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Topic Page No.

INORGANIC CHEMISTRY

PERIODIC PROPERTIES

1. Classification of Elementsand Periodicity in properties 01

2. Mendeleev’s Periodic Law 03

3. Modern Periodic Law 07

4. Effective Nuclear Charge 11

5. Atomic Size 12

6. Ionisation Potential 15

7. ElectronAffinity 18

8. Electronegativity 20

9. Density 27

10. Diagonal Relationship 28

11. Metallic Character 28

12. Non-metallic character 29

PERIO I PROPERTIES


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13. Reactivity, Oxidation state, Valency 29

14. Acid-base behaviour of oxides and hydroxides 30

15. Atomic volume 30

16. Graphs of periodic properties 30

17. General Trend of different properties in the period and groups 35

18. Important facts of remember 36

19. Solved Examples 38

20 Exercise - 1 41

Exercise - 2 46

Exercise - 3 51

Exercise - 4 55

21. Answer Key 58

22. Hints/Solution 60


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ACC CH PERIODIC PROPERTIES

1

PERIODIC PROPERTIES

CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

Need for classification:

It is verydifficulttostudyindividuallythe chemistryofalltheelementsand millionsoftheir compounds,

hence to simplify and systematize the study of chemistry of the elements and theircompounds, theyare

classified intogroupsandperiods. Earlyattempt to classifytheelements:

Classification of Lavoiser

Elements had been classified into two majorgroups byLavoiser

1. Metals 2. Non–metals

This classification wasbased on thedifferencesin their properties.

The metal word itself can beconsideredanacronym in which each alphabet signifiesa characteristicof

thesesubstances:

1. M = malleability. The mostmalleable beingAu andPt.

s–block < p–block < d–block > f–block.

M = melting point (alsoboilingpoint) of themetals in general ishigh.Metal having thehighest melting

point isTungsten(W) and the highest boilingpoint isRenium(Re).The lowestmeltingpointisofMercury

(Hg)and it is one of the two elements in the periodic table which are liquid at 25oC. (The other being

Bromine).

Theelement having highest mp isW.

Element havinglowest mp and bpis He.

2. E standsfor elasticity(Young’s modulus)

E is also forelectropositive.

E also forgood conductors ofElectricity. Best isAg > Cu >Au. Metals aregoodconductors ofelectricity

and heat due to thepresence of free mobile delocalized electrons.

Graphite andblack phosphorous are also conductors of electricity butnot of heat.

Whenever a compound of metalis electrolysed metal always deposited at cathode while non metal at

both anodeas well as cathode.

3. T =Tenacity(Tensile/ductilemeans elongation) Most tensile/ductile–Au/Pt.

Least – Pb

T = Toughness/ Hardness. Hardest is W. Exception : diamond is thehardest material ever known.

Synthetic hard materials are called abrasive.These are

CorundumAl2O3, CarborundumSiC, Borazone BN, Boron carbideB4C (non stoichiometric),Tantalum

Carbide andTungsten Carbide as hard as diamond.

4. A=Allotropy/polymorphism, shown mainlybynonmetals. Metals suchasSnshowpolymorphism.

=Atomicity, Numberofatomsinthemoleculeofanelement.Atomictyis verysignificantfor non–metals becauseit reveals the shape of themolecules andphysical properties (state) of the molecules.

A=Alloys formation.


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Alloys are homogeneous mixture of twoormore metals. It may also have some non metals as well.

EveryalloyisasolidsolutionexceptthealloysofHgwhichisasolutionofasolidintoaliquid.Thealloys

ofmercuryarecalledamalgum.Alloysofmetalswithatleast a non–metalarecalled interstitial compounds.

Alloys of only metals of nearly similar character are called substitutional compounds. Steel is a

homogeneous mixture.Italso is aninterstitialcompound.Purpose ofalloying a metal are:

a. To increase the utilityof the metalseg. fusewire (Pb+Sn + Bi). Itmelts in boiling water.

b. To increase the hardness.

Alloying is propertyofmetalsnotofnonmetals.

5. L = lustre, Shine propertyofmetals. Eachmetal havinga characteristic.

6. S = sonorous. Eachmetal givesspecialsoundonbeating.

Dobereiner’s law of traids:

It was first attempt towards classification. He arranged similar elements in a group of three elements

called triad and the atomic mass of the middle elements of the traid is approximately the arithmetic

mean of the other two.

e.g. Ca40 Sr 87.5 Ba137

At. wt. of Sr = 5.882

40137

88.5 is nearly similar to 87.5 of atomic wt. of Sr.Such a group of elements is called Dobereiner’s triad.

Triad of atoms Mean of first and last element

Li Na K 7 39

232

7 23 39

Be Mg Ca

8 4 0

242

8 24 40

Dobereiner could arrange onlya few elements as triads and there are some such elements present in a

triad, whose atomic weights areapproximatelyequal, e.g.

Fe Co Ni

Ru Rh Pd

There fore, this hypothesiswas not acceptablefor allelements.

Newland’s law of octaves:

When the lighter elements are arranged in order of their increasing atomic weights, then every eighth

element is similar to the first element in its properties, similarly as the eighth node of a musical scaleis similar to 1st one. e.g. Na 8th element resembles in their properties with Li. Similarly K the 8th

element with Na and so on.


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Name ofelement Li Be B C N O F

7 9 11 12 14 16 19

Name ofelement Na Mg Al Si P S Cl

23 24 27 28 31 32 35.5

It is clear from theabovetablethat sodium is theeighth element from lithium,whosepropertiesresemblethat oflithium.

This typeofclassification was limited up toonly20 elements.

Inert gas element were not discovered till then.

Lother Meyer arrangement:

A t o m i c V o l u m e

Metalloid and transition metals

Atomic Weight

Li

BeF

Na

Mg

Cl

K

CaBr

Rb

Sr I

Cs

Ba

Thegraphs plottingthe atomic volumesagainst atomicweights areknown as LotharMayervolume

curves.

Thealkali metals have highest atomicvolumes.

Alkalineearthmetals (Be, Mg Ca, Sr, Ba, etc.) which are relativelya little less electropositive. Occupy

positions on the descendingpart of the curve.

Halogens and thenoblegases (except helium) occupypositions on theascendingpart of thecurve.

Transition elements have very small volumes and thereforethese arepresent at thebottoms of thecurve

Exercise

1. Lother Meyer attempt was based on plotting atomic mass vs

(A)Atomicsize (B)Atomicvolume (C) Density (D)Meltingpoint

Ans. (B)

MENDELEEV’S PERIODIC LAW

(i) Mendeleev’s Periodic Law - The physical and chemical properties of elements are the periodicfunction of their atomic weight

(ii) Characteristic of Mendeleev’s Periodic Table -

(a) It is based on atomic weight (b) 63 elements were known, noble gases were not discovered.

(c) 12 Horizontal rows are called periods. (d) Vertical columns are called groups and there were 8groups in mendeleev’s Periodic table. (e) Each group upto VIIth is divided into A& B subgroups.‘A’sub groups element arecallednormalelements and ‘B’subgroupselements are called transition

elements. (f)TheVIIIth group was consists of 9 elements in three rows (Transitionmetals group).(g) Theelements belonging to same groupexhibit similar properties.


