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XI..CH..P-BLOCK ELEMENTS

Elements  in  which  the  last  electron  enters  in  the  any  one  of  the  three  p -  orbital  of
their outermost shell s  – p-block ele ments
•
Gen. electronic configuration  of  outer shell  is  ns
2
np
1-6
The  inner  core  of  e-config.may  differ  which  greatly  influences  their  physical  &  to
some extent chemical properties.
GROUP 13  :  The boron group
Outer Electronic Configuration:-ns
2
np
1
•  group   members:   boron   (B),   aluminum   (Al),   gallium   (Ga),   indium   (In)&
thallium (Tl) .  All, except boron, are metals.
•  Boron  show  diagonal  relationship  with  Silicon;  both  are   semiconductors
metalloids & forms covalent compounds.
•    Boron compounds  are  electron  deficient,  they are  lack of an octet  of electrons
about the B atom .
•  diborane  B2H6
,  is  simplest boron hydride

•  Structure: three-center two-electron: the  H atoms are  simultaneously bonded to
two B atoms the B-H bridging bond  lengths are greater than B-H terminal.
•  - Boron  oxide is acidic (it reacts readily with water to f orm boric acid)
•  aluminium compounds:aluminium oxide is amphoteric
•  aluminum   hali des,    e.g.,    AlCl
3
is   dimer,    an   important   catalyst   in   organic
chemistry  have  anincomplete  octet,  acts  as  Lewic  acid  by  acc epting  lone  pairs
from Lewic bases, forming adduct
•  aluminum hydride, e.g., LiAlH4
,  a reducing agen

. Atomic and  ionic radii
•  The  atomic  and  ionic  radii  of  group  13  elements  are  compared  to
corresponding  elements  of  group  2.  From  left  to  right  in  the  period,  the
magnitude  of  nuclear  charge  increases  but  the  electrons  are  added  to,  the  same
shell.    These  electrons  do   not   screen   each  other,   theref ore,   the  electrons
experience greater nuclear charge.
•  In  other  words,  effective  nuclear  charge  increases  and  thus,  size  decreases.
Therefore,  the  elements of this group  have smaller size than the  corresponding
elements of  second group.
•  On   moving  down  the   group  both  atomic  and   ionic   radii  are  expected  to
increase  due  to  the  addition    of  new  shell s.  However,  the  observed  atomic
radius of  Al (143 pm) is slightly more than that of  Ga (l35 pm).
Ionization energies
The  first  ionization  energies  of  group  13  elements  are  less  than  the  corresponding
me mbers of  the alkaline earths.

The sharp decrease in I.E.  from B to  Al is due to  increase in size. In case of Ga, there
are ten d-electrons in its inner electronic configuration.
The  very  high  value  of  3
rd
I.  E.  of  thallium  indicates  that  +3  O.N.  state  is  not  stable,
rather +1 is more stable  f or thallium  .
Electropositive (or metallic) character
the elements of group 13 are  less electropositive as compared to ele ments of group  2.
On  moving  down the  group  the  electropositive  (metallic)  character  increases because
ionization  energy  decreases.  For  e.g.,  Boron  is  a  non -metal  white  the  other  elements
are typical metals.
Oxidation states
The  common  oxidation  states  of  group  13  elements  are  +3  and  +  l  .The  stability  of
the  +  1  oxidation  state  increases  in  the  sequence  Al  <Ga<  In  <Tl,  Due  to  Inert  pair
effect.

Hydrides
•  None  of  the  group  13  ele ments  reacts  directly  with  hydrogen.  However,  a  no.
of  hydrides  of  these  elements  have  been  prepared  by  indirect  methods.  The
boron  hydrides  are  called  boranes&  classified  in  two  s eries:  (a)  BnHn+4
called nidoboranes  (b)  BnHn+6
called arachnoboranes
•  INUDUSTRIAL PREPERATION  :-2BF
3
(g) + 6LiH(s) → B2H6
(g)  + 6LiF(s)
•  Laboratory method:
(i)  By  the  reaction  of  iodine  with  sodium  borohydride  in  a  high  boiling
solvent.
2NaBH4
+  I
2    →  B2H6
+  2NaI  + H2
(ii)  By reduction of BCl
3
with LiAlH4
4BCl
3
+  3LiAlH4   →  2 B2H6
+  3AlCl
3
+  3 LiCl


Some important characteristics of  boranes:
i) Lower boranes are colourless gases while higher boranes are volatile liquids
or solids.
ii) They undergo spontaneous combustion in  air due to strong affinity of  boron
for oxygen.
B2H6
+  3O2    →  B2O3
+  3H2O  + Heat
iii)   Boranes   react   with   alkali   metal   hydrides  in   diethyl   ether   to  form
borohydride complexes.
B2H6
+  2MH →2M
+
[BH4
]
-
(M= Li  or Na)
Metal borohydride
•  (iv) Diborane reacts with ammonia to give borazine at 450 K.
B2H6
+  6NH3  →  3B3N3H6
+  12H2
•  Borazine  has a  cyclic  structure similar  to benzene and thus  is   called  inorganic
benzene
•  The other ele ments of this  group  form only a  few  stable  hydrides. The  thermal
stability decreases as we move down  the group.

•  AlH3
is a colourless soli d polymerized via Al - H - Al  bridging units. These
hydrides are weak Lewis acids and readily form adducts with strong Lewis
base (B:) to give compounds of  the type MH3
(M = Al  or Ga). They also  form
complex-tetrahydrido anions, [MH4]-. The most important tetrahydrido
compound  is  Li[AlH4
]
ether
4LiH  + AlCl
3 ―――→  LiAlH4
+  3LiCl
Dimeric structure of  aluminium chloride
–  Boron halides do  not  form dimers because  the  size of boron
is  so  small  that  it  is   unable  to  coordinate  four  large -sized
halide ions.

