XII chemistry..

Explain how electrical neutrality is maintained in compounds showing Frenkel and Schottky defect.

 In compound showing Frenkel defect, ions just get displaced within the lattice. While in compounds showing Schottky defect, equal number of anions and Cations are removed from the lattice. Thus, electrical neutrality is maintained in both cases

Classify each of the following as either a p-type or n-type semi-conductor. a) Ge doped with In b) Si doped with P

(a) Ge is group 14 element and In is group 13 element. Therefore, an electron deficient hole is created. Thus semi-conductor is p-type. (b) Since P is a group 15 element and Si is group 14 element, there will be a free electron, thus it is n-type semi-conducto

CdCl2 will introduce impurity defect if added to AgCl crystal. Explain.

Ans.For every Cd++ ion dopped to the AgCl crystal, one cation vacancy is created.

The electrical conductivity of a metal decreases with rise in temperature while that of a semiconductor increases. Explain.

In metals with increase of temperature, the kernels start vibrating and thus offer resistance to the flow of electrons. Hence conductivity decreases. In case of semi-conductors, with increase of temperature, more electrons can shift from valence band to conduction band. Hence conductivity increases. 

some examples of 3rd law of motion in our daily life ..for school students

Jumping...you exert a force on the ground, the ground exerts a force on you which causes you to move upwards.

you can begin walking across the floor because you push on the floor and the floor pushes back on you.

 An airplane pushes back on the air and the air pushes forward on the plane. 

Jump off a skateboard - it is pushed one way and you are pushed the other way.


Rockets and jets push fuel and other gas out their exhaust nozzles and those exhaust streams push back on the rockets and jets, propelling them forward.

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CLASS XII CHEMISTRY.........SOLID

General Characteristics of Solid State:
Solids have definite mass, volume and shape.
Intermolecular forces are strong.
The constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate about their mean positions.
They are incompressible and rigid. Solids are classified as crystalline and amorphous on the basis of order of arrangement of constituent particles.

Crystalline solid

1.Solids which have regular orderly arrangements of constituent particles (Long range order) 2. They have sharp melting point. 3. They give regular structure on cleavage. 4. They are anisotropic, i.e. ; they have different optical and electrical properties in different directions due to different arrangements of particles in different directions. 5. They have high and fixed heat of fusion. 6. Diamond, Graphite, NaCl, Metal (Fe, Cu, Ag etc) ice.

Amorphous solid

1.Solids which have irregular arrangement of constituent particles Short range order 2. They melt over a range of temperature. 3. They give irregular structure on cleavage. 4. They are isotropic, i.e. ; the value of physical properties is same in all directions due to irregular arrangement in all directions. 5. They do not have fixed heat of fusion. 6.Glass,rubber,plastics, Quartz glass


Crystal lattice: A regular arrangement of atoms, molecules or ions in the three dimensional space. Unit Cell: The smallest repeating portion of a crystal lattice which, when repeated in different direction generates the entire lattice. Lattice sites (points): The positions which are occupied by the constituent particles in the crystal lattice.

Crystalline solids are classified on the basis of nature of intermolecular forces (bonding) as

(SEE THE TABLE FROM YOUR TEXT)

Types of Unit cells Primitive Unit Cell: The unit cell which contains constituent particles at its corner positions.
Centred unit cells: The unit cells which contains constituent particles at other positions in addition to the corner positions.
Centred Unit cells are of three types:
Body centered unit cells: it contains constituent particles at its body centre in addition to the corner positions.
Face centered unit cell: it contains constituent particles in all the corner positions and also at the centre of each face.
 End centered unit cell: it contains constituent particles at the centre of any two opposite faces (end faces) in addition to the corner positions.


The seven crystal systems are: cubic tetragonal orthorhombic, hexagonal, trigonal, monoclinic, triclinic

 Bravais Lattices: there are 14 types of crystal lattices (space lattices) corresponding to seven crystals systems.
UNIT CELL          CALCULATION OF No. ATOMS                   No. OF ATOMS PER UNIT CELL

Primitive              8 (corner atoms)× 1/8(atom per unit cell)               1


Body centred    8 (corner atoms)× 1/8 (atom per unit cell) + 1           2

Face centred   8 (corner atoms) × 1/8 (atom per unit cell)                                                                                                                      
                       + 6 (face centred atoms) × 1/2 (atom per unit cell)       4


Close packed structures: various types of close packing of  constituent particles. Close packing in three dimensions Simple Cubic Lattice or Structure:  AAA….type arrangement generates simple cubic lattice in three dimension  Its unit cell is primitive cubic unit cell.  Packing efficiency = 52.4%

Hexagonal Close Packing Structure (HCP):  ABAB….Type arrangement  Example: Mg, Zn.  Packing efficiency = 74% o Coordination number = 12  It has N no of octahedral voids and 2N no. of tetrahedral voids if it has N no. of spheres.

Cubic Close Packing Structure (CCP):  ABCABC….Type arrangement  CCP structure is also called FCC structure. Example: Ag, Cu.  Packing efficiency = 74%  Coordination number = 12.  If there is N no. of close packed spheres then it has 2N no. of tetrahedral voids and N no. of octahedral voids.

 Voids (Interstitial Voids or Sites or Hole): Free space or vacant between close packed constituent particles. Types of voids: 
Tetrahedral void
Octahedral void

 Coordination number: the no of nearest neighbors of a particle or the no of spheres which are touching a given sphere.

Packing efficiency: the percentage of total space filled by the particles in a crystal.  Example: Packing efficiency of BCC structure = 68%

Density of the crystal: Density = mass of unit cell /Volume of unit cell

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FOR AIPMT, CCE(MEDICAL ENTR) BIOLOGY QUESTIONS

1.  Formation of gametophyte directly from
sporophyte without meiosis is  [1988]
(a)  Apospory  (b)  Apogamy
(c)  Parthenogenesis  (d)  Amphimixis
2.  Parthenogenesis is  [1988]
(a)  development of embr yo without
fertilization
(b)  development of fr uit without
fertilization
(c)  development of fruit without hormones
(d)  development of embryo from egg without
fertilization
3.  Male gametophyte of angiosperms is shed as
[1988]
(a)  four celled pollen grain
(b)  three celled pollen grain
(c)  microspore mother cell
(d)  anther
4.  Total number of meiotic division required for
forming 100 zygotes/100 grains of wheat is
[1988]
(a)  100  (b)  75
(c)  125  (d)  50
5.  Double fertilization and triple fusion were
discovered by  [1988, 93]
(a)  Hofmeister
(b)  Nawaschin and Guignard
(c)  Leeuwenhoek
(d)  Strasburger
6.  Development of an organism from female
gamete/egg without involving fertilization
is    [1989]
(a)  Adventitive embryony
(b)  Polyembryony
(c)  Parthenocarpy
(d)  Parthenogenesis
7.  Nucellar embryo is  [1989]
(a)  Amphimictic haploid
(b)  Amphimictic diploid
(c)  Apomictic haploid
(d)  Apomictic diploid
8.  Generative cell was destroyed by laser but a
normal pollen tube was still formed because
(a)  vegetative cell is not damaged  [1989]
(b)  contents of killed generative cell stimulate
pollen growth
(c)  laser beam stimulates growth of pollen
tube
(d)  the region of emergence of pollen tube
is not harmed
9.  Which is correct?  [1989]
(a)  Gametes are invariably haploid
(b)  Spores are invariably haploid
(c)  Gametes are generally haploid
(d)  Both spores and gametes are invariably
haploid
10.  A diploid female plant and a tetraploid male
plant are crossed. The ploidy of endosperm
shall be    [1989, 2004]
(a)  tetraploid  (b)  triploid
(c)  diploid  (d)  pentaploid
11.  Which ones produces androgenic haploids in
anther cultures?  [1990]
(a)  Anther wall
(b)  Tapetal layer of anther wall
(c)  Connective tissue
(d)  Young pollen grains
12.  Male gametophyte of angiosperms/monocots
is    [1990]
(a)  Microsporangium (b)  Nucellus
(c)  Microspore  (d)  Stamen
13.  Female gametophyte of angiosperms is
represented by  [1990]
(a)  Ovule
(b)  Megaspore mother cell
(c)  Embryo sac
d)nucleus.

