Periodicity

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SirReconstruct Your Chemistry With Prince SirPage 1

PERIODIC PROPERTIES OF ELEMENTSDOBEREINERS TRIADS

According to Dobereiner when elements of same properties are kept in theincreasing order of atomic weights, the atomic weight of middle element isequal to the mean atomic weight of remaining two elements. Such a group ofelements is called Dobereiners triad.

Mean of first and last elementLi Na K7 23 39Be Mg Ca8 24 40P As Sb31 75 120

Cl Br I35.5 80 123

Dobereiner could arrange only a few elements as triads and there are someelements present in a triad, whose atomic weights are approximately equal

Fe Co NiRu Rh Pd

There fore, this hypothesis was not acceptable for all elements.NEWLANDS RULE OF OCTAVE

As in music, the eighth node is same as the first node. If the elements arearranged in the increasing order of atomic weights, on starting with anelement, the first element will exhibit similarities with the eighth element e.g.Name of element Li Be B C N O F

7 9 11 12 14 16 19Name of element Na Mg Al Si P S Cl

23 24 27 28 31 32 35.5It is clear from the above table that sodium is the eighth element fromlithium, whose properties resemble that of lithium.This type of classification was limited up to only 20 elements.

CHEMISTRY

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MENDELEEFS PERIODIC LAWAccording to Mendeleefs periodic law, the physical and chemical propertiesof elements are periodic function of their atomic weights.

MENDELEVS PERIODIC TABLEMendeleves Periodic table is based on atomic weight. In the periodic table, the horizontal lines are called periods and the verticallines are called groups.The periodic table consists of seven periods and nine groups (The earlierperiodic table had only 8 groups.The noble gases were added later in the zero group because these were notdiscovered when Mendeleef put forward his periodic table.All the groups (except VIII and Zero groups) are divided into subgroups Aand B.2, 8, 18 and 32 are called magic numbers.

ORIGINAL MENDELEEVS PERIODIC TABLE

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MERITS OF MENDELEEFS PERIODIC TABLE Classification of elements then known, was done for the first time and the

elements having similar properties were kept in the same group. It encouraged research and led to discovery of newer elements. Mendeleef had even predicted the properties of many elements not

discovered at that time. This helped in the discovery of these elements.For example. Mendeleef predicted the properties of the following elements.(a) Eka-boron- This was later called scandium (Sc)(b) Eka-aluminium - This was later called gallium (Ga)(c) Eka-silicon - This was later called germanium (Ge)

DEFECTS OF MENDELEEFS PERIODIC TABLE

(1) Position of Hydrogen Like alkali metals hydrogen has one electron in itsoutermost level and like halogens, it has one electron less than the next inertgas. Therefore, placing hydrogen with alkali metals is equally appropriate asplacing it with halogens.

(2) Position of Isotopes The isotopes have different atomic weights and theperiodic table is based on atomic weights. Therefore, isotopes should get dif-ferent places in the periodic table on the basis of atomic weights.

(3) The periodic table is not fully based on increasing order of atomic weights.For example :Ar(39.94) was placed before Ca (39.098)Te (127.6) was placed before I(126.9)Co(58.93) was placed before Ni (58.69)

(4) It is not proper to place together the elements having differing properties,such as coinage metals (Cu, Ag and Au) with alkali metals; Zn, Cd and Hgwith alkaline earth metals and metal like Mn with halogens. Similarly. Pt andAu having similar properties have been placed in different groups.

(5) There is no indication whether lanthanides and actinides are associated withgroup IIIA or group IIIB.

(6) Position of Isobars These elements have different groups when mass remains same.

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SirReconstruct Your Chemistry With Prince SirPage 4MODERN PERIODIC LAW AND MODERN PERIODIC TABLE

Mosley proved that the square root of frequency () of the rays, which areobtained from a metal on showering high velocity electrons is proportional tothe atomic number of the atom. This can be represented by the followingexpression.

= a (Zb)where Z is nuclear charge on the atom and a and b are constants.The nuclear charge on an atom is equal to the atomic number.According to modern periodic law.

The properties of elements are the periodic functions of their atomic numbersMODERN PERIODIC TABLE

One the basic of the modern periodic law, Bohr proposed a long form ofperiodic table that was prepared by Rang and Warner.In the periodic table the horizontal lines are periods and the vertical lines aregroups.The periodic table has a total of seven periods and 18 groups. But the numberof groups is 16, because the eighth group has been divided into three groups.There are two elements in the first period eight elements in each of thesecond and third periods, eighteen elements in each of the fourth and fifthperiod thirty two elements in the sixth period and only nineteen elements tillnow in the seventh period.The first period is very short period, second and third are short periodsfourth and fifth are long periods sixth is very long period, while the seventhis incomplete period.The lanthanides (Elements from atomic numbers 58 and 71) and actinides(elements from atomic numbers 90 to 103) are included in the sixth andseventh groups through these have been kept outside the periodic table.Period - The details about the seven periods are as follows.

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Period Atomic number Number of elementsFrom to

First H (1) He (2) 2Second Li (3) Ne (10) 8Third Na (11) Ar (18) 8Fourth K (19) Kr (36) 18Fifth Rb (37) Xe (54) 18Sixth Cs (55) Rn (86) 32 (including lanthanides)

Seventh Fr (87) Ha (105) 19 (including actinides)

Group - The modern periodic table has 18 vertical columns and according to CAS system there are 16groups having the following number of elements.

Group Number of Elements(a) IA or group1 7 (H, Li, Na, K, Rb, Cs Fr) Alkali metals(b) IIA or group2 6 (Be, Mg, Ca , Sr, Ba, Ra) Alkaline earth metals(c) IIIA or group13 5 (B, Al , Ga , In, Tl) Boron family(d) IV A group14 5 (C, Si, Ge , Sn , Pb) Carbon family(e) V A group 15 5 (N, P , As, Sb, Bi) Nitrogen family(f) VI A group16 5 (O, S , Se, Te , Po) Oxygen family (chalcogen)(g) VII A group17 5 (F, Cl, Br, I, At) - Halogen family(h) Zero group18 6 (He , Ne , Ar, Kr, Xe, Rn) Inert elements(i) III B group3 32(Sc, Y, La, Ac amd 14 lanthanide elements and 14 actinide

elements.These are elements of IIIB group, whichwhich could not be accommodated in one column andtherefore written separately outside the periodic table.

(j) IV B group4 4 (Ti, Zr, Hf, Ku)(k) V B group5 4 (V, Nb, Ta, Ha)(l) VI B group6 3 (Cr, Mo, W)(m) VII B group7 3 (Mn, Tc, Re)(n) VIII (3) group8,9,10 9 (Fe, Co, Ni, Ru, Rh, Pd, Os, Ir, Pt)(o) I B group11 3 (Cu, Ag, Au)(p) II B group12 3 (Zn, Cd, Hg)

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MERITS OF LONG FORM OF PERIODIC TABLE OVER MANDELEEFS PERIODIC TABLE

Positions of Isotopes and Isobars - Isotopes have same atomic number and theperiodic table is based on atomic numbers. Therefore, various isotopes of thesame elements have to be provided the same position in the periodic table.Isobars gave same atomic weights but different atomic numbers and thereforethey have to be placed at different positions.The positions of actinides and lanthanides is more clear now because thesehave been placed in IIIB groups and due to paucity of space, these are writtenat the bottom of the periodic table.In the periodic table a diagonal line from B to At separates the metals,nonmetals and metalloids from one another. The elements like B, Si As. Te At,that are situated near this line are metalloids i.e. these have characteristic ofboth metals and nonmetals.The general electronic configurations of the elements remains same in group.