Anomalous properties of  boron
1. Boron is a non-metal & bad conductor of electricity whereas aluminium is a metal
&  good conductor. B is  hard but  Al is a soft metal.
2.  Boron exists in  two forms-crystalline and amorphous. But Al does not exist in
different  f orms.
3.  The melting  and boiling  point of  boron are much higher  than that of Al .
4.  Boron forms only covalent compounds whereas Al forms even some  ionic
compounds.
5.  The hydroxides and oxides of  boron are acidic in nature whereas those of
aluminium are amphoteric.
6.  The trihalides of  boron exist as monomers. On  the other hand, aluminium halides
exist as dimers .
7.  The hydrides of  boron are quite stable whil e those of  aluminium are unstable


Group 14 Ele ments:-The Carbon Family
Group 14 includes carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb).
General electronic configuration of  carbon fa mily  is ns
2
np
2
.
Covalent radius:-Covalent radius expected to increase from Cto Si,
From Si to Pb small increase is  found.
Ionization  Enthalpy:-The  first  ionization  enthalpies  of  group  14  elements  are  higher
than those of  the corresponding group 13 elements.
Electronegativi ty:-Group  14  ele ments  are  smaller  in  size  as  compared  to  group  13
elements that‘s why this group ele ments  are slightly  more electronegative than group
13
Chemical  properties:-Carbon   and   silicon   mostly   show   +4   oxidation   state.   Germanium   forms   stable
compounds  in +4 state and only few compounds in +2 state.
Tin f orms compounds in  both oxidation states. Lead compounds  in +2 state are stable
and in +4 state are strong oxidizing agents.
Exception:-Pb
4
and SnF
4
are ionic in nature.
Except CCl
4
other tetrachlorides are easily hydrolysed by water.

Allotropes of  Carbon:-The three types of  all otropes are –

1-Diamond  2-Graphite  3-Full erence
Dia mond:-In  diamond each carbon atom under goes SP
3
hybridisation. Each carbon is
tetrahedrally linked to four other carbon atoms.
Graphite:-In graphite,  carbon is SP
2
-hybridized  graphite has a two-dimensional sheet
like structure consisting  of  a number of  hexagonal rings fused together.
Graphite conducts electricity along the sheet. It is very soft and Slippery
Fullerence  was  discovered  collectively  by  three   scientists  namely  R.E
Smalley,R.F Curl and H.W  Kroto

1.  Why is boron used in nuclear reactions?
Ans:-Because Boron can absorb neutrons.
2.  By giving  a balanced equation  show how B(OH)
3
behaves as an acid in water.
Ans:-B(OH)
3
+2H2O  [B(OH)4
]
-+H3O
+
3.Name   the   ele ment   of   group   14   which   exhibits   maximum   tendency   for
catenation?
Ans:-Carbon
4.  What is the basic building unit of  all  silicates?
Ans:-SiO4
4-is the basic unit of all  silicates.
5.  What happens when NaBH4
reacts with iodine?

6.  What happens when  boric acid  is heated
Ans:-4H3BO3
4HBO2  H2B4O7
.
7.  What is producer gas?
Ans:-Producer gas is a mixture of  CO and N2
in  the ratio of  2:1.
8.Write the state of hybridization  of  ‗B‘ in BF
3
.
ANS:-Hybridisation of  ‗B‘ in BF
3
is Sp
2
.
9.Mention the state of hybridization in B in BH4
-.
Ans:-Sp
3
.
10.  Which  oxide of  carbon is regarded as anhydride of  carbonic acid.
Ans:-CO2
is regarded as a hydride of  carbonic acid .

What happens when
(i)  Quick lime  is heated with coke?
(ii)  Carbon monoxide reacts with Cl
2
Ans:- (i)  Cao +3C  →  CaC2
+CO
(iii)  CO +Cl
2→    COCl
2

Give reason
(i) C and Si are always tetravalent but Ge,Sn,Pb show divalency.
(ii) Gallium has higher  ionization enthalpy than Al. Explain.
Ans:-(i)  Ge,  Sn,  Pb  show  divalency  due  to  inert  pair  effect,  Pb
2+
is  more  stable  than
Pb
4+
.
(ii)  Due  to  poor  shielding  effect  of  d-electrons  in  Ga  effect ive  nuclear
charge  increases as compared to Al thus the  I.E is higher  than Al.


If  B-Cl  bond  has  a  dipole  moment,  Explain  why  BCl
3    molecule  has  zero  dipolemo ment.

Ans:- B-Cl bond has dipole mo ment because of  polarity.In BCl
3
since the molecule issymmetrical thus the polarities cancel out.

What do you understand by-(a) Inert pair effect:-The pair of electron in the valence shell does not take part in
bond  formation it is called inert pair effect.
(b) Allotropy:-It is the property of the  element by which an element can exists in
two  f orms  which have  same chemical properties  but different physical properties due
to their structures.

Give reason for the  f ollowing observations:-(a)  The tendency for catenation decreases down  the group in Group 14.
(b)  The  decreasing  stability  of  +3  oxidations  state  with  increasing  atomic
number in group 13.
(c)  PbO2
is a stronger oxidizing agent than  SnO2.
(d)  Molten aluminium bromide  is a poor conductor of  electricity.
Ans:- (i)(a) It  is due  to decrease  in bond dissociation energy which is  due to  increase
in atomic size.
C-C > Si-Si >Ge-Ge>Sn-Sn>Pb-Pb.
(b) It is due to inert pair effect.
(c)  PbO2
is  stronger  oxidizing  agent  than  SnO2
because  Pb
2+
is  more
stable than Pb
4+
whereas Sn
4+
is more stable than Sn
2+
.
compound.
(d)   Molten  AlBr
3
is  poor  conductor of electricity  because  it  is  covalent.

CHEM..ORGANIC..XI

Resonance  Effect  :  The  polarity  produced  in  the  molecule  by  the  interaction of  two
pi  bonds or between a pi bond and lone pair of  electron present on an adjacent atom.
There are two types of  resonance effect:
1)  Positive  resonance  effect  :  In  this  effect  the  transfer  of electr ons  is  away  fro m
an atom or substituent group attached to the conjugated system.
The  atoms  or  groups  which  shows  +R  effect  are  halogens, -OH  ,  -OR,-NH2
2)  Negative  resonance  effect  :  In  this  effect  the  transfer  of  electrons  is  towards
the  atom or substituent group attached to the conjugated system.
The atoms or groups which  shows  -R effect are –COOH , -CHO , -CN

Q.1. What effect does branching of an alkane chain has on its boiling point? 

Ans. As the branching increases, the surface area of an alkane approaches that of a sphere. Since a sphere has minimum surface area, therefore, Vander Waal forces of attraction are minimum and hence the boiling point of the alkane decreases with branching.

Q2. Among Cis and Trans structure of Hex-2-ene which has higher boiling point and why?

Ans: The dipole moment of a molecule depends upon dipole-dipole interactions. Since cis-isomer has higher dipole moment, therefore it has higher boiling point

Q3. Write IUPAC names of the following compounds )

(a) CH₃CH=C(CH₃)₂ (b) CH₃CH(CH₃)CH=CH₂

Ans. (a) 2-Methylbut-2-ene                  (b) 3-Methylbut-1-ene

Q.4. Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?(L-II)

Ans. -CH₃ group is electron donating while –NO₂ group is electron withdrawing. Therefore, maximum electron density will be in toluene, followed by benzene and at least in m-dinitrobenzene. Therefore, the ease of nitration decreases in the order:

                  

Toluene> benzene> m-dinitrobenzene.

