. e.g: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2. H2O 21. Explain the formation of [CoF6]3- . Is this complex paramagnetic? 3 Marks Co, Z=27 [Ar]3d7 4s2 4p0 4d0 Co3+ [Ar] 3d6 4s0 4p0 4d0 It undergoes sp3d2 hybridisation. When F - ligand attacks the central metal ion, Co3+ This complex uses outer d orbital (4d) for hybridisation. It is an outer orbital complex. It has unpaired electrons. [CoF6]3- is paramagnetic. This complex is called high spin or spin free complex.
- 48. 22. Give thegeometry, hybridization and magnetic property of [Co(NH3)6]3+ based on VBT. 3 Marks Co, Z=27 [Ar]3d7 4s2 4p0 Co3+ [Ar] 3d6 4s0 4p0 When NH3 ligand attacks the central metal ion Co3+ , pairing of electrons in 3d orbital occurs against Hund’s rule. Co3+ can undergo d2sp3 hybridization. This complex uses inner d orbital (3d) for hybridisation. It is an inner orbital complex. It has no unpaired electrons. [Co(NH3)6]3+ is diamagnetic. This complex is called low spin or spin paired complex. It has octahedral geometry. 23. Using VBT, explain the type of hybridization, geometry and magnetic property of [NiCl4]2-. 3 Marks Ni, Z =28 [Ar] 3d8 4s2 4p0 Ni2+, [Ar] 3d8 4s0 4p0 It undergoes sp3 hybridisation. Four pair of eelctrons from 4 Cl- ions This complex has unpaired electrons. It is paramagnetic. [NiCl4]2- has tetrahedral structure. 24. Explain the hybridization, geometry and magnetic property in the complex compound [Ni(CN)4]2-. 3 Marks Ni, Z =28 [Ar] 3d8 4s2 4p0 Ni2+, [Ar] 3d8 4s0 4p0 When CN- ligand attacks the central metal ion Ni2+, unpaired electrons in 3d orbital are paired up against Hund’s rule. It undergoes dsp2 hybridisation. Four pair of electrons from 4 CN- ligands. This complex has no unpaired electrons. It is diamagnetic. [Ni(CN)4]2- has square planar structure. 25. What are the limitations of Valence Bond Theory? 2 Marks This theory, a) has a number of assumptions b) does not give quantitative interpretation of magnetic data c) does not explain colour shown by complexes. d) could not predict tetrahedral or square planar shape for the coordination number 4 e) could not distinguish between strong and weak ligands. 26. Why [CoF6]3- is called an outer orbital complex? 1 Mark In this complex, Co3+ uses outer d orbital (4d) for hybridization. Therefore it is called an outer orbital complex.
- 49. 27. What areinner orbital complexes? Give an example. 2 Marks Inner orbital complex is one where the central metal ion uses inner d orbital {(n-1)d orbital} for hybridsation. e.g: [Co(NH3)6]3+ 28. Explain the salient features of crystal field theory. 2 Marks This theory considers ligands as point charges in case of anionic ligands and dipoles in case of neutral molecules. The bond formed between central metal ion and the ligands is purely ionic. 29. What is crystal field splitting? Explain crystal field splitting in octahedral entities using energy level diagram. 3 Marks In an isolated gaseous central metal atom or ion, all the five d orbitals are having same energy. i.e they are degenerated. In the presence of attacking ligands, it becomes asymmetric and the d orbitals lose degeneracy, resulting in splitting of d orbitals. This is called crystal field splitting. In an octahedral complex, six ligands surround the central metal ion. dx2 – y2 and dz2 orbitals (called eg set) are directed along the direction of ligands and experience more repulsion. They have more energy. dxy, dyz and dzx orbitals (called t2g set) are directed between the axes of attacking ligands and experience lesser repulsion by the ligands. They have lesser energy. The energy separation between two split sets is denoted as ∆o . The energy of eg orbitals increase by 3/5 ∆o and that of t2g set decrease by 2/5 ∆o . 30. What is spectrochemical series? Arrange the following ligands in the increasing order of their field strength. Br-, I-, H2O, CO, F- 2 Marks Spectrochemiccal series is the arrangement of ligands in the order of increasing field strength. Correct order for the given set is, I- < Br- < F- < H2O < CO 31. Draw a figure to show the splitting of d orbitals in a tetrahedral crystal field. 2 Marks
- 50. UNIT.10 HALOALKANES AND HALOARENES ONEMARKS QUESTIONS 1. What are haloalkanes? [1] A: Haloalkane is a derivative obtained by replacing hydrogen atom of alkane by halogen atom. 2. What is the hybridization of the carbon attached with vinylic halides (or) aryl halides? [1] A: sp2hybridisation. 3. Among phosphorus trihalides which halides are generated insitu ? [1] A: PBr3, PI3 4. Free radical halogenation of hydrocarbons is not a best method to prepare haloalkanes. Why? [1] A: Because this method gives mixture of isomeric mono and poly haloalkanes, which is difficult to separate as pure compounds. 5. Why fluoro compounds cannot be prepared from electrophilic substitution reaction? [1] A: Due to high reactivity of fluorine. 6. What happens to the boiling point of isomeric haloalkanes with increase in branching?[1] A: decreases. 7. How density and atomic mass of halogen atoms in haloalkanes are related? [1] A: Directly 8. Why tertiary alkyl halide undergoes SN1 reaction very fast? [1] A: Because of the high stability of tertiary carbocation 9. What is the order of reactivity of alkyl halides towards SN1 and SN2 reaction [1] A: Towards SN1 reaction order of reactivity is 30> 20 > 10 Towards SN2 reaction order of reactivity is 10> 20> 30. 10.Allylic and benzylic halides are highly reactive towards SN1 reaction. Why? [1] A: Because the carbocation formed from allylic and benzylic halides is more stable due to resonance. 11.Between SN 1and SN2 reaction which one proceeds with complete stereo chemical inversion? [1] A: SN2 reaction. 12.Between SN1 and SN2 reactions which one proceeds with racemisation? [1] A: SN1 reaction. 13.What are optically active compounds? [1] A: Certain compounds rotate the plane polarized light when it is passed through their solutions are called as optically active compounds. 14.What are dextro rotatory compounds? [1] A: The compound which rotates plane polarized light in clockwise direction is called as dextro rotatory compound. 15.What is a laevo rotatory compound? [1] A: The compound which rotates plane polarized light in anticlockwise direction is called as laevo rotatory compound. 16.What are optical isomers? [1] A: The dextro and laevo rotatory isomers of a compound are called as optical isomers. 17.What is asymmetric carbon (or) stereocentre [1] A: A carbon atom attached with four different substituent groups is called as asymmetric carbon (or) stereocentre. 18.What are chirals? [1] A: The objects which are non-superimposable on their mirror image are said to be chirals. 19.What are achiral molecules? [1] A: The molecules which are, superimposable on their mirror images are called achiral molecules. 20.Between propan – 2- ol and butan – 2 – ol, identify the chiral molecule? [1] A: Butan – 2- ol. 21.What are enantiomers? [1] A: The stereo isomers related to each other as non-superimposable mirror images are called enantiomers. 22.What are racemic mixtures? [1] A: A mixture containing two enantiomers in equal proportions will have zero optical rotation, are called as racemic mixture (or) racemic modification.