14.  Sperm and egg nuclei fusedue to  [1990]
(a)  base pairing of their DNA and RNA
(b)  formation of hydrogen bonds
(c)  mutual attraction
(d)  attraction of their protoplasts
15.  Entry of pollen tube through micropyle is
[1990]
(a)  Chalazogamy  (b)  Mesogamy
(c)  Porogamy  (d)  Pseudogamy
16.  Cellular totipotency was demonstrated by
[1991]
(a)  Theodore Schwann
(b)  A.V. Leeuwenhoek
(c)  F.C. Steward
(d)  Robert Hooke
17.  Pollination occurs in  [1991]
(a)  Bryophytes and angiosperms
(b)  Pteridophytes and angiosperms
(c)  Angiosperms and gymnosperms
(d)  Angiosperms and fungi
18.  Embryo sac occurs in    [1991]
(a)  Embryo
(b)  Axis part of embryo
(c)  Ovule
(d)  Endosperm
19.  Which of the following pair has haploid
structures?  [1991]
(a)  Nucellus and antipodal cells
(b)  Antipodal cells and egg cell
(c)  Antipodal cells and megaspore mother
cell
(d)  Nucellus and primar y endosper m
nucleus
20.  Point out the odd one?  [1991]
(a)  Nucellus  (b)  Embryo sac
(c)  Micropyle  (d)  Pollen grain
21.  Syngamy means  [1991]
(a)  fusion of gametes
(b)  fusion of cytoplasms
(c)  fusion of two similar spores
(d)  fusion of two dissimilar spores
22.  Double fertilization is fusion of   [1991]
(a)  two eggs
(b)  two eggs and polar nuclei with pollen
nuclei
(c)  one male gamete with egg and other with
synergid
(d)  one male gamete with egg and other with
secondary nucleus
23.  Meiosis is best observed in dividing  [1992]
(a)  cells of apical meristem
(b)  cells of lateral meristem
(c)  microspores and anther wall
(d)  microsporocytes
24.  A population of genetically identical
individuals, obtained from asexual
reproduction is  [1993]
(a)  Callus  (b)  Clone
(c)  Deme  (d)  Aggregate
25.  Which of the following plant cells will show
totipotency?  [1993]
(a)  Sieve tubes  (b)  Xylem vessels
(c)  Meristem  (d)  Cork cells
26.  Study of formation, growth and development
of new individual from an egg is  [1993]
(a)  Apomixis  (b)  Embryology
(c)  Embryogeny  (d)  Cytology
27.  Ovule is straight with funiculus, embryo sac,
chalaza and micropyle lying on one straight
line. It is    [1993]
(a)  Orthotropous  (b)  Anatropous
(c)  Campylotropous  (d)  Amphitropous
28.  Double fertilization is characteristic of
[1993]
(a)  Angiosperms  (b)  Pteridophytes
(c)  Gymnosperms  (d)  Bryophytes
29.  Number of meiotic divisions required to
produce 200/400 seeds of Pea would be
[1993]
(a)  200/400  (b)  400/800
(c)  300/600  (d)  250/500
30.  Haploid plant cultures are got from  [1994]
(a)  leaves  (b)  root tip
(c)  pollen grain  (d)  buds
31.  Chief pollinators of agricultural crops are
[1994]
(a)  butterflies  (b)  bees
(c)  moths  (d)  beetles
32.  Transfer of pollen to the stigma of another
flower of the same plant is  [1994]
(a)  Autogamy  (b)  Allogamy
(c)  Xenogamy  (d)  Geitonogamy
33.  Fertilization involving carrying of male
gametes by pollen tube is  [1994]
(a)  Porogamy  (b)  Siphonogamy
(c)  Chalazogamy  (d)  Syngonogamy

34.  One of the most resistant biological material
is    [1994]
(a)  lignin  (b)  hemicellulose
(c)  lignocellulose  (d)  sporopollenin
35.  In an angiosperm, how many microspore
mother cells are required to produce 100
pollen grains  [1995]
(a)  25  (b)  50
(c)  75  (d)  100
36  The polyembryony commonly occurs in
[1995]
(a)  citrus  (b)  turmeric
(c)  tomato  (d)  potato
37.  Reproducing new plants by cells instead of
seeds is known as  [1995]
(a)  mutation  (b)  tissue culture
(c)  antibiotics  (d)  biofertilizer
38.  How many pollen grains will be formed after
meiotic division in ten microspore mother
cells?  [1996]
(a)  10  (b)  20
(c)  40  (d)  80
39.  In angiosperms, triple fusion is required for
the formation of  [1996]
(a)  embryo  (b)  endosperm
(c)  seed coat  (d)  fruit wall
40.  If an angiospermic male plant is diploid and
female plant tetraploid, the ploidy level of
endosperm will be  [1997]
(a)  haploid  (b)  triploid
(c)  tetraploid  (d)  pentaploid
41.  The endosperm of gymnosperm is  [1999]
(a)  triploid  (b)  haploid
(c)  diploid  (d)  polyploid
42.  Flowers showing ornithophily show few
characteristic like  [1999]
(a)  blue flower with nectaries at base of
corolla
(b)  red sweet scented flower with nectaries
(c)  bright red f lower into thick
inflorescence
(d)  white flowers with fragrance
43.  Double fertilisation leading to initiation of
endosperm in Angiosperms require  [2000]
(a)  fusion of one polar nucleus and the
second male gamete only
(b)  fusion of two polar nuclei and the second
male gamete
(c)  fusion of four or more polar nuclei and
the second male gamete only
(d)  all the above kinds of fusion in different
angiosperms
44.  Eight nucleate embryo sacs are   [2000]
(a)  always tetrasporic
(b)  always monosporic
(c)  always bisporic
(d)  sometime monosporic, sometimes
bisporic and sometimes tetrasporic
45.  Anemophily type of pollination is found in
[2001]
(a)  Salvia  (b)  Bottle brush
(c)  Vallisneria  (d)  Coconut
46.  Adventive polyembryony in citrus is due to
[2001, 05]
(a)  nucellus  (b)  integuments
(c)  zygotic embryo  (d)  fertilised egg
47.  In angiosperms pollen tubes liberate their
male gametes into the     [2002]
(a)  central cell  (b)  antipodal cell
(c)  egg cell  (d)  synergids
48.  What is the direction of micropyle in
anatropous ovule ?    [2002]
(a)  upward  (b)  downward
(c)  right  (d)  left
49.  Which type of association is found in between
entomophilous flower and pollinating agent
[2002]
(a)  mutualism  (b)  commensalism
(c)  cooperation  (d)  co-evolution
50.  In angiosperms all the four microspores of
tetrad are covered by a layer which is formed
by    [2002]
(a)  pectocellulose  (b)  callose
(c)  cellulose  (d)  sporopollenin
51.  In a flowering plant, archesporium gives rise
to    [2003]
(a)  only tapetum and sporogenous cells
(b)  only the wall of the sporangium
(c)  both wall and the sporogenous cells
(d)   wall and the tapetum
52.  An ovule which becomes curved so that the
nucellus and embryo sac lie at right angles to
the funicle is     [2004]
(a)  Hemitropous
(b)  Campylotropous
(c)  Anatropous  (d)  Orthotropous

TRY THESE FOR CHEMISTRY EXM

ALAAP TUTORIALS.

Himan Kr. Barman

9706984610/ hbtutor11.blogspot.com

PART-1

1.Define limiting reagent? Give example.

2.State Hunds rule of maximum multiplicity? Write E.C. of Ca.

3.write Aufbau principle ? arrange in increasing order of energy :  2s,3d,4f,2p

4.which of the following orbital does not exist ? give reason-: 2s,2d,3f,4f

5.write Bohrs postulate ? mention any two limitations.

6.Write Rutherfords atomic models drawbacks.

7.write any  two A. Heisenburgs uncertainty principle.

B.Paulis exclusion principle.      C.De Broglie hypothesis

8.define quantum no. ? write significance.

9. which q.n gives shape & orientation of orbital.

10.distinguish orbit & orbital.

11.what is degenerated orbital ? give example.

12.write E.C. of Cu & Cr ? justify your answer.

13state modern periodic law ? write gen E.C of S-block & P-block elements.

14.what is shielding effect ?(screening)

15why Cations are smaller than parent atoms ?

16. Define isoelectron species? Give example.

17.find the iso electronic species  O^2- , F^-    Na⁺ ,

18.Define Ionisation Enthalpy ? how it is changed across a period ?

19.Define diagonal relationship ? Write the diagonal relationship between Mg & Li ?

20.Explain 1^st ionization enthalpy is less than 2^nd I.E. ?

Q1.Write 1^st law of thermodynamics ?  prve that ΔH=ΔU+ΔnRT .

Q.2.Write Hess`s law of constant heat summation ?

Q.3. what is bond enthalpy ? write the 2^nd law of thermodynamics ?

Q.5. define entropy ? write criteria of spontaneity w.r.t. entropy ?