DEFECTS OF LONG FORM OF PERIODIC TABLEThe position of hydrogen is still disputable as it was there in Mendeleef peri-odic table in group I A as well as IVA & VIIA.Helium is an inert gas but its configuration is different from that of the otherinert gas elementsLanthanide and actinide series could not be adjusted in the main periodictable and therefore they had to be provided with a place separately below thetable.

S-BLOCK ELEMENTS The elements of the periodic table in which the last electron enters in s

orbital, are called sblock elements. s-Orbital can accommodate a maximum of two electrons.

Their general formulae are ns1 and ns2 respectively, where n = (1 to 7). I A group elements are known as alkali metals because they react with water

to form alkali. II A group elements are known as alkaline earth metals becausetheir oxides react with water to form alkali and these are found in the soil orearth.

The total number of s block elements are 14. Fr87and Ra88 are radioactive elements while H and He are gaseous elements Cs and Fr are liquid elements belonging to s-block.

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P - BLOCK ELEMENTS The elements of the periodic table in which the last electron gets filled up in

the p-orbital, called p-block elements. A p-orbital can accommodate a maximum of six electrons. Therefore, p-block

elements are divided into six groups which are III A, IV A, V A, VI, A VII A andzero groups.

The general formulae of p block elements are ns2np6 (where n = 2 to 6) The zero group elements having general formula ns2 np6 are inert, because

their energy levels are fully filled. The total number of p block elements in the periodic table is 30 (excluding

He) There are nine gaseous elements (Ne, Ar, Kr, Xe, Rn, F2, Cl2, O2 and N2)belonging to p-block. Gallium (Ga) and bromine (Br) are liquids. The step-like thick lines drawn in the periodic table in the p-block divides

elements into metals nonmetals and metalloids. B to At, which are presenton this line are metalloids.

D-BLOCK ELEMENTS The elements of the periodic table in which the last electron gets filled up in

the d orbital, called d block elements. The d block elements are placed in groups named IIIB, IV B, V B, VI B, VII B,

VIII, I B and II B. In d block elements the electron gets filled up in the d orbital of the penultimate

shell. That is why these elements are known as transition elements. Through the total number of d block elements is 33 in the periodic table but

there are only 30 transition elements. Becuase only those elements aretransition in which d orbital partilly filled.

The general formula of these elements is (n1)s2, p6, d110 ns12 wheren = 4 to 7.

All of these elements are metals Out of all the d block elements mercury is the only liquid element.

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F-BLOCK ELEMENTS The elements of the periodic table in which the last electron gets filled up in

the f orbital, called f block elements. The f block elements are from atomic number 58 to 71 and from 90 to 103. The lanthanides occur in nature in low abundance and therefore, these are

called rare earth elements. There are 28 f block elements in the periodic table. The elements from atomic number 58 to 71 are called lanthanides because

comes after lanthanum (57). The elements from 90 to 103 are called ac-tinides because comes after actinium (89).

All the actinide elements are radioactive All the elements after atomic number 82 (i.e., Pb82) are radioactive. All the elements after atomic number 92 (i.e.U92 ) are transuranic elements. The general formula of these elements is

(n2)s2p6 d10 f(114) (n1)s2 p6 d01 ns2where n = 6 & 7.

EFFECTIVE NUCLEAR CHARGE In a polyelectronic atom, the internal electrons repel the electrons of the

outermost orbit. This results in decrease the nuclear attraction on theelectrons of the outermost orbit.

Therefore, only a part of the nuclear charge is effective on the electrons of theoutermost orbit. Thus, the inner electrons protect or shield the nucleus andthereby decrease the effect of nuclear charge towards the electrons of theoutermost orbit.

Thus the part of the nuclear charge works against outer electrons, is knownaseffective nuclear charge (Z*) = (Z - ) Where is screening constant .Shielding (Screening) effect : The repulsion of valency electrons by the elec-trons in peunltimate shell to reduce effective nuclear charge (thereby reduc-ing ionization enthalpy value).

Zeff. = Z (where Z = atomic number and = shielding constant)

+

1 2

Nucleus Attraction

Repulsion1e01e0

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Calculation of Shielding Constant ()In order to estimate extent of shielding, Slater proposed a set of empiricalrules. These rules are simplified generalizations based upon averagebehaviour of various electrons. These calculations are useful in explainingatomic size and electronegativity. To calculate the shielding constant foran electron in the following rules are used :A. Write the electronic configuration of element in the following orderand grouping(1s), (2s, 3p), (3s, 3p) (3d), (4s, 4p) (4d) (4f), (5s, 5p) etc.B. Electrons in any group to the right of the (ns, np) group contributenothing to shielding constant.C. All the electrons in the (ns, np) group shield the valence electron tothe extent of 0.35 each except for the 1s (i.e., n= 1) for 1s s is takenas 0.3.D. All the electrons in the (n 1) shell shield to an extent of 0.85 each.E. All the electrons in (n 2) or lower shield completely and theircontribution is 1.00 each.F. When electron being shielded is in an nd and nf group, rules (B) and(C) are same but rule (D) and (E) are changed. Here all the electronsin groups lying to the left of the nd or nf group contribute 1.00 each.Thus, = [0.35 Number of electrons in nth shell excluding test (valence)electron] + [0.85 Number of electrons in (n 1)th shell] +[1.0 Number of electrons ininner shell]e.g., (i)7N : Consider the valence electron in 2p orbital, then excluding it 4electrons are present in (2s,2p) group.

7N : (1s)2 (2s, 2p)5s = [0.35 4] + [0.85 2] = 3.10(ii) 30Zn : (1s)2, (2s, 2p)8, (3s, 3p)8, (3d)10, (4s)2[Valence electron in 4s, then excluding it one electron is present in4s group]s = [0.35 1] + [0.85 18] + [1.00 10] = 25.65

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Cationic RadiusAn atom forms a cation on loss of electron/s. The cationic radius can bedefined as the distance between the nucleus and the limit of the electroncloud scattered around the nucleus.The size of a cation is smaller in comparison to the size of its correspondingatom. This is because of the fact that an atom on losing electrons/s form acation, which has lesser number of electrons/s than the number of proton/s.This results in increase in the effective nuclear charge.Size of cationExamples - (1) Mn > Mn+2 > Mn+3 > Mn+4 > Mn+6 > Mn+7

(2) Pb+2 > Pb+4

Anionic RadiusWhen a neutral atom gains electron/s it becomes a negatively charged ioncalled an anion. The distance between the nucleus of an anion and the limitof the electron cloud scattered around the nucleus, is called its anionic radius.The size of an anion is greater than the size of its corresponding atom, becausethe number of electrons present in the anion gets larger than the number ofprotons due to gain of electron/s. This results in decrease in the effectivenuclear charge.Size of anion O0 < O1 < O2

. Atomic Radius Problem in calculating actual size of atom and hence distance between nucleiis calculated giving rise to three type of radii for atoms.

(a) Covalent radius : Cr = 2

d

Cr < actual atom size [Slight difference]

[Used for H2, Cl2 and such molecules]Half of the internuclear distance between two atoms of the element held by a single covalentbond.

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SirReconstruct Your Chemistry With Prince SirPage 11Calculation for covalent radius can be made as :

A. In homodiatomic molecule (AA) :r =

dA A2

(dAA = Bond length in molecule)B. In heterodiatomic molecule (AB) :

(i) When (A B) is very small :dAB = rA + rB

dAB = Bond lengthwhere rA & rB = Covalent radii of A & B respectively.A & B = Electronegativities of A & B.