- 51. 23.What is racemization?[1] A: The process of conversion of enantiomer into a racemic mixture is known as racemization. 24.Between α and β- hydrogen which one is removed during dehydrohalogenation of alkyl halides? [1] A: β - Hydrogen. 25.What are organo-metallic compounds? [1] A: Most organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds are known as organo – metallic compounds. 26.Write the general formula of Grignard reagent? [1] A: RMgX. 27.Name the product formed when Grignard reagent treated with water? [1] A: Alkanes 28.What is the hybridization of carbon atom in C-X bond of alkyl halides? [1] A: sp3hybridisation 29.Mention the hybridization of carbon atom in C-X bond of aryl halides? [1] A: sp2hybridisation 30.Why SN1mechanism is ruled out in haloarenes? [1] A: In case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilized by resonance and therefore, SN1mechanism is ruled out. 31.What happens to the reactivity of haloarenes towards nucleophilic substitution when electron withdrawing group present at ortho or para position? [1] A: Increases. 32.What is wurtz-fitting reaction? [1] A: A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether and is called as wurtz-fitting reaction. 33.Expand DDT [1] A: Dichlorodiphenyltrichloro ethane 34.Among chloral, chloroform, DDT, carbontetrachlorides , which has more number of chlorides ? A;DDT [1] Cl Cl H Cl Cl Cl O2N OH NO2 NO2 Br CH3 H3CH2C H CH3 H CH2CH3 + Br Θ 2-bromobutane secondary haloalkane C R H R X primary haloalkane C H H R X 35. Identify the product. A: 36.Write the structure of DDT [1] A: TWO MARKS QUESTIONS 37.Explain the classification of monohaloalkanes containing Sp3C – X bond [2] A: These are classified into three types
- 52. X X X H3C X Tertiary haloalkaneC R R R X X CH2 X CH2 X 38.What is the difference between allylic acid and benzylic halides [2] A: 39.What are vinylic halides? Give example? [2] A: These are the compounds in which the halogen atom is bonded to a Sp2-hybridised carbon atom of a carbon-carbon double bond. 40.What are aryl halides? Give example? [2] A: These are the compounds in which the halogen atom is bonded to the Sp2hybridised carbon atom of an aromatic ring. Allylichalide These are the compounds where the halogen atom is bonded to a Sp3-hybridised carbon atom next to carbon-carbon double bond. Benzylic halides These are the compounds in which the halogen atom is bonded to an Sp3-hybridised carbon atomnext to an aromatic ring. Br Br Br C H3C CH3 H3C CH2 Cl CH2Cl C + X - CH3 CH Cl Cl CH2 CH2 ClCl 41.Write the IUPAC name of the following compounds [2] a) b) c) d) CH2=CH-Cl A: a) 1, 3, 5 – tribro,obenzece b) 1, chloro 2, 2 – dimethyl propane c) Chloro phenyl methane d) Chloroethene 42.Explain nature of C-X bond in haloalkanes? [2] A: Haloalkanes are polar in nature because halogen atoms are more electronegative than carbon as a result the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge 43.What are geminal halides? Give example? [2] A: These are the dihalo compounds where two same halogen atoms are present on the same carbon atom Ex: 44.What are vicinal halides? Given example? [2] A: These are the dihalo compounds where two same halogen atoms are present on the adjacent carbon atoms Ex:
- 53. CH3 + Cl2 Fe dark CH3 Cl Toluene O-chloro toluene + Cl CH3 P-chlorotoluene N2 + X - X Cu2X2 + N2 Bebzene diazonium halide haloarene X = Cl, Br 45.R-OH + A R – Cl + H2O Identify A and B in the given reaction [2] A: R – OH + HCl R-Cl + H2O A = HCl, B = ZnCl2 46.How do you prepare chloroalkane using reaction between alcohol and phosphorus pentachloride? [2] A: R – O – H + PCl5 R – Cl + HCl + POCl3 47.How do you prepare chloroalkane using alcohol and thionylchloride? Mention advantage of this reaction? [2] A: R – OH +SOCl2 R – Cl + SO2↑ + HCl↑ 48.Explain the reaction between toluene and chlorine? [2] A: 49.Why iodination of arenes by electrophilic substitution requires an oxidizing agent? [2] A: Because iodination of arenes is a reversible reaction due to formation of biproduct HI and presence of an oxidizing agent oxidises the HI formed, there by prevents reversible reaction. 50.Explain sandmeyer’s reaction to prepare haloarenes (or) how do you convert benzene diazonium salt into haloarenes? [2] A: ZnCl2 B N2 + Cl - I KI + N2 + CH3CH3 CH3 I In this reaction iodination of benzene diazonium halide does not requires cuprous halide 51.Identify major product in the given reaction and give reason? [2] CH3 – CH = CH2 + H – I CH3 – CH2– CH2– I Iodopropane 2- Iodo propane A: 2- Iodo propane Because 2- Iodo propane involves stable 2° - carbocation. 52.What happens when ethane reacts with bromine in presence of CCl4. Write the reaction [2] A: This reaction gives 1.2- dibromoethane product. CH2 = CH2 + Br2 BrCH2 – CH2Br 53.What is finkelstein reaction? Give its general reaction? [2] A: The reaction in which an alkyl chloride (or) bromides reacts with sodium iodide in dry acetone gives alkyl iodides is called finkelstein reaction. R – X + NaI R – I + NaX X= Cl, Br 54.How do you get fluoro methane from chloro (or) bromo methane and name the reaction? [2] A: CH3 – Br + AgF CH3 – F + AgBr Bromomethane. Silver fluoride fluoro methane This reaction is called as swart’s reaction. 55.Why the boiling point of halides are higher than hudrocarbons of comparable molecular mass [2] A: Because of greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the inter molecular forces of attraction are stronger in the halogen derivatives. CCl4
- 54. Nu C X C Nu + X OH + Cl OH Cl H HO H H HH H H H + Cl H (Slow step)r [OH ] [CH 3 - Cl] order = 1 + 1 = 2 56.Arrange the halides R-Br, R-Cl, R-F, R-I containing same alkyl group in the increasing order of their boiling point [1] A: R-F < R – Cl< R-Br < R – I . 57.In isomeric dihalobenzenes, the para - isomers has high melting point than ortho and meta – isomers. Why? [2] A: Because para – isomers are symmetric in nature and fits in crystal lattice better as compared to ortho and meta – isomers. 58.Haloalkanes are less soluble in water even though they are polar in nature. Why? [2] A: For a haloalkane to dissolve in water, energy is required to overcome and break the hydrogen bond between water molecules. Less energy is released when new attractions are set up between the haloalkane and water molecules as these are not as strong as the original hydrogen bonds in water. 59.Whyhaloalkanes undergoes nucleophilic substitution reaction? [2] A: In haloalkanes due to difference in electronegativity of carbon and halogen, carbon bears partial positive charge and attracts nucleophile and halogen atom departs as halide ion 60.Write SN2 mechanism [2] 61.In SN2 reaction transition state cannot be isolated. Why? [2] CH3 - CH2 - CH = CH - CH3 CH3 - CH2 - CH2 - CH - CH2 HC3 - CH2 - CH2 - CH = CH2 OH OH Θ ΘBr HPent - 2 - ene (81%) 2 - Bromo pentane Pent - 1 - ene (19%) CH3 H CH2 CH2 OH CH3 + HCl heat CH3 H CH2 CH2 Cl CH3 + H - OH (-) - 2 methylbutan -1-of (+) -1-chloro-2-methylbutane A: In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group; as a result carbon atom in transition state is simultaneously bonded to five atoms and therefore is unstable. 62.Why tertiary haloalkanes are less reactive towards SN2 reaction? [2] A: Because SN2 reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituent on (or) near the carbon atom like in 30 – haloalkane have a dramatic inhibiting effect. 63.What is retention of configuration? Give example [2] A: Retention of configuration is the preservation of integrity of the spatial arrangement of bonds to an asymmetric centre during a chemical reaction. 64.Write the reaction between 2-bromopentane with alcoholic solution of potassium hydroxide and mention the major product in thereaction? [2] A: Pent-2-ene is the major product. 65.During dehydrohalogenation of 2-bromopentane, Pent-2-ene is the major product. Why? [2] A: According to saytzeff rule alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms are stable. 66.How do you prepare ethyl magnesium bromide from ethyl bromide? [2] A: CH3CH2Br + Mg CH3CH2Mg Br Dry Ether
- 55. :X: x + X + X +.. Θ Θ Θ Cl Cl NO2 Cl NO2 + Conc.HNO3 CONC. H2SO4 Chlorobenzene 1 - chloro - 2 - nitro - benzene (minor) 1 - chloro - 4 - nitrobenzene (major) Cl Cl CH3 Cl CH3 + anhydrous Chlorobenzene 1 - chloro - 2 - methyl - benzene (minor) 1 - chloro - 4 - methylbenzene AlCl3+ CH3 - Cl Chloromethane Cl (i) NaoH, 623 K, 300 atm (ii) H + OH :Cl: Cl + Cl + Cl Θ Θ Θ .. 67.Aryl halides are extreamely less reactive towards nucleophilic substitution reaction than alkyl halides. Why? [2] A: In haloarenes, the electron pairs on halogen atom are in conjugation with π- electrons of the ring and C-Cl bond aquires partial double bond character due to resonance. As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive towardsnucleophilic substitution reaction. 68.How do you convert chlorobenzene to phenol? [2] A: 69.Write the resonance structure to show the halogenatom present on the haloarenes areortho and para directing group? [2] A: 70.Write the reaction involved in the nitration of chlorobenzene? [2] A: 71.Explain methylation of chlobenzene and name the reaction? [2] X + 2Na + 2NaXdry ether2 Diphenyl X + 2Na + R - X R + 2NaX dry ether haloarene haloalkane alkylarene (CH3)3 CBr CH3 CH3 CH3 + + Br (slow) Θ CH3 CH3 CH3 + + OH Θ (CH3)3 COH (fast) 72.Write general equation of wurtz-fitting reaction? [2] A: 73.Explain fitting reaction with general equation? [2] A: Two arylhalides reacts with sodium in presence of dry ether gives diphenyl. This reaction is called as fitting reaction. 74.What are freons? Give an example [2] A: The fluro, chloro compounds of methane or ethane collectively called as freons Ex: CCl2 F2 THREE MARKS QUESTIONS 75.Explain SN1machanism [3] A: (CH3)3CBr + ŌH (CH3)3 C OH + BrΘ 2-bromo-2-methylpropane 2-methylpropan – 2 -ol Step (1): formation of carbocation Step (2): attack of nucleophile on carbocation