Q5. Explain why (CH₃)₃C⁺ is more stable than CH₃CH₂⁺ and ⁺CH_3.  (L-I)

Ans. Greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation interaction and +I effect of methyl group greater is the stabilisation of the cation.Thus, CH₃)₃C⁺ is more stable than CH₃CH₂⁺ and ⁺CH_3.

Q6. Explain electromeric effect, resonance effect and hyperconjugation.

Ans: Electromeric effect is defined as the complete-electrons totransferoneof  o

the atoms joined by a multiple bond on the demand of an attacking reagent. It is represented by E

and the shifting of the electrons is shown by a curved arrow.

The resonance effect is defined as ‘the polarity produced in the molecule by the interaction of two-bndsπ or between-bondandlone pairaof electronsπ present on an adjacent atom’. The effect is transmitted through the chain.

Hyperconjugation involves delocalisation of σ electrons—Hbondof an ofalkyl groupC directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. The—H σbondelectronsofthealkylgroupenterofinto partialC conjugation with the attached unsaturated system or with the unshared p orbital. Hyperconjugation is a permanent effect.

CHEM..CLASS XI..ORGANIC..QUESTIONS..

Q.1. Write structures of different chain isomers of alkanes corresponding to the molecular
formula C4H10 and C5 H12. Also write their IUPAC names.
Q.2. Write IUPAC names of the following compounds :
(i) (CH3)3CCH2C(CH3) 3
(ii) (CH3) 2 C(C2H5) 2
(iii) tetra – tert-butylmethane
Q.3. Write structural formulas of the following compounds:
(i) 3-Methylhexane
(ii) 3-ethyl-2, 2–dimethylpentane
Q.4. Define conformers/rotamers.
Ans.The spatial arrangements of atoms which can be converted into one another by
rotation around a C-C single bond are called conformations or conformers or rotamers

Q.5. Write IUPAC names of the following compounds:
(i) CH3 – CH = CH2
(ii) CH3 – CH2 – CH = CH2
(iii) CH3 – CH = CH–CH3

Q.6.  Out of pentane, 2-methylbutane and 2,2-dimethylpropane which has the highest
boiling point and why?
Ans. Pentane having a continuous chain of five carbon atoms has the highest boiling point
(309.1K) whereas 2,2 –  dimethylpropane boils at 282.5K. With increase in number of
branched chains, the molecule attains the shape of a sphere. This results in smaller area of
contact and therefore weak intermolecular forces between spherical molecules, which are
overcome at relatively lower temperatures.

Q.7. Calculate number of sigma (σ) and pi (π) bonds in the following
(i)  CH2 = C (CH2CH2CH3) 2
(ii)   CH2 = CH – CH – CH3
|
CH3
(iii) CH3 – CH = CH–CH3

Q.8.  Write structures and IUPAC names of different structural isomers of alkenes
corresponding to C5H10

Ans.pent-1-ene
pent-2-ene
2-methylbut-2-ene
3-methylbut-1-ene
2-methyl-but-1-ene

Q.9. Discuss about structural isomerism and stereoisomerism.
Ans. Structural Isomerism: Compounds having the same molecular formula but different
structure i.e. different arrangement of atoms within the molecule are called structural
isomers and the phenomenon is called as structural isomerism.
Types:
•   Chain isomerism
•   Position isomerism
•   Functional isomerism
•   Metamerism
•   Tautomerism
    Stereoisomerism: Isomers which have the same structural formula but have
different relative arrangement of atoms or groups in space are called stereoisomers and the
phenomenon is called as stereoisomerism.

Q.10. Draw cis and trans isomers of the following compounds. Also write their IUPAC names
:(i) CHCl = CHCl
(ii) C2H5CCH3 = CCH3C2H5

Q.11. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-ene
(i) in the absence of peroxide and
(ii) in the presence of peroxide.

Q.12. Define Hückel rule.
Ans.  Huckel Rule gives information about the aromaticity. According to this rule, the
necessary and sufficient conditions for a molecule to be aromatic are:
   Planarity
   Complete decolisation of the pi electrons in the ring.
   Presence of (4n+2) pi electrons in the ring where n is an integer (n= 0,1,2…)

Q.13How would you convert the following into benzene?
(i)Ethyne   (ii) Ethene   (iii) Hexane

Q.14. Out of benzene, m–dinitrobenzene and toluene which will undergo
nitration most easily and why?
Ans. -CH3 group is electron donating while –NO2 group is electron withdrawing. Therefore,
maximum electron density will be in toluene , followed by benzene and at least in m-dinitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene> benzene> m-dinitrobenzene

Q.15. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also
give reason for this behaviour.
Benzene Hexane    Ethyne
sp
3  
sp
2
sp
Since s-electrons are close to the nucleus, therefore as the s-character of the orbital making
the C-H bond increases, the electrons of C-H bond lie closer and closer to the carbon atom.
In other words, the partial positive charge on the H-atom and hence the acidic character
increases as the s character of the orbital increases.
Thus the acidic character decreases in the order:
Ethyne> Benzene> Hexane

Q.16.Why does benzene undergo electrophilic substitution reactions easily and nucleophilic
substitutions with difficulty?

Ans. Due to the presence of an electron cloud containing 6 pi electrons above and below the
plane of the ring, benzene is a source of electrons. Consequently, it attracts the
electrophiles towards it and repels nucleophiles. As a result, benzene undergoes
electrophilic substitution reactions easily and nucleophilic substitution with difficulty.

Q.17. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and
why?

Ans. cis-Hex-2-ene & Trans-Hex-2-ene

The dipole moment of a molecule depends upon dipole-dipole interactions. Since cis-isomer
has higher dipole moment, therefore it has higher boiling point.

Q.18.  Explain the extra ordinarily stability of benzene though it contains three double
bonds?
Ans. Resonance or delocalisation of electrons usually leads to stability. Since in benzene all
the six pi electrons of the three double bonds are completely delocalised to form one lowest
molecular orbital which surrounds all the carbon atoms of the ring, therefore, it is extra-ordinarily stable.
Q.19. What are the necessary conditions for any system to be aromatic?
Ans. The necessary conditions for a molecule to be aromatic are:
(i)It should have a single cyclic cloud of delocalised pi electrons above and below the plane of
the molecule.
(ii)It should be planar. This is because complete delocalisation of pi electrons is possible only if
the ring is planar to allow cyclic overlap of p-orbitals.
(iii)It should contain Huckel number of electrons, i.e. (4n+2) pi electrons where n=0,1,2,3….etc.