(ii) When (A B) is large :dAB = rA + rB 0.09 (A B)

(b) Metallic Radius :Half of the internuclear distance between two nearest atoms in the metallic lattice.

d

Mr = 2

d

[Used for metals]

(c) Vanderwaal radius :Half of the internuclear distance between two nearest atoms belonging totwo adjacent molecules of noble gases.

VVr = 2

dV

r >> actual size [very large difference]

In general for an elementVander waal radius > Metallic radius > Covalent Radius

(d) Ionic Radius : A cation is smaller than parent atom . An anion is largerthan parent atom.SIZE OF ISOELECTRONIC SERIES

The species, which have same number of electrons but different nuclear charges,constitute anisoelectronic series.Size in an isoelectronic seriesC4->N3->O2->F-1>Ne>Na+>Mg2+>Al3+>Si4+>P5+>S6+(left to right size decreases)

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IONISATION ENERGIESThe energy required to remove the most loosely bound electron from theoutermost orbit of one mole of isolated gaseous atoms of an element, is calledionisation energy or ionisation potential (IP). This ionisation is anendothermic or energy-absorbing process.

An electron cannot be removed directly from an atom in solid state. For thispurpose, the solid state is converted to gaseous state and the energy requiredfor this is called sublimation energy.

Sucssesive Ionisation Energies The energy required to remove one electron from a neutral gaseous atom to

convert it to monopositive cation, is called first ionisation energy (IE1). Theenergy required to convert a monopositive cation to a diapositive cation iscalled second ionisation energy (IE2)M(g) M+(g) + e IE1

M+(g) M2+(g) + e IE2 IE1 < IE2 < IE3 because as the electrons go out of the atom, the ionic size goeson decreasing and the amount of positive charge goes on increasing.

Factors Affecting Ionisation Potential(i) Atomic size : When the size of an atom is very large the electron of the

outermost orbit bound to the nucleus by weaker attractive forces. Such anelectron will be readily removed from the atom. Therefore. The value ofionisation potential will be low.

radiiAtomicEnergiesIonisation 1

(ii) Effective Nuclear Charge : Atomic size decreases with increase in effectivenuclear charge because, higher the effective nuclear charge stronger will bethe attraction of the nucleus towards the electron of the outermost orbit andhigher will be the ionisation potential

effective ZEnergiesIonisation

(iii) Shielding Effect : The electrons of internal orbits repel the electrons of theelectron of the outermost orbit due to which the attraction of the nucleustowards the electron of the outermost orbit decreases and thus atomic sizeincreases and the value of ionisation potential decreases.

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effectSheildingEnergiesIonisation1

(iv) Stability of half filled and fully filled orbitals : The atoms whose orbitals arehalf-filled (p3,d5,f7) or fully-filled (s2, p6 , d10, f14) have greater stability thanthe others. Therefore, they required greater energy to for removing out elec-tron. However stability of fully filled orbitals is greater than that of the halffilled orbitals

(v) Penetration power : In any atom the s orbital is nearer to the nucleus incomparison to p, d and f orbitals. Therefor, greater energy is required toremove out electron from s orbital than from p, d and f orbitals. Thus thedecreasing order of ionisation potential of s, p, d and f orbitals is as followss > p > d > f

Periodic Table & Ionisation Potential(a) In a Period :- The value of Ionisation potential normally increase on going

from left to right in a period, because effective nuclear charge increasesand atomic size decreases.

Exceptions In second period ionisation potential of Be is greater than that of B, and

in the third period ionisation potential of Mg is greater than that of Aldue to high stability of fully filled orbitals.

In second period ionisation potential of N is greater than O and in thethird period ionisation potential of P is greater than that of S, due tostability of half filled orbitals.

The increasing order of the values of ionisation potential of the secondperiod elements isLi < B < Be < C < O < N < F < NeThe increasing order of the values of ionisation potential of the third pe-riod elements isNa < Al < Mg < Si < S < P < Cl < Ar

Ionisation Potential of Transition ElementsIn transition elements, the value of ionisation potential has very little in-crease on going from left to right in a period because the outermost orbitremains the same but electrons get filled up in the (n1)d orbitalsresulting in very little increase in the values of ionisation potential.

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SirReconstruct Your Chemistry With Prince SirPage 14 The value of ionisation potential of transition elements depends on the

following two important factors.(a) The value of ionisation potential increases with increase in effective nuclearcharge.(b) The value of ionisation potential decreases with increase in shielding effectwhen the number of electrons increases in (n-1)d orbitals.In the first transition element series the first ionisation potential normallyincreases on going from left to right from Sc to Cr because shielding effect ismuch weaker in comparison to effective nuclear charge. The value of firstionisation potential of Fe, Co and Ni remains constant, because shieldingeffect and effective nuclear charge balance one another. The value of ionisationpotential shows slight increase from Cu to Zn because they have fully filled sand d orbitals. The value of first ionisation potential of Mn is maximumbecause it has maximum stability due to fully filled s and d orbitals.

Inner Transition ElementsThe size of inner transition elements is greater than that of d block ele-ments. Therefore the value of ionisation potential of f block elements issmaller than that of d block elements and due to almost constant atomicsize of f block elements in a period the value of their ionisation potentialremains more constant than that of d block elements.In a GroupThe value of ionisation potential normally decreases on going from top tobottom in a group because both atomic size and shielding effect increase.Exception :The value of ionisation potential remains almost constant from Al to Ga inthe III A group.The actual variation for this group is

(B > Tl > Ga> Al > In)In the periodic table the element having highest value of ionisationpotential is He.The values of ionisation potential of noble gases are extremely high, becausethe orbitals of outermost orbit are fully-filled (ns2 np6) and provide greatstability.

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SirReconstruct Your Chemistry With Prince SirPage 15 In a period, the element having least value of ionisation potential is an

alkali metal (group I A ) and that having highest value is inert gas (Group0)

Applications of Ionisation PotentialThe elements having high values of ionisation potential have low reactivity,e.g. inert gases.The value of ionisation potential decreases more on going from top tobottom in a group in comparison to a period. Therefore, reactivity increases and the atom forms a cation by loss of electron.The elements having low value of ionisation potential readily looseelectron and thus behave as strong reducing agents. The elements having low value of ionization potential readily looseelectron and thus exhibit greater metallic property.The elements having low value of ionisation potential readily looseelectron and thus have basic property.

ELECTRON AFFINITYThe energy released on adding up one mole of electron to one mole ofneutral atom (A) in its gaseous state to form an anion (A) is called elec-tron affinity of that atom. Since the electron adds up in the outermostorbit, energy is given out. Therefore, electron affinity is associated with anexothermic process.

A(g) + e A (g) , H = EnWhen one electron adds up to a neutral atom, it gets converted to a unitnegative ion and energy is released. On adding one more electron to themononegative anion, there is a repulsion between the negatively chargedelectron and anion. In order to counteract the repulsive forces, energy hasto be provided to the system. Therefore, the value of the second electronaffinity is positive.

A (g) + e A2 (g) , H = + En

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SirReconstruct Your Chemistry With Prince SirPage 16 Factors Affecting Electron Affinity

1.Effective Nuclear chargeWhen effective nuclear charge is more then, the atomic size less.

ergyNuclear EnEffectiveAE .. 2. Atomic Size or Atomic RadiusWhen the size or radius of an atom increases, the electron entering theoutermost orbit is more weakly attracted by the nucleus and the value ofelectron affinity is lower.

eAtomic sizAE 1..