- 56. H Br CH3 C6H13 + OHOH CH3 H C6H13 + Br Θ Θ Br CH3 H3CH2C H CH3 H CH2CH3 + Br Θ 2-bromobutane CH3 CH2CH3CH3 Θ OHHO H CH2CH3 CH3 OH Θ OH CH3 H H3CH2C (+) - Butan - 2 - ol (-) Butan - 2 - ol Rate of the reaction depends on slow step of the reaction r α [(CH3)3C Br]1 Order = 1 76.With example explain SN2reactions of optically active halides areaccompanied by inversion of configuration. [3] A: SN2 mechanism of optically active halides has the inverted configuration because nucleophile attaches itself on the side opposite to the one where the halogen atom is present. For example when (-) – 2 – bromooctane is allowed to react with sodium hydroxide, (+) – octan – 2 – ol is formed with the – OH group occupying the position opposite to what bromide had occupied. 77.SN1 reaction of optically active halides accompanied by racemisation. Explain with example?[3] A: SN1 reaction of optically active halides accompanied by racemization (but not 100% racemization) because carbocation formed in the slow step being sp2hybridised is planar. As a result the attack of the nucleophile may be accomplished from either side resulting in a mixture of products, one having the same configuration and the other having opposite configuration. Ex: (Please make correction of CH3 as H in the transition state) 1 Unit 11. Alcohols, Phenols and Ethers One mark questions 1. Name the alcohol which is used for polishing wooden furniture. Ans: Ethanol 2. What are alcohols? Ans: Hydroxyl derivatives of aliphatic compounds are called alcohols. 3. What is the IUPAC name of ? Ans: Ethane-1, 2-diol 4. Write the structure of 2-methyl cyclopentanol. Ans: 5. Name the simplest hydroxyl derivative of benzene. Ans: Phenol 6. What is the IUPAC name of Resorcinol? Ans: Benzene-1, 3-diol 7. What is the common name of CH3OC2H5? Ans: Ethylmethyl ether 8. Write the formula of anisole. Ans: C6H5OCH3 or 9. What is the IUPAC name of anisole? Ans: Methoxybenzene. 10. Write the IUPAC name of CH2 = CH CH2OH Ans: prop-2-en-1-ol
- 57. 11. Why isthe bond angle in alcohols is slightly less than the tetrahedral angle? Ans: It is due to the repulsion between the unshared electron pairs of oxygen atom. 12. Why is the bond angle slightly greater than the tetrahedral angle in ethers? Ans: It is due to the repulsive interaction between the two bulky R groups or alkyl groups. 13. Name the product obtained when propene is subjected to acid catalysed hydration. Ans: Propan-2-ol or 2-propanol 14. In the reaction, H 2 2 2H C CH H O X Identify X. Ans: Ethanol 15. In a reaction, 2 2 diporane 3 2 H O /NaOH CH CH CH X Name the product X formed in the reaction. Ans: Propan-1-ol. 16. Write the chemical name of cumene. Ans: Isopropyl benzene. 17. The boiling point of alcohols is much higher than ethers and other classes of compounds with similar molecular masses. Give reason. Ans: Due to intermolecular hydrogen bonding in alcohols. 18. Give reason: Lower alcohols are soluble in water. Ans: Due to the formation of hydrogen bonds with water molecules. 19. Name the compound which is also known as carbolic acid. Ans: Phenol 20. Name the method by which O-nitrophenol and p-nitrophenol are separated. Ans: By steam distillation the two isomers are separated. 21. Ether is soluble in water. Give reason. Ans: Ether is soluble in water because oxygen of ether form hydrogen bonds with water molecule. Two Mark Questions 1. What happens when an aldehyde is reduced? Write the general reaction OR explain the reduction of aldehydes. Ans: Aldehydes on reduction by hydrogen in presence of catalyst like finely divided Nickel or platinum give the respective primary alcohols. Ni 2 2RCHO H RCH OH Or Aldehydes on reduction in presence of sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4) yield the respective primary alcohols. 4 4 NaBH or 2LiAlH RCHO 2(H) RCH OH 2. An aldehyde reacts with Grignards reagent forming an inter immediate product which on hydrolysis gives primary alcohol. Name the aldehyde and write the chemical equation. Ans: The aldehyde is methanal or formaldehyde. Reaction: HCHO + RMgX RCH2OMgX 2H O RCH2OH+MgX(OH) 3. How is phenol prepared from aniline? Write the equation. Ans: Aniline is treated with nitrous acid in presence of HCl at 273-278 K, when benzene diazonium chloride is obtained. Which on warming with water or treating with dilute acids gives phenol. 2 2NaNO H O 6 5 2 6 5 6 5 2HCl warm C H NH C H N NCl C H OH N HCl
- 58. 4. What ismeant by hydroboration – oxidation reaction. Illustrate with an example. Ans: Diborane reacts with alkenes to give trialkyl boranes which is oxidized to alcohol by hydrogen peroxide in presence of sodium hydroxide. Reaction 2 2 3 2 3 2 3 2 2 3 OH 3 2 2 33H O 3CH CH CH (BH ) (CH CH CH ) B 3CH CH CH OH B(OH) 5. Give two reactions that show acidic nature of phenol. Ans: Reaction (1) C6H5OH + Na C6H5ONa + H2 (2) C6H5OH + NaOH C6H5ONa + H2O These two reactions prove that phenol is acidic. 6. Name the following reaction and predict the product X obtained. H2SO4 R’COOH + RO-H X+ H2O Ans: The name of the reaction is esterification and product X is an ester with the formula RCOOR. 7. When phenol is treated with acid chloride in presence of pyridine base, what is the product obtained. Write the equation. Ans: The reaction is pyridine 6 5 6 5C H OH RCOCl C H OCOR HCl The product is an ester. 8. Explain the dehydration of ethanol with equation. Ans: Ethanol undergoes dehydration by heating it with conc. H2SO4 at 443 K. forming ethene. 2 4conc H SO 3 2 2 2 2443 K CH CH OH CH CH H O 9. Explain the dehydration of a secondary alcohol with equation. OR How is isopropyl alcohol converted to propene by dehydration reaction? Ans: Secondary alcohols like isopropyl alcohol undergo dehydration on heating with 85% phosphoric acid at 440 K. forming an alkene (propene) Reaction 3 485% H PO 3 3 3 2 2440 K CH CH CH CH CH CH H O | OH 10. Explain the dehydration of tertiary alcohols. Ans: Tertiary alcohols undergo dehydration when heated with 20% H3PO4 at 358 K forming the respective alkene. Reaction: 11. Complete the following reactions: (a) Cu 2 573 RCH OH X (b) Cu 573 X R C R || O Name X in both the reactions. Ans: (a) Cu 2 573 RCH OH RCHO X-Aldehyde (b) Cu 573 R CH OH R C R ||| OR X = Secondary alcohol 12. Explain the reaction of phenol with dil. nitric acid at 298 K. Write equation. Ans: Phenol reacts with dil. HNO3 at 298 K forming O-nitrophenol and p- nitrophenol respectively. 13. How do you convert phenol to picric acid? Explain with equation. Ans: Phenol reacts with concentrated nitric acid forming picric acid or 2, 4, 6- trinitro phenol.
- 59. 14. Explain thebromination of phenol forming ortho and para bromophenols with equation. Ans: Phenol reacts with bromine in CS2 at 273 K forming ortho – and para bromophenols respectively. 15. How is phenol converted to 2, 4, 6-tribromophenol? Explain with equation. Ans: Phenol reacts with bromine water forming a white ppt of 2, 4, 6-trinitrophenol 16. Explain Kolbe’s reaction with equation. OR What happens when sodium phenate is treated with carbon dioxide? Write equation and name the reaction. Ans: Sodium phenate is treated with carbon dioxide and the product on acidification forms salicylic acid. This reaction is called Koble’s reaction. 17. How is phenol converted to benzene? Write the equation. Ans: Phenol is converted to benzene on heating with zinc dust. 18. Explain Reimer – Tiemann reaction with equation. Ans: Phenol is treated with chloroform and sodium hydroxide solution. The product on acidification forms salicyl aldehyde. 19. Explain the oxidation of phenol with equation. Ans: Phenol undergoes oxidation with acidified sodium dichromate forming benzoquinone. 20. How is diethyl ether or ethoxy ethane prepared from ethanol? Write equation. Ans: Ethanol is heated with conc. H2SO4 to 413 K when ethoxy ethane is obtained. 2C2H5OH C2H5OC2H5+H2O
- 60. 21. Explain Wilhamsonsynthesis with equation. Ans: An alkyl halide reacts with sodium alkoxide forming the respective ethers. By this method both symmetrical and unsymmetrical ethers can be prepared. R X R ONa R O R NaX 22. Identify A and B in the following reactions and name the product obtained. (A) (B) Ans: (A) (B) 23. Explain the reaction of anisole with HI. Write the equation. Ans: Anisole reacts with HI forming phenol and methyl iodide. 6 5 3 6 5 3C H O CH HI C H OH CH I 24. Explain the bromination of anisole with equation. Ans: Anisole (methoxy benzene) undergoes bromination with bromine in ethanoic acid in absence of FeBr3 catalyst forming O-bromoanisole and p-bromoanisole respectively. 25. Explain the Friedel crafts reaction of anisole with equation. Ans: Anisole reacts with chloromethane in presence of anhydrous aluminium chloride as catalyst forming 2-methoxy toluene and 4-methoxy toluene. OR Anisole reacts with acetyl chloride in presence of anhydrous aluminium chloride forming 2-methoxy acetophenone and 4–methoxy acetophenone. 26. Explain the reaction of anisole with a mixture of conc. H2SO4 and conc. HNO3 or Explain the nitration of anisole with equation. Ans: Anisole reacts with a mixture of conc. Sulphuric acid and conc. Nitric acid forming ortho nitro anisole and paranitroanisole.