CLASS XII CHEMISTRY

Smallest repeating unit in a space lattice is called unit cell.
Thereare4 typesofunitcells,7crystalsystemsand14bravaislattices.
. Types of unit cell No.ofatomsperunitcell
 Simple cubicunitcell 8x1/8=1
 FCC(Facecenteredcubic) 8x1/8+6x1/2=4
BCC(Body centeredcubic) 8x1/8+1x1=2
 End centered 8x1/8+1/2 x 2 = 2 10.
Hexagonalclosepackingand cubicclosepackinghaveequal packing efficiencyi.e.74% 11. Packingefficiency= Volume occupied by the sphere*100/Volume of unit cell

Radius ratiois the ratio of radius of void to the radius of sphere .
For tetrahedral void radius ratio = 0.225
Foroctahedralvoidradiusratio = 0.414
.No.oftetrahedralvoid=2x N (N=No.ofparticles)
.No.ofoctahedral void=N

Simplecubicunitcell a=2r
. FCC a=4r/√ 2
 BCC a=4r/√ 3

Point defect or Atomic defect‐>it is the deviation from ideal arrangement of constituent atom
.Point defects are two types
(a)Vacancy defect
(b)Interstitial defect

Frenkel defect
 is the combination of vacancy and interstitial defects.Cations leave their  actual
lattice sites and come to occupy the interstitial spaces.Density remains the same
eg.AgCl.

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CLASS XI CHEM PERIODIC TABLE...SOME IMP Q&A

Q.State Mendeleev Periodic Law.
Ans: The chemical and physical properties of the elements are periodic functions of their atomic
masses.
Q State modern periodic law.
Ans-The chemical and physical properties of the elements are periodic functions of their atomic
numbers.Mosley state this law.
Q.Why are isotopes not considered in Mendeleev’s periodic table?Name the elements which are
predicted by Mendeleev as eka –silicon and eka –aluminum ?
Ans.Because Mendeleev gave priority to similarities in properties.
Germanium and gallium.
Q. What are Dobereiner’s triads? Give example of 2 triads ?
Ans: he arranged the elements in several sets of 3 elements .these sets were called triads which
resembled with each other in chemical properties and atomic mass of the middle element was
approx the mean of the atomic masses of the remaining two.


Example (i) chlorine, bromine and iodine A.T of cl =35.5 and iodine =127 .middle element bromine
has (35.5 +127)/2 =81.25
(ii) Calcium (40) ,strontium(88.5) and barium (137)

Q Write the general E.C of s, p, d and f Block elements.(L2)
Ans:   (i) s block- ns
1-2
(ii) p block – ns
2
np
1-6
(iii) d block – (n-1)d
1-10
ns
1-2
(iv) f block -- (n-2) f
1-14
(n-1)d
0-1
ns

Q Assign the position of the element having outer electronic configuration
(i)ns
2
np
4
for n=3 , {L3]
(ii) (n-1) d
2
ns
2
for n=4
(iii) (n-2)f
7
(n-1)d
1
ns
2
for n=6
Ans: (i) n= principal no. which shows no of periods. N=3 so E.C =3s23p4=elements having position 3
rd
period 16
th
group.
(ii) n=4 so elements belong in 4
th
period .it is d block element .d2= 2 electron in d orbital it belongs to
4
th
group in d block.
(iii) n=6 and electron goes in fshell, so it is f block element having 6
th
period and 7
th
element of
lanthanides series.

Q. Which out of F or Cl has a more negative electron gain enthalpy?
Ans. Cl has more negative electron gain enthalpy. It is because there is more inter electronic
repulsion between valence electrons of F, due to smaller size than Cl.

Q.1 Consider the following species
N
-3
, O
-2
, F-, Na
+
, Mg
+2
and Al
+3
(a). what is common in them?
(b). arrange them in the order of increasing ionic radii.
Ans. (a). They have same number of electrons.
(b). Al
3+
<Mg
2+
<Na
+
<F
-<O
2-<N
3

Q. Explain Screening effect ?

Ans..Screening effect. The inner electrons between valence electron and nucleus shield the
valence electrons from nucleus; it is called shielding effect.
Q.2 (a). Which is largest in size- Cu
+
, Cu
+2
, Cu and why?
Ans.. Cu is largest due to less effective nuclear charge. It has 29 electrons, 29 protons, Cu
+
has 28
electrons and 29 protons, Cu
2+
has 27 electrons and 29 protons.

Q. How does electro negativity vary down the group 17 and why? How does it vary from left to
right in the period? Name an element having highest electro negativity.
Ans. Electro negativity decreases down the group due to increase in atomic size. It increases along
the period from left to right due to decrease in atomic size.
Fluorine (F) has highest electro negativity.

Q.1 Would you expect the second electron gain enthalpy of O as positive, more negative or less
negative than the first? Justify your answer.
Ans. Second electron affinity of O is largely +ve because of repulsion between negatively charged
ions and second electron to be added. Energy required to overcome repulsion is more than the
energy released in gaining electron, so net energy is absorbed.

Q.Explain why Halogens have very high negative electron gain enthalpy?
Ans.The electron gain enthalpy for halogens is highly negative because they can acquire the nearest
stable noble gas configuration by accepting an extra electron.

Q. How would you explain the fact that the first ionization of Na is lower than that of Mg but its
second ionization enthalpy is higher than that of Mg? (L-3)
Ans. The electronic configurations of Na and Mg are: Na:1s
2
2s
2
2p
6
3s
1
And Mg: 1s
2
2s
2
2p
6
3s
2
thus, the
first electron in both the cases has to be removed from the 3s orbital, but the nuclear charge of Na is
lower than that of Mg, therefore, the first ionization energy of Na is lower than that of Mg. But after
the removal of the first electron, Na acquire the nearest noble gas configuration (Ne) which makes
Na highly stable, but an electron is still left in the p orbital of Mg. Hence second ionization energy of
Na is higher than that of Mg.

Q. Explain why Electron gain enthalpy of noble gases is positive. (L-1)
Ans. Noble gases have large positive electron gain enthalpies because the extra electron has to be
placed in the next higher principal quantum energy level, thereby producing highly unstable
electronic configuration.