3.Shielding EffectShielding effect is directly proportional to atomic size and atomic size isinversely proportional toelectron affinity.4.Stability of Fully-Filled and Half-Filled OrbitalsThe stability of the configuration having fully-filled orbitals (p6,d10,f14) andhalf-filled orbital (p3, d5, f7 ) is relatively higher than that of other con-figurations.

Periodic Table and Electron AffinityIn a period, atomic size decreases with increase in effective nuclear chargeand hence increase in electron affinity.

Exception :On going from C6 to N7 in the second period, the values of electronaffinity decreases in stead of increasing. This is because there arehalf-filled (2p3) orbitals in the outermost orbit of N, which are morestable. On the other hand, the outermost orbit in C has 2p2 configuration.In the third period, the value of electron affinity of Si is greater thanthat of P. This is because electronic configuration of the outermost orbit inP atom is 3p3 , which being half-filled, is relatively more stableThe values of electron affinity of inert gases are zero or +ive, becausethere outermost orbit has fully-filled p orbitals, So they donot havetendency to accept electrons or energy has to be given to add an extraelectron.In a period, the value of electron affinity goes on decreases on going fromgroup IA to group IIA. The value of electron affinity of the elements ofgroup IIA is zero because ns orbitals are fully-filled and such orbitals haveno tendency to accept electrons.

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In a GroupThe values of electron affinity normally decrease on going from top tobottom in a group because the atomic size increases which decreases theactual force of attraction by the nucleus.

ExceptionsThe value of electron affinity of F is lower than that of Cl, because thesize of F is very small and compact and the charge density is high on thesurface. Therefore, the incoming electron experiences more repulsion incomparison to Cl . That is why the value of electron affinity of Cl is highestin the periodic table. The order among the halogens is

Cl > F > Br >IFor the elements of oxygen family the order of electronegativity is

S > Se > Te > Po > OThe values of electron affinity of alkali metals and alkaline earth metalscan be regarded as zero, because they do not have tendency to form anionsby accepting electron.

Imortant NoteWhile comparing electron affinity we will only consider magnitude butwhile comparing Electron Gain Enthalpy we will consider sign also.For e.g. following E.A. values

O = -144 kj/mole, S = -200 kj/mole,Se = -195 kj/moleTe = -190 kj/molePo = -174 kj/mole

Order of E. A. is : S > Se >Te > Po > O (consider magnitude only )Order of E.G.E. is : S < Se

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SirReconstruct Your Chemistry With Prince SirPage 18 (i) Paulings approach :Pauling based his scale on thermo-chemical data. He concluded that thebond formed between the two atoms A and B must be stronger than theaverage of single bond energies AA and BB molecules. According tohim the electronegativity difference between two atoms A and B (A - B)is given by :

(A - B) = 0.101 In SI Units .....(1)(A - B) = 0.208 Energy in k cal Units .....(2)where = = EAB E EAA B BThe electronegativity of atoms on Pauling scale are obtained by assumingH = 2.05.Thus A - B = 0.1017 = 0.1017 [EAB (EAA EBB)1/2]1/2

Where EAB, EAA and EBB are in kJ mol1

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(ii) Mullikens approach :According to Mulliken, electronegativity of an atom is average of I.E.and E.A.M = I.E. E.A.2

I.E. and E.A. are ionization energy and electron affinity inelectron volts.Mulliken values are 2.8 times greater than Pauling values.Thus, P = I.E. E.A.2 2.8 I.E. E.A.5.6 If I.E. and E.A. are being measured in kJ mol1 then equation becomes as

M = I.E. E.A.2 96

.48 ( 1 eV/molecule = 96.48 kJ mol1)

Also P = I.E. E.A.2 96.48 2.8

P = I.E. E.A.540

(iii) Allerd and Rochow approach :In this method, electronegativity is calculated as(A)AR = 0.744 + 3.59Zeffectiver 2

A = Electronegativity of atom A on Allred & Rochow scale.Zeff. = Effective nuclear charge at periphery of element A.r = Radius of an element in A

Heat of formation of a molecule can be calculated if electronegativities ofbonded atoms areknown.Hf = 23 S(A B)2 55.4 nN 26.0 nO

Hf = Approximate Heat of formation of molecule (in kcal mol1).A & B = Electronegativities of bonded atoms.nN & nO = number of nitrogen and oxygen atoms in molecule.

Factors Affecting Electronegativity1.Atomic size Electronegativity of a bonded atom decreases withincrease in its size. When effective nuclear charge is high the nucleus will attract the sharedelectrons with greater strength and the electronegativity will be high.This effect increases the atomic size which decreases the electronegativityvalue.

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2.Hybridisation state of atom Electronegativity increases with increasesin the s character of the hybrid orbital. This is because the s orbital is nearer to thenucleus and thus suffers greater attraction resulting in increase in electronegativity.

or

The number of covalent bonds present between two bonded atoms isknown as bond order. With increases in the bond order, the bond distancedecreases, effective nuclear charge increases and thus electronegativityincreases.The increasing order of electronegativity is as follows :

CC < C = C < C Csp3 < sp2 < sp

3. Oxidation number The electronegativity value increases with increasein oxidation number because radius decreases with increase in oxidationnumber.The increasing order of electronegativity is as follows : Fe < Fe2+

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The elements of group IV A, show no change in electronegativity value ongoing from top to bottom in the group from Si onwords.

APPLICATIONS OF ELECTRONEGATIVITY(i) Nomenclature

Compounds formed from two nonmetals are called binary compounds. Nameof moreelectronegative element is written at the end and (-ide) is suffixed to it. Thename of lesselectronegative element is written before the name of more electronegativeelement of the formula.

(ii) Nature of BondIf difference of electronegativities of the two elements is 1.7 or more, thenionic bond is formed between them whereas if it is less than 1.7, then cova-lent bond is formed. (HF is exception in which bond is covalent althoughdifference of electronegativity is 1.9)

(iii) Metallic and Nonmetallic NatureGenerally values of electronegativity of metallic elements are low, whereaselectronegativity values of nonmetals are high.

Partial Ionic Character in Covalent bondsPartial ionic characters are generated in covalent compounds by the differ-ence of electronegativities. Hanny and smith calculated percentage of ioniccharacter from the difference of electronegativity.Percentage of ionic character = 16(XA XB) + 3.5(XA XB)2

= 16 + 3.52

= (0.16 + 0.0352) 100(Here XA is electronegativity of element A XB is electronegativity of element B) = XA XBCalculation of Heat of Formation

Heat of formation = (XA XB)2 100 Kcal/Mole

(iv) Bond LengthWhen difference of electronegativities of atoms present in a molecule in-creases, then bond lengthdecreases. Shoemaker and stephensen determined.

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SirReconstruct Your Chemistry With Prince SirPage 22Bond length dAB = rA + rB 0.09 (XA XB)

or dAB = (DAA + DBB) 0.09 (XA XB) Here XA > XBBond EnergyBond energy is inversely proportional to bond lengthBond Strength & StabilityBond strength and stability of AB increases on increases in difference ofelectronegativities of atoms A and B bonded AB.Therefore HF > HCl > HBr > HIReactivityBond energy is proportional to stability and stability is inversely proportionalto reactivity.HF < H Cl < H Br < HI

(v) Nature of OxidesIf difference of the two electronegativities (XOXA) is 2.3 or more then oxidewill be basic in nature. Similarly if value of XOXA is lower than 2.3 then thecompound will be first amphoteric then acidic in nature.