- 61. III. Three MarkQuestions 1. Give three reasons that phenols are more acidic than alcohols. Ans: (1) In phenol, the OH group is attached to sp2 hybridised carbon which is more electronegative, hence the OH bond becomes more polar. (2) Due to resonance is phenol, oxygen gets a positive charge and this increases the polarity of the O H bond. (3) Delocalisation of negative charge in phenoxide ion makes phenoxide ion more stable than phenol favouring the ionization of phenol. 2. Explain the mechanism of dehydration of ethanol to ethane. Ans: The dehydration of ethanol to ethane occurs in the following three steps, when heated with conc. H2SO4 at 443 K. 2 4conc H SO 3 2 2 2 2443 K CH CH OH CH CH H O Page 1 UNIT 12 ALDEHYDES KETONES AND CARBOXYLIC ACIDS 1) What are aldehydes ? 1 Aldehydes are the organic compounds containing carbonyl group,linked with one hydrogen and one alkyl /aryl group. 2) What are carboxylic acids? 1 Carboxylic acids are the organic compounds containing carboxyl(-COOH) group/s 3) Between aldehyde and ketones which one is confirmed using Tollen’s reagent. 1 Aldehyde. 4) Between aldehyde and ketones which one is confirmed using Fehling’s solution.. 1 Aldehyde. 5) Write the IUPAC name of the compound.CHO-CH2-CH(CHO)-CH2-CHO. 1 Propane-1,2,3-tricarbaldehyde. 6) The boiling point of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular mass. Why. 1 Because in aldehydes and ketones there is a weak molecular association arising out of dipole- dipole interaction. 7) Arrange the following compounds in the increasing order of their acidic strength. HCOOH, CH3COOH, CH3CH2COOH. 1 CH3CH2COOH <CH3COOH< HCOOH. 8) Arrange the following compounds in the decreasing order of their acidic strength. HCOOH, CH3COOH, C6H5COOH. 1 HCOOH> C6H5COOH> CH3COOH. 9) Arrange the following compounds in the increasing order of their acidic strength. Cl-CH2COOH, Br- CH2COOH ,F-CH2COOH 1 Br-CH2COOH <Cl-CH2COOH<F-CH2COOH . 10) Name the reagent used in the Stephen reaction. 1 Stannous chloride in presence of HCl. 11) Explain the nature of carbonyl group in aldehydes and ketones. 2 The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen relative to carbon. Hence, the carbonyl carbon is an electrrophilic and carbonyl oxygen is a nucleophilic centre. 12) Identify the product and name of the reaction. 2 Benzaldehyde Rosenmunds reduction
- 62. Page 2 13) Howdo you prepare aldehydes from alkane nitrile? Write the general reaction and name of the reaction. 2 Stephen reaction 14) What happens when toluene treated with chromyl chloride in CS2 solvent upon hydrolysis? Write chemical equation. 2 Benzaldehyde is obtained 15) Write the general reaction to prepare ketones from acyl chloride. 2 16) Name the functional group obtained when benzene reacts with acetyl chloride in presence of anhydrous AlCl3 2 Functional group is Ketone 17) Explain the mechanism of addition of HCN to aldehyde . 3 Aldehydes and ketones reacts very slowly with pure HCN. Hences,it is catalysed by base and generated cyanide adds to carbonyl compound to yield corresponding cyanohydrins. 18) How do aldehydes and ketones react with ammonia/ hydroxylamine/ hydrazine/ phenyl hydrazine /semicarbazide? (Each one carries 2 marks.) 19) Identify the following reaction. 1 Page 3 Clemmensons reduction 20) Name the products obtained when aldehydes are oxidized. 1 Carboxylic acid 21) Write Wolff-Kishner reduction equation. 2 22) What is Tollen’s reagent. 1 Tollen’s reagent is ammonical silver nitrate solution. 23) What is Fehling’s solution. 1 Fehling’s solution is a mixture of alkaline copper sulphate solution and soium-potasium tartarate solution. 24) How do aldehydes and ketones react with sodium bisulphite? 2 25) Why are α- hydrogen of aldehyde are acidic in nature? 2 26) What is Aldol condensation reaction and explain this reaction by taking ethanol as example. 3
- 63. Page 4 27) Howis propanone converted into 4-methylpent-3-en-2-one? 2 28) Explain the reaction between benzaldehyde and acetophenone in presence of dilute base and identify the name of the reaction. 3 C6H5CHO + C6H5COCH3 C6H5CH=CH-CO-C6H5 Benzaldehyde acetophenone 1,3-diphenylprop-2-en-1-one. This reaction is called as cross-aldol condensation reaction. 29) Write the reaction involved when two molecules of methanal reacted each other in presence of concentrated base. Name the reaction. 3 Cannizzaro’s reaction 30) How is toluene/ propyl benzene converted into benzoic acid? 2 Toluene benzoic acid. Page 5 31) How are carboxylic acids obtained from alkyl nitrile? Give example. 2 32) How do you prepare carboxylic acid obtained from Grignard reagent? 2 33) How is benzoic acid obtained from ethyl benzoate? 2 34) Identify the product in the following reaction. 1 35) Mention any two uses of acetic acid 2 36) Identify the product in the following reaction. 1 m-Nitrobenzaldehyde 37) What is formalin? Mention its uses. 2 40 % aqueous solution of formaldehyde is called as formalin.It is used for preservation of biological specimen
- 64. Page 6 39) Whathappenes when carboxylic acid reacts with PCL5/PCl3/SOCl2. 2 40) How do acetic acid react with ammonia. 2 41) Write the equation involving the reaction between benzoic acid and ammonia. 2 42) How do you convert benzene-1,2-dicarboxylic acid into phthalimide. 3 43) Name the product obtained when sodium acetate treated with sodalime. 1 Methane 44) Write the general reaction of Hell-Volhard-Zelinsky reaction. 2 45) Explain nitration reaction of benzoic acid. 2 46) Write the name of the following reaction. 1 38) How are carboxylic acids obtained from alcohols? 2 Page 7 Gutterman-koch reaction
- 65. 1 Unit 13-NITROGEN CONTAININGORGANIC COMPOUNDS Two marks: 1. Name the product obtained when a nitrile is reduced by H2/Ni, . Give the equation. Ans. Primary amine: RCN 2H /Ni, RCH2NH2. 2. How is nitrobenzene converted into aniline. Give the equation. Ans. By reduction using Sn/ HCl 3. What are A and B? Ans. A is RCH2NH2, B is RNH2 4. Give the equation which will be an example for Hofmann bromamide reaction. Ans. CH3CONH2 + Br2 + 4NaOH CH3NH2 + 2NaBr + Na2CO3 + 2H2O Acetamide methanamine 5. Gabriel phthalimide synthesis is used to prepare which class of organic compound? Aniline cannot be prepared by this method. Give reason. Ans. 1° aliphatic amine Aryl halides are not reactive towards nucleophilic substitution reaction. 6. Name the reaction by which a 1° amine is prepared from an amide having one carbon atom more than 1° amine. Give the general equation. Ans. Hofmann bromamide degradation reaction RCONH2 + Br2 + 4KOH 2NH2 + 2KBr + K2CO3 + 2H2O 7. Between CH3CH2CH2NH2 and (CH3)3N, which has higher boiling point and why? Ans. CH3CH2CH2NH2 has higher boiling point. CH3CH2CH2NH2 has more H atoms on N to form intermolecular hydrogen bonding. 8. Give reason: i) Amines have lower boiling point than alcohol of same molar mass. ii) Amines are insoluble in water. Ans. i) Nitrogen in amines is less electronegative than oxygen in alcohol. Hence amines do not form H-bonds among them. ii) Amines do not form H bonds with water. 9. Amines are both Bronsted base and Lewis base. How? Ans. Amines can accept H+ , hence Bronsted bases. Amines can donate a pair of electrons hence Lewis bases. 2 10.Give reason: aniline is a weaker base than ammonia but methanamine is a stronger base than ammonia. Ans. Aniline is weaker base because the pair of electron on nitrogen gets delocalized towards benzene ring. Methanamine is stronger base, because CH3 group is electron releasing group and makes pair of electrons on nitrogen more available for protonation. 11.Arrange 1°, 2°, 3°methylamines in decreasing order of their base strength i) in gaseous phase ii) in aqueous medium. Ans. i) (CH3)3N > (CH3)2NH > CH3NH2 ii) (CH3)2NH > CH3NH2 > (CH3)3N 12.Name two factors that effect the basic strengths of 1°, 2°, 3° methyl amines in water. Ans. i) Solvation (hydration) ii) steric hindrance 13.What is the final product obtained when 1° amine is alkylated? Give its general formula. Ans. Quaternary ammonium salt: 4R NX 14.Give equation for the reaction between ethanamine and acetylchloride. Name the product obtained. Ans. C2H5NH2 + CH3COCl CH3CONHC2H5 + HCl N-ethylacetamide 15.What is benzoylation of 1° amine? Give the equation with methanamine. Ans. Reaction of amine with benzoyl chloride is benzoylation. CH3NH2 + C6H5COCl C6H5CONHCH3 + HCl 16.Name the family of compounds that answers carbylamine test. Give the equation. Ans. 1° amine RNH2 + CHCl3 + alc. 3KOH RNC + 3KCl + 3H2O 17.How does a 1° aliphatic amine react with nitrous acid? Give the equation. Ans. 1° aliphatic amine reacts with nitrous acid to form respective alcohol. RNH2 + HNO2 2NaNO /HCl 2R N Cl 2H O ROH + HCl + N2. 18.Name the reaction by which aniline is converted into phenyl isocyanide. Give the equation. Ans. Carbylamine reaction C6H5NH2 + CHCl3 + Alc.3KOH C6H5NC + 3KCl + 3H2O 19.Complete the following equations: i) CH3NH2 + CH3COCl HCl + ____________ ii) RNH2 1 mole R-X HX __________ Ans. i) CH3CONHCH3 ii) RNHR