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MODIFIED MENDELEEF’S PERIODIC TABLE OF ELEMENTS


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(iii) Merits or advantages of Mendeleev’s periodic table -

(a) Study of elements - First time all known elements were classified in groups according to their similar properties. So studyof theproperties becomeeasierof elements.

(b) Prediction of new elements - It gave encouragement to the discovery of new elementsas somegapswereleftinit.Sc(Scandium),Ga(Gallium),Ge(Germanium),Tc(Technetium)were theelementsfor whom position and properties were defined byMendeleev even before their discoveries and heleft the blank spacesfor themin his table.

e.g.- Blank space at atomicweight 72 in silicongroup was called Eka silicon (meansproperties like silicon)andelement discoveredlaterwasnamed Germanium.Similarlyotherelements discovered after mendeleev periodic table were.Ekaaluminium- Gallium(Ga) Eka Boron - Scandium (Sc)Eka Silicon - Germanium (Ge) Eka Manganese- Technetium (Tc)

(c) Correction of doubtful atomic weights - Correction weredone in atomicweight ofsome elements.Atomic Weight =Valency× Equivalent weight.

Initially, it was found thatequivalentweightofBeis4.5 and it is trivalent (V= 3), sothe weight ofBewas 13.5 and there is no space in Mendeleev’s table for this element. So, after correction, it wasfound thatBeis actuallydivalent (V = 2). So, the weight of Be became 2 × 4.5 = 9 and there was aspace between Liand B for this element in Mendeleev’s table. – Correctionswere done in atomicweight of elements are – U, Be, In,Au, Pt.

(iv) Demerits of Mendeleev’s periodic table -

(a) Position of hydrogen- Hydrogen resemblesboth, thealkalimetals(IA)andthehalogens(VIIA) in properties so Mendeleev couldnot decide where to place it.

(b) Position of isotopes - As atomic weight of isotopes differs, they should have placed in different positionin Mendeleev’s periodictable. But there wereno suchplaces for isotopes in Mendeleev’s

table.(c) Anomalous pairs of elements - There were some pair of elements which did not follow the

increasing orderof atomic wts.eg. Ar and Cowere placed before K and Ni respectively in the periodic table, but havinghigher atomic

weights.

1.399.39

K Ar

1275.127

ITe

6.589.58

NiCo

231232

PaTh

(d) Like elements were placed in different groups.

Thereweresome elements like Platinum(Pt) and Gold (Au) whichhave similar properties butwere placedin different groups in mendeleev’s table.

Pt – VIII Au – IB(e) Unlike elements were placed in same group.

I groupst

IA IB

Li

Na

K More reactive Cu Less reactive

RbAlkali metal Ag Coin metal

Cs Normal elements Au Transition element

Fr

– Cu,Ag andAu placedin Ist group alongwith Na, K etc.Whiletheydifferin their properties (Onlysimilar inhavingns1 electronicconfiguration)


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MODERN PERIODIC LAW

Mosley proved that the square root of frequency() of the rays, which are obtained from a metal on

showeringhigh velocityelectronsisproportional tothenuclear chargeoftheatom.This can berepresented bythefollowing expression.

= a (Z–b) where Z is nuclear charge on the atom and a and b are constants.

The nuclear chargeon an atom is equal to the atomic number.

Accordingtomodern periodiclaw. “Thepropertiesof elements arethe periodicfunctions oftheiratomic

numbers”

Exercise

1. Which of thefollowingexpressions is correct ?

(A) ν = a(z–b) (B) ν2 = a (z–b)2 (C) ν = a2 (z–b)2 (D) ν2 = a2 (z–b)2

Ans. (C)

Cause of periodicity: It is due to the repetition of similar outer shell electronic configuration at a

certain regular intervals.

Structural features of the long form of the periodic table.

On the basic of the modern periodic law, a scientist named Bohr proposed a long form of periodic

table that was prepared by Rang and Warner.

(i) It consists of 18 vertial columns called groups and 7 horizontal rows called periods.

(ii) Elements of groups, 1, 2, 13 – 17 are called normal or representative elements.

(iii) Elements of groups 3 – 11 are called transition elements.

(iv) The 14 elements with atomic numbers (Z) = 58 – 71 (occurring after lanthanum 57La in the periodic

table) are called lanthanides or rare earth elements and are placed at the bottom of the periodic

table. The 14 elements with atomic numbers (Z) = 90 – 103 (Occurring after actinium 89Ac in the

periodic table) are called actinides and are placed at the bottom of the periodic table.(v) The Eleven elements with Z = 93 – 103 (93 Np – 103Lr) which occur in the periodic table after uranium

and have been prepared from it by artificial means are called transuranics. These are all radioactive

elements.

(vi) The elements belonging to a particular group are said to constitute a chemical family which is usually

named after the name of the first element. For example, Boron family (group 13), carbon family (group

14), nitrogen family (group 15), and oxygen family (group 16). In addition to this, some groups

have typical names. For example,

Elements of group 1 are called alkali metals

Elements of group 2 are called alkaline earth metalsElements of group 16 are called chalcogens

Elements of group 17 are called halogens


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Elements of group 18 are called zero group or noble gases.

The long form of the periodic table contains seven periods. These are :

1st period (1H – 2He) contains only two elements. This is the shortest period.

2nd period (3Li – 10 Ne) and third period (11 Na – 18Ar) contain 8 elements each and are called

short periods.

4th period (19K – 36Kr) and 5th period (37Rb – 54Xe) contain 18 elements each and are called long

periods.

6th period (55Cs – 86Rn) contains 32 elements and is the longest period.

7th period (87Li –) is, however, incomplete and contains at present only 24 elements.

In yet another classification, the long form of the periodic table has been divided into four blocks (i.e.,

s, p, d and f ), depending upon the subshell to which the last electron enters.

MERITS OF LONG FORM OF PERIODIC TABLE OVER MANDELEEF’S PERIODIC TABLE

Positions of Isotopes and Isobars - Isotopes have same atomic number and the periodictable is

based on atomic numbers.Therefore, various isotopesof the sameelementshave to be provided the

same position in theperiodic table. Isobars gave same atomicweights butdifferent atomicnumbers and

therefore they have to be placed at different positions.

Thepositions of actinides andlanthanides is more clear now becausethese have been placed in IIIB

groups and due to paucityof space, these are written at the bottom of the periodic table.

In the periodictable a diagonal line from B toAt separates the metals, nonmetalsand metalloidsfrom

one another. The elements like B,Si,As, Te,At, that are situated near this line are metalloids i.e. these

have characteristic ofboth metalsandnonmetals.

Thegeneral electronic configurationof theelementsremainssame in group.

DEFECTS OF LONG FORM OF PERIODIC TABLE

Thepositionofhydrogen is still disputableasit was therein Mendeleef periodictablein group IA aswellas IVA & VIIA.

