Inorganic Chemistry - Periodic Properties

7 inch

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2.5 inch2.5 inchPERIODIC PROPERTIES

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Topic Page No.

INORGANIC CHEMISTRY

PERIODIC PROPERTIES

1. Classification of Elements and Periodicity in properties 01

2. Mendeleev’s Periodic Law 03

3. Modern Periodic Law 07

4. Effective Nuclear Charge 11

5. AtomicSize 12

6. IonisationPotential 15

7. ElectronAffinity 18

8. Electronegativity 20

9. Density 27

10. DiagonalRelationship 28

11. Metallic Character 28

12. Non-metallic character 29

PERIODIC PROPERTIES

Topic Page No.

13. Reactivity, Oxidation state,Valency 29

14. Acid-base behaviour of oxides and hydroxides 30

15. Atomicvolume 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 verydifficult to studyindividuallythe chemistryof all the elements and millionsof their compounds,

hence to simplifyand systematize the studyof chemistryof the elements and their compounds, theyare

classified into groups and periods. Earlyattempt to classify the elements:

Classification of Lavoiser

Elements had been classified into two major groups by Lavoiser

1. Metals 2. Non–metals

This classification was based on the differences in their properties.

The metal word itself can be considered an acronym in which each alphabet signifies a characteristic of

these substances:

1. M = malleability.The most malleable beingAu and Pt.

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

M = meltingpoint (also boiling point)of the metals in general is high. Metal having the highest melting

point isTungsten(W) and thehighestboilingpoint isRenium(Re).The lowestmeltingpoint isof Mercury

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

Bromine).

The element havinghighest mp isW.

Element having lowest mp and bp is He.

2. E stands for elasticity(Young’s modulus)

E is also for electropositive.

E also for good conductors of Electricity. Best isAg> Cu >Au. Metals are good conductors of electricity

and heat due to the presence of free mobile delocalized electrons.

Graphite and black phosphorous are also conductors of electricity but not of heat.

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

both anode as well as cathode.

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

Least – Pb

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

Synthetic hard materials are called abrasive. These are

CorundumAl2O3, Carborundum SiC, Borazone BN, Boron carbide B4C (non stoichiometric),Tantalum

Carbide and Tungsten Carbide as hard as diamond.

4. A=Allotropy/polymorphism, shown mainlybynonmetals. Metals such as Sn show polymorphism.

=Atomicity,Number of atoms in the molecule of anelement.Atomictyis verysignificant for non–metals

because it reveals the shape of the molecules and physical properties (state) of the molecules.

A=Alloys formation.

2 ACC- CH-PERIODIC PROPERTIES


Alloys are homogeneous mixture of two or more metals. It mayalso have some non metals as well.

Everyalloyisa solid solution except the alloys of Hgwhich is a solution ofa solid into a liquid. The alloys

ofmercuryarecalledamalgum.Alloysofmetalswithat leastanon–metal arecalled interstitialcompounds.

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

homogeneous mixture. It also is an interstitial compound.

Purpose of alloying a metal are:

a. To increase the utilityof the metals eg. fuse wire (Pb+ Sn + Bi). It melts in boiling water.

b. To increase the hardness.

Alloying is propertyof metals not of nonmetals.

5. L = lustre, Shine property of metals. Each metal having a characteristic.

6. S = sonorous. Each metal gives special sound on beating.

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 Sr87.5 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 K7 39

232

7 23 39

Be Mg Ca

8 40

242

8 24 40

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

triad, whose atomic weights are approximatelyequal, e.g.

Fe Co Ni

Ru Rh Pd

There fore, this hypothesis was not acceptable for all elements.

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 scale

is 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.



Name of element Li Be B C N O F

7 9 11 12 14 16 19

Name of element Na Mg Al Si P S Cl

23 24 27 28 31 32 35.5

It is clear from the above table that sodium is the eighth element from lithium, whose properties resemble

thatoflithium.

This type of classification was limited up to only20 elements.

Inert gas element were not discovered till then.

Lother Meyer arrangement:

Ato

mic

Vo

lum

e

Metalloid and transition metals

Atomic Weight

Li

BeF

Na

Mg

Cl

K

CaBr

Rb

Sr I

Cs

Ba

The graphs plotting the atomic volumes against atomic weights are known as Lothar Mayer volume

curves.

The alkali metalshave highest atomic volumes.

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

positions on the descending part of the curve.

Halogens and the noble gases (except helium) occupy positions on the ascending part of the curve.