- 66. 3 20.Identify the mainorganic product in the following reactions: i) C6H5NH2 + HNO2 2NaNO /HCl 0 C ________ ii) C6H5CONH2 2Br /NaOH __________ Ans. i) C6H5N2Cl ii) C6H5NH2 21.What is benzene sulphonyl chloride also known as? An amine with benzene sulphonyl chloride forms a compound insoluble in an alkali. Identify the class of the amine. Ans. Hinsberg’s reagent. 2° amine. 22.How does Hinsberg’s reagent help to distinguish 1° amine and a 2° amine? Explain. Ans. The given amine is treated with Hinsberg’s reagent. If the product formed is soluble in an alkali, the amine is 1°. If the product formed is insoluble in an alkali, the amine is 2°. 23.Complete the following equations: 24.What is the significance of acetylation of aniline before nitrating it? Ans. When aniline is treated with concentrated HNO3, much of the aniline gets oxidized, aniline gets protonated and the major product is meta-nitroaniline. Hence to avoid all this aniline is acetylated. Acetylated aniline, avoids oxidation of aniline and controlled nitration yields p-nitro aniline as the major product. 25.Give reasons: i) aniline does not undergo Friedel-Crafts reaction. ii) aniline with concentrated HNO3 forms meta nitro compound in significant amounts. Ans. i) Aniline reacts with AlCl3 to form a salt, which makes nitrogen of aniline to get a positive charge, which becomes a strongly deactivating group. ii) Aniline with conc. HNO3 forms anilinium ion which is meta directing. 26.What is diazotization? Give the general formula of a diazonium salt. Ans. Conversion of 1° aromatic amine into diazonium salt is diazotization. General formula: 2Ar N X or 2R N X , where R = Ar Ans. 4 27.How is benzene diazonium chloride prepared from aniline? Give the equation. Ans. It is prepared by the reaction of aniline with nitrous acid (NaNO2/ HCl) at 0°C C6H5NH2 + NaNO2 + 2HCl 0 C 6 5 2C H N Cl + NaCl + 2H2O 28.What is Sandmeyer’s reaction? Give an example. Ans. Replacement of diazonium group by Cl / Br in presence of Cu(I) ion. E.g.: 2Ar N X 2 2Cu Cl /HCl ArCl + N2 29.Name the organic products obtained in the following reactions: i) 2Ar N X CuCN/KCN __________ + N2 ii) 2Ar N Cl + H3PO2 + H2O _____ + N2 + CH3CHO + HCl Ans. i) Aryl cyanide ii) Benzene 30.How is a diazonium salt converted into iodobenzene? Give the equation. Ans. By treating diazonium salt with potassium iodide. 2Ar N X + KI ArI + KCl + N2 iodobenzene 31.Give an example for a coupling reaction with an equation. Ans. Benzene diazonium chloride reacts with phenol to form p-hydroxyazobenze. This is an example for coupling reaction. 32.How is benzene diazonium chloride converted into an azo dye? Give an example for an azo dye. Ans. Azo dyes are the products obtained when reaction of benzene diazonium chloride with phenol or aniline takes place with retention of diazo group. E.g.: benzene diazonium chloride couples with aniline to form an azo dye p-amino azo benzene (yellow dye) 33.Mention the importance of diazonium salt in synthetic organic chemistry. Ans. i)Aryl fluoride and iodides that cannot be prepared by direct halogenation can be synthesized. ii) It helps to introduce many functional groups into aromatic ring, which cannot be done by direct methods. THREE MARKS: 34.Identify the X, Y, Z in the following: Ans. X is aniline, Y is benzene diazonium chloride, Z is iodobenzene
- 67. 5 35.Give equations forthe preparation of methylamine (methanamine) by Gabriel- phthalimide synthesis. 36.RCN 2H /Ni, X 3CHCl /Alc.KOH Y. Y is a three carbon compound. What is R in RCN, X and Y? Ans. R = CH3, X = CH3CH2NH2, Y = CH3CH2NC 37.Give equation for the conversion of aniline into 4-bromoaniline. 38.An organic compound with formula C2H7N does not answer carbylamine test, but give a product that is insoluble in an alkali, with Hinsberg reagent. Give the IUPAC name of X and to what class of organic compound does it belong to? Ans. X is CH3NHCH3. IUPAC name : N-methylmethanamine. It is a 2° amine. 39.X 2NaNO /HCl 0 C Y warm Z. Y + Z orange dye (p-hydroxyazobenzene). What are X, Y and Z? Ans. Ans. Ans. Unit 14 BIOMOLECULES 1 What are carbohydrates? Give examples 2 Carbohydrates are polyhydroxy aldehydes or ketones or the substances which gives these upon hydrolysis. Example: glucose fructose maltose lactose sucrose starch cellulose glycogen etc. 2 How are carbohydrates classified? 3 Carbohydrates Reducing sugars Sugars Non sugarsNon reducing sugars Aldoses Mono saccharaides Oligo saccharides Poly sacchrides Ketoses Trioses Tetroses Pentoses Hexoses di tri tetra penta hexa hepta octa nano deca 3 What are sugars and non-sugars? 2 Sugars are the carbohydrates; soluble in water crystalline in nature and sweet in taste Example glucose fructose maltose lactose etc. non-sugars are carbohydrates; insoluble in water, amorphous in nature and tasteless. Example :starch cellulose glycogen etc. 4 What are reducing sugars? Give example 2 The sugars which can reduce Tollen’s reagent, Benedict’s reagent and Fehling’s reagent are reducing sugars. These contain a free hydroxyl group on anomeric carbon. Example glucose fructose maltose lactose 5 What are non-reducing sugars? Give example (Is sucrose a reducing sugar or not? Give reason.) 2 The sugars which cannot reduce Tollen’s reagent, Benedict’s reagent and Fehling’s reagent are non- reducing sugars. These do not contain a free aldehydic group(aldehydic groups are bonded). Example : sucrose 6 What are monosaccharaides? Give examples 2 Monosaccharaides are the simple sugars which do not undergo hydrolysis. Example : glucose fructose, Galactose 7 What are oligosaccharides? Give examples 2 Oligosaccharides are the sugars which undergo hydrolysis to give 2 to 10 monosaccharaide units. Example: maltose lactose sucrose etc. 8 What are disaccharides? Give examples 2 Disaccharides are the sugars which undergo hydrolysis to give 2 monosaccharaide units. Example: maltose lactose sucrose etc. 9 What are polysaccharides? Give examples Polysaccharides are the carbohydrates which undergo hydrolysis to give more than 10 (many) monosaccharaide units. Example: starch, cellulose, glycogen etc. 10 Give an example of aldohexose 1 Glucose or Galactose
- 68. 11 Give exampleof ketohexose 1 Fructose 12 How is glucose prepared? 2 13 Elucidate the structure of glucose 5 (i) Molecular formula − C6H12O6 (ii) Suggestion of straight chain (iii) Confirmation of carbonyl (> C = O) group (iv) Confirmation of the presence of carbonyl group as aldehydic group (v) Confirmation of the presence of five −OH groups (vi) Indication of the presence of a primary alcohol The correct configuration of glucose is given by Kiliyanissyntesis 14 Gluconic acid on oxidation with HNO3 gives saccharic acid. What does it indicate about the structure 1 of glucose? Confirmation of the presence of primary alcoholic group 15 Mention the structural features of open chain structure of glucose 2 It has 1 aldehyde group, 1 primary alcohol group and 4 secondary alcoholic groups 16 Mention the structural features of open chain structure of fructose 2 It has 1 ketone group, 2 primary alcohol group and 3 secondary alcoholic groups 17 Mention demerits of open chain structure of glucose 3 The following reactions of glucose cannot be explained by its open-chain structure. 1. Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions. 2. The penta-acetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose. 3. Glucose exists in two crystalline forms, α and β. 18 How do you explain the absence of aldehyde group of the pentaacetate of D – glucose? 2 The aldehyde group is involved in formation of cyclic hemiacetal with secondary alcoholic group of 5th carbon. In pentaacetate of D – glucose, all 5 -OH groups are acetylated, therefore, it does not form an open chain structure, and does not react with NH2OH. This fact indicates absence of aldehyde group in glucose. But, D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. 19 What is glycosidic bond / linkage? 1 Glycosidic linkage − Linkage between two monosaccharide units through oxygen atom 20 Name the sugar present in cane sugar 1 Sucrose 21 What are the expected products of hydrolysis of sucrose 1 α –glucose and β - fructose 22 What are the expected products of hydrolysis of lactose 1 Β – Galactose and β- glucose 23 Name the sugar present in milk sugar 1 Lactose 24 Name the components of starch 1 Amylose and amylopectin 25 Name water soluble component of starch 1 Amylose 26 Name water insoluble component of starch 1 Amylopectin 27 Name the storage polysaccharide in plants 1