Helium is aninert gas but itsconfiguration is different fromthatof the other inert gas elements

Lanthanideand actinide series could not beadjusted in themain periodic table and therefore theyhad to

be placed in a separate block below the periodic table.

CLASSIFICATION IN BLOCKS

s-block elements.

Elements of groups 1 and 2 including He in which the last electron enters the s-orbital of the valenceshell are called s-block elements. There are only 14 s-block elements in the periodic table.


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p-block elements.

Elements of groups 13–18 in which the last electron enters the p-orbitals of the valence shell are called

p-block elements.

d-block elements.

There are three complete series and one incomplete series of d-block elements. These are: 1st or 3d-

transition series which contains ten elements which atomic numbers 21–30 (21Sc – 30Zn).

2nd or 4d-transition series which contains ten elements with atomic numbers 39 – 48 (39Y – 48Cd).

3rd or 5d transition series which also contains ten elements which atomic numbers 57 and 72 – 80

(57La, 72Hf – 80Hg).

4th or 6d transition series which is incomplete at present and contains only nine elements. These are

89Ac, 104Rf, 105Ha, Unh (Unnihexium, Z = 106), 107 Ns (Neilsobohrium), 108Hs (Hassium), 109Mt

(Meitherium), Uun (Ununnilium, Z = 110) and Uud (Unundium, Z = 112) or Ekamercury. The element,

Z = 111 has not been discovered so far. Thus, in all there are 39 d-block elements.

f-Block elements are called inner-transition elements. In these elements, the f-subshell of the

anti-penultimate is being progressively filled up. There are two series of f-block elements each

containing 14 elements. The fourteen elements from 58Ce – 71Lu in which 4 f-subshell is being

progressively filled up are called lanthanides or rare elements. Similarly, the fourteen elements from

90Th – 103Lr in which 5 f-subshell is being progressively filled up are called actinides.

Illustrati on

1. Which ofthe followingis the period numberof the element whose atomic numberis 98

(A) 4 (B) 7 (C) 5 (D) 6

Ans. B

Sol. Theelectronic configuration of theelement with atomicnumber 98is as follow

1s2 , 2s2 , 2p6, 3s2, 3p6, 4s2 , 3d10, 4p6, 5s2, 4d10 , 5p6, 6s2, 4f 14, 5d10, 6p6, 7s2, 5f 10

The last electron enters in f orbital, so it belongs to f block in the period.

2. Thenuclei ofelements X,Yand Z have same numberof protons, but different numbers of neutrons.

Accordingto Mendeleef periodictable, theelements X,Y and Z

(A) belong to same group and same period

(B)belongto different groupsanddifferent periods

(C)belongto same group and different periods

(D)are isotopes, whichdo not have different positions

Ans. D

Sol. Isotopeshave same number of protons (i.e. same atomicnumber). So they occupy same position

in theperiodictable. However, due to different numbers of neutrons their atomic weightsare different.


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Exercise

1. Which ofthefollowing has the samenumberofelectrons inits outermost shell and penultimateshell ?

(1)Al3+ (2) Ca2+ (3) F – (4) N3–

Ans. 2

2. Whichofthe following statement isnot correct about theelectronic configurationofchromium atom (Cr

with atomic number= 24)?

(1) It has five electrons in 3d - sub - shell

(2) It has one electrons in 4s-orbital

(3)The principal quantum numbers of its valence electrons are 3 and 4.

(4)It has six electrons in 3d-sub-shell

Ans. 4

TEMPORARY NOMENCLATURE

The IUPAC proposed a system for naming elements with Z > 100.1. The names are derivedbyusing roots for the three digits in the atomic number ofthe elements and

adding theending-ium. The rootsfor the numbers are:0 1 2 3 4 5 6 7 8 9nil un bi tri quad pent hex sept oct enn

2. Incertain cases the names are shortened; for example, bi ium andtri ium are shortened to bium andtrium, and enn nil is shortened to ennil.

3. The symbolfor the elementismade upfromthefirst lettersfromtheroots which makeup the name.Thestrange mixture ofLatin and Greek rootshave been chosen to ensure that thesymbols areall different.

Illustrati on

1. According to theIUPAC system ofnaming elements, the symbol of theelement of atomic number 121willbe(A) Unu (B) Ubn (C) Ubu (D) Bus

Ans. (C)

Exercise

1. The IUPAC name of the element with atomic numberZ = 109 is(A) U np (B) U ns (C) U no (D) U ne

Ans. (D)

Anomalous behaviour of the first element of a group:

The first element of a group differs considerably from its congeners (i.e. the rest of the elements of its group). This is due to (i) small size (ii) high electronegativity and (iii) non availability of d-orbitals

for bonding. Anomalous behaviour is observed among the second row elements (i.e., Li to F).


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EFFECTIVE NUCLEAR CHARGE

Ina polyelectronic atom, the internal electronsrepel the electronsof the outermost orbit. This results in

decrease thenuclear attraction on theelectrons of theoutermost orbit.

Therefore,onlya part ofthe nuclear chargeis effective onthe electrons ofthe outermost orbit. Thus, the

inner electrons protect or shield thenucleus andthereby decrease theeffect of nuclear charge towards

theelectrons of theoutermost orbit.

Thus the part of the nuclear charge works against outer electrons, is known as effective nuclear

charge

Z * Z

Z* = effectivenuclear charge

s = shielding constant and Z = nuclear charge

Ascientist named slater, determined the valueofshielding constant and put forward some rules asfollowing.

(1) The shielding effectorscreening effect ofeachelectron of1sorbital is 0.30.

(2) The shielding effect ofeach electronsofns and npi.e.electronof the outermost orbit, is 0.35.

(3) The shielding effect ofeachelectronof s orp orbitals of the penultimateorbit (n – 1)is 0.85.

(4) The shieldingeffectof eachelectronof s orp orbital ofthe prepenultimateorbit(n– 2)andbelow thisis1.0.

Illustrati on

1. Thescreeningeffect of inner electronsof the nucleus causes

(A) Decreaseinthe ionisationenergy (B) Increase in the ionisation energy(C) No effect on the ionisation energy (D) Increases in the attraction of the nucleus to the electrons

Ans. (A)

Sol. Theattraction of thenucleus with the electron decreases so it becomes easy to extract the electron.

Exercise

1. The screening effect of d - electrons is -

(A) equal to the p - electrons (B)muchmorethan p- electron(C) same as f - electrons (D) less than p - electrons

Ans. (D)

2. Theincreasingorder of effective nuclear chargein Na,Al,Mg and Si atoms

(A) Na < Mg < Si < Al (B) Na < Mg < Al < Si (C) Mg < Na < Al< Si (D) Na = Mg = Al = Si

Ans. (B)

Periodic Properties: Properties which are directly or indirectly related to their electronic configuration and

show a regular gradation when we move from left to right in a period or from top to bottom in a groupare called periodic properties. Some important periodic properties are atomic size, ionization energy

electron affinity, electronegativity, valency, density, atomic volume, melting and boiling points etc.