Oxide Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7(XOXA) 2.6 2.3 2.1 1.8 1.5 1.1 0.5Nature Strong Basic Amphoteric weak Acidic Strong Strongestbasic acidic acidicBasic character of oxides decreases with increasing difference of

electronegativities of central atom and oxideTherefore basic character decreases in the period and acidic character in-creases.

NATURE OF HYDROXIDESAccording to Gallis if electronegativity of A in a hydroxide (AOH) is morethan 1.7 then it will be acidic in nature whereas it will be basic in nature ifelectronegativity is less than 1.7For example NaOH and ClOHElectronegativity (XA) 0.9 3.00Nature Basic AcidicIf the value is more than XOXH , then that hydroxide will be basic otherwiseit will be acidic.

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SirReconstruct Your Chemistry With Prince SirPage 23 Acidic and Basic Nature of Hydroxides of Elements

Acidic and basic nature of hydroxide of an element AOH depends on ionisationpotential of A. If ionisation potential of A is low then it will give its electronto oxygen easily thus AOH will be basic.

NATURE OF OXY ACIDSThose inorganic acids in which XOH group is present are called oxy acids.(Here X = a non metal) like HOCl is an oxy acid.

In a PeriodMoving from left to right in a period strength of oxy acids increases likeOrder of strength of oxy acid of elements of second periodH3BO3 < H2CO3 < HNO3Order of strength of oxy acids of elements of third periodH2SiO3 < H2SO4 < HClO4

In a GroupMoving upwards to downward in a group, strength of oxy acids decreasesLike - in VA group

HNO3 > H3PO4When amount of oxygen increases in an oxy acid of an element (i.e. onincreasing oxidation state of element) strength of acid increases like(a) H2SO3 < H2SO4 (b) HClO < HClO2 < HClO3 < HClO4Stability of acids also increases in the same order

NATURE OF HYDRIDESMoving from left of right in a period, nature of hydrides of nonmetal ele-ments becomes basic to acidic. LikeIn second period NH3 H2O HFWeak base Neutral Acidic (Amphoteric)

In third period PH3 H2S HCl Very weak acid Weak acid Strong acidMoving upwards to downwards in a group, acidic nature and reducing power

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SirReconstruct Your Chemistry With Prince SirPage 24 of hydrides of nonmetal elements both increases but stability decreases.

LikeAcid strength and reducing strength : HF < HCl < HBr < HI

stability : HF > HCl > HBr > HIDENSITY

Density in solid state changes on change in atomic number of elements.Moving left to right in the period densities of solid and liquid elements firstincreases, it is maximum in the middle and then decreases.Moving upwards to downwards in a group, density generally increases butthere is irregularity in this order.Density decreases instead of increasing on moving towards K in IA group andmoving from Mg to Ca in IIA group. Because penultimate value of K and Caincreases, due to which density gets reduced.Os (22.6 gm/cc) is most dense in solid elements and in liquids Hg is mostdense (13.6 gm/cc)

DIAGONAL RELATIONSHIPSome elements of second period Li, Be, B shows dissimilarities with otherelements of this group but shows similarities with elements of third group likeMg,. Al, Si situated diagonally to them. It is called diagonal relationship.

Similarities between properties of Li and Mg are as follows(a) Li and Mg both reacts directly with nitrogen to form lithium nitride (Li3N)and magnesium nitride (Mg3N2) whereas other alkali metals of IA group doesnot form nitride.(b) Fluoride carbonate and phosphate of Li and Mg are insoluble in water whereas

these compounds of other alkali metals are soluble.(c) Li and Mg both are hard metals, whereas other metals of IA group are soft.(d) LiOH and Mg(OH)2 both are weak base, whereas hydroxides of other elementsof IA group are strong base.(f) Metallic bond in Li and Mg both are strong compare to other alkali metals .(g) Their melting and boiling points are high.(h) By thermal disintegration of LiNO3 and Mg (NO3)2 Li2O and MgO isobtained respectively.(i) Thermal stability of Li2CO3 and Mg CO3 is very less compare to other alkalimetals and they liberates CO2 gas easily.

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SirReconstruct Your Chemistry With Prince SirPage 25Similarly Be shows similarity to Al of IIIA group

(a) They both elements does not provide colour to Bunsen burner.(b) They both are comparatively stable in air.(c) Both are insoluble in NH3 therefore does not form blue coloured solution.(d) There is not tendency of making peroxide and superoxide in them.(e) Reducing power is very less due to low value of standard electrode potential

in the form of oxidation potential.(f) Be and Al both forms halogen bridge halides.

Some Important Points(1) Triad rule Dobereiner(2) Octet rule Newland(3) Study of atomic volume Lothar Mayer(4) Inventor of atomic number Moseley(5) Godfather of periodic table Mandeleef(6) Maker of modern periodic table Bohr(7) Mg is bridge element, which joins metals of IIA and II B groups.(8) Elements after atomic number 92 are transuranic elements.(9) Artificial element is Tc43.(10) Liquid non-metal Br(11) Liquid metal Hg, Ga, Cs, Fr(12) Solid volatile non-metal Iodine(13) Lightest metal Li(14) Heaviest metal Ir(15) Hardest metal W(16) Nobel metals Pc, Pt, Au, Ag(17) Element most found on earth Al(18) Gaseous elements 11 (He, Ne, Ar, Kr, Xe, Rn, H2, N2, O2, Cl2, F2)(19) Liquid elements 5(Br, Hg, Ga, Cs, Fr)(20) Submetals 5(B, Si, As, Te, At)(21) Inert gases 6(22) Lowest electronegativity : Cs(23) Highest electronegativity : F(24) Highest ionisation potential : He(25) Lowest ionisation potential : Cs(26) Highest electron affinity : Chlorine (Cl)(27) Least electropositive element : Fluorine (F)

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(28) Most reactive solid element : Li(29) Most reactive liquid element : Cs(30) Most stable element : Te(31) Largest atomic size : Cs(32) Most electropositive element : Cs ; Fr

(in stable element) (In all element)(33) Group containing maximum no. : Zero gp ; next to zero gp is VII gp or of gaseous elements in periodic table halogen gp (F2 and Cl2)(34) Total number of gaseous : 11 (H2, He, N2, O2, F2 Ne, Cl2

elements in periodic table Ar, Kr, Xe, Rn)(35) Total number of liquid elements : 4 (Ga, Br, Cs, Hg) in periodic table (Fr and Eka are also liquid)(36) Volatile d-block elements : Zn, Cd, Hg(37) Most abundant element on earth : Oxygen followed with Si(38) Most stable carbonate : Cs2CO3(39) Strongest alkali : Cs(OH)(40) Element kept in water : P(41) Elements kept in kerosene oil : Na, K, I, Cs(42) Liquid non metal : Br2(43) Bridge metals : Na, Mg(44) Noble metals : Au, Pt(45) Lightest element : H(46) Poorest conductor of current : Pb(metal), S (non metal)(47) Most abundant gas : N2(48) Lightest solid metal : Li(49) Heaviest solid metal : Os(highest density 22.6 g/cm3)(50) Natural explosive : NCl3(51) Dry ice : CO2(52) First Nobel prize of chemistry was given to : vant Hoff* Core charge - Atomic number Kernel of electron* Penultimate shell Shell present inside one shell (n 1) from outermost shell, is called penultimate shell.* Prepenultimate shell Shell present inside two shells (n 2) from outermost shell, is called prepenultimate shellAbout 75% of known elements are metals. The non-metals (20 ) are located at the top right of theperiodic table.Elements at the border line of metals and non-metals are called metalloids or semimetals e.g., B,Ge, As.Total number of solid elements are about 90.Total number of radioactive elements are about 42.Liquid radioactive element : Francium.Rarest element in the earths crust : Astatine.Most poisonous metal : Plutonium.Element having maximum number of natural isotopes (10) : Tin.First man made element : Teachnetium (At. no. = 43).van der Waals radius > Metallic radius > Covalent radius. (for an atom)