Transition elements have verysmall volumes and therefore these are present at the bottoms of the curve

Exercise

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

(A)Atomic size (B)Atomic volume (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 8

groups in mendeleev’s Periodic table. (e) Each group upto VIIth is divided intoA& B subgroups.

‘A’sub groups element are called normal elements and ‘B’sub groups elements are called transition

elements. (f) The VIIIth group was consists of 9 elements in three rows (Transition metals group).

(g) The elements belonging to same group exhibit similar properties.



MODIFIED MENDELEEF’S PERIODIC TABLE OF ELEMENTS



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

(a) Study of elements - First time all known elements were classified in groups according to theirsimilar properties. So study of the properties become easier of elements.

(b) Prediction of new elements - It gave encouragement to the discovery of new elements as somegapswereleft in it.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 spaces for them in his table.

e.g.- Blank space at atomic weight 72 in silicon group was called Eka silicon (means properties like silicon)and element discovered later was named Germanium.

Similarlyother elements 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 were done in atomic weight of some elements.

AtomicWeight =Valency× Equivalent weight.

Initially, it was found that equivalent weight of Be is 4.5and it is trivalent (V= 3), so the weight of Bewas 13.5 and there is no space in Mendeleev’s table for this element. So, after correction, it wasfound that Be is actually divalent (V = 2). So, the weight of Be became 2 × 4.5 = 9 and there was aspace between Li and B for this element in Mendeleev’s table.

– Corrections were done in atomic weight of elements are – U, Be, In,Au, Pt.

(iv) Demerits of Mendeleev’s periodic table -

(a) Position of hydrogen - Hydrogen resembles both, the alkali metals (IA) and the halogens (VIIA) inproperties so Mendeleev could not decide where to place it.

(b) Position of isotopes -As atomic weight of isotopes differs, they should have placed in differentposition in Mendeleev’s periodic table. But there were no such places for isotopes in Mendeleev’stable.

(c) Anomalous pairs of elements - There were some pair of elements which did not follow theincreasing order of atomic wts.

eg. Ar and Co were placed before K and Ni respectively in the periodic table, but having higher atomicweights.

1.399.39

KAr

1275.127

ITe

6.589.58

NiCo

231232

PaTh

(d) Like elements were placed in different groups.

There were some elements like Platinum (Pt) and Gold (Au) which have similar properties but wereplaced in different groups in mendeleev’s table.

Pt – VIII Au – IB

(e) Unlike elements were placed in same group.

I groupst

IA IBLi

Na

KMore reactive Cu Less reactive

RbAlkali metal Ag Coin metal

CsNormal elements Au Transition element

Fr

– Cu,Ag andAu placed in Ist group along with Na, K etc. While theydiffer in their properties (Only

similar in havingns1 electronic configuration)





MODERN PERIODIC LAW

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

showeringhighvelocityelectrons isproportional to thenuclearchargeof theatom.Thiscan berepresented

bythefollowing expression.

= a (Z–b) 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

numbers”

Exercise

1. Which of the following expressions 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 (93Np – 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 metals

Elements of group 16 are called chalcogens

Elements of group 17 are called halogens



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 – 10Ne) and third period (11Na – 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 periodic table is

based on atomic numbers. Therefore, various isotopes of the same elements have to be provided the

same position in the periodic table. Isobars gave same atomic weights but different atomic numbers and

therefore theyhave to be placed at different positions.

The positions of actinides and lanthanides is more clear now because these have been placed in IIIB

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

In the periodic table a diagonal line from B toAt 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 of both metals and nonmetals.

The general electronic configuration of the elements remains same in group.

DEFECTS OF LONG FORM OF PERIODIC TABLE

The position of hydrogen is still disputable as it was there in Mendeleef periodic table in group IAas well

as IVA & VIIA.

Helium is an inert gas but its configuration is different from that of the other inert gas elements

Lanthanide and actinide series could not be adjusted in the main 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 valence

shell are called s-block elements. There are only 14 s-block elements in the periodic table.



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), 107Ns (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.