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(a) Atomic size : It refers to the distance between the centre of the nucleus of the atom to the outermost

shell containing electrons. Since absolute value of the atomic size cannot be determined, it is usually

expressed in terms of the following operational definitions.

A

X X

B

1/2 AB = r covalent(of element X)

FE G H

XH XHXH

1/2 EF = r van der Waals of hydrogen in HX molecule

1/2 GH = r van der Waals of X in HX molecule

*0

2

nZ

anr

(i) Covalent radius. It is defined as one-half of the distance between the nuclei of two covalently bonded

atoms of the same element in a molecule.

Single Bond Covalent Radius, SBCR - (a)ForHomolatomic molecules

dA–A = r A + r A or 2r A

r A = A Ad

2

(b)Forhetrodiatomic moleculeswhile electronegativityis approx same.dA–B = r A + r B

Forheteronuclear diatomic molecule.A–B,whiledifference betweenthe electronegativityvalues of atomA and atom B is relatively larger, (XA andXB) arethe electron negativityin Pauling Scale.

dA–B = r A + r B – 0.09 |(XA –XB)| [Bond length or radius expressed in Å]whereXA andXB areelectronegativityvalues ofhighelectronegative elementA andlesselectronegativeelement B, respectively. This formula was given byStevenson & Schomaker.

Note : Covalent radius is slightly smaller then actual radius.

(ii) Van der Waals’radius. It is defined as one-half of thedistance between the nuclei oftwo non-bondedisolatedatomsor twoadjacent atomsbelongingtotwoneighbouringmoleculesofanelement inthesolidstate.By definition, van der Waals’radius of an element is always larger than itscovalent radius.Variation of atomic radii :

(i) Across the period atomic radii decreases(ii) Where we move from 17th group to 18th atomic radii increases the period decreases


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Illustrati on

1. Calculate the bond length of C–X bond, if C–C bond length is 1.54 Å,X–X bond length is 1.00 Å and

electronegativityvalues of C and X are2.0 and 3.0 respectively

Sol. (1) C–C bond length = 1.54 Å

r C = 1.54

2 = 0.77 Å

r X = 1.00

2 = 0.50 Å

(2) C–X bond length

dC–X = r C + r X – 0.09 |XX – XC|

= 0.77 + 0.50 – 0.09 |3–2| = 0.77 + 0.50 – 0.09 × 1 = 1.27 – 0.09 = 1.18 Å

Thus C–X bondlength is 1.18Å

2. Which of the followingshould bethe longest bond ?

(A) S–H (B) O–H (C) N–H (D) P–H

Ans. (D)

Sol. The atomic radius of P is largest out of O.S.N. and P. Therefore. P–H bond will be thelongest one.

Exercise

1. The van der Waal’s radii of O, N, Cl, F and Ne increase in the order (A) F, O, N, Ne, Cl (B) N, O, F, Ne, Cl (C) Ne, F, O, N, Cl (D) F, Cl, O, N, Ne

Ans. (C)

2. Whenevera list ofradii isgiven, wefindthat the size of the noble gases is larger than the size of their

adjacent halogens.The reason is

(A) Noblegases have a complete octet

(B)Theyhave a higherinterelectronic repulsion

(C) Inhalogensit is covalent radii and in noble gases it is vanderwaals radii

(D)Noble gases cannot be liquifiedAns. (C)

Variation in Period :

Li Be B C N O F Ne

Z 3 4 5 6 7 8 9 10

Z* 1.30 1.95 2.60 3.25 3.90 4.55 5.20 5.85

n 2 2 2 2 2 2 2 2

rn

(pm) 123 90 80 77 75 74 72 160

(covalent) (Van der Waals)


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Ionic size. An atom can be changed to a cation by loss of electrons and to an anion by gain of

electrons. A cation is always smaller than the parent atom because during its formation ef fe ct ive

nuclear charge increases and sometimes a shell may also decrease. On the other hand, the size of

an anion is always larger than the parent atom because during its formation effective nuclear charge

decreases e.g..Mg2+ < Mg, Cl – > Cl

The sizeofa cation issmaller in comparisontothe sizeof itscorresponding atom. This is because of the

fact that anatom on losing electrons/s form a cation, which has lesser number of electrons/s than the

numberofproton/s. This resultsin increase in theeffectivenuclear charge.

Size ofcationeff

1

Amount of positive ch arg e or Z

Examples- (1) Mn > Mn+2

> Mn+3

> Mn+4

> Mn+6

> Mn+7

(size)(2) Pb+2 > Pb+4

Thesize ofan anion is greater than thesize of itscorresponding atom,because thenumber of electrons

present in the anion gets larger than the number of protons due to gain of electron/s. This results in

decrease in theeffective nuclear charge.

Size ofanion Amount of negative charge

O0 < O –1 < O –2

Size of Isoelectronic Series

Isoelectronic ions or species are the neutral atoms, cations or anions of different elements which have

the same number of electrons but different nuclear charge e.g..

The size of the isoelectronic species depends upon their nuclear charge Greater the nuclear charge,

lesser the radii.

The ionic radii decrease moving from left to right across any period in the periodic table

Na+ Mg2+ Al3+

102 pm 72 pm 53.5 pm

Sizein anisoelectronic series

1Nuclear charge

No. of protons

Illustrati on

1. What should be the order of size of H –1 , H+1 andH?

H –1 H+1 H

1p 1p 1p

2e 0e 1e

Sol. H+1 < H < H –1


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Exercise

1. Whichofthefollowing has thelargest size

(A) N –3 (B) O –2 (C) K +1 (D) Ca+2

Ans. (A)

2. Which oneof thefollowing is correct order of the size of iodine species?

(A) I > I – > I+ (B) I > I+ > I – (C) I+ > I – > I (D) I – > I > I+

Ans. (D)

IONISATION POTENTIAL

The energyrequired to remove the mostlooselybound electron fromthe outermost orbit ofone moleof isolatedgaseous atomsofanelement,iscalled ionisation potential (IP). This ionisation isanendergonic

or energy-absorbing process.

An electron cannot be removed directlyfrom an atom in solid state.For this purpose, the solid state is

converted to gaseousstate andtheenergyrequired forthis is called sublimation energy.

A stI IP A+1 ndII IP AA

+2 rdI II IP A+3

The energyrequired to remove one electron from a neutral gaseous atom to convert it to monopositive

cation, is called first ionisation potential(Ist IP). Theenergyrequired to convert a monopositive cation to

a diapositivecation is called second ionisation potential (IInd IP)ofanatom

Ist IP


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(iv) Stability of half filled and fully filled orbitals : The atoms whose orbitals arehalf-filled(p3, d5, f 7) or

fully-filled (s2 , p6 , d10 , f 14) have greater stability than the others. Therefore, they required greater

energyto for removingout electron. However stabilityof fully filledorbitals is greater thanthatof the

halffilledorbitals

I.P.of fully filled orbitals I.P. of half filled orbitals

(v) Penetration power : In any atom the s orbital is nearer to the nucleus in comparison to p, d and f

orbitals.Therefor, greaterenergyis required to remove out electron from s orbital than from p, d and f

orbitals. Thus the decreasing order of ionisation potential of s, p, d and f orbitals is as follows

s > p > d > f

I o n is a tio n p o t en t ia l p e n e tr a tio n p o w e r

Periodic Table & Ionisation Potential(a) In a Period : The valueof Ionisation potential normallyincrease on goingfrom left to right in a period,

becauseeffective nuclear chargeincreases andatomic sizedecreases.