Q.6. what is Gibbs free energy ? when a reaction become

 A. spontaneous

B.equillibrium

Q.7.define

1. reversible process

2.state function

3.intensive property.

Q.8.Define rate constant ? for a reaction show that  Kc=[C]^c [D]^d/[A]^a[B]^b

Q.9.for a gaseous reaction  : aA+bB ↔cC+dD show that Kp = Kc(RT)^Δn

Q.10.Write Le Chatelier principle . for the reaction N_2(g) +3H_2(g)↔2NH_3(g)  ΔH=-93.6 KJ write the direction of reaction if

1.temparature is increased

2.pressure is increased

3.N₂ is removed

Q.11.BCl₃ has zero dipole moment-explain?

Q.12.graphite good conductor of electricity where as diamond-Explain ?

Q.13.what are sillicons ?how they are formed ?

Q.14.What is catenation ? how it is changed in a group ?

Q.15.KO₂  is paramagnetic –explain ?

Q.16. Find Lewis acid & bases

Br ^_ ,Ag⁺ , NCl₃,NH₃

Q.17.F has less electron gain enthalpy than Cl     , explain why ?

Q.17.What is isomerism ? write different types isomerism ?

Q.18.what is hybridization ? write the conditions for hybridization ?

Q.19.δ bond are stronger than ᴨ bond ?

Q.20.what is free radicals ? explain free radicals are very reactive ?

Q.21.what are carbocations ? write their stability order ?

Q.22.what is Markownikoffs rule ? give example ?

Q.23. some important topics from Environmental chemistry …

A.ozone layer, depletion & adverse effect

B.green house effect & global warming.

C.acid rain & its effect

D. fog

E.effect of- carbon monoxide, oxides of sulpher.

F.B.O.D , eutrophication

Q.24.what is resonance ? explain resonance in O_{3 ,} CO₃^2- ,benzene

Q.25 define : functional group,homologous series,inductive effect.

Q.26. define coefficient of viscosity ? why liquid drops are spherical ?

Q.27. explain HF is liquid but HCl is gas.

Q.28.define H- bonding ? how viscosity changes with temperature ?

Q.29. define  bond order ? Draw molecular orbital diagram of Oxygen ?

Q.30. He ₂ doesn’t exist- why ?

Q.31.prove  Cp-Cv=R. what is gas constant ?

Q.32.Write the real gas equation ? give significance of a & b ?

Q.33.Interprete absolute zero in terms of K.T. of gases ?

Q.34.what is diborane ?

Q.35. what is isotope ? which one isotope of hydrogen is radioactive ?

Q.36.white is heavy water ? write preparation & uses ?

Q.37.how hydrogen peroxide is preapared ? why it is sored in coloured bottle ?

Q.38. describe a process to remove permanent hardness of water ?

Q.39.write difference betn oxidation no. & valency ?

Q.40.what is common ion effect ? what is abuffer soln. ?give exm ?

Q.41. fid O.N.

A) Cr in Cr₂ O₇^-  (B) Mn in  MnO₄^-

© P in H₄P₂O₇

Q.42. balance by ion electron method 

(A) MnO₄^- +H⁺ + C₂O₄^2-  →  Mn²⁺ +H₂O+ CO₂

(B) Cr₂ O₇^-  + H⁺+ Cl ^-   →  Cr³⁺ + Cl₂ +H₂O.

Q.43.calculate P^H of 10^-8 M HCl ?

Q.44. determine the empirical formula of an oxide of iron which has 69.9 % Iron & 30.10 % hydrogen by mass ?

Q45. If 2 moles of Carbon  are burnt in 16g dioxygen calculate the CO₂ produced ?

Q.46.at 300K Kp =2.0 Pa. Find Kc for the reaction : 2A_(g) ↔2B_(g) + C_(g)

_Q.47. complete the reaction :

(A) CH₂=CH₂ +Br₂       →(CCl₄)→?

B) CH₃CH=CH₂ + HBr →peroxide→ ?

© benzene +Zn dust → ?

(D) Ethylbromide + alc KOH→ ?

Q.47. Sorry I am unable to give some question due to my typing  inefficiency…..

CHEMISTRY PART II

ALAAP TUTORIALS
H.K.BARMAN

Q1.Write 1^st law of thermodynamics ?  prve that ΔH=ΔU+ΔnRT .

Q.2.Write Hess`s law of constant heat summation ?

Q.3. what is bond enthalpy ? write the 2^nd law of thermodynamics ?

Q.5. define entropy ? write criteria of spontaneity w.r.t. entropy ?

Q.6. what is Gibbs free energy ? when a reaction become

 A. spontaneous

B.equillibrium

C.not possible

Q.7.define

1. reversible process

2.state function

3.intensive property.

Q.8.Define rate constant ? for the reaction show that  Kc=[C]^c [D]^d/[A]^a[B]^b

Q.9.for a gaseous reaction show that Kp = Kc(RT)^Δn

Q.10.Write Le Chatelier principle . for the reaction N_2(g) +3H_2(g)↔2NH_3(g)  ΔH=-93.6 KJ

write the direction of reaction if

1.temparature is increased

2.pressure is increased

3.N₂ is removed

Q.11.BCl₃ has zero dipole moment-explain?

Q.12.graphite good conductor of electricity where as diamond-Explain ?

Q.13.what are sillicons ?how they are formed ?

Q.14.What is catenation ? how it is changed in a group ?

q.15.KO₂  is paramagnetic –explain ?

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SUB: G.Sc, G.Math, English, S.Science
 PHYSICS,CHEMISTRY,BIOLOGY,MATHS.

     

CLASS XI CHEMISTRY IMP QUESTION..SET 1

PART-1

1.Define limiting reagent? Give example.

2.State Hunds rule of maximum multiplicity? Write E.C. of Ca.

3.write Aufbau principle ? arrange in increasing order of energy : 

2s,3d,4f,2p

4.which of the following orbital does not exist ? give reason-: 

2s,2d,3f,4f

5.write Bohrs postulate ? mention any two limitations.

6.Write Rutherfords atomic models drawbacks.

7.write any  two

 A. Heisenburgs uncertainty principle.

B.Paulis exclusion principle. 

C.De Broglie hypothesis

8.define quantum no. ? write significance.

9. which q.n gives shape & orientation of orbital.

10.distinguish orbit & orbital.

11.what is degenerated orbital ? give example.

12.write E.C. of Cu & Cr ? justify your answer.

13state modern periodic law ? write gen E.C of F-block & P-block elements.

14.what is shielding effect ?(screening)

15why Cations are smaller than parent atoms ?

16. Define isoelectron species? Give example.

17.find the iso electronic species  

O^2- , F^-    Na⁺ ,

18.Define Ionisation Enthalpy ? how it is changed across a period ?

19.Define diagonal relationship ? Write the diagonal relationship between Mg & Li ?

20.Explain 1^st ionization enthalpy is less than 2^nd I.E. ?

PHYSICS IMPORTANT QUESTIONS

1..DEEINE INST. VELOCITY & ACCN ?
2..DEFINE CEFFICIENT OF FRICTION & COEFF. OF RESTITUTION
3..DEFINE MOMENTUM. STATE LAW OF CONSERVATION OF MOMENTUM AND PROVE IT.
4..WRITE NEWTONS 2ND LAW. SHOW F= ma..
5..SHOW THAT INST POWER IS DOT PRODUCT OF FORCE AND VELOCITY.
6..STATE KEPLERS LAW OF PLANETARY MOTION.
7.DEFINE ESCAPE VELOCITY. DEDUCE THE EXPRESSION
8.STATE BERNOULIS THEOREM..PROVE IT.
9..DEFINE COEFF OF VISCOSITY..WRITE ITS DIMENSION
10.DEFINE ACCN DUE TO GRAVITY. STATE & EXPLAIN NEWTONS LAW OF GRAVITATION.
11. WRIT NEWTONS LAW OF COOLING
12. DEFINE BEAT. FIND THE EXPRESSION.
13..WRITE LAPLACES CORRECTION OF NEWTONS FORMULA.
14..WHAT IA GEO STATIONARY SATELLITE. WRITE USES.
15..DEFINE ANGLE OF FRICTION & REPOSE.
16..WHAT IS ELASTIC COLLISION..
17..SHOW THE ENERGY OF A FALLING BODY IS CONSERVED..
18..DEFINE TERMINAL VELOCITY.. DEDUCE
19..WRITE HOOKES LAW OF ELASTICITY..DEFINE YOUNG MODULUS OF ELASTICITY.
20.WRITE DIMENSION OF GRAVTATIONAL CONSTANT.
21..FIND THE RELATIONSHIP OF COEFFICIENT OF LINEAR, VOLUMETRIC EXPANSION.
22..STATE PASCALS LAW..
23..FIND THE EXPRESSION FOR ACCN DUE TO GRAVITY AT A HEIGHT h
24..FIND EXPRESSION FOR TIME PERIOD OF A SIMPLE PENDULUM..
25..DEFINE KINETIC ENERGY..DEDUCE THE EXPRESSION FOR I
26..DEFINE CONSERVATIVE & NON CONSERVATIVE FORCE.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