- 69. Starch 28 Name thestorage polysaccharide in animals 1 Glycogen( animal starch) 29 Name the structural polysaccharide in plants 1 cellulose 30 Write Haworth structure for α glucose / monomer in cellulose. (β glucose) / α fructose /β fructose 2 31 Write Haworth structure of sucrose/ maltose / lactose 2 Structure of sucrose: Structure of Maltose: Structure of Lactose 32 Why cellulose cannot be used as food by human beings? 1 Human saliva do not contain the enzyme that can hydrolyses β 1-4 linkages present in cellulose 33 What is glycogen? How does it differ from starch 3 Glycogen is a polymer of α – glucose linked by α 1-4 glycosidic bond and α 1-6 glycosidic bond at the point of branching starch glycogen Storage polysaccharide in plants Storage polysaccharides in animals Made of two compenents 1) amylose 2) Made of one component amylopectin Amylopectin has branched structure. The frequency of branching is at every 30 glucose units Glycogen has branched structure. The frequency of branching is at every 10 glucose units 34 Mention two differences between starch and cellulose 2 starch cellulose Storage polysaccharide in plants Structural polysaccharides in plants Made of two compenents 1) amylose 2) amylopectin Made of one component Amylose is linear chain of α – glucose linked by α 1-4 glycosidic bond Amylopectin has branched structure. The frequency of branching is at every 30 glucose units cellulose is linear chain of β – glucose linked by β 1-4 glycosidic bond 35 Name the products obtained when proteins are hydrolysed? What do you understand by this reaction? 2 Proteins upon hydrolysis form amino acids. This indicates that proteins are made of amino acids 36 What are amino acids? How many naturally occurring amino acids are present in proteins 2 These are the organic compounds containing both amino and carboxyl group on α carbon atom. These are the building blocks(monomers) of proteins. There are 20 naturally occurring amino acids 37 Write the general structure of amino acids 1 38 Write the structure of an optically inactive aminoacid 1 39 Name an amino acid containing sulphur 1 Cysteine ,methionine 40 Name an amino acid which is acidic 1 Aspateric acid, Glutamic acid 41 Name an amino acid which is basic 1 Glutamine,Lysine 42 Name an amino acid which contains heterocyclic nucleus 1 Proline,histidine 43 How amino acids are classified based on dietary requirement? 2 Based of dietary requirement they are classified into essential and Non-essential amino acids:
- 70. Essential amino acids:Amino acids that cannot be synthesised in the body, and must be obtained through diet Example − Valine, leucine, isoleucine Non-essential amino acids: Amino acids that can be synthesised in the body Example − Glycine, alanine, glutamic acid Non-essential amino acids: 44 What is zwitter ion? Write its general structure 2 These are the amino acid dipolar ions, carrying both positive and negative charges. These moves neither towards cathode nor towards anode in electric field 45 What is isoelectric point 1 The pH at which amino acids acts as zwitter ions in aqueous solution is called isoelectric pH / point 46 What is peptide bond? How is it formed? 2 It is the amide bond present between two amino acids units in peptides and protein. It is formedby eliminating on molecule of water from α−COOH group and α −NH2 group of two amino acid 47 What is poly peptide? 1 Poly peptides are the polymers of (n)amino acids containing 10 to 50 amino acids in chain linked by (n-1) peptide bonds 48 How many peptide bonds are present in a pentapeptide? 4 49 What are proteins? 1 proteins’ are the polymers of (n)amino acids containing more than 50 amino acids in chain linked by (n-1) peptide bonds 50 Name a hormone which controls the carbohydrate metabolism. 1 insulin 51 How are proteins classified based on their molecular shape and solubility? 3 Based on the molecular shape, proteins are classified into two types Fibrous proteins, polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. These are insoluble in water. These are also called structural proteins Example: keratin (hair and nail), actin and myosin ( muscles) and collagen( cartilage) Globular proteins In Polypeptide chains coil around, giving a spherical shape. These are soluble in water. These are also called functional proteins. Example: albumin, globulin etc 52 Write a note on structure of proteins 4 Structures and shapes of proteins are studied at four different levels: primary, secondary, tertiary and quaternary. Primary structure of proteins: Contains one or more polypeptide chains, and each chain has amino acids linked with each other in a specific sequence. This sequence of amino acids represents the primary structure of proteins. Secondary structure of proteins: Shape in which a long polypeptide chain can exist; two types of secondary structures: α-helix, β-pleated sheet, stabilised by hydrogen bonds α- Tertiary structure of proteins: Overall folding of the polypeptide chains; results in fibrous and globular proteins; secondary and tertiary structures of proteins are stabilised by hydrogen bonds, disulphide linkages, van der Waals forces and electrostatic forces. Quaternary structure of proteins: Spatial arrangement of subunits, each containing two or more polypeptide chains 53 What is denaturation of proteins? 2 Denaturation means loss of biological activity of proteins due to the unfolding of globules and uncoiling of helix. Denaturation takes place due to action of heat, addition of electrolytes etc Example − Coagulation of egg white on boiling, curdling of milk 54 What are enzymes? Give example 2 Enzymes are biocatalysts. Specific for a particular reaction and for a particular substrate For example, maltase catalyses hydrolysis of maltose 55 What are vitamins? How are these classified? 3 Vitamins are micronutrients that take part in metabolic process, to produce energy and growth. These are classified as water soluble vitamins (vitamin B complex and vitamin C). fat soluble vitamins ( vitamin A,D,K,E,) 56 Mention the source and deficiency syndrome of vitaminA 1 Sources Deficiency diseases Fish liver oil, carrots, butter and milk Xerophthalmia, night blindness 57 Mention the source and deficiency syndrome of Vitamin B12 1 Sources Deficiency diseases Meat, fish, egg and curd Pernicious anaemia 58 Mention the source and deficiency syndrome of Vitamin C 1 Sources Deficiency diseases Citrus fruits, amla and green leafy vegetables Scurvy 59 Mention the source and deficiency syndrome of Vitamin D 1 Sources Deficiency diseases
- 71. Exposure to sunlight,fish and egg yolk Rickets and osteomalacia 60 Name the products when nucleic acids are hydrolysed step wise 2 Nucleic acids → nucleotides Nucleotides → nucleosides + phosphoric acid Nucleosides → pentose sugar + heterocyclic bases (purine and pyrimidine) 61 How nucleoside and nucleotide are formed? 2 1) Nucleoside is formed when N-base gets attached to 1 position of pentose sugar. N-base + Pentose sugar nucleoside 2) Nucleotide is formed when nucleoside is linked to phosphoric acid at 5th position of pentose sugar. Nucleoside + H3PO4 nucleotide 62 What are nucleic acids? 1 Nucleic acids are the polymers of nucleotides linked by 3-5 phosphodiester bond 63 What are the differences between DNA and RNA 3 DNA RNA Contains de- oxy ribose sugar Contains ribose sugar Bases are A,G,C,T Bases are A.G.C.U Has double helical structure Has single stranded structure Present in nucleus of the cell Present in cytoplasm Hereditary material Involved in protein synthesis Messenger RNA (m-RNA) Ribosomal RNA (r-RNA) Transfer RNA (t-RNA) 64 Name a. The sugar moiety present in DNA b. Nitrogenous base present only in DNA, but not in RNA. a) de- oxy ribose sugar b)Thymine 65 Write the structure of ribose sugar / deoxy-ribose sugar 2 66 Name any 3 Biological functions of nucleic acids 3 1. DNA is chief chemical as reserve genetic information. 2. DNA is chiely responsible for identity of a species. 3. DNA is capable of self replication during cell division. 4. Important function of RNA is in protein synthesis in the cells. Message for the protein synthesis is in DNA but various RNAs take part in protein synthesis. 67 What are hormones? Give an example for each type of hormone a) Polypeptide hormones b) Amino acid derivatives c) Steroid hormones Hormones are biochemical messengers produced by endocrine glands. a) Polypeptide hormones ----- insulin/ glucagons b) Amino acid derivatives----- Thyroxine/Epinephrine c) Steroid hormones--- Testosterone/Estradiol/progesterone 68 Write the function of the following hormones : a) Insulin b) Thyroxine c) Estrogen and androgen a) Insulin: Maintains blood sugar level b) Thyroxine: Growth and development c) Estrogen and androgen: Development of secondary sex characters