Exceptions :

In second period ionisation potential of Beis greater than that of B, and in thethird period ionisation

potential ofMgis greaterthan that ofAl dueto high stabilityoffullyfilled orbitals.

Insecondperiodionisation potential ofN is greater thanO and in the third periodionisation potential of

P is greaterthanthat ofS,due to stability ofhalf filledorbitals.

The increasing order of thevalues of ionisationpotentialof thesecondperiod elements is

Li < B < Be < C < O < N < F < NeThe increasingorder of the valuesof ionisation potential of the thirdperiod elements is

Na < Al < Mg < Si < S < P < Cl < Ar

InnerTransition Elements : The size ofinnertransitionelementsisgreater thanthatofd blockelements.

Therefore thevalue of ionisation potential off block elements is smaller than that ofd block elements

and due to almost constant atomic size of f block elements in a period the value of their ionisation

potential remainsmore constant thanthat of d block elements.

In a Group

Thevalueof ionisation potential normallydecreases on goingfrom topto bottom in a group because both atomicsize and shieldingeffect increase.

Exception :

The valueof ionisation potential remains almost constant fromAl to Ga in theIIIA group.

(B>Al , Ga > In)

IP1(Sn) < IP1(Pb)

IP1(In) < IP1(Tl)

In theperiodic table theelement havinghighest valueof ionisation potentialis He.

Thevaluesof ionisationpotential of noble gasesareextremelyhigh, because theorbitals of outermost

orbit arefully-filled (ns2 , np6) andprovidegreat stability. In a period, the element havingleast valueof ionisation potential is an alkali metal (group IA) and that

having highest valueis inert gas (Group0)


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1 7

Applications of Ionisation Potential

Theelementshavinghigh values of ionisation potential havelowreactivity, e.g. inert gases.

Thevalue of ionisation potential decreases moreon going from top to bottom in a group in comparison

to a period. Therefore, reactivityof metal increases and the atom forms a cation byloss of electron.

Ionisation potentialmetalof activityRe

1

Theelementshavinglowvalueof ionisationpotential readilylose electron andthus behave as strong

reducing agents.

Ionisation potential propertyducingRe

1

Theelements havinglowvalueof ionization potential readilylose electron and thus exhibit greater

metallicproperty.

Ionisation potential 1

Metallic property

Theelementshaving lowvalueofionisation potentialreadilylose electron and thus have basicproperty.

Ionisation potential 1

Basic property

Illustrati on

1. Which of thefollowingshould be theorderof increasingvalues of second ionisation potential of C6, N7,

O8 and F9(A) C > N > F > O (b) C < F < N < O (C) C < F < N < O (D) C < N < F< O

Ans. (D)

Sol. Thesecondionisationpotential means removal ofelectron from a cation

C+1 (5e) = 1s2 , 2s2 , 2p1

N+1 (6e) = 1s2, 2s2, 2p2

O+1 (7e) = 1s2, 2s2, 2p3

F

+1

(8e) = 1s

2

, 2s

2

, 2p

4

Therefore C < N < F < O

2. Which of the following should be correct for Z1 andZ2 in thefollowingtwo processes

M+ + Z1 M+2 + e –

M+2 + Z2 M+3 + e –

(A) 1

2 Z1 = Z2 (B) Z1 = Z2 (C) Z1 =

1

2 Z2 (D) Z1 < Z2

Ans. (D)

Sol. Z1 = secondionisation potential and Z2 = Third ionisationpotential.

Second ionisation potential is always less than thethird ionisation potential.


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Exercise

1. Onemole ofmagnesiumin thevapourstateabsorbed1200kJ ofenergy. If thefirstand secondionization

enthalpies ofmagnesium are 750 and 1450kJ mol –1respectively, thefinal compositionof themixtureis

(A)69%Mg+,31%Mg2+

(B) 59% Mg+,41%Mg2+

(C) 49% Mg+,51%Mg2+

(D)29%Mg+,71%Mg2+

Ans. (A)

2. Theincorrectstatement in thefollowingis

(A) The thirdionisation potential ofMgis greater thanthe third ionisation potential ofAl

(B) The first ionisationpotential ofNais less thanfirst I.P ofMg

(C) The first I.P. ofAl is less than the first I.P. of Mg

(D) The second I.P. of Mg is greater than the second I.P. of Na

Ans. (D)

ELECTRON AFFINITY (EA)

The energyreleased on adding upone mole of electron to one mole ofneutral atom (A) in its gaseous

state to form an anion (A – ) is called electron affinity of that atom. In general, electron affinity is

associated withan exothermic process.

A(g) + e – A – (g) , Heg = –En

When one electron adds up to a neutral atom, it gets converted to a unit negative ion and energy is

released. On adding one more electron to the mononegative anion, there is a repulsion between the

negatively charged electron and anion. In order to counteract the repulsive forces, energy has to be

provided to thesystem. Therefore, thevalue of the second electron affinityis positive.

A – (g) + e – A –2 (g) ,Heg = + En Heg : Electron gain enthalpy

Factors Affecting Electron Affinity

Effective Nuclear charge : When effective nuclear charge is more, then the atomic size less. Hence

EA increases.

Electron affinity Effective nuclear charge

Atomic Size orAtomicRadius : Whenthe sizeorradiusofanatomincreases, the electron enteringthe

outermost orbit is more weakly attracted bythe nucleus and the value of electron affinityis lower.

1Electron affinity Atomic size


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1 9

Shielding Effect : Shieldingeffect is directlyproportional to atomic size andatomic sizeis inversely

proportional to electron affinity.

1Electron affinity

Shielding effect

Stability of Fully-Filled and Half-Filled Orbitals :

Thestabilityof theconfigurationhavingfully-filled orbitals (p6, d10, f 14) and half-filled orbital (p3 , d5 , f 7)

is relativelyhigherthan thatofother configurations.

Periodic Table and Electron Affinity :

In a period, atomic size decreases with increase in effective nuclear charge and hence increase in

electronaffinity.

Exception :

On going from C6 to N7 in the second period, the values of electron affinity decreases in stead of

increasing. This is because there arehalf-filled (2p3) orbitals in the outermost orbit ofN,which are more

stable. On the other hand, the outermost orbit in C has 2p2 configuration.

In the thirdperiod, thevalue of electron affinityof Si is greater than that of P. This is because electronic

configurationoftheoutermost orbit in P atom is3p3 , whichbeinghalf-filled, is relativelymore stable

The values of electron affinityof inert gases are zero, because there outermost orbit has fully-filled p

orbitals.

In a period, the value of electron affinitygoes on decreases on going from group IA to group IIA. The

valueofelectronaffinityof theelementsofgroupIIA is zero because ns orbitals arefully-filled and such

orbitals have no tendencyto accept electrons.