OZONE LAYER.........SOME IMAGES FOR YOUR STUDY HELP

ENVIRONMENTAL CHEMISTRY

Environmental chemistry is the branch of chemistry which is concerned with the chemical
phenomenon occurring in the environment.
Classification of Environment
1.Atmosphere
Atmosphere is Ii gaseous mixture of air that surrounds the earth. Its different layers are as
(1) Troposphere It is the lowest region of the atmosphere extending from earth’s surface to the
lower boundary of the stratosphere. It contains water vapours and is greatly affected by air
pollution.
(ii) Stratosphere The layer of the earth’s atmosphere above the troposphere and below the
mesosphere, is called stratosphere. Ozone layer to; present in this region.
(iii) Mesosphere It is the region of the earth’s atmosphere above the stratosphere and below
the thermosphere. It is the coldest region (temperature – 2 to 92°C) of atmosphere.
(iv) Thermosphere The upper region of the atmosphere above the mesosphere is called
thermosphere It is the hottest region (temperature up to 1200°C).
(v) Exosphere It is the uppermost region of atmosphere. It contains atomic and ionic O
2
,
H2
and He.
2. Hydrosphere
It is the aqueous envelop of the earth e.g., oceans. lakes etc.
3. Lithosphere
The solid rocky portion of the earth constitute the lithosphere.
4. Biosphere
The biological envelop which supports the life is called biosphere. e.g., animal, human beings.

Environmental Pollution

It may be described as contamination of environment with harmful wastes mainly arising from
certain human activities. These activities release materials which pollute atmosphere, water and
soil.
Types of Pollutions
(i) Natural pollution This type of pollution is caused by the natural sources e.g., volcanic
eruptions. release of methane by paddy fields and cattles, forest fires etc.
(ii) Man-made pollution This type of pollution is resulting from human activities like burning
of the fuels, deforestation, industrial effluents, pesticides etc.
Pollutants
Any substance produced either by a natural source or by human activity which causes adverse
effect on the environment is called pollutant.
Pollutants can be of the following types depending upon the following factors:
Classification on the Basis of Their Degradation
(i) Biodegradable pollutants Pollutants capable of being degraded by biological or microbial
actions are called biodegradable pollutants, e.g., domestic sewage.
(ii) Non-biodegradable pollutants The substances which are normally not acted upon by
microbes are called non-biodegradable pollutants. These undergo biological magnification.
They can further be of two types
(i) Wastes, e.g., glass, plastic, phenols
(ii) Poisons, e.g., radioactive substances, Hg salts, pesticides. heavy metals.
Classification on the Basis of Their Occurrence in Nature
(i) Primary pollutants These are present in same form in which these are added by man e.g.,
DDT. pesticides. fertilizers etc.
(ii) Secondary pollutants These occur in different forms and are formed by the reaction
between the primary pollutants in the presence of sunlight e.g., HNO
3, H
2SO
4
PAN, ozone etc.
Classification on the Basis of Their Existence in Nature
(i) Quantitative pollutants These are naturally present in nature and also added by man. These
become pollutants when their concentration reaches beyond a threshold value in the
environment, e.g., CO2, nitrogen oxide etc.

(ii) Qualitative pollutants These arc not present in the nature but are added by nature only due
to human activities. e.g., pesticides. fungicides. herbicides etc.
Tropospheric Pollution
It is caused by gaseous pollutants and particulate matter.
Gaseous air pollutants Oxides of sulphur (SOx), oxides of nitrogen (NO
x
), oxides of carbon
(CO, CO
2), hydrogen sulphide (H
2
S), hydrocarbons etc.
Particulate pollutants Dust, fumes. mist, smoke etc.
Air Pollution
Air pollution occurs when the concentration of a normal component of the air or a new
chemical substance added or formed in air, build up to
undesirable proportions causing harm to humans, animals, vegetation and materials. The
chemical substances and particles causing pollution are called air pollutants.
Air Pollutants
The major air pollutants are
(i) Carbon monoxide (CO) It is produced by incomplete combustion of gasoline in motor
vehicles, wood. coal, incineration and forest fires. It induces headache, visual difficulty, coma
or death. It blocks the normal transport of oxygen from the lungs to other parts of the body, by
combining with haemoglobin of the blood. (Its affinity towards haemoglobin is about 200
times more than the oxygen.)
(ii) Sulphur dioxide (SO
2
) It is produced by petrol combustion, coal combustion, petrol
refining and smelting operation.
It obstruct the movement of air in and out of lungs. It is particularly poisonous to trees causing
chlorosis and dwarfing. In the
presence of air. it is oxidised to SO
3
which is also an irritant.
2SO
2 + O
2 (air) → 2SO
3
Taj Mahal is reported to be affected by SO2
and other pollutants released by oil refinery of
Mathura.
(iii) Oxides of nitrogen NO2
and NO are obtained by combustion of coal, gasoline. natural gas.
petroleum refining, chemical industries and tobacco smoke. In upper atmosphere. these are
emitted by high flying jets and rockets.


Breathing NO2
causes chlorosis to plants and chronic lung conditions leading to death in
human beings . These oxides destroy ozone layer.
(iv) Smoke, dust These are obtained in cement works, iron and steel works. gas works, power
generating stations. Coal miners
suffer from black lung disease and textile workers suffer from white lung disease.
(v) Ammonia It is produced by fertilizer works.
(vi) Mercaptans These are obtained from oil refineries. coke ovens etc.
(vii) Zn and Cd These are obtained from zinc industries.
(viii) Freon (or CFC’8) Their source is refrigerator.
Smog
It is a mixture of smoke (composed of tiny particles of carbon, ash and oil etc from coal
combustion) and fog in suspended droplet form. It is of
two types:
1. London smog or classical smog
It IS coal smoke plus fog The fog part is mainly SO
2 and SO
3
. It has sulphuric acid aerosol. It
causes bronchial irritation and acid rain. It is reducing in nature and occurs in cool humid
climate.
2. Photochemical smog or Los Angeles smog
The oxidised hydrocarbons and ozone In a warm. dry and sunny climate cause photochemical
smog. Its brown colour is due to the presence of NO
2
.
The nitrogen dioxide by absorbing sunlight in blue and UV region decomposes into nitric oxide
and atomic oxygen followed by a series of the other reactions producing O
3
, formaldehyde,
acrolein and peroxyacetyl nitrates.


(WRITE REACTION FROM TEXT)


Hydrocarbons + O2, NO
2 NO, O, O3
Peroxides, formaldehyde, peroxyacetyl nitrate (PAN),
acrolein etc.
It is oxidising in nature and causes irritation to eyes, lungs, nose, asthmatic attack and damage
to plants.
Green House Effect and Global Warming
The phenomenon in which atmosphere of earth traps the heat coming from the sun and
prevents it from escaping into the outer space is called green house effect. Certain gases, called
green house gases [carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFCs)
and water vapour] in the atmosphere absorb the heat given by earth and radiate back it to the
surface of the earth. Thus, warming of the earth led to the warming of air due to green house
gases. which is called global warming.
Consequences of Green House Effect (or Global Warming)
1. The green house gases are useful in keeping the earth warm with an average temperature of
about 15° to 20°C.
2. There may be less rainfall in this temperature zone and more rainfall in the dried areas of the
world.
3. Increase in the concentration of CO
2
in the atmosphere leads to increase in the temperature
of the earth’s surface. As a result evaporation of surface water will increase which further help
in the rise of temperature and results in the melting of glaciers and polar ice caps and hence,
level of sea water may rise

ACid Rain
The pH of normal rain water is 5.6 due to the dissolution of CO
2
from atmosphere.
(REACTION FROM TEXT)

when the pH of rain water drops below 5ppm, it is called acid rain (by Robert Augus.) Oxides
of N ans S are responsible for making rain
water acidic, Much of the NO
x and SO
x
entering in the atmosphere are converted into
HNO3 and H2SO
4
respectively. The detailed photochemical reactions occurring in the
atmosphere are given as


HNO3 is removed as a precipitate or as particulate nitrates after reaction with bases (like NH
3
,
particulate lime etc).
The presence of hydrocarbons and NOx
step up the oxidation rate of the reaction. Soot particles
are also known to be strongly involved in catalysing the oxidation of SO
2
Acid rain causes extensive damage to building and sculptural materials of marble, limestone,
slate. mortal’ etc


Ozone is a light bluish gas and absorbs UV radiations of the sun which are harmful to living
beings, But nowadays ozone layer is being depleted by CFCs (chlorofluorocarbons).
UV radiations cause the chlorofluorocarbons to dissociate to  highly reactive chlorine
free radical which reacts with ozone to form chlorine monoxide.