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Size of anion > Size of atom > Size of cation e.g., I > I >I +.Size of largest atom (Cs) is approximately 4.4 times greater than that of H-atom.Helium has the highest value of I.E. and the lowest value of electron gain enthalpy among all theelements.Fluorine is the most electronegative element whereas cesium is the least electronegative element.Size of isoelectronics decreases with increase in atomic number.Hydrogen is the most abundant element in universe.Oxygen is the most abundant metal on earth.Al is the most abundant metal on earth.The abundance of first five element on earth is O > Si > Al > Fe > Ca.Metallic nature decreases along the period but increases down the group.Non-metallic nature increases along the period but decreases down the group.Out of 17 non-metals, 11 are gases, one is liquid (Br) and five are solid C, P, S, Si & 1.The most electronegative element is F(EN = 4.0 on the Pauling scale) while the leastelectronegative elements are Cs and Fr with EN = 0.7.Metalloids have electronegativity values closer to 2.0.Lanthanoids were also called rare earth elements since their oxides were rarely found on earth inearlier days.

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Exerxise 1Only one choice is correct

Q.1. Among O, C, F, Cl, Br the correct order of increasing atomic radii is :(A) F < O < C < Cl < Br (B) F < C < O < Br < Cl (C) F < Cl < Br < O < C (D) C < O < F < Cl < Br

Q.2. Atomic radii of fluorine and neon in Angstrom units are respectively given by :(A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) None of these

Q.3. The first ionisation energy of Na, Mg, Al and Si are in the order :(A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg < Al > Si (D) Na > Mg > Al < Si

Q.4. The first ionization energy will be maximum for :(A) Uranium (B) He (C) Lithium (D) Iron

Q.5. Which has maximum value for proton :(A) IE (B) Radius (C) Charge (D) Hydration energy

Q.6. Ionisation potential of Na would be numerically the same as :(A) Electron affinity of Na+ (B) Electronegativity of Na+

(C) Electron affinity of He (D) Ionisation potential of Mg.Q.7. The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is

likely to be :(A) Na (B) Si (C) F (D) Ca

Q.8. Ionic radius are :(A) Inversely proportional to effective nuclear charge(B) Inversely proportional to sqaure of effective nuclear charge(C) Directly proportional to effective nuclear charge(D) Directly proportional to sqaure of effective nuclear charge.

Q.9. Which one is incorrect statement :(A) IE1 of He is maximum among all elements (B) EA1 for noble gases is zero(C) Electronegativity is maximum for fluorine (D) IE1 for N < IE1 for O

Q.10. Which set of elements have nearly the same atomic radii :(A) F, Cl, Br, I (B) Na, K, Rb, Cs (C) Li, Be, B, C (D) Fe, Co, Ni, Cu

Q.11. In which pair, the second atom is larger than first :(A) Br, Cl (B) Na, Mg (C) Sr, Ca (D) N, P

Q.12. An element X occurs in short period having configuration ns2np1. The formula and nature of its oxide is :(A) XO3, basic (B) XO3, acidic (C) X2O3, amphoteric (D) X2O3, basic

Q.13. An example of a non-stoichiometric compound is(A) Al2O3 (B) Fe3O4 (C) NiO (D) PbO

Q.14. Mendeleeff corrected the atomic weight of :(A) Be (B) ln (C) Os (D) All of these.

Q.15. The ionic radii of N3, O2 and F are respectively given by :(A) 1.36, 1.40, 1.71, (B) 1.36, 1.71, 1.40 (C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40

Q.16. Most of the known elements are :(A) Metals (B) Non-metals (C) Transition metals (D) Inner transitional metals

Q.17. Zn and Cd do not show variable valency like d block elements due to :(A) Softness (B) Completely filled d orbital(C) Two electrons in outermost orbit (D) Low m.pt.

Q.18. Which of the following is paramagnetic :(A) O2

(B) CN (C) CO (D) NO+

Q.19. The element having 18 electrons in its outermost shell is :(A) 28Ni (B) 46Pd (C) 29Cu (D) None of these.

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Q.20. The order of first electron affinity of O, S and Se is :(A) O > S > Se (B) S > O > Se (C) Se > O > S (D) Se > S > O

Q.21. Which of the following sequence regarding the first ionisation potential of coinage metal is correct :(A) Cu > Ag > Au (B) Cu < Ag < Au (C) Au > Cu > Ag (D) Ag > Cu < Au

Q.22. Which is not arranged in the correct sequence :(A) d5, d3, d1, d4 increasing magnetic moment(B) MO, M2O3, MO2, M2O5 decreasing basic strength(C) Sc, V, Cr, Mn increasing number of oxidation states(D) Co2+, Fe3+, Cr3+, Sc3+ increasing stability

Q.23. Na+, Mg2+, Al3+, Si4+ are isoelectronics. Their ionic size follows the order :(A) Na+ < Mg2+ < Al3+ < Si4+ (B) Na+ > Mg2+ < Al3+ < Si4+

(C) Na+ < Mg2+ > Al3+ > Si4+ (D) Na+ > Mg2+ > Al3+ > Si4+.Q.24. Which transition involves maximum amount of energy :

(A) M(g) M(g) + e (B) M(g) M+(g) + e (C) M+(g) M2+(g) + e (D) M2+(g) M3+(g) + e.Q.25. Which set of metals becomes passive in conc. HNO3 :

(A) Fe, Co, Ni, Al, Cr (B) Na, Mg, K, Cs (C) Mn, Ba, Sr, Ca, Al (D) Pb, Ag, Hg, CuQ.26. Low melting point of manganese in the 1st transition series is due to :

(A) Strong metallic bond due to d10 configuration (B) Weak metallic bond due to d5 configuration(C) Weak metallic bond due to d7 configuration (D) None of these.

Q.27. The order of basic character of given oxides is :(A) Na2O > MgO > Al2O3 > CuO (B) MgO > Al2O3 > CuO > Na2O(C) Al2O3 > MgO > CuO > Na2O (D) CuO > Na2O > MgO > Al2O3.

Q.28. Which pair of elements is chemically most similar :(A) Na, Al (B) Cu, S (C) Ti, Zr (D) Zr, Hf

Q.29. Decreasing order of atomic weight of the elements given below is :(A) Fe > Co > Ni (B) Ni > Co > Fe (C) Co > Ni > Fe (D) Co > Fe > Ni.

Q.30. The largest number of known elements are placed in period :(A) IV (B) V (C) VI (D) VII

Q.31. In the transition elements, the incoming electron occupies (n 1)d sublevel in preference to :(A) np (B) ns (C) (n 1) d (D) (n + 1)s

Q.32. The correct increasing order of bond energy of the following is :(A) N2 < O2 < Cl2 < F2 (B) Cl2 < F2 < N2 < O2 (C) F2 < Cl2 < O2 < N2 (D) O2 < Cl2 < F2 < N2.

Q.33. Which of the following is paramagnetic molecule :(A) ClO2 (B) SO2 (C) SiO2 (D) CO2.