In a Group

Thevalues ofelectron affinitynormallydecreaseon goingfrom top to bottom in a group because the

atomicsize increases whichdecreases theactual force of attractionbythenucleus.

Exceptions (E.A. 2nd period p-block element < E.A. 3rdperiod p-block element)

ThevalueofelectronaffinityofFislowerthanthatofCl,becausethesizeofFisverysmallandcompact

andthechargedensityishigh onthesurface.Therefore,theincomingelectronexperiencesmore repulsion

in comparisonto Cl . Thatis why the value ofelectronaffinityofCl is highest in the periodic table.

Thevalues ofelectron affinityofalkalimetalsand alkalineearthmetalscan be regarded aszero, because

they do not have tendency to form anions byaccepting electron.


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Illustrati on

1. O(g) + 2e –1 O –2 (g) –E = + 744.7

ThereasonforthepositivevalueofEis

(A) endothermicreaction (B)exothermicreaction(C) both 1 and 2 (D)All of the above are wrong

Ans. (A)

Sol. When electron is brought near O –1 there will be repulsion between them, and therefore theenergy

will be positive i.e therewill be absorptionof energyduring theprocess.

2. Theincreasingorder ofelectron affinityvalues of O,Sand Se is

(A) O < S < Se (B) S < O < Se (C) O < S > Se (D) Se < O > S

Ans. (C)

Sol. Atomic sizeofSe is verylarge.

Exercise

1. Theleast electron affinityis foundin

(A) Kr (B) O (C) N (D) B

Ans. (A)

2. Of thefollowingelement ofwhichelectronicconfigurationwillhave thehighest electron affinity

(A) 1s2 2s2 2p3 (B) 1s2 2s2 2p5 (C) 1s2 2s2 2p6 3s2 3p5(D) 1s2 2s2 2p6 3s2 3p3

Ans. (C)

ELECTRONEGATIVITY

The measure of the capacityortendencyof anatomto attract thesharedpair ofelectrons ofthecovalent

bond towards itself is called electronegativity ofthat atom.

Electronegativityis a relative value that indicates thetendency ofan atom to attract shared electronsmorethan the other atombondedto it.Therefore it doesnot haveany unit.Paulingwas the first scientist

to put forward theconcept of electronegativity.

Thenumericalvalueof electronegativityof anatom depends on its ionisation potentialandelectron

affinityvalues.

FactorsAffecting Electronegativity

Atomic size – Electronegativity of a bonded atom decreases with increase in its size.

1Electronegativity Atomic size


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SOLVED EXAMPLES

Q.1 For the element X,student Surbhi measureditsradius as102 nm, Mr.Guptaas113nm and Mr.Agarwalas 100 nmusingsame apparatus. Theirteacherexplained that measurements were correct bysayingthat

recorded values bythree students were(A)Crystal, vanderWaal and covalent radii (B) Covalent, crystal and vanderWaal radii

(C) Vander Waal, ionic and covalent (D) None is correct

Ans. (A)

Sol. Order of radius is – VanderWaal’s radius > Metallic radius > Covalent radius

Q.2 WhichoxideofN is isoelectronic with CO2 -

(A) NO2 (B) NO (C) N2O (D) N2O3Ans. (C)

Sol. N2OTotal electrons 1 4 + 8 = 2 2

CO2Total electrons 6 + 1 6 = 2 2

Q.3 Givethe correct order of initials T (true) orF (false) for following statements.

(I) Top positions of Lother-Mayer’s atomicvolume curve are occupied byAlkali metals.

(II) Number of elements presents in the fifth period of the periodic table are 32.

(III) 2nd I.P. of Mg is less than the 2nd I.P. of Na.

(IV) A p-orbital can takemaximum ofsix electrons.(A) TFTF (B) TFFT (C) FFTF (D) TTFF

Ans. (A)

Sol. (II) 5s 4d 5p

Orbital 1 5 3

Total orbitals = 9

Total element = 18

(III) Mg+ 1s2 2s2 2p6 3s1

Na+ 1s2 2s2 2p6

Na+ attainedinertgasconfiguration IE2 of Na is greater than Mg.

(IV) p-orbital can have2 electrons but p-subshell can have 6 electrons.

Q.4 The electronic configurationofanelement is 1s2, 2s22p6,3s13p33d2. The atomic numberof the element present just below this element in the periodictable -

(A) 36 (B) 34 (C) 33 (D) 32

Ans. (B)

Sol. At. No. = 16(S)

Next element below this element has atomic number = 16 + 18 = 34


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4 1

EXERCISE-1 (Exercise for JEE Mains)

[SINGLE CORRECT CHOICE TYPE]

Q.1 The correct order of ionic radius is -(A) Ce > Sm > Tb > Lu (B) Lu > Tb > Sm > Ce

(C) Tb > Lu > Sm > Ce (D) Sm > Tb > Lu > Ce [2010110651]

Q.2 TheZeff for 3d electron of Cr 4s electron of Cr 3d electron of Cr 3+

3s electron of Cr 3+ arein the order respectively(A) 4.6, 2.95, 4.95, 8.05 (B) 4.95, 2.95, 4.6, 8.05

(C) 4.6, 2.95, 5.3, 12.75 (D) none of these [2010110575]

Q.3 According to the PeriodicLaw of elements,theVariation in properties of elements is related to their ?

(A) Nuclear masses (B)Atomic numbers

(C) Nuclear neutron-proton numberratio (D)Atomic masses [2010112297]

Q.4 Atomic radius decreases in a period, but after halogens, the atomic radius suddenlyincreases. Thus,inert gases has almost highest radiusin a period. The explanation forsuch an increase is-(A) Inert gaseshasmoststable configuration(B) Inert gases do not takepart in bonding

(C)Vander Wall’s radius is reported in case of inert gases(D) None of these

[2010110748]

Q.5 Which one of thefollowing groups represent a collection of isoelectronic species ?

(At. no. Cs = 55, Br = 35)

(A) N3– , F – , Na+ (B) Be, Al3+, Cl – (C) Ca2+, Cs+, Br (D) Na+, Ca2+, Mg2+

[2010110847]

Q.6 Consider thefollowingstatements:

I. The radius of ananion is larger thanthat ofparent atomII. TheI.E. increases from left to right in a period generally

III. The electronegativityof an element is thetendency of an isolated atom to attract an electron

The correct statements are -

(A) I alone (B) II alone (C) I and II (D) II and III

[2010110542]

Q.7 Theatomicnumbers ofvanadium(V). Chromium (Cr), manganese (Mn) and iron (Fe) respectively23,24, 25 and 26. Which one of these may be expected to have the highersecondionization enthalpy?