(REACTION FROM TEXT)

Type I Clouds They contain some solidified nitric acid trihydrate (HNO3 * 3H
2
O) formed at
about -77°C.
Type II Clouds They contain some ice formed at about – 85°C. These clouds play important
role in ozone depletion by hydrolysing
chlorine nitra.


Consequences of Depletion of Ozone Layer
(a) Loss of sight The UV radiation damage the cornea and lens of the eyes.
(b) Effect on immune system The UV radiations are also likely to suppress immune system.
(c) Skin cancer This type of radiation is known to be cancer causing agent.
Water Pollution
The contamination of water by foreign substances which would Constitute a health hazard and
make It unfit for all purposes (domestic, industrial or agriculture etc) is known as water
pollution. The polluted Water may have foul odour. bad taste, unpleasant colour etc.
 Sources of Water Pollution
(i) Domestic sewage Discharge from kitchens, baths, etc.
(ii) Industrial water Wastes from manufacturing processes which includes acids, alkalies,
pesticides, insecticides, metals. fungicides etc.
(iii) Oil From oil spills or washings of automobiles.
(iv) Atomic explosion Processing of radioactive materials.
(v) Suspended particles (organic or inorganic) Viruses, bacteria, algae, protozoa etc.
(vi) Wastes from fertilizer Industries such as phosphates, nitrates, ammonia etc.
(vii) Clay Ores, minerals, fine particles of soil.
Effects of Impurities in Water
(a) Fluorides Mottling of teeth enamel, above 1 mg/L fluoride causes fluorosis.
(b) Sulphates Sulphates of Na, K, Mg cause diarrhoea.
(c) Lead It damages kidney, liver, brain and central nervous system.
(d) Cadmium and mercury They causes kidney damage.
(e) Zn It causes dizziness and diarrhoea. .
(f) Arsenic It can cause cramps and paralysis.
(g) Phosphates from fertilizers They promote algae growth and reduce dissolved oxygen
concentration of water. This process is known as eutrophication.

Aerobic and Anaerobic Oxidation

The oxidation of organic compounds present in sewage in the presence of good amount of
dissolved or free oxygen (approx, 8.5 mlJL) by aerobic bacteria is called aerobic oxidation.
When dissolved or free oxygen is below a certain value, the sewage is called stale.
Anaerobic bacteria bring out putrefaction by producing H2S, NH
3, CH
4, (NH
4
)
2
S etc. This type
of oxidation is called anaerobic oxidation.
The optimum value of dissolved oxygen for good quality of water is 4·6 ppm (4-6 mg/L). The
lower the concentration of dissolved oxygen, the more polluted is the water.
Biological oxygen demand (BOD) It is defined as the amount of free oxygen required for
biological oxidation of the organic matter under aerobic conditions at 20°C for a period of five
days. Its unit is mg/L or ppm.
An average sewage has BOD of 100 to 150 mg/L.
Chemical oxygen demand (COD) It is the measure of all types of oxidisable impurities
(biologically oxidisable and biologically inert organic matter such as cellulose) present in the
sewage. COD values are higher than BOD values.
Control of Water Pollution
(i) Recycling of waste water
(ii) Use of chemicals Lead poisoning can be cured by g iving the patient an aqueous solution of
calcium complex of EDTA. Lead ions displace calcium in the EDTA complex to form chelated
lead and Ca
2+
. The soluble lead chelate is excreted with the urine.
Ca – EDTA + Pb
2+
→ Pb – EDTA + Ca
2+
(iii) Special techniques such as adsorption, ion exchangers, reverse osmosis, electrodialysis etc.
(iv) Waste water reclamation
Sewage Treatment
It involves the following steps
(i) Preliminary process Passing sewage through screens to remove large suspended matter and
then through mesh screens to remove solids, gravels, silt etc.
(ii) Settling process (sedimentation) The residual water when allowed to stand in tanks, the
oils and grease, float on the surface and skimmed off and solids settle down. The colloidal
material is removed by adding alum, ferrous sulphate etc. Primary sludge can be separated.

(iii) Secondary treatment or biological treatment It is aerobic chemical oxidation or aeration
which converts carbon of the organic matter to CO2
, nitrogen into NHJ and finally into nitrite
and nitrates, dissolved bases form salts such as NH
4O2, NH
4NO3 and Ca(NO
3
)
2
etc., and
secondary sludge is obtained.
(iv) Tertiary treatment It is treatment of waste water with time for removal of phosphate
which is then coagulated by adding alum and ferric chloride and removed by filtration.
Water is disinfected by adding chlorine.
Secondary sludge forms a good fertilizer for soil as it contains nitrogen and phosphorus
compounds.

Soil or Land Pollution
The addition of substances in an indefinite proportion changes the productivity of the soil. This
is known as soil or land pollution.
Sources of Soil Pollution
(i) Agricultural pollutants e.g., chemicals like pesticides, fertilizers, bactericides, fumigants.
insecticides, herbicides, fungicides.
(ii) Domestic refuge and industrial wastes.
(iii) Radioactive wastes from research centres, and hospitals.
(iv) Soil conditioners containing toxic metals like Hg, Pb, As. Cd etc.
(v) Farm wastes from poultries, dairies and piggery farms.
Control of Soil Pollution
(i) Use of manures Manures prepared from animal dung is much better than the commonly
used fertilizers.
(ii) Use of bio- fertilizers These are the organisms which are inoculated in order to bring about
nutrient enrichment of the soil. e.g., nitrogen fixing bacteria and blue-green algae.
(iii) Proper sewerage system A proper sewerage system must be employed and sewage
recycling plants must be installed.
(iv) Salvage and recycling Rag pickers remove a large number of waste articles such as paper,
polythene, card board. rags. empty bottles and metallic articles. These are subjected to
recycling and this helps in checking soil pollution.

Radioactive Pollution
Cosmic rays that reach the parth from outer space and terrestrial radiation from radioactive
elements are natural radiations. This natural or background radiation is not a health hazard due
to its low concentration.
Man made sources of radiations include mining; and refining of plutonium and thorium, atomic
reactors and nuclear fuel. These are produced during preparation of radio-isotopes. These are of
two types: electromagnetic (radio waves UV, IR, α-rays) and particulate.
Other Sources of Radioactive Pollution
(i) Atomic explosions Atomic explosions produce radioactive particles which are thrown high
up into the air as huge clouds.
The process releases large amount of energy as heat. Due to atomic -explosion nuclear fallout.
These radioactive elements may reach the human beings through food chain.
(ii) Radioactive wastes Wastes from atomic power plants come in the form of spent fuels of
uranium, and plutonium. People
working in such power plants, nuclear reactors, fuel processors etc., are vulnerable to their
exposure.
(iii) Radio isotopes Many radioactive isotopes like C
14
, I
125
, p
32
and their compounds are used
in scientific researches. The waste water of these research centres contains the radioactive
elements which may reach the human beings through water and food chains.
Effects of Radiations
1. Strontium-90 accumulates in the bones to cause bone cancer and tissue degeneration in
number of organs.
2. 1-131 damages WBCs, bone marrow, lymph nodes and causes skin cancer, sterility and
defective eye sight.
3. These may cause ionisation of various body fluids, chromosomal aberrations and gene
mutations.
4. Radioactive iodine may also cause cancer of thyroid glands.
5. Cesium-137 brings about nervous, muscular and genetic change.
6. Uranium causes skin cancers and tumours in the miners.
7. Radon-222 causes leukemia, brain tumours and kidney cancers.