Q. Explain why cations are smaller and anions are larger in size than their parent atom. (L-2)
Ans.The ionic radius of cation is always smaller than the parent atom because the loss of one or
more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus
for the electrons increases and hence, the ionic radii decreases. In contrast, the ionic radius of an
anion is always larger than its parent atom because the addition of one or more electrons decreases
the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons
decreases and hence the ionic radii increases.

Q Arrange the following order of the property indicated:
(a). F, Cl, Br and I (negative electron gain enthalpy)
(b). Mg, Al, Si and Na (ionization enthalpy)
(c). C, N, O and F (second ionization enthalpy)

Q. Arrange the following elements in increasing order of metallic character: B, Al, Mg and K.

Q. Why is fluorine (F2) more reactive than chlorine (Cl2)?

Q.Explain why
(i). Be has higher first ionization enthalpy than B?
(ii). O has lower first ionization enthalpy than N and F?

Q.3 ∆iH1 value of Mg is more as compared to that of Na while its ∆
iH2 value is less. Explain.
Ans. ∆iH1 value of Mg is more than that of Na due to greater symmetry and smaller size. But ∆iH2
value of Na is higher because Na
+
ion has the configuration of noble gas element neon while Mg
+
ion
does not have a symmetrical configuration.

Q. The increasing order of reactivity among group 1 elements is Li<Na<K<Rb<Cs whereas among
the group 17 elements, it is F>Cl>Br>I. Explain

CHEMISTRY..STRUCTURE OF ATOM..Q&A

Q1 State Heisenberg Uncertainty Principle ?

Ans. It is impossible to determine the exact position & exact momentum or velocity of microscopic particle.

Q2-Write E.C. for Oxygen ?

Q3. Which orbital has quantum no(a) n= 3 & l=1 

(b) n= 2, l= 0  

Ans: (a)3p orbitals.  (b) 2s

Q4.Find the wave length of Yellow emission of sodium whose frequency 5.09 x 10

14

per sec ?

Ans:(hint: wave length = C/V0)

Q5. How was Neutron discovered? Write equation only?

Q6. Calculate the wave length of an electron whose KE is 3x 10

-25 J ?

Q7. An element with mass number 81 contains 31.7% more neutrons as compare to protons find out atomic symbol of element(Br)

Q8. What is threshold frequency? 

Ans –The minimum frequency required to eject out electron from an atom.

Q9.Write the draw back of Rutherford Atomic Model ?

Ans   (1). He was unable to explain the stability of Atom.

(2). He could not explain the line spectrum.

Q10.Give the points of Rutherford Model ?

Ans. 1. Atom is neutral.

2.Whole mass is concentrated in nucleus.

3.Electrons are revolving in orbits.

Q10-A 10 g Table Tennis Ball is moving with a speed of 90 m/s which is measured with an accuracy of 4% find out uncertainty in it speed as well as position.

Ans:

m = 10g

m = 0.01kg

∆v = 4% of 90 m/s

= 3.6 m/s

∆x = h/4π m∆v

Put the values

=1.46 x 10

-33

m

Q.11 How many electrons with  l=2 are there in an atom having atomic number 54 ?

Ans. 3d

10 

& 4d

10 

= 20.

Q. What is electronic configuration of Cu (z=29) and Cr (z=24) by aufbau principle?

Q.Define Pauli’s exclusion principle ?  

Ans. No two electrons in an atom can have same set ofall four quantum numbers. If three are same one Quantum number must have different value.         

Q.Define Aufbau Principle. Give a Suitable example ?

Q. a) Where is the probability of finding an electron is zero?  

b) How many maximum electrons can be found in 3d orbital with spin quantum

number -1/2? 

c) Which quantum no. determines the orientation of the orbital? 

Ans.a) at Node.                

b) 5-electrons.

c) magnetic quantum no

Q. Why is s-orbital orbital considered non directional while p-orbital directional?

Ans. Electron density in s-orbital is equal in all direction while it dumble shaped along one axis in p-orbitals.

Q4.Which of the following can exhibit maximum Para magnetism: Ca, Fe

+2

,Cu

+2

?

Ans.Fe

2+ 

due to presence of 4 unpaired electrons in 3d orbitals.  

Q..What will be the increasing order of energy of 2s, 3s, 3d, 4f in atom? 

Ans. 2s< 3s < 3d < 4f 

Q.Is the energy of an electron in 2s orbital same as of that in 2p orbital? Why? 

Ans. No, the energy depends upon (n+l) values in an atom. Hence 2p orbital has greater energy than 

2s

Q.“In reality all the s-orbitals are spherically symmetric.” What does this represent?

Ans. The probable density of an electron in s-orbital is equally distributed at all coordinates in space. 

Q.3  Give three differences between orbit and orbital.

Ans

ORBIT                                                                                             

1 It is a well defined circular path around the 

nucleus in which electron move .

 2 Maximum no. of electrons in an orbit is 

2n

2

. 

3.  Do not have directional characteristics.     

  ORBITAL           

 1 It is a 3dimensional region around the nucleus where 

the probability of finding electron is 

maximum.  

2 Maximum 2 electrons in an orbital.

3. Except s all have directional characteristics

QExplain the significance of Azimuthal quantum number ?

Ans. It refers as sub-energy level of an atom.

Q. How many of the following are possible: 1p, 2s, 3p, 3f, 3d. ?

Ans.2s , 3p & 3d are possible.

Q. n = 3, l = 3, m

l 

= -3, ms

= +1/2. Is the given set of quantum numbers possible? If not, explain by 

giving reasons. 

Ans. It is not possible because 3f does not exist in an atom

Q.Explain why half-filled orbitals have extra stability ?

Ans. It has symmetrical electron density as well as greater exchange energy.

Q. Explain the significance of spin quantum number ?

Ans.The direction of spinning of an electron in an orbit as well as the net magnetic moment of an orbital.

Q..   What will be the value of n & l of the followings? (L1)

1.  4s

1

2.  5d

1

3.  4f

1

Q.  What is Hund’s rule of maximum multiplicity? What is its significance?

Ans Pairing of electrons in an orbital takes place only when all orbitals of same sub-shell become singly occupied. It helps to equalize energy of a sub-shell

CHEMISTRY..SOME SIMPLE Q&A..

Q1. Define the term precision and accuracy?

Ans. The term precision refers for the closeness of set of values obtained from identical
measurement of a quantity, whereas accuracy is a related term, refers to the closeness of a single
measurement to its true value.

Q2-What do you mean by significant figure?

Ans. Total number of digits in a number including the last digit whose value is uncertain is called the
number of significant figure.