(A) Cr (B) Mn (C) Fe (D) V [2010111695]


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EXERCISE-2 (Exercise for JEE Advanced)

[PARAGRAPH TYPE]

Paragraph for Question Nos. 1 to 2Four elementsP, Q, R & S have ground state electronic configuration as:

P 1s2 2s2 2p6 3s2 3p3 Q 1s2 2s2 2p6 3s2 3p1

R 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 S 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

[2010111148]

Q.1 Comment whichof thefollowingoption represent thecorrect order of true (T)& false (F) statement

I size of P < size of Q II sizeofR R > S Q (B) P < R < S < Q (C) R > S > P > Q (D) P > S > R > Q

Paragraph for Question Nos. 3 to 4

Theelectron affinityis a inherent propertyof theatom and it depends upon several factors.

[2010111642]

Q.3 Thecorrect electron affinity order is

(A) F > Cl (B) Cl > F (C) S < P (D) N > O

Q.4 Choose the incorrect statement.

(A)1st I.E. ofA¯ is equal to electron affinityof A

(B)2nd electron affinityis always greater than1st electron affinity.

(C)O O2– process is endothermic

(D) Li Li+ process is endothermic.

Paragraph for Question Nos. 5 to 7

Atoms of metals have onlya few electronsin their valence shells while atoms of non-metalsgenerallyhave more electrons in theirvalence shells. Metallic character is closely related to atomic radius andionization enthalpy. Metalliccharacter increasesfrom top to bottom ina group and decreasesfrom left toright in a period. Metallic character is inverselyrelated to electronegativity.

[2010111495]

Q.5 Which of thefollowinggroupscontainsmetals, non-metals and metalloids -(A) Group 1 (B) Group 17 (C) Group 14 (D) Group 2

Q.6 Non-metals belong to -

(A) s-blockelements (B) p-block elements (C)d-blockelements (D) f-block elementsQ.7 Consideringthe elements B, C, N, and F the correct order of their non-metallic character is -

(A) B > C > N > F (B) C > B > N > F (C) F > N > C > B (D) F > N > C > B


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5 1

EXERCISE-3 (Miscellaneous Exercise)

Subjective Type

Q.1 Usethefollowingsystem ofnamingelements inwhich first alphabetsofthedigitsarewrittencollectively,0 1 2 3 4 5 6 7 8 9nil uni bi tri quad pent hex sept oct ennto write three-letter symbols for theelementswith atomicnumber 101 to 109.[Example : 101 is Unu....] [2010110901]

Q.2 Mg2+, O2– , Na+, F – , N3– (Arrange in decreasing order of ionic size) [2010110877]

Q.3 WhyCa2+ has a smaller ionic radius than K +. [2010110043]

Q.4 Arrange in decreasing order of atomic size : Na, Cs, Mg, Si, Cl. [2010110195]

Q.5 Why the first ionisation energyof carbonatom is greater than thatofboron atom whereas, the reverse istruefor thesecondionisation energy. [2010110601]

Q.6 The IE donot follow a regular trend in II & III periods with increasingatomic number. Why?[2010110943]

Q.7 Explainwhya fewelements such asBe(+0.6), N(+0.3) & He(+0.6)have positiveelectron gain enthalpieswhile majorityof elements do havenegative values. [2010110599]

Q.8 Which bondin each pair is more polar (a) P – Cl or P – Br (b) S – Cl or S – O (c) N – O or N – F

[2010110441]

Q.9 From among the elements, choose the following: Cl, Br, F,Al, C, Li, Cs & Xe.(i)Theelement withhighest electrongainenthalpy.(ii)Theelementwithlowest ionisationpotential.(iii) Theelement whoseoxideis amphoteric.(iv) The elementwhichhas smallest radii.(v)The element whose atom has 8 electrons in theoutermost shell. [2010110597]

Q.10 In the ionic compound KF, the K + andF – ions are found to have practically radii, about 1.34 Å each.What do you predict about the relative covalent radii of K and F? [2010110350]

Q.11 Which oxide is more basic, MgO or BaO? Why? [2010110301]

Q.12 Thebasic natureof hydroxides of group 13 (III-A) decreasesprogressivelydown thegroup. Comment.[2010110596]

Q.13 Based on locationin P.T., whichof thefollowingwouldyou expect to be acidic & whichbasic.(a) CsOH (b) IOH (c) Sr(OH)2 (d) Se(OH)2 (e) FrOH (f)BrOH

[2010110868]


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5 5

EXERCISE-4

SECTION-A

(IIT JEE Previous Year's Questions)

Q.1 Movingfrom right to left in a periodictable, theatomic size is: [JEE 1995](A) increased (B) decreased (C) remains constant (D) none of these

[2010112001]

Q.2 Theincreasing orderofelectronegativityinthefollowing elements: [JEE 1995](A) C, N, Si, P (B) N, Si, C, P (C) Si, P, C, N (D) P, Si, N, C

[2010110537]

Q.3 One element has atomic weight 39. Its electronic configuration is 1s2, 2s2 2p6, 3s2 3p6 4s1. The true

statement forthat element is:(A)Highest value ofIE (B)Transition element(C) Isotone with

18Ar 38 (D) None [JEE 1995]

[2010111632]

Q.4 The numberof paired electronsin oxygen atom is: [JEE 1995](A) 6 (B) 16 (C) 8 (D) 32

[2010112648]

Q.5 The decreasing size of K +, Ca2+, Cl – & S2– followsthe order: [REE 1995]

(A) K +

> Ca+2

> S –2

> Cl –

(B) K +

> Ca+2

> Cl –

> S –2

(C) Ca+2 >K + > Cl – > S –2 (D) S –2 > Cl – > K + > Ca+2

[2010111675]

Q.6 Whichofthefollowing oxideis neutral? [JEE 1996](A) CO (B) SnO

2 (C) ZnO (D) SiO

2

[2010111977]

Q.7 Whichofthefollowinghas themaximumnumber ofunpairedelectrons [JEE 1996](A) Mg2+ (B)Ti3+ (C) V3+ (D) Fe2+

[2010111613]

Q.8 The followingacids have been arranged in the order of decreasing acid strength. Identify thecorrectorder [JEE 1996]

ClOH(I) BrOH(II) IOH(III)(A) I > II > III (B) II > I > III (C) III > II > I (D) I > III > II

[2010110900]

Q.9 Theincorrect statement amongthefollowing is: [JEE 1997](A) the first ionisationpotential ofAl is less thanthe first ionisationpotentialofMg(B)thesecond ionisation potential ofMg is greater than thesecond ionisation potential of Na

(C)thefirst ionisation potential ofNais less thanthefirst ionisation potential ofMg(D) the thirdionisation potential ofMgis greater thanthe third ionisation potential ofAl

[2010112041]


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5 7

Q.20 Identify the correct order of acidic strengths of CO2, CuO,CaO, H

2O: [JEE 2002]

(A*) CaO < CuO < H2O < CO

2 (B) H

2O < CuO < CaO < CO

2

(C) CaO < H2O < CuO < CO

2 (D) H

2O < CO

2 < CaO < CuO

[2010111496]

SECTION-B

(AIEEE Previous Year's Questions)

Q.21 In which of the followingarrangements, the sequence is notstrictly according to the propertywrittenagainstit? [AIEEE-2009]

(A)H2O < H2S < H2Se < H2Te : Increasing acidic strength(B) HF< HCl < HBr < HI: Increasingacidic strength