Q.34. Which arrangement represents the correct order of electron gain enthalpy (with negative sign) of the givenatomic species :(A) S < O < Cl < F (B) O < S < F < Cl (C) Cl < F < S < O (D) F < Cl < O < S

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Exerxise 2Passage1

Ionisation energies of five elements are given below :Atom I.E. (in kcal/mol)

I II IIIA 300 549 920B 99 734 1100C 118 1091 1652D 176 347 1848E 497 947 1500

Q.1. Which element is a noble gas :(A) A (B) E (C) C (D) D

Q.2. Which element form stable unipositive ion :(A) A (B) B (C) C (D) D

Q.3. The element whose msot stable oxidation state is + 2 :(A) B (B) C (C) D (D) E

Q.4. Which is a non-metal :(A) A (B) B (C) C (D) D

Q.5. Most reactive metal is :(A) A (B) B (C) C (D) D

Q.6. If B reacts with fluorine and oxygen, the molecular formula of fluoride and oxide will be respectively :(A) BF3, B2O3 (B) BF, B2O (C) BF2, BO (D) None of these.

Q.7. Which of the following pair represents elements of same group :(A) B, C (B) A, B (C) A, D (D) B, D

Passage2Read the following paragraph and answer the questions given below :Bohrs periodic table has seven horizontal rows called periods and eighteen vertical columns calledgroups. Alternatively, the elements of the Bohrs table are divided into four blocks namely s-, p-d- and f-blocks. Group, period and block of given element can be predicted with the help of its electronic configuration.(i) The principle quantum number of valence shell refers to period of element.(ii) The nature of subshell (s, p or d) having last electron refers the block to which the element

belongs.(iii) The number of electrons present in outermost or penultimate shell predicts the group of element.

For s-block, the group number = No. of valence electronFor p-block, the group number = 10 + numbers of valence electronFor d-block, the group number = 2 + (n 1)d electrons

Q.8. The element of atomic number 36 belongs to(A) 3rd period and p-block (B) 4th period and p-block(C) 4th period and s-block (D) 3rd period and d-block

Q.9. The element with atomic number 58 belongs to(A) s-block (B) p-block (C) d-block (D) f-block.

Q.10. An element with atomic number 24 is an example of(A) Alkali metal (B) Alkaline earth metal (C) Inert gas (D) Transition element

Q.11 Number of groups of s-block element, possible in a period is/are(A) 1 (B) 2 (C) 3 (D) 4

Q.12. Number of groups possible for s + p + d block elements in a period are(A) 6 (B) 2 (C) 10 (D) 18

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Passage3The amount of energy required to remove an electron from the last orbit of an isolated (free) atom in

gaseous state is known as ionization potential or first ionization potential of the element. Similarly, the energyrequired for the removal of the electron from the unipositive ion (produced above) is reffered to as second ionizationpotential and thus the third, fourth etc., Ionization potentials may be defined in general as the number of ionization

potentials of an element may be as much as its number of electrons.Q.13. The relationship between IE3 and IE1 of an element is :

(A) IE3 > IE1 (B) IE3 < IE1 (C) IE3 = IE1 (D) None of the aboveQ.14. Which of the following statement is correct ?

(A) IE of elements increases along the period.(B) IE of the 3rd group elements is less than that of the second group elements.(C) IE of group 16 elements is less than that of group 15 elements.(D) All of the above.

Q.15. IE1, IE2 and IE3 of an element are 9.32, 18.21 and 153.83 eV. What information these data convey ?(A) The element has two s electrons in the valence shell.(B) The element has two p-electrons in the valence shell.(C) Both of the above(D) None of the above

Q.16. Ionization energy increases on moving down a group due to(A) increase in the number of electrons (B) Decrease in size of elements(C) Screening effect (D) None of the above.

Q.17. Ionization energy of nitrogen is more than that of oxygen which can be explained by(A) High-effective nuclear charge (B) Low screening(C) Half-filled stability (D) None of the above

Assertion & Reason Type QuestionsIn each sub question below a statement S and an explanation E is given. Choose the correct answersfrom the codes, A, B, C, D given for each question.(A) S is correct but E is wrong.(B) S is wrong but E is correct.(C) Both S and E are correct and E is correct explanation of S.(D) Both S and E are correct but E is not correct explanation of S.

Q.18. S : Iodine possess metallic nature.E : It has violet crystalline nature and shows ionic compounds IPO4.

Q.19. S : The third period contains only 8 electrons and not 18 like 4th period.E : In III period filling starts from 3s1 and complete at 3p6 whereas in IV period it starts from 4s1 andcomplete atfter 3d10 and 4s2.

Q.20. S : Transition elements show horizontal as well as vertical relationship.E : This is due to shielding effect and similar electronic configuration.

Q.21. S : Cs and F2 combines violently to form CsF.E : Cs is most electropositive and F is most electronegative.

Q.22. S : Each transiting series possesses ten molecule.E : In transition elements filling of electrons occurs in d sub-shells.

Q.23. S : A jump in 3rd ionisation energy is noticed in case of alkaline earth metals.E : The jump in ionisation energy is due to change in major energy shell during successive removal ofelectron.

Q.24. S : First ionisation energy of Be is more than that of IE1 of B.E : Removal of electron in Be occurs from 2s sub-shell whereas in B from 2p sub-shell.

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Q.25. S : Formation of Cl is exothermic whereas formation of O2 is endothermic.E : EA2 of oxygen is endothermic and greater than its exothermic value EA1 of oxygen.

Q.26. S : BiCl5 does not exist.E : In Bi inert pair effect is more prediominant.

Q.27. S : Second EA for halogens is zero.E : Fluorine has maximum value of electron affinity.

Q.28. S : Pure NaCl is hygroscopic in nature.E : Pure NaCl is non-hygroscopic in nature.

Q.29. S : Sulphate is estimated as BaSO4 and not MgSO4.E : Ionic radii of Mg2+ is smaller than that of Ba2+.

Q.30. S : First ionization enthalpy of Be is greater than that of B.E : 2p orbital is lower in energy than 2s orbital.

Q.31. S : Nobel gases have highest ionization enthalpies in their respective periods.E : Nobel gases have stable closed shell electronic configuration.

More than One CorrectQ.32. The elements which are radioactive and have been named after the names of planet are

(A) Hg (B) Np (C) Pu (D) RaQ.33. Which of the process do not involve absorption of energy ?

(A) S(g) + e S(g) (B) O + e O2 (g) (C) Cl(g) + e Cl(g) (D) O(g) + e O (g)Q.34. Which of the following pairs of elements have almost similar atomic radii ?

(A) Zr, Hf (B) Mo, W (C) Co, Ni (D) Nb, TaQ.35. Which of the following elements have similar value of electronegativity ?

(A) H (B) S (C) Te (D) PQ.36. Mark the correct statements out of the following

(A) He has highest IE1 in the periodic table.(B) Cl has the highest EA out of all elements in the periodic table.(C) Hg and Br are liquid at room temperature.(D) In any period, the atomic radius of the noble gas is lowest.

Q.37. Which of the following show amphoteric behaviour ?(A) Zn(OH)2 (B) BeO (C) Al2O3 (D) Pb(OH)2

Q.38. Ionization energy is influenced by(A) Size of atom (B) Charge of nucleus(C) Electrons present in inner shells (D) None of the above

Q.39. Which is correct in increasing order of ionic character ?(A) AlCl3 < MgCl2 < NaCl (B) LiF < LiCl < LiBr < Lil (C) NaCl > MgCl2 < AlCl3 (D) None of the above.

Q.40. Which of the following is correct in order of increasing size ?(A) I+ < I < I (B) Fe < Fe2+ < Fe3+ (C) Fe3+ < Fe2+ < Fe (D) All of the above.