- 72. UNIT 15 POLYMERS Polymers inGreek means, poly means many and mer means unit or part. Polymers means many units or parts. 1. What are polymers? 1M A large number of simple repeating units linked together through covalent bond are called polymers. They are also called as macromolecules 2. What is a monomer? 1M The simple molecule which combine to form polymer are called monomers. 3. What is polymerisation ? 1M The process by which monomers are converted into polymer is called polymerisation. Classification of polymers: Classification based on Source: 4. What are natural polymers? Give example. 2 M The polymers which are found in nature i.e in plants and animals are called natural polymers. Ex: proteins, Nucleic acid , starch, cellulose, rubber 5. What are semi synthetic polymers? Give examples. 2M Chemically modified natural polymers are called semi synthetic polymers. Ex: Cellulose acetate (rayon), cellulose nitrate, valcanised rubber. 6. What are synthetic polymers? Give examples. 2M Synthetic polymers are man –made polymers synthesized in the Laboratories or industries used in daily life. Ex: Polythene, poly vinyl chloride, nylon, terylene, Teflon bakelite Classification based on structure of polymer: 7. What is Linear polymer? Give example. In Linear polymer, the monomer units are linked together to form Long straight chains of polymer molecule Ex: polythene, p v c, nylon, polyester, poly styrene 8. What is branched chain polymer? Give example In branched chain polymer, the monomer unit combines to produce the Linear chains having some branches. Ex: Low density poly then, starch, glycogen etc. 9. What are cross linked or network polymer ? Give examples. Cross- linked polymers are formed from monomer units containing two or more functional Group. They contain strong covalent bond between various linear polymer chains. Ex: Bakelite, melamine, urea –formaldehyde etc. Classification based on mode of polymerization 10. What is addition polymerization? Give examples A polymer formed by the addition of repeating monomer units possessing double or triple bond without elimination of by product molecule during polymerization is called addition polymer. Ex: polythene , poly propene Low density polyethene (LDPE) n(CH2=CH2) (CH2CH2)n Electrical insulator, toys, squeeze bottles HDPE (high density polyethene) n(CH2=CH2) (CH2CH2)n Buckets, dustbin, pipes Teflon (polytetra fluroethene) nCF2 = CF2 (CF2CF2)n Non-stick cookware, oil seals, gaskets Polyacrylonitrile (orlon) Substitute for wool (Any one example) 11. What are homo polymer? Give example Addition polymers formed by the polymerization of one type of monomers are called homo polymer 2 2000 atm, 200°C peroxide or O 4 2 5 3 Ziegler-Natta catalyst TiCl -Al(C H ) ,6 atm, 60°C free radical initiator
- 73. Ex: Polythene (monomerunit in ethene) 12. What are co polymers? Give on example Addition polymers formed by the polymerization of two different monomer units are called co- polymer. Ex: Buna-S, Buna-N, Nylon 6,6 etc. 13. What is Co- polymerization It is polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerize and form a co polymer. 14. What is condensation polymerization? Give examples A polymer formed by the condensation of two different bifunctional or trifunctional monomers with the elimination of simple molecules like water, methanol ammonia is called condensation polymerization. Ex: Nylon 6,6 Polymer polymerization uses Terylene/ Dacron (a polyester Terywoo l, terycot fabrics, safety glass Nylon 6, 6 (a polyamide) Textiles, bristles for brushes Nylon-6 Tyre cords, fabrics, ropes Bakelite (phenol formaldehyd e) Or Combs, handles of utensils, electrical switches Melamine formaldehyd e Crockery Classification based on molecular forces: 15. What are elastomers? Give examples Elastomers are rubber like solid with elastic properties. In these the polymer chains are held by weakest intermolecular forces. The weak binding forces permit the polymer to be stretched. Ex: vulcanized rubber, Buna-S, Buna-N, neoprene etc. 16. What are fibers? Give examples Fibres are thread- like polymer possessing high tensile strength and high modulus. These characterization are due to strong intermolecular forces like hydrogen bonding which result in close packing of chain impart crystalline structure to the polymer. Ex: Nylon 6, 6, terylene, Nylon 6, silk etc 17. What is thermoplastic polymer? Give example Thermoplastic are linear or slightly branched polymers which can be repeatedly softened on heating and hardened on cooling. Ex: polythene, polypropene , pvc, polystyrene, Teflon etc.
- 74. 18. What arethermosetting polymers? Give Examples Thermosetting polymers are cross linked and heavily branched molecules. On heating they undergo extensive cross linking and become hard and infusible. These cannot be reused. Ex: Bakelite, urea formaldehyde resin, etc. 19. Name the monomer and write the partial structure of polythene? Monomer of polythene - Ethene or Ethylene Partial structure - (CH2CH2)n 20. Name the monomer and write the partial structure of Nylon-6 ? Monomer of Nylon-6 - caprolactum Partial structure - [-CO-(CH2)5-NH-]n 21. Name the monomers and write the partial structure of Nylon- 6,6 Monomer of Nylon-6,6 - Hexamethylene diamine and Adipic-acid Partial structure - [-OC-(CH2)4-CONH-(CH2)6-NH-]n 22. Name the monomers and write the partial structure of terylene (Dacron) Monomer of terylene - Ethylene glycol and terephthatic-acid Partial structure - 23. Name the monomer and write the partial structure of Bakalite ? Monomer of Bakalite - Phenol and formaldehyde Partial structure - RUBBER: 24. Name the monomer present in natural rubber.Write the partial structure Natural rubber is a polymer of cis-2-methyl-1,3-butadiene (isoprene). ( cis-poly-isoprene). Its partial structure is 25. Define Synthetic rubber? Give one example Synthetic rubber is defined as any valcanisable rubber like polymer capable of getting stretched to twice its length and returns to its original length, size and shape when the stretching force is withdrawn Ex: Neoprene, Buna-S, Buna-N 26. What is Valcanisation ? The process of heating natural rubber with sulphur or sulphur containing compounds at about 415k for a few hours in order to give strength and elasticity to natural rubber is called valcanisation. 27. Explain the preparation of Neoprene? Write the equation. When chloroprene (2-chloro-1,3-butadiene) is heated in the presence of peroxide catalyst, polychloroprene or neoprene is formed
- 75. 28. Explain thepreparation of Buna-N? When 1,3-butadiene and acrylonitrile are heated in the presence of peroxide catalyst, Buna-N is formed 29. What is bio-degradable polymer? Give example Bio-degradable polymer are those which contain functional groups similar to the functional groups present in bio-polymers Ex: 1. Polyhydroxybutyrate-co-hydroxyvalerate (PHBV ) 2. Nylon-2-Nylon-6 H2N CH2 COOH + NH2 (CH2)5 COOH (HNCH2 CO NH (CH2)5 CO)n glycine aminocaproic acid Polyamide 30. What is non bio-degradable polymer? Give example A large number of synthetic polymers are resistant to the environmental degradation processes and responsible for the accumulation of polymers solid waste materials and cause environmental problems are called Non-biodegradable polymers. Ex: polythene, Nylon, terylene etc
- 76. Chapter-16 Chemistry in EverydayLife 1. Sleeping pills are recommended by doctors to the patients suffering from sleepness but it is not advisable to take their doses with out consultation with the doctor. Why? Ans. Sleeping pills contain drugs which may be tranquilizers or anti-depressants . They affect the nervous system and induce sleep. However, if these doses are not properly controlled, they may create havoc. They even adversely affect the vital organs of the body. It is advisable to take these sleeping pills under the supervision of a doctor. 2. “Ranitidine is an antacid” With reference to which classification, has this statement been given? Ans. Ranitidine is labelled as antacid since it is quite effective in neutralizing the excess of acidity in the stomach. It is sold in the market under trade name Zintac. 3. Why do we require artificial sweetening agents? Ans. The commonly used sweetening agent i.e., sucrose is a carbohydrate with molecular formula C12H22O11. Since it has high calorific value, it is not recommended to the patients, diabetics in particular which require low calorie diet. Most of the artificial sweeteners are better than sucrose but hardly provide any calories to the body. These are being used as substitutes of sugar. 4. Write Chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulas of these compounds are given: (i) (C15H31COO)3 C3H5(Glyceryl palmitate) (ii) (C17H33COO)3C3H5 (Glyceryl oleate) Ans. CH2 OCOC15H31 CH2OH CHOCOC15H31 + 3NaOH CHOH + 3C15H31COONa CH2OCOC15H31 CH2OH Sod. Palmitate (soap)) (C15H31COO) 3C3H5 Glycerol Glyceryl palmitate CH2OCOC17H33 CH2OH CHOCOC17H33 + 3 NaOH CHOH + 3C173 COONa CH2OCOC17H33 CH2OH Sod. oleate (Soap) (C17H33COO)3C3H5 Glycerol Glyceryl oleate 5. Label the hydrophilic and hydrophobic parts in the following molecule which is a detergent. Also identify the functional groups present. C9H19 O(CH2CH2O)xCH2CH2 OH (x = 5to 10) Ans: C9H19 O(CH2CH2O)xCH2CH2OH Hydrophobic part Hydrophilic part Detergents are esters formed by the combination between carboxylic acid and polyethylene glycol. 6. Why do we need to classify the drugs in different ways? Ans Drugs are to attack different targets which are the biomolecules from which our body is made.Moreover, the drugs also differ in action. Therefore, there is a genuine necessity to classify the drugs in different ways. 7. Explain the following as used in medicinal chemistry (a) Lead compounds (b) Target molecules or drug targets. Ans. (a) Lead compounds are the compounds which are effective in different drugs. They have specific chemical formulas and may be extracted either from natural sources (plants and animals) or may be synthesized in the laboratory. (b) Target molecules or drug targets.