Q4. Define the law of conservation of mass ?

Ans. Matter can neither be created nor destroyed in the course of a Physical or chemical process
although it may change from one form to another.

Q5- How many significant figure are present in the following ?

(I)  0.0025 (II) 208 (III) 5005   (IV) 126000

Ans.- (I) 2    (II) 3      (III) 4    (IV) 3

Q6  Define Molarity ?

Ans.  It is the number of moles of the solute dissolved per litre of the solution.
Q.7  Define Molality ?

Ans.    It is the number of moles of the solute dissolved per 1000 g of the solvent.

Q8  Define mole?

Ans.  The amount of substance that contains the same number of entities(atoms, molecules, ions
or other particles) as the number of atoms present in 12g of carbon-12 isotope.

Q9   How are 0.50 m Na2CO3 and 0.50 M Na2CO3 different?

Ans.   0.50 m Na2CO3 solution means that 0.5 mol (or 53g) of Na2CO3 are present in 1000 g of
solvent.
0.50 M Na2CO3 solution means that 0.5 mol (or 53g) of Na2CO3 are present in 1L of solution

Q10.   a) Why Molarity varies with temperature?
b) Which concentration term is suitable when solute is present in very minute quantities or
traces?
c) What is the sum of mole fraction of all the components of a solution?

Ans.  a) Because it is number of moles of solute per litre of solution, and volume varies with
temperature.
b) ppm(Parts per million)
c) 1

Q11.   Calculate the mass of an atom of silver (atomic mass= 108)

Ans.  Mass of 6.022x10
23
atoms of Ag = 108 g
Mass of one atom of Silver
108
6.022X10
23
= 1.79x10
-22
g

Q12.   Calculate the number of moles of iodine in a sample containing 1.0x10
22
molecules?

Ans.  6.022x10
23
molecules of iodine = 1 mol of iodine
1.0x10
22
molecules of iodine =
1
6.022X10
23X1.0 X10
22
= 0.0166 mol of iodine

Q13. Calculate the volume occupied by 10 mole of CO2 at STP. ?

Ans.  1 mol of CO2 atSTP = 22.7 L
10 mole of CO2 at STP = 22.7 x 10 =227 L

Q14 A solution is prepared by dissolving 2g of the substance A in 18 g of water. Calculate the mass
percentage of solute ?

Ans.  Mass of solute A = 2g
Mass of water = 18g
Mass of solution = 2+18 = 20g

Mass percentage =
Mass of A/
Mass of solution
*100g
=
2/
20
* 100

Q15   In three moles of ethane (C2H6) calculate the following:
(a)  Number of moles of carbon
(b)  Number of moles of hydrogen atom
(c)  Number of moles of ethane

Ans.  (a)   1 mole of C2H6 contain = 2 moles of carbon
3 moles of C2H6 contain = 6 moles of carbon
b)    1 mole of C2H6 contain = 6 moles of hydrogen atom
3 moles of C2H6 contain = 18 moles of hydrogen atom
c)  1 moles of C2H6 contain = 6.022 x 10
23
molecules
3 moles of C2H6 contain = 6.022 x 10
23
x 3 = 1.0807 x 10
24
molecules.

Q17  A solution is 25% H2O, 25% C2 H5 OH and 50% acetic acid by mass. Calculate mole fraction of each component.

Q18  Calculate the molarity of pure water (density of water = 1 g mL
-1
) . (L 3)

Ans.  Density of water = 1 g mL
-1
Mass of 1000 mL of water = Volume x Density
1000 x 1 = 1000 g
Moles of water =
1000/18
= 55.55

Q18  Calculate the weight of carbon monoxide having same number of oxygen atoms as are
present in 88 g of carbon dioxide.

KINEMATICS..PHYSICS..PART 1

Rest and Motion are relative terms, nobody can exist in a state of absolute rest or of
absolute motion.

Speed:- It is rate of change of distance covered by the body with respect to time.
Speed = Distance travelled /time taken
Speed is a scalar quantity
S.I. unit is m /s
dimensional formula is[M
0
L
1
T
-1
]

*Uniform Speed:-  If an object covers equal distances in equal intervals of time than the
speed of the moving object is called uniform speed.
In this type of motion, position –time
graph is always a straight line.

*Instantaneous speed:-The speed of an object at any particular instant of time is called
instantaneous speed. In this measurement, the time ∆t→0.
When a body is moving with uniform speed its instantaneous speed = Average speed =
uniform speed.

*Velocity:- It is defined as the displacement covered by an object per unit time.
                                             Velocity =Displacement /Time
                                               Velocity is a vector quantity
                                             SI unit is m /s


Its  dimensional formula  is (see speed)

difference between speed & velocity :
1.Speed is scalar but velocity is vector quantity.
2.speed may be +veor zero where as velocity may be zero,+ve or -ve.

*Acceleration:-  The rate of change of velocity of an object  with respect to time  is called 

its acceleration.
Acceleration = Change in velocity /time taken
unit-m/s(2)
It may be positive ,negative or zero.
*Positive Acceleration:- If the velocity of an object increases with time , its acceleration is 
positive .
*Negative Acceleration :-If the velocity of an object decreases with time ,  its acceleration 
is negative . The negative acceleration is also called retardation or deacceleration.

PHYSICS..Unit - I Physical World And Measurement

There are four fundamental forces which govern both macroscopic and microscopic 
phenomena. These are 
(i)  Gravitational force 
(iii) Electromagnetic force 
(ii)  Nuclear force
 (iv) Weak force

All those quantities which can be measured directly or indirectly and in terms of which 
the laws of physics can be expressed are called physical quantities

The units of the fundamental quantities called fundamental units , and the units of 

derived quantities called derived units.

DIMENSION & DIMENSIONAL FORMULA:
The equation which expresses a physical quantity in terms of the fundamental 

units of mass, length and time, is called dimensional equation.

According to the principle  of homogeneity  a physical equation will be 
dimensionally correct if the dimensions of all the terms in the all the terms 

occurring on both sides of the equation are the same.

There are three main uses of the dimensional analysis-
1.To check the correctness of a given physical relation.
2.To derive a relationship between different physical quantities.

  Significant figures:  -  The significant figures are normally those digits in a 
measured quantity which are known reliably plus  one additional digit that is 
uncertain.
RULES OR FINDING SIGNIFICANT FIG.:

(i)  All non- zero digits are significant figure.
(ii)  All zero between two non-zero digits are significant figure.
(iii)  All zeros to the right of a non-zero digit but to the left  of an understood decimal 
point are not significant. But such zeros are significant if they come from a 
measurement.
(iv)  All zeros to the right of a non-zero digit but to the left of a decimal point are 
significant.
(v)  All zeros to the right of a decimal point are significant.
(vi)  All zeros to the right of a decimal point but to the left of a non-zero digit are not 
significant. Single zero conventionally placed to the left of the decimal point is 
not significant.