Q.41. Which is the correct increasing order of ionization energy ?(A) Li < B < Be (B) Be < B < Li (C) Li < Na < K (D) O < N < F

Q.42. The first eight ionization energies for a particular neutral atom is as given below. All values are expressedin MJ mol1. Which oxidation state(s) is/are not possible for the atom ?1st 2nd 3rd 4th 5th 6th 7th 8th

1.31 3.39 5.30 7.47 10.99 13.33 71.33 94.01(A) 2 (B) 3 (C) 6 (D) 6

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Exerxise 3Previous Years IIT Based QuestionsOnly one option is correct :Q.1. The hydrogen energy of Mg2+ is larger than the of : (1984)

(A) Al3+ (B) Na+ (C) Be2+ (D) Mg3+

Q.2. The first ionization potential in electron volts of nitrogen and oxygen atoms are respectively given by :(A) 14.6, 13.6 (B) 13.6, 14.6 (C) 13.6, 14.6 (D) 14.6, 14.6 (1987)

Q.3. Atomic radii of fluorine and neon in Angstrom units are respectively given by : (1987)(A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) None of these

Q.4. The electronegativity of the following elements increases in the order : (1987)(A) C, N, Si, P (B) Na, Si, C, P (C) Si, P, C, N (D) P, Si, N, C

Q.5. The first ionization potential of Na, Mg, Al and Si are in the order : (1988)(A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg < Al > Si (D) Na > Mg > Al < Si

Q.6. Which of the following is the smallest is size ? (1989)(A) N3 (B) O2 (C) F (D) Na+

Q.7. Amongst the following elements (whose electronic configurations are given below), the one having thehighest ionization energy is : (1990)(A) [Ne] 3s2 3p1 (B) [Ne] 3s2 3 p3 (C) [Ne] 3s2 3p2 (D) [Ar] 3d10 4s2 4p3.

Q.8. The statement that is not correct for the periodic classification of elements is : (1992)(A) The properties of elements are the periodic functions of their atomic numbers.(B) Non-metallic elements are lesser in number than metallic elements.(C) The first ionization energies of elements along a period do not vary in a regular manner with increase inatomic number.(D) For transition elements the d-sub shells are filled with electrons monotonically with increase in atomicnumber.

Q.9. Which has most stable +2 oxidation state : (1995)(A) Sn (B) Pb (C) Fe (D) Ag

Q.10. Which of the following has the maximum number of unpared electrons : (1996)(A) Mg2+ (B) Ti3+ (C) V3+ (D) Fe2+.

Q.11. The incorrect statement among the following is : (1997)(A) The first ionisation potential of Al is less than the first ionisation potential of Mg(B) The second ionsation potential of Mg is greater than the second ionisation potential of Na(C) The first ionisation potential of Na is less than the first ionisation potential of Mg(D) The third ionisation of Mg is greater than third ionisation potential of Mg

Q.12. Which of the following compounds is expected to be coloured ? (1997)(A) Ag2SO4 (B) CuF2 (C) MgF2 (D) CuCl.

Q.13. The following question of an Assertion in column 1 and the Reason in column 2 : (1998)Column I (Assertion) Column -2 (Reason)F atom has a less negative electron affinity Additional electrons are repelled more effectivelythan Cl atom by 3p electrons in Cl atom than by 2p electrons in F atom.Which of the following is correct ?(A) If both assertion and Reason are correct, and reason is the correct explanation of the assertion(B) If both assertion and reason are correct, but reason is not the correct explanation of the assertion(C) If assertion is correct but reason is incorrect(D) If assertion is incorrect but reason is correct

Q.14. The correct order of radii is : (2000)(A) N < Be < B (B) F < O2 < N3 (C) Na < Li < K (D) Fe3+ < Fe2+ < Fe4+.

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Q.15. This question contain an Assertion in column 1 and the Reason in column 2. Use the following key tochoose the appropriate answer : (2000)Column I (Assertion) Column -2 (Reason)The first ionisation energy of Be is 2p orbital is lower in energy than 2s.greater than that of B(A) If both assertion and Reason are correct, and reason is the correct explanation of the assertion(B) If both assertion and reason are correct, but reason is not the correct explanation of the assertion(C) If assertion is correct but reason is incorrect(D) If assertion is incorrect but reason is correct

Q.16. The set representing the correct order of first ionization potential is : (2001)(A) K > Na > Li (B) Be > Mg > Ca (C) B > C > N (D) Ge > Si > C

More than one is correctQ.17. The statements that are true for the long form of the periodic table are : (1988)

(A) It reflects the sequence of filling the electrons in the order of sub-energy level s, p, d and f.(B) It helps to predict the stable valency states of the elements(C) It reflects tends in physical and chemical properties of the elements(D) It helps to predict the relative ionicity of the bond between any two elements

Q.18. Sodium sulphate is soluble in water whereas barium sulphate is sparingly soluble because :(1989)(A) The hydration energy of sodium sulphate is more than its lattice energy(B) The lattice energy of barium sulphate is more than its hydration energy(C) The lattice energy has no role to play ion solubility(D) The hydration energy of sodium sulphate is less than its lattice energy

Q.19. The first ionization potential of nitrogen and oxygen atoms are related as follows : (1989)(A) The ionization potential of oxygen is less than the ionization potential of nitrogen.(B) The ionization-potential of nitrogen is greater than the ionization potential of oxygen(C) The two ionization-potential values are comparable.(D) The difference between the two ionzation-potential too large.

Q.20. Ionic radii of : (1999)(A) Ti4+ < Mn7+ (B) 35Cl < 37Cl (C) K+ > Cl (D) P3+ > P5+.

Subjective Type Questions:Q.21. Arrange the following in order of their (1985)

(i) Decreasing ionic sizeMg2+, O2, Na+, F

(ii) Increasing first ionization energy Mg, Al, Si, Na(iii) Increasing bond length F2, N2, Cl2, O2.

Q.22. Arrange the following in the order of their increasing size : (1986)Cl, S2 , Ca2+, Ar

Q.23. Explain the following : (1989)The first ionzation energy of carbon atom is greater than that of boron atom whereas, the reverse is truefor the second ionzation energy .

Q.24. Arrange the following as stated : (1991)Increasing order of ionic size

N3, Na+, F, O2, Mg2+

Q.25. Compare qualitatively the first and second potentials of copper and zinc. Explain the observation.(1996)

Q.26. Arrange the following ions in order of their increasing radii (1993)Li+, Mg2+, K+, Al3+.

By Prince


ANSWERSExercise - 11. A 2. A 3. A 4. B 5. D6. A 7. B 8. A 9. D 10. D11. D 12. C 13. B 14. D 15. C16. A 17. B 18. A 19. B 20. B21. C 22. A 23. D 24. D 25. A26. D 27. A 28. D 29. C 30. C31. A 32. C 33. A 34. B

Exercise - 21. B 2. B 3. C 4. A 5. B6. B 7. A 8. B 9. D 10. D11. B 12. D 13. A 14. D 15. C16. C 17. C 18. C 19. C 20. C21. C 22. C 23. C 24. C 25. C26. C 27. A 28. A 29. D 30. A31. C 32. B, C 33. A, C, D 34. A, B, C, D 35. A, C, D36. A, B, C 37. A, B, C, D 38. A, B, C 39. A, B 40. A, C41. A, C, D 42. B, C

Exercise - 31. B 2. A 3. A 4. C 5. A6. D 7. B 8. D 9. B 10. D11. B 12. B 13. C 14. B 15. C16. B 17. A, C, D 18. A, B 19. A, B, C 20. D

By Prince


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Periodicity

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