- 77. Ans. The differentmacromolecules or biomolecules , which are drug targets are carbonates, proteins, enzymes, nucleic acids. Out of these , enzymes are the most significant because their deficiency leads to many disorders in the body. 8 Why the medicines should not be taken without consulting doctors? Ans. No doubt medicines are panacea for most of the body ailments. But their wrong choice and overdose can cause havoc and may even prove to be fatal. Therefore, it is of utmost importance that the medicines should not be given without consulting doctors. 9. Define the term chemotherapy. Ans: Chemotherapy means the treatment of the disease by means of chemicals that have specific effect upon the disease causing micro-organisms without harming the friendly micro-organisms or bacterias which the body needs. 10. Which forces are involved in holding the drugs to the active sites of enzymes? Ans. These are different inter-molecular forces like dipolar forces, Hydrogen bonding , van der Waals’ forces etc.. 11. Antacids and antiallegic drugs interfere with the function of histamines but donot interfere with the function of each other . Explain. Ans They donot interfere with the functioning of each other because they work on different receptors in the body . 12. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Na,me two drugs. Ans: Low level of noradrenaline which acts as a neurotransmitter reduces the signal sending ability to the nerves and the patient suffers from depression. Antidepressants are needed to give relief from depression. These are also called tranquilizers or neurologically active drugs. The two specific drugs are iproniazid and phenelzine. 13. What is meant by the term ‘broad spectrum antibiotic’? Explain Ans. Broad spectrum antibiotics are drugs which are effective against a large number of harmful micro-organisms causing diseases. 14. Why are cimetidine and ranitidine better antacids than sodium bicarbonate or magnesium or aluminium hydroxides ? Ans. Both sodium bicarbonate and hydroxides of magnesium or aluminium are very good antacids since they neutralise the acidity in the stomach. But their prolong use can cause the secretion of excessive acid in the stomach . This may be quite harmful and may lead to the forrmation of ulcers Both cimetidine and ranitidine are better salts without any side effect. 15. Name a substance which can be used as an antiseptic as well as disinfectant. Ans. About 0.2 percent solution of phenol can act as antiseptic whereas about 1.0 percent solution of the same can act as disinfectant. 16. What are the main constituents of dettol? Ans. The main constituents of antiseptic dettol are chloroxylenol and terpenol. 17. What is tincture of iodine? What is its use? Ans: Tincture of iodine is a dilute solution of iodine ( 2 to 3 percent ) prepared in ethanol. It is a powerful antiseptic particularly in case of fresh wounds. 18. Why is use of aspartame restricted to cold foods and drinks? Ans: Aspartame is a very good sweetener for foods and drinks. But its use is restricted to cold stuff only. In case these are hot, the sweetener may decompose and it may not be effective any more. 19. Name the sweetening agent used in the preparation of sweets for a diabetic patient. Ans: Saccharine is the well known sweetening agent which is more than 550 times sweet as compared to sucrose ( or sugar). It is commonly used in the preparation of sweets for diabetic patients. Actually, it is not a carbohydrate. Now better sweetening agents are also available. 20. What problem arises by using alitame as artificial sweetener? Ans: Alitame is no doubt, a very potent sweetener. Its sweetening capacity is more than 2000 times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to control the sweetness level in the food which is actually desired. 21. Why are detergents called soapless soaps? Ans: Detergents are called soapless soaps becauses they resemble soaps in their cleansing action but they donot contain the usual chemical contents of soaps i.e., sodium or potassium salts of long chain fatty acids. In other words, we can say that they behave as soaps without being actually soaps. 22. What are biodegradable and non-biodegradable detergents? Give an example of each. Ans: Detergents are non-biodegradable in the sense that they cannot be degraded or decomposed by the micro-organisms. They mix with water present in rivers, ponds, lakes etc. as such without getting decomposed and thus cause pollution problems. The biodegradable detergents are the ones which can be degraded. These are being synthesised by reducing the branching of the chain. Sodium n-dodecylbenzene sulphonate is a biodegradable detergent. Even soaps act as biodegradable detergents. 23. Why do soaps not work in hard water? Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid
- 78. ( C 15H31 COOH), oleic acide ( C17 H33COOH ) and stearic acid ( C17 H35 COOH). Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds These being insoluble, get separated as curdy white precipitates resulting in wastage of soap. 24. Can you use soaps and synthetic detergents to check the hardness of water? Ans: Soaps can be used to check hardness of water as they will form insoluble precipitates of calcium and magnesium salts on reacting with hard water. Since detergents donot form any precipitate they cannot check hardness of water. 25. If water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents, which one will you use for cleansing clothes? Ans: Calcium bicarbonate makes water hard. Soap (RCOONa) will react with the salt to form corresponding calcium salt which will be precipitated and wasted. The synthetic detergents are chemically different from soaps. They will not react with the calcium bicarbonate and can be used for cleaning dirty clothes without being precipitated. In other words, there will be no wastage when the detergents are used. 3RCOONa + Ca(HCO3 )2 (RCOO)2Ca + 2NaHCO3 (Soap) 26. Label the hydrophilic and hydrophobic parts in the following compounds. a) CH3(CH2)10 CH2OSO3 Na # (b) CH3(CH2) 15 N + (CH 3)3 Br C) CH3 (CH2)16 – COO(CH2CH2O)n CH2CH20H. Ans: (a) CH3(CH2)10CH2OSO3-Na + B) CH3(CH2)15 – N + ( CH3 )3 Br – (Hydrophobic ) ( Hydrophilic ) (Hydrophobic ) (Hydrophilic) O (c) CH3 (CH2) 16 – C – O(CH2 CH2O)n CH2CH2OH (Hydrophobic ) (Hydrophilic) 13. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Na,me two drugs. Low level of noradrenaline which acts as a neurotransmitter reduces the singal sending ability to the nerves and the patient suffers from depression. Antidepressants are needed to give relief from depression. These are also called tranquilizers or neurologically active drugs. The two specific drugs are ipronaizine and phenelzine. 14. What is meant by the term ‘broad spectrum antibiotic’? Explain Ans. Broad spectrum antibiotics are druga which are effective against a large number of harmful micro-organisms causing disease.
- 79. 15. Why are cimetidineand ranitidine better antacids then sodium bicarbonate or magnesium or aluminium hydroxides? Ans. Both sodium bicarbonate and hydroxides of aluminium are very good antacids since they neutralize the acidity in the stomach. But their prolong use can cause the secretion of excessive acid in the stomach . This may be quite harmful and may lead to the foirmation of ulcers Both cimetidine and rantidine are better salts without any side effect. 16. Name a substance which can be used as an antiseptic as well as disinfectant. Ans About 0.2 percent solution of phenol can act as antiseptic whereas about 1.0 percent solution of the same can act as disinfectant. 17. What are the main constituents of dettol? Ans. The main constituencnts of antiseptic dettol are chloroxylenol and terpenol. 18. What is tincture of iodine? What is its use? Ans: Tincture of iodine is a dilute solution of iodine ( 2 to 3 percent ) prepared in ethanol. It is a powerful antiseptic particularly in case of fresh wounds. 19. Why is use of aspartame restricted to cold foods and drinks? Ans: Asparatame is a very good sweetener for foods and drinks. But its use is restricted to cold stuff only. In case these are hot, the sweetener may decompose and it may not be effective any more. For more details, consult section 16.16. 20. Name the sweetening agent used in the preparation of sweets for a diabetic patient. Ans: Saccharine is the well known sweetening agent which is more than 550 times sweet as compared to sucro9se ( or sugar). It is commonly used in the preparation of sweets for diabetic patients. Actually, it is not a carbohydrate. Now better sweetening agents are also available. 21. What problem arises by using alitame as artificial sweetener? Ans: Alitame is no doubt, a ver4y potent sweetener, Its sweetening capacity is more than 2000 times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to control the sweetness level in the food which is actually desired. 22. Why are detergents called soapless soaps? Ans: Detergents are called soapless soaps becauses they resemble soaps in their cleanising action but they donot contain the usual chemical contents of soaps i.e., sodium or potassium salts of long chain fatty acids. In other words, we can say that they behave as soaps without being actually soaps. 23. What are biodegradable and non-biodegradable detergents? Give an example of each Ans: Detergents are non-biodegradable in the sense that they cannot be degraded or decomposed by the micro-organisms. They mix with water present in rivers, ponds, lakes etc. as such without getting decomposed and thus cause pollution problems. The biodegradable detergents are the ones which can be degraded. These are being synthesized by reducing the branching of the chain. Sodium n-didecylbenzene sulphonate is a biodegradable detergent. Even soaps act as biodegradable detergents. 24. Why do soaps not work in hard water? Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid ( C H COOH) oleic acide ( C17 H33COOH ) abnd stearic acid ( C17 H35 COOH) Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds These being insoluble, get separated as curdy white precipitates resulting in wastage of soap. 25. Can you use soaps and synthetic detergents to check the hardness of water? Ans: Soaps can be used to check hardness of water as they will form insoluble precipitates of calcium and magnesium salts on reacting with hard water. Since detergents donot form any precipitate they cannot check hardness of water. 26. If water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents, which one will you use for cleansing clothes? Ans: Calcium bicarbonate makes water hard. Soap (RCOONa) will react with the salt to form corresponding calcium salt which will be precipitated and wasted. The synthetic detergents are chemically different from soaps. They will not react with the calcium bicarbonate and can be used for cleaning dirty clothes without being precipitated. In other words, there will be no wastage when the detergents are used. 3RCOONa + Ca(HCO3 )2 (RCOO)2Ca + 2NaHCO3 27. Label the hydrophilic and hydrophobic parts in the following compounds. a) CH2(CH2)10 CH2OSO3 Na # (B) CH3(CH2) 15 N + (CH 3)3 Br – C) CH3 (CH2)16 – COO(CH2CH2O)NCH2CH20H. Ans: (a) CJH3(CH2)10CH20SO3-Na + B) CH3(CH2)15 – N + ( CH3 )3 Br –
- 80. (Hydrophobic ) (Hydrophilic ) (Hydrophobic ) (Hydrophilic) (c) CH3 (CH2) 16 – C – O(CH2 CH2O)N CH2CH2OH (Hydrophobic ) (Hydrophilic)