(vii)  The number of significant figures does not depend on the system of units.

PHYSICS..SOME IMPORTANT PHYSICAL CONSTANTS

Mass of an electron (Me) = 9.1x10
-31
kg.
2.  Mass of a proton ( Mp) = 1.6725 x 10
-27
kg.
3.  Mass of a neutron (Mn) = 1.6746 x 10
-27
kg.
4.  Charge of an electron (e) =-1.6 x 10
-19
c
5.  Speed of light in vacuum (c) = 3 x10
8
m/sec.
6.  Planck Constant (h) = 6.6256 x 10
-34
J x sec .
7.  Universal Gravitation constant (G) = 6.67 x 10
-11
Nm
2
/ kg
2.
8.  Avogadro Number (NA) = 6.023 x10
23
mol
-1
.
9.  Boltzmann constant (K) = 1.38 x 10
-23
J/K

10.Radius of the earth (Re) = 6400 Km. = 6.4x10
6
m

PHYSICS..GRAVITATION..SOME IMP Q&A

Q1.What are the factors which determine ; Why some bodies in solar system have 
atmosphere and others don’t have?
Ans. The ability of a body (planet) to hold the atmosphere depends  on acceleration due to gravity.
Q2.Should the speed of two artificial satellites of the earth having different masses but the same orbital radius, be the same?

Ans.Yes it is so because the orbital speed of a satellite is independent of the mass of a satellite. Therefore the speeds of the artificial satellite will be of different masses but of the same orbital radius will be the same.

Q3.Can a pendulum vibrate in an artificial satellite?
Ans. No, this is because inside the satellite, there is no gravity ,
i.e., g=0.

As t = 2π     , hence, for g=0  , t =    . Thus, the pendulum will not vibrate.

Q4.Why do different planets have different escape speed?
Ans. As, escape speed =        ,  therefore its value  are different for different
planets which are of different masses and different sizes.
Q5.Show that weight of all body is zero at Centre of earth?
Ans. The value of acceleration due to gravity at a depth d below the surface of earth
of radius R is given by  ɠ=g(1-d/R).At the center of earth,  (dept)d=R; so,  ɠ  =0.The
weight of a body of mass m at the centre of earth =mg’=m x 0=0.

Q6.Why does moon have no atmosphere?
Ans. Moon has no  atmosphere because the value of acceleration due to gravity ‘g’
on surface of moon is small. Therefore, the value of escape speed on the surface of
moon is small. The molecules of atmospheric gases on the surface of the moon
have thermal speeds greater than the escape speed. That is why all the molecules
of gases have escaped and there is no atmosphere on moon

Q7.What are the conditions under which a rocket fired from earth, launches
an artificial satellite of earth?
Ans. Following are the basic conditions: (i) The rocket must take the satellite to
a suitable height above the surface of earth for ease of propulsion.
(ii)From the desired height, the satellite must be projected with a suitable
speed, called orbital speed.
(iii)In the orbital path of satellite, the air resistance should be negligible so that
its speed does not decrease and it does not burn due to the heat produced.

Q1.State Kepler’s laws of planetary motion. Prove second Kepler’s law using
concept of conservation of angular motion.
Q2.State universal law of gravitation. What is the significance  of this law. Find the
expression for acceleration due to gravity.
Q4.  Define gravitational potential energy. Derive the expression for gravitational
potential energy. What is the maximum value of gravitational potential energy?
Q5.What is escape speed? Derive the expressions for it. Calculate escape speed for
the Earth

ORGANIC CHEM...

Organic compounds are the hydrocarbons and their derivatives and organic chemistry is that branch of chemistry that deals with the study of these compounds

Tetravalency of carbon The atomic number of Carbon is 6 and its electronic configuration is 2,4 i.e. it has 4 valence electrons. Thus carbon is always tetracovalent, i.e. it forms 4 covalent bonds with other atoms C Due to tetravalency of carbon it has a tetrahedron shape.

Catenation- The self linking property of carbon is known as catenation. This is the main reason of existence of such large number of compounds 

Classification of organic compounds 

Organic compounds

 Acyclic                                             Cyclic

                                   (SEE TEXT TABLE)

Functional groups:A functional group may be defined as an atom or a group of atoms present in a molecule which largely determines the chemical properties. CLASS OF ORGANIC NAME OF FUNCTIONAL STRUCTURE COMPOUNDS GROUP Alkenes double bond -C=C

Alkynes triple bond - C Ξ C - 

Halogens halogen - X ( F,Cl,Br,I ) 

Alcohols hydroxyl -OH

 Aldehydes aldehydic(formyl) -CHO

 Carboxylic acids carboxyl -COOH 

Acid amides Primary amines amides amino -CONH2 - NH2

HOMOLOGOUS SERIES Homologous series is defined as a family or group of structurally similar organic compounds all members of which contain the same functional group, show a gradation in physical and similarity in chemical properties and any two adjacent members of which differ by -CH2 group. The individual members of this group are called homologues and the phenomenon is called homology.

NOMENCLATURE OF ORGANIC COMPOUNDS

Straight chain alkanes: 

The names of such compounds is based on their chain structure,and end with suffix ‗-ane‘ and carry a prefix indicating the number of carbon atoms present in the chain. Branched chain hydrocarbons: 

1.) The longest carbon chain in the molecule is identified. 

2.) The numbering is done in such a way that the branched carbon atoms get the lowest possible value. 

3.) The names of the alkyl groups attached as a branch are then prefixed to the name of the parent alkane and its position is indicated by numbers.

 4.) The lower number is given to the first in alphabetical order. 

5.) The carbon atom of the branch that attaches to the root alkane is numbered 

1. Organic compounds having Functional Groups: The longest chain of carbon atoms containing the functional groups is numbered in such a way that the functional group attached to the carbon atom gets the lowest possible number in the chain. 

When there are more functional groups then a priority order is followed as:

 -COOH, -SO3H, -COOR, COCl, -CONH2, -CN, -HC=O, =C=O, -OH, -NH2, =C=C=, -CΞ C-.