(C*) NH3 > PH3 Na+ > Mg2+ (B) Na+ > Li+ > Mg2+ > Be2+

(C) Li2+ > Na+ > Mg2+ > Be2+ (D) Mg2+ > Be2+ > Li+ > Na+

[2010113569]

Q.23 Thecorrect sequence which shows decreasing order of theionic radii of the elements is

[AIEEE-2010]

(A)Al3+ > Mg2+ > Na+ > F – > O2– (B) Na+ > Mg2+ > Al3+ > O2 > F –

(C) Na+ > F – > Mg2+ > O2– > Al3+ (D) O2– > F – > Na+ > Mg2+ > Al3+

[2010113620]

Q.24 Which oneof thefollowingorderedpresents thecorrect sequenceof theincreasingbasicnature of thegiven oxides ? [AIEEE-2011]

(A) Al2O3 < MgO < Na2O < K 2O (B) MgO < K 2O < Al2O3 < Na2O(C) Na2O < K 2O < MgO < Al2O3 (D) K 2O < Na2O < Al2O3 < MgO

[2010113671]

Q.25 The increasing order of theionic radii of the given isoelectronicspecies is [AIEEE-2012](A) Ca2+, K +, Cl – , S2– (B) K +, S2– , Ca2+, Cl –

(C) Cl – , Ca2+, K +, S2– (D) S2– , Cl – , Ca2+, K + [2010113722]

Q.26 The first ionisationpotential ofNa is 5.1 eV. The value of electron gain enthalpyof Na+ will be:(A*) –5.1 eV (B) –10.2 eV (C) +2.55 eV (D) –2.55 eV

[JEE MAIN-2013]

[2010113967]

Q.27 Which ofthe followingrepresents the correct order of increasingfirst ionization enthalpyfor Ca, Ba, S,Se and Ar?(A) S < Se < Ca < Ba < Ar (B*) Ba < Ca < Se < S < Ar

(C) Ca < Ba < S < Se < Ar (D) Ca < S < Ba < Se < Ar [JEE MAIN-2013][2010114018]


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EXERCISE 1

Q.1 (A) Q.2 (C) Q.3 (B) Q.4 (C)Q.5 (A) Q.6 (C) Q.7 (A) Q.8 (D)

Q.9 (D) Q.10 (A) Q.11 (B) Q.12 (B)

Q.13 (B) Q.14 (C) Q.15 (B) Q.16 (B)

Q.17 (C) Q.18 (D) Q.19 (D) Q.20 (D)

Q.21 (D) Q.22 (C) Q.23 (B) Q.24 (C)

Q.25 (B) Q.26 (A) Q.27 (D) Q.28 (A)

Q.29 (A) Q.30 (B) Q.31 (C) Q.32 (A)

Q.33 (A)

EXERCISE 2

Q.1 (B) Q.2 (A) Q.3 (B) Q.4 (B)

Q.5 (C) Q.6 (B) Q.7 (C) Q.8 (B)Q.9 (A) Q.10 (C) Q.11 (C) Q.12 (B)Q.13 (D) Q.14 (A) Q.15 (A) Q.16 (C)Q.17 (A) Q.18 (C) Q.19 (C) Q.20 (C)Q.21 (D) Q.22 (A) Q.23 (D) Q.24 (A), (C)

Q.25 (A), (B), (C) Q.26 (B), (D) Q.27 (A), (B), (C) Q.28 (A), (C), (D)Q.29 (A), (C), (D) Q.30 (A), (C)

Q.31 [(A) S (B) R (C) P (D) Q ] Q.32 [(A) Q, (B) S, (C) P, Q]Q.33 [(A) Q,R; (B) P,S; (C) S; (D) Q,R ] Q.34 [(A) Q (B) P,Q,R (C) P (D) R,S ]Q.35 [(A) Q,R (B) R (C) P (D) S] Q.36 [(A) P,Q,R (B) R,S (C) Q,R (D) P,Q ]

EXERCISE 3

Q.1 [101 102 103 104 105 106 107 108 109

Unu Unb Unt Unq Unp Unh Uns Uno Une]Q.2 [N3– > O2– > F – > Na+ > Mg2+] Q.3 [Isoelectronic Ca+2(Value of Zeff is higher)]Q.4 [Cs > Na > Mg > Si > Cl] Q.5 [Zeff&half filledconfig.]Q.6 [halffilled&fullyfilledorbitals] Q.7 [halffilledandfullyfilled orbitals]Q.8 [(a) P–Cl (b) S–O, (C) N–F] Q.9 [(i) Cl (ii) Cs (iii) Al (iv) F (v) Xe]Q.10 [r k > 1.34Å > r F] Q.11 [BaO]Q.12 [False]Q.13 [(a) basic (b) acidic (c) basic (d) acidic (e) basic (f) acidic ]Q.14 [CaO < CO < CO2 < N2O5 < SO3] Q.15 [EN1 > EN2]Q.16 [5.919 Å] Q.17 [1.21 Å]

Q.18 [4.64 Å ; b = 3.28 Å] Q.19 [IE2 = 1825 kJ/mole, IE3 = 2737.5 kJ/mol]Q.20 [3.476 eV] Q.21 [3.03(Pauling)]Q.22 [3.8752] Q.23 [3.2]


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6


EXERCISE 1

Q.1 [Due to lanthanide contraction; left to right atomicsize decreases. ]Q.2 [Cr = 1s

2 2s2 2p6 3s2 3p6 4s1 3d5

Zeff =Atomic no s for 4s electron (valance shell e¯ –1) × 0.35 +

Penultimate shell e¯ × 0.85 + remaininge¯ = (1 – 1) × 0.35 + 13 × 0.85 + 10= 0 + 11.05 + 10 = 21.05Zeff = 24 – 21.05 = 2.95

Zeff for 3d electron for 3de¯ = (valance shell e¯ –1) × 0.35 + remaining e¯

= (5 – 1) × 0.35 + 18= 1.4 + 18 = 19.4= 24 – 19.4 = 4.6 ]

Q.3 [Atomic no.]

Q.5 [Isoelectronic species = Total no. of electronsare same ]

Q.6 [After theremoval of oneelectron Cr acquiredhalf filled stabled-orbital configuration]Q.7 [Basiccharacters electropositive/ metallic character ]

Q.8 [Metals oxidearebasicas well as they areelectropositivelyin nature ]

Q.10 [(Heg)2 is an endothermic process ]Q.11 [Dueto thehalffilledstableconfigurationofN itsionisationpotential is maximum]

Q.12 [Correct order is B < C < O < N]Q.13 [Correct order of IE1

Li < B < Be < C]

Q.14 [Poorer shielding of 5d electrons by4f electron. ]

Q.15 [e

zAtomic size]

Q.22 [r c + r F = 1.33 Å

r si + r F = 1.54 Å----------------------r c + r si + 2r F = 2.87 Å

r c + r si = 1.87 Å

1.87Å + 2 r F = 2.87 Å2 r F = 1.00 Å

r F = .5 Å]

Q.23 [Electronegativity sizeatomic1

]

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