Bhopal Gas Tragedy
In Dec. 2, 1984 a dense cloud of methyl isocyanate gas (Mlq leaked from a storage tank of the
Union Carbide ltd plant in Bhopal. It caused a great loss of life to people and animals. Methyl
Isocyanate was prepared by the reaction of methyl amine with phosgene and stored in
abundance
Green Chemistry-An Alternative Tool for Reducing Pollution
Green chemistry may be called chemistry involved in the design, development, and
implementation of chemical products and processes to reduce or eliminate the use and
generation of substances hazardous to human health and the environment.
Thus, the goal of green chemistry is ‘to promote the development of products and processes
that reduce or eliminate the use or generation of toxic substances associated with the design,
manufacture, and use of hazardous chemicals. Some important principles and method of green
chemistry are
1. It is better to prevent waste than to treat or clean up waste after it is formed.
2. Synthetic methods should be designed to maximize the incorporation of all materials used in
the process into the final product. .
3. Whenever possible, synthetic methodologies should be designed to use and generate
substances that possess little or no toxicity to human health and the environment.
4. Chemical products should be designed to preserve efficiency of function while reducing
toxicity.
5. The use of auxiliary substance (e.g., solvents, separation agents etc.) should be avoided as
far as possible.
6. Energy requirements should be recognised for their environmental and economic impacts
and should be minimized.
7. Synthetic methods should be conducted at ambient temperature and pressure.

XI.........BIOLOGY.........BOTANY

The Root:The root is underground part of the plant and develops from elongationof radicle of the embryo.
Various types of root
1. Tap root: Originates from radicle. Dicotyledonous plants e.g., mustard,gram, mango.
2. Fibrous root: Originates from base of the stem. Monocotyledonous plants e.g., wheat, paddy.
3. Adventitious root:  Originates from parts of the plant  other than radicle.  Banyan tree (Prop roots)Maize
(Stilt roots)
Root Cap:The root is covered at the apex by the thimble-like structure which protects the tender apical part of
the root. It is positively Geotropic, hydrotropic and negatively phototropic.
Regions of the root:
1. Region of meristematicactivity:Cells of this region have the capabilityto divide.
2. Region of elongation:Cells of this region are elongated and enlarged, responsible for root growth.
3. Region of Maturation:This region has differentiated and matured cells.  Some of the epidermal cells of this region form thread-like root hairs for absorption of water and minerals

Modifications of Root:
Roots are modified for support, storage of food, respiration.......

• For support:Prop roots in Banyan tree, stilt roots in maize andsugarcane.
• For respiration:Pneumatophores in Rhizophora(Mangrove).
• For storage of food:Fusiform (radish), Napiform (turnip), Conical (carrot).

The Stem:Stem is the aerial part of the plant and develops from plumule of theembryo.It bears nodes and
internodes.
Modifications of Stem:
In some plants the stems are modified to perform the function of storage of food, support, protection
and vegetative propagation.

Modifications of Stem......

• For food storage:Rhizome (ginger), Tuber (potato), Bulb (onion), and Corm(colocasia).
• For support:Stem tendrils of watermelon, pumpkin, cucumber.
• For protection:Axillary buds of stem of Citrus, Bougainvillea get modified into pointed thorns.
• For vegetative propagation:Underground stems of grass, strawberry, lateral branches of mint and
jasmine.
• For assimilation of food:Flattened stem of Opuntia contains chlorophyll and performs
photosynthesis.


The Leaf: Develops from shoot apical meristem, flattened, green structure, manufacture the
food by photosynthesis. It has bud in axil. A typical leaf has leaf base, petiole and lamina.
Venation:The arrangement of veins and veinlets in the lamina of leaf.


Types of Venation:
1. Reticulate:Veinlets form a network as in leaves of dicotyledonous plants(China rose, Peepal).
2. Parallel:Veins run parallel to each other as in leaves of monocotyledonous plants (grass, maize).
Types of Leaves
Simple  Compound
(Single leaf blade)(Leaf has number of leaflets)
e.g., mango, peepal
Pinnately Compound      Palmately Compound
(Neem, rose)             (Silk cotton)


Phyllotaxy: The pattern of arrangement of leaves on the stem or branch.


Types of phyllotaxy
Alternate         Opposite         Whorled
(Single leaf at a node)     (Two leaves at a node) (More than two leaves in a whorl at a node)
e.g., China rose, Mustard     e.g., Calotropis, guava     e.g., Nerium,Alstonia

Modifications of Leaves:
• Tendrils: (Climbing) −Sweet wild pea
• Spines (Protection) −Aloe, Opuntia, Argemone
• Pitcher: (Nitrogen Nutrition) −Nepenthes
• Fleshy: (Storage) −Onion

The Inflorescence:The arrangement of flowers on the floral axis.
Main types of Inflorescence:
1. Racemose:Main axis is unlimited in growth-Radish, Mustard, Amaranthus. Flowers in Acropetal order.
2. Cymose:Main axis is limited in growth-Cotton, Jasmine, Calotropis. Flowers in Basipetal order.
3. Special type:Ficus, Salvia, Euphorbia.

The Flower:A flower is modified shoot and reproductive unit in angiosperms.
Flowers may be unisexual or bisexual, bracteate or ebracteate. Some features of flower are:

Symmetry of flower ............
Actinomorphic (radial
symmetry)
Zygomorphic (bilateral
symmetry)
Asymmetric (irregular)

On the basis of no. of
floral appendage................

Trimerous

Tetramerous
Pentamerous

On the basis of position of calyx,corolla,
androecium with respect to ovary

Hypogynous (superior ovary)

Perigynous (half inferior ovary)

Epigynous (inferior ovary)


Parts of aflower......( DRAW DIAGRAM )

1. Calyx:Sepals, green in colour, leaf like.Gamosepalous− (Sepals united)Polyseppalous− (Sepals free)
2. Corolla:Petals, usually brightly coloured to attract insects forpollinationGamopetalous−  (Petals
united)Polypetalous − (Petals free)


Aestivation:The mode of arrangement of sepals or petals in floral bud  with respect to other members of the same whore.

Types of aestivation:

1. Valvate:Sepals or petals do not overlap the sepal or petal at margins as in Calotropis.
2. Twisted:Sepals or petals overlap the next sepal or petal as in China rose.
3. Imbricate:The margins of sepals or petals overlap one another but not  in any definite direction  as in
Gulmohar.
4. Vexillary:The largest petal overlaps the two lateral petals which in turn  overlap two smallest anterior
petals as in Pea. (Papilionaceous)
Perianth:If calyx and corolla are not distinguishable (tepals), they are called perianth
3. Androecium:Stamens (filament, anther),  male reproductive organ  and produce pollengrains. Stamens
may be  epipetalous  (attach to petals) or  epiphyllous  (attachto perianth).  Stamens may be  monoadelphous
(united into one bundle-china rose),  diadelphous(two bundles-pea) or  polyadelphous
(more than two bundles-citrus).
4. Gynoecium:Made up of one or more carpels, female reproductive part, consists of
stigma,  style  and  ovary, ovary bears one or more  ovules. Carpels maybe  apocarpous
(free) or  syncarpous  (united). After fertilisation, ovules developinto seeds and ovary into fruit.


Placentation:The arrangement of ovules within the ovary.
Types of Placentation:
1. Marginal:Placenta forms a ridge along the ventral suture of ovary as in pea.
2. Axile:Margins of carpels fuse to form central axis as in China rose.
3. Parietal:Ovules develop on inner wall of ovary as in mustard.
4. Free central:Ovules borne on central axis, lacking septa as in Dianthus.
5. Basal:Placenta develops at the base of ovary as in sunflower.

The fruit:
After fertilization, the mature ovary develops into fruit. Theparthenocarpic
fruits are formed from ovary without fertilization.

.......................................................
SOME IMPORTANT QUESTIONS:

1.  Define placentation. Describe various types of placentation.

STATES OF MATTER.............Q&A.

Q 1.Define Boyle’s law. How is it represented mathematically?
Ans.The volume of a given mass of a gas is inversely proportional to its pressure at constant
temperature
Mathematically V α 1/P. PV = constant

Q 2 .Define (a) Dipole-Dipole Forces (b) Avogadro law (c) Intermolecular forces .

Ans.(a) Dipole-dipole forces : The force of attraction which act between two molecules having
permanent dipoles.
(b) Avogadro Law : Equal volumes of all gases under the similar condition of temperature and
pressure contain equal number of molecules . V α n
(c) Intermolecular Forces: The forces of attractions and repulsion between interacting particles.


Q 3 .Define dispersion forces. Write characteristic features of dispersion forces.
Ans. Dispersion forces: The forces of attraction between two temporary dipoles.
Characteristics of dispersion forces:
(i)  These forces are always attractive.
(ii)  These forces are effective only at short distance.
(iii)  Their magnitude depends on the polarisability of the particles

Q 4 .Describe Charle’s Law. In terms of Charle’s Law explain why -273
0
C is the lowest temperature?
Ans. The volume of a fixed mass of a gas is directly proportional to its absolute temperature at
constant Pressure. V α T , constant P.
At -273
0
C, volume of the gas approaches to zero i.e the gas ceases to exist.