ISOMERISM Two or more compounds having the same molecular formula but different physical

and chemical properties are called isomers and this phenomenon is called isomerism. Chain isomerism: When two or more compounds having same molecular formula but different carbon skeletons are referred to as chain isomers.

pentane,isopentane,neopentane(SEE STRUCTURE FROM TEXT)

Position Isomerism :

 Compounds which have the same structure of carbon chain but differ in position of double or triple bonds or functional group are called position isomers and this phenomenon is called Position Isomerism. e g CH3-CH2-CH=CH2 CH3-CH = CH – CH3

 Functional Isomerism 

Compounds which have the same molecular formula but different functional group are called functional isomers and this phenomenon is called functional Isomerism. e g CH3 – CH2 – OH CH3 – O – CH3 

Metamerism:

It is due to the presence of different alkyl groups on either side of functional group in the molecule. Ex. C4H10O represents C2H5OC2H5 and CH3OC3H7. 

FISSION OF COVALENT BOND 

Heterolytic cleavage: In this cleavage the bond breaks in such a way that the shared pair of electron remains with one of the fragments. 

H3C – Br +CH3 + Br - 

Homolytic Cleavage: In this cleavage the shared pair of electron goes with each of the bonded atom. R – X R . + X . 

Alkyl free radical 

Nucleophiles : A reagent that brings an electron pair is called nucleophile ie nucleus seeking e g -OH , -CN 

Electrophiles: A reagent that takes away electron pair is called electrophile I e electron seeking e g > C= O , R3C– X 

Inductive Effect: The displacement of the electron along the chain of the carbon atoms due to presence of an atom or group at the end of the chain

ɗ+++ ɗ ++ ɗ+ 

CH3- C H2 CH2 Cl

Resonance Effect : 

The polarity produced in the molecule by the interaction of two pi bonds or between a pi bond and lone pair of electron present on an adjacent atom. There are two types of resonance effect:

 1) Positive resonance effect : In this effect the transfer of electrons is away from an atom or substituent group attached to the conjugated system. The atoms or groups which shows +R effect are halogens,-OH , -OR,- NH2

 2) Negative resonance effect : In this effect the transfer of electrons is towards the atom or substituent group attached to the conjugated system. The atoms or groups which shows -R effect are –COOH , -CHO , -CN

Sublimation : This method is used to separate the sublimable compounds from non sublimable compounds. 

Crystallisation: This method is based on the difference in the solubilities of compound and impurities in a suitable solvent. The impure compound is dissolved in solvent and heated at higher temp .On cooling, from the hot and conc solution pure compounds crystallizes out. 

Distillation: This method is used to separate volatile liquids from non volatile liquids and liquids having sufficient difference in their boiling points.

 Fractional distillation: If the boiling points of two liquids is not much , they are separated by this method. Distillation under reduced pressure : This method is used to purify liquids having high boiling points and decomposes at or below their boiling points. S

team distillation : This method is used to separate substances which are steam volatile and are immiscible with water.

 Differential Extraction: When an organic compound is present in an aqueous medium it is separated by shaking it with organic solvent in which it is more soluble than in water. The aqueous solution is mixed with organic solvent in a separating funnel and shaken for sometimes and then allowed to stand for some time .when organic solvent and water form two separate layers the lower layer is run out by opening the tap of funnel and organic layer is separated. the process is repeated several times and pure organic compound is separated. 

Chromatography :This technique is used to separate mixtures in to their components ,purify the compounds and test the purity of compounds.It is classified 

 Adsorption Chromatography : It is based on the fact that different compounds are adsorbed on an adsorbent to different degrees. Silica jel or alumina is used as adsorbents.

 Partition Chromatography : It is based on the continuous differential portioning of components of a mixture between stationary and mobile phase. 

 QUANTITIVE ANALYSIS(Carbon and Hydrogen)

Let the mass of organic compound be m g. Mass of water and carbon dioxide produced be m1 and m2 g respectively;

 % of carbon = 12 x m2 x 100 /44 x m

 % of hydrogen = 2 x m1 x 100 /18 x m

Nitrogen DUMAS METHOD: A known mass of organic compound is heated with excess of CuO in an atmosphere of CO2, when nitrogen of the organic compound is converted into N2 gas. 

The volume of N2 thus obtained is converted into STP and the percentage of nitrogen determined by applying the equation: 

Volume of Nitrogen at STP = P1V1 x 273 /760 x T1 

%N = 28 x vol of N2 at STP x 100 /22400 x mass of the substance taken

 

Physics Notes Class.... GRAVITATION

Every object in the universe attracts every other object with a force which is called the force of gravitation

Newton’s                      Law   of   Gravitation

Gravitational force is a attractive force between two masses m₁ and m₂ separated by a distance r.

The gravitational force acting between two point objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Gravitational force.

where G is universal gravitational constant.

Dimensional formula of Gis [M^-1L³T^-2].

The value of G is 6.67 X 10^-11 Nm² kg^-2 and is same throughout the universe.

The value of G is independent of the nature and size of the bodies well as the nature of the medium between them.

(i)   Gravitational force is a central as well as conservative force.

(ii) It is the weakest force in nature.

(iii)                                         It   is   1036   times   smaller   than   elect

(iv)  The law of gravitational is applicable for all bodies, irrespective of their size, shape and

position.

(v)  Gravitational force acting between sun and planet provide it centripetal force for orbital motion.

(vi)  Gravitational pull of the earth is called gravity.

(vii)                                      Newton’s third law of motionmeansthe holds gravitation forces between two bodies are action-reaction pairs.

Following three points are important regarding the gravitational force

(i)   Unlike the electrostatic force, it is independent of the medium between the particles.

(ii)  It is conservative in nature. 

(i)       It expresses the force between two point masses (of negligible volume). However, for external points of spherical bodies the whole mass can be assumed to be concentrated at its centre of mass.

Acceleration Due to Gravity

The uniform acceleration produced in a freely falling object due to the gravitational pull of the earth is known as acceleration due to gravity.

It is denoted by g and its unit is m/s². It is a vector quantity and its direction is towards the centre of the earth.

The value of g is independent of the mass of the object which is falling freely under gravity.

The value of g changes slightly from place to place. The value of g is taken to be 9.8 m/s² for all practical purposes.

The value of acceleration due to gravity on the moon is about. one sixth of that On the earth and on the sun is about 27 times of that on the earth.

Among the planets, the acceleration due to gravity is minimum on the mercury. Relation between g and a is given by

g = Gm / R²

where M = mass of the earth = 6.0 * 10²⁴ kg and R = radius of the earth = 6.38 * 10⁶ m. Acceleration due to gravity at a height h above the surface of the earth is given by

g_h = Gm / (R+h)² = g (1 –2h / R)