Q5 Give one example for each of the following types of inter-particle forces .
(a) London dispersion forces
(b)Dipole-Dipole forces
(c)  Hydrogen bond
(d) Ion-dipole forces  
Ans.   (a) noble gases.
(b) between HCl molecules.
(c) Between water molecules or HF molecules.
(d)Between NO3
-and water molecules.

Q 6 What are isochores? From the three isochores I, II, III given below, for a certain amount of ideal
Gas,What is the correct arrangement of volumes V1,V2 and V3.

Ans. Isochore is the plot beween P and T for a definite amount of a gas at a constant volume.
From different isochors at different volumes draw a line parallel to Temperature axis represent a
constant P and cutting the three isochors at T1,T
2 and T3 respectively.Fr om the graph we find T1,>T
2
>T3. Since V α T , at constant P. Thus V1 > V2 > V3.

(draw  from text)


Q7. In a hospital an oxygen cylinder holds 10 L of oxygen at 200atm pressure. If a patient breathes
in 0.50ml of oxygen at 1atm with each breath, for how many breaths the cylinder will be sufficient.
(Assume that all the data is at 30
0
C.)................( 4x10
6 )

Q 8. Explain the absolute zero (in terms of volume) with the help of isobar.

Ans: At absolute zero ,volume approach to zero and below this temperature, the volume will be
negative, which is meaningless. Hence absolute zero is the lowest possible temperature .

.....................................................................................................................................................
•  Stronger intermolecular forces result in higher boiling point
•  Strength of London forces increases with the number of electrons in the molecule
•  Boiling point of HF ,HCl, HBr and HI are 293K,189K,206K and 238K respectively
.....................................................................................................................................................

Q1.What is Boyle’s point?
Ans. The temperature at which the real gas obeys ideal gas law for an appreciable range of
pressure is called Boyle’s point.

Q2.Why liquid drops are spherical in shape?

Ans. Because of surface tension, the molecules tend to minimize the surface area and sphere has
minimum surface area.

Q3.State with expression Dalton’s law of partial pressure?
Ans. The total pressure exerted by the mixture of non reactive gases is equal to the sum of the
partial pressures of individual gases.
Ptotal = p1 + p2 + p3 +…………………

Q4.What are the two faulty assumptions of kinetic theory of gases?
Ans. i) There is no force of attraction and repulsion between the molecules of a gas.
ii) Volume of the molecules of a gas is negligibly small in comparison to the empty space
between them.

Q5. Give the units of Vander Waal’s constants a and b.

Q6. Give Vander Waal’s equation of state for real gases? What is the significance of constants a
and b in this equation?

Q7.What is the effect of temperature on surface tension and viscosity of a liquid?
Ans. Both decrease by increasing temperature.


Q8. Account for the following properties of gases on the basis of kinetic molecular theory of gases-a) high compressibility b) gases occupy whole of the volume available to them.
Ans. a) High compressibility is due to large empty space between the gas molecules.
b) Due to absence of attractive forces between the molecules, they can easily separate from
one another.

Q9. Explain the pressure and volume corrections in ideal gas equation?
Ans. At low temperature and high pressure intermolecular attractions cannot be neglected, the
observed pressure P is smaller than the ideal pressure Pi
.
Therefore, Pi = P + p = P + an
2
/V
2
As volume of gas molecules is not negligible as compared to total volume of the gas
the ideal volume Vi
is smaller than the observed volume.
Therefore, Vi
= V + v = V - nb


Q10. What is meant by compressibility factor of gases? How does its value deviate from that of an
ideal gas in case of real gases and what does it indicate?
Ans. It is the ratio of the product PV to nRT.
Z = PV/nRT. For ideal gas its value is unity. Z < 1 indicates negative deviation i.e. gas is more
compressible due predominance of attractive forces. Z > 1 indicates positive deviation i.e.
gas is less compressible due predominance of repulsive forces.

.....................................................................................................................................
Critical pressure(Pc) – it is the pressure required to liquefy the gas at critical temperature.
Critical volume( Vc) – it is the volume of one mole of the gas at T
cand Pc
........................................................................................................................................

Q1.Explain:
i)Vapour pressure increases with increase in temperature.
ii)Giycerine is more viscous than water.
Ans. i) At higher temperature inter particle attraction weakens and more number of mole cules
escape to vapour.
ii) As interparticle forces are stronger in glycerine.


Q. Derive relation between density and pressure of a gas


Q.At what temperature 128 g of SO2 confined in vessel of 5 dm
3
capacity will exhibit a pressure of
10.0 bar? Given a = 6.7 bar L
2
mol
-2
and b= 0.0564 L mol
-

Q. Calculate the pressure exerted by 8.5 g of NH3 contained in a 0.5 L vessel at 300 K. For ammonia
a = 4.0 atm L
2
mol
-2
and b = 0.036 L mol
-1

Q5.  Derive ideal gas equation.

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                                    ........................
                                          ...........

HYBRIDIZATION & MOT....Q&A

Q1 What is hybridization?
Ans: The process of intermixing of atomic orbitals of slightly different energies of same atom to get
same number of new orbitals of equivalent energies and shape is called hybridization
Q2 Which hybrid orbital are used by carbon atoms in the following molecules (L-II)
(a) CH3CH2OH b) CH3COOH .            
Ans a)sp
3
,sp
3
b)sp
3
sp
2
Q3. Give an example for sp
2
and sp
3
hybrid molecules also give their shape.
Ans sp
2
-Trigonal planar e.g, BH3
sp
3
- Tetrahadral eg CH4
Q4 . Draw the shapes of following hybrid orbitals. sp, sp
2
(L_II)
B
Ans sp B A B sp
2
B  A
B
Q5. Draw the shapes of H2O and CH4 molecule.

Q6 Arrange the following orbitals in the increasing order of s- character: sp, sp
2
, sp
3
(L-I)
Ans. sp
3
< sp
2
<sp
Q7 How is paramagnetic character of a compound is related to the no. of unpaired electrons?
Ans. Greater the number of unpaired electrons greater will be the paramagnetic character
Q3 . Calculate bond order of N2 ? (L-II)
Ans. 3
Q8 Give electronic configuration for C2?
Ans.
1s
2
*1s
2
2s
2
*2s
2
2py
2
2pz

Q9 Calculate the number of unpaired electrons in the B2 molecule.
Ans 2

Q10 Define bond order and give the relationship between bond order and bond length .
Ans Bond order (b.o.) is defined as one half the difference between the number of electrons present
in the bonding and the antibonding orbitals i.e., Bond order (b.o.) = ½ (Nb–Na)
Bond order is inversely proportional to bond length

Q11 Explain the shape of NH3 molecule using hybridization.

Q12 Arrange the following in order of decreasing bond angles.
i)  CH4, NH3, H2O, BF3, C2H2
ii)  NH3, NH2
-,NH4
+
Ans (i)C2H2(180) > BF3(120)>CH4 (109.28)>NH3(107) > H2O(104.5)
(ii)NH4
+
> NH3  >NH2
-,

Q13 Compare the relative stability of following species and indicate their magnetic properties .
O2, O2
+
O2

Ans. Stability increases as O2
-< O2, <O2
+
Bond order 1
1
2
� ,2, 2
1
2
�
Magnetic properties -Paramagnetic


Q15 Show the non-existence of helium molecule based on molecular orbital theory.
Ans Helium molecule (He2 ): The electronic configuration of helium atom is 1s2. Each helium atom
contains 2 electrons, therefore, in He2 molecule there would be 4 electrons. These electrons will be
accommodated in σ1s and σ*1s molecular orbitals leading to electronic configuration: He2 : (σ1s)
2
(σ*1s)
2
.Bond order of He
2
is ½(2 – 2) = 0. He2 molecule is therefore unstable and does not exist.
.....................................................................................................................................................................................................................................................................................................................................
.................................................................................................................................................................

BONDING...PART-2

One Mark Questions
Q 1.Define polarizing power and polalrisability.
Ans . The ability of an atom to deform the electron cloud of an atom is called polarizing power .
while the ease with which any atom can be polarized by its neighbouring bonded atom is called
polarizability.
Q 2.Why PCl5  is more covalent than PCl3 ?
Ans . Because P
5+
has more polarising power than P
3+
.
Q 3.Draw Resonating structure of O3.
Q 4.Why is SF4 more reactive than SF6 ?
Ans . SF4 has two unpaired electrons which it can donate further to extend its valency , hence it is
reactive .’S’ in SF6 has maximum 6 0.S.
Q 5.Explain the following order of bond angle.
NO2
+
>  NO2   >   NO2
—
Ans. This is because NO2
+
has no lone pair of e-, and hence its linear.NO2 has one unshared e- while
NO2 has one unshared e- pair.Greater the repulsion on N-O bond in case of NO2
-than in case of NO2.

Q6.Suggest the expected shape of the following molecules with reasons.
a.SO2
b.NH3
Ans.(a) In SO2 molecule there are three electron pairs .the three electron pairs should acquire
trigonal planar arrangement with bond angle 120 ° .Since one of the position is occupied by lone pair
the geometry is described as V-shaped or bent shaped.
(b) In NH3 molecule there are are four electron pairs (three bond pairs and one lone pair) .These
four electron pairs adopt tetrahedral geometry but due to the repulsion by lone pair the bond
decreases to 107
0
from 109.5
0
and hence the geometry of ammonia is regarded as pryamidal.
Q 7. Why is CO2 a linear molecule while SO2 a non-linear molecule?
Ans.  In SO2 molecule there are two bond pairs and one lone pair of electron .Due to repulsion by
lone pair its bond angle decreases from 120° to 119.5° hence it is not linear whereas in CO
2 molecule
there is no lone pair of electron due to which it has linear geometry.

Q8. i) Calculate formal charges of the atoms in nitrite ion.
Ans. i) Formal charge = V-L- ½ S
V--Total no. of valence electrons in free atom
L-- Total no.of non-bonding electrons
S-- ½ Total no. of bonding electrons

For N= 5 – 2 – ½ 6 = 5 – 3= 0
For O= 6 – 4 – ½ 4 = 2 – 2 = 0
For O(-vely charged)= 6 – 6 – 1/2 x 2 = -1
Total charge = -1

Japanese Scientists Become First From Asia to Name a New Element

On New Year's Eve, a team of researchers at RIKEN Nishina Center for Accelerator-Based Science in Japan were notified by the International Union of Pure and Applied Chemistry (IUPAC) that they would have the honor of naming element 113. The news came almost 12 years after the RIKEN team first synthesized the element in a lab, and three years after they conclusively demonstrated its decay chain. It will be the first element named by an Asian research institution.
Element 113, provisionally known as ununtrium (Latin for ‘one one three’, signifying its atomic number) is a highly radioactive element that isn’t found in nature. It is located in the wild west of the periodic table, home to recently-synthesizedsuperheavy synthetic elements which have no real practical purpose due to their incredibly short half-lives. Here, ununtrium can be found nestled between elements copernicium and flerovium.

CHEMICAL.........................................BONDING

Q. On the basis of VSEPR theory predict the shape of Hydronium (H₃O⁺) ion 
Ans. pyramidal

Q.Define formal charge of an atom.

Ans.The formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a LEWIS STRUCTURE,

Q.Define Bond length& Bond dissociation enthalpy.

Ans. The equilibrium distance between the centre s of the nuclei of two bonded atoms is called bond length.

The energy required to break a bond is called bond dissociation enthalpy.

Q.On the basis of VSEPR theory give the geometry & bond angle of 

(i) BF₃ (ii) SiH₄ 

Ans. Trigonal planar 120°

Tetrahedral 109.5°

Q .What is the geometry of molecule of type AB₄E₂ & AB₃E.

Ans. Square planar & Pyramidal shape

Q (a)State one significance of Formal Charge.

Answer. It helps us to select the most stable structure out of all the different Lewis structures. (b)In nitrate ion what should be total of Formal Charge on each atom? Why?

Ans. The sum of formal charge is always equal to the charge present on the ion. Hence sum of charges in NO₃^- will be -1.

a)        NaCl gives a white ppt. with AgNO₃ but CCl₄ does not. Why?.

Answer. Being an ionic compound NaCl gives Cl negative ion and thus gives White ppt. with silver nitrate but CCl₄ is a covalent compound and thus do not give Cl negative ions.

Q.Dipole moment of NH₃ is greater than of NF₃.?

Ans. Resultant moment of N-H bond adds up to the bond moment of lone pair, that of 3 N-F bonds partly cancels the resultant moment of lone pair. Thus the dipole moment of NH₃ is greater than of NF₃.

Q. Explain why PCl₅ is trigonal bipyramidal and IF₅is square pyramidal?

Ans. PCl₅ has 5 bond pairs of electrons around central atom while in IF₅ there are 5 bond pair & 1 lone pair of electrons around central atom.

Q .Give Reason

(i)N₂ is inert at room temperature.

(ii) Overlapping of S orbital with any other orbital forms sigma bonds.

Ans .Due to high bond order (3) , it has high bond dissociation energy ,which is not possible at room temperature

(ii)Because s orbital has same electron distribution in all axis/ directions

Q Out of NH₃ & BF₃ Which is polar & Why?

Ans. NH₃ has net dipole moment because resultant of 3N-H bond & lone pair lie in the same direction. While BF3 being trigonal planar has zero dipole moment.

Q.When does ionic character develop in covalent bond?

Ans. When the difference in electro negativity of the combining atom is more than 1.9 it develops 50 % ionic character . 

Q  .Account for the following :

(i) AlF₃ is a high melting point solid whereas SiF₄ is a gas?

Ans.AlF₃ is an ionic solid due to large difference in electro negativity of Al and F whereas SiF₄ is a covalent compound and hence only weak Vander Waals force exists between the molecules.

(ii) Out of peroxide ion and superoxide ion which has larger bond length and why?

Ans. The bond orders of(superoxide ion) O₂^- is 1.5 while that of(peroxide ion ) O₂^2- is 1.0 .Smaller the bond order, greater the bond length. Hence, O₂^2- has larger bond length.

Q. What do you mean by canonical structures? Draw canonical structures of Carbonate ion & Benzene. [1+2]

Ans. Sometimes a molecule possess several structures and the actual one is in between of them. This is called Canonical structure.

Q .Define polarizing power and polalrisability,

Ans . The ability of an atom to deform the electron cloud of an atom is called polarizing power . while the ease with which any atom can be polarized by its neighbouring bonded atom is called polarizability.

Q .Why PCl₅  is more covalent than PCl₃ ?

Ans . Because P⁵⁺ has more polarising power than P³⁺.

Q .Why is SF₄ more reactive than SF₆ ?

Ans . SF₄ has two unpaired electrons which it can donate further to extend its valency , hence it is reactive .’S’ in SF₆ has maximum 6 0.S.

Q .Explain the following order of bond angle ,
NO2+		NO2		NO2—
Ans. This is because NO + has no lone pair of e-, and hence its linear.NO						has one unshared e- while
	>	2	>		2		-	than in case of NO2.
NO2 has one unshared e- pair.Greater the repulsion on N-O bond in case of NO2

Q . Why is CO₂ a linear molecule while SO₂ a non-linear molecule?

Ans. In SO₂ molecule there are two bond pairs and one lone pair of electron .Due to repulsion by lone pair its bond angle decreases from 120° to 119.5° hence it is not linear whereas in CO₂ molecule there is no lone pair of electron due to which it has linear geometry.

Q . Why He₂ molecule does not exist?

Ans. It has been found that when two helium atoms approach each other four new attractive forces and five new repulsive forces come into play. Therefore repulsive forces predominate and the potential energy of the system increases which leads to instability. Or Bond order is Zero.

Q Explain the formation of H₂ molecule on the basis of valence bond theory.

Ans. When two H atoms come closer the following forces operate between them: a. Attractive :

i.nucleus of atom and its electron

ii.nucleus of one atom and electron of other. b.repulsive :

i.electrons of two atoms. ii.nucleus of two atoms.

It has been found experimentally that magnitude of attractive forces is more than repulsive forces hence two atoms come closer to each other and the potential energy of the system decreases. A stage is reached when attractive forces balance repulsive forces and the system acquires minimum energy .At this stage two hydrogen atoms are said to be bonded together

Q. i) Why is bond angle in PH₄⁺ higher than that in PH₃?

ii) Arrange the bonds in order of increasing ionic character in the molecules: (L3) LiF, K₂O, N₂, SO₂ and ClF₃.

Ans- N₂ < SO₂ < ClF₃ < K₂O < LiF.

Q4 i) Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.

 Ans. The Lewis structure for BeH2 is as follows:

There is no lone pair at the central atom (Be) and there are two bond pair. It has a linear structure. Dipole moments of each H–Be bond are equal and are in opposite directions.

Therefore, they nullify each other and the BeH₂ molecule has zero dipole moment.