Illustration

1. Which of the following is the period number of the element whose atomic number is 98

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

Ans. B

Sol. The electronic configuration of the element with atomic number 98 is as follow

1s2 , 2s2 , 2p6, 3s2, 3p6, 4s2 , 3d10, 4p6, 5s2, 4d10 , 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f 10

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

2. The nuclei of elements X,Yand Z have same number of protons, but different numbers of neutrons.

According to Mendeleef periodic table, the elements X,Y and Z

(A) belong to same group and same period

(B) belong to different groups and different periods

(C) belong to same group and different periods

(D) are isotopes, which do not have different positions

Ans. D

Sol. Isotopes have same number of protons (i.e. same atomic number). So theyoccupy same position

in the periodic table. However, due to different numbers of neutrons their atomic weights are different.

1 0 ACC- CH-PERIODIC PROPERTIES


Exercise

1. Which of the following has the same number of electrons in its outermost shell and penultimate shell ?

(1)Al3+ (2) Ca2+ (3) F– (4) N3–

Ans. 2

2. Which of the following statement is not correct about the electronic configuration of chromium 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 derived byusing roots for the three digits in the atomic number of the elements and

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

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

3. The symbol for the element is made up from the first letters from the roots which make up the name. Thestrange mixture of Latin and Greek roots have been chosen to ensure that the symbols are all different.

Illustration

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

Ans. (C)

Exercise

1. The IUPAC name of the element with atomic number Z = 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).


ACC- CH-PERIODIC PROPERTIES 1 1

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 the electrons of the outermost orbit.

Therefore, onlya part of the nuclear charge is effective on the electrons of the outermost orbit. Thus, the

inner electrons protect or shield the nucleus and thereby decrease the effect of nuclear charge towards

the electrons of the outermost orbit.

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

charge

Z * Z

Z* = effective nuclear charge

s = shielding constant and Z = nuclear charge

Ascientist named slater, determined the value of shielding constant and put forward some rules as

following.

(1) The shielding effect or screening effect of each electron of 1s orbital is 0.30.

(2) The shielding effect of each electrons of ns and np i.e. electron of the outermost orbit, is 0.35.

(3) The shielding effect of each electron of s or p orbitals of the penultimate orbit (n – 1) is 0.85.

(4) Theshieldingeffectofeachelectronofsorporbitalof the prepenultimateorbit (n– 2)andbelowthis is1.0.

Illustration

1. The screening effect of inner electrons of the nucleus causes

(A) Decrease in the ionisation energy (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. The attraction of the nucleus 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) much more than p - electron

(C) same as f - electrons (D) less than p - electrons

Ans. (D)

2. The increasing order of effective nuclear charge in 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 group

are called periodic properties. Some important periodic properties are atomic size, ionization energy

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



(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 = rcovalent

(of element X)

FEG H

XH XHXH

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

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

*

02

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) For Homolatomic molecules

dA–A = rA + rA or 2rA

rA = A Ad

2

(b) For hetrodiatomic molecules while electronegativity is approx same.dA–B = rA + rB

For heteronuclear diatomic molecule.A–B, while difference between the electronegativityvalues ofatomAand atom B is relatively larger, (XA and XB) are the electron negativity in Pauling Scale.

dA–B = rA + rB – 0.09 |(XA –XB)| [Bond length or radius expressed in Å]where XA and XBare electronegativityvalues of high electronegative elementAand less electronegativeelement B, respectively. This formula was given by Stevenson & Schomaker.

Note : Covalent radius is slightly smaller then actual radius.

(ii) Van derWaals’radius. It is defined as one-half of the distance between the nuclei of two non-bondedisolated atomsor two adjacent atomsbelonging to two neighbouringmolecules of an element in the solidstate.By definition, van der Waals’radius of an element is always larger than its covalent 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



Illustration

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

electronegativity values of C and X are 2.0 and 3.0 respectively

Sol. (1) C–C bond length = 1.54 Å

rC =1.54

2= 0.77 Å

rX =1.00

2= 0.50 Å

(2) C–X bond length

dC–X = rC + rX – 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 bond length is 1.18 Å

2. Which of the following should be the 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 the longest 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. Whenever a list of radii is given, we find that the size of the noble gases is larger than the size of their

adjacent halogens. The reason is

(A) Noble gases have a complete octet

(B) Theyhave a higher inter electronic repulsion

(C) In halogens it is covalent radii and in noble gases it is vander waals radii

(D) Noble gases cannot be liquified

Ans. (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)



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 effective

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 size of a cation is smaller in comparison to the size of its corresponding atom. This is because of the

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

number of proton/s. This results in increase in the effective nuclear charge.

Size of cationeff

1

Amount of positive charge or Z

Examples - (1) Mn > Mn+2 > Mn+3 > Mn+4 > Mn+6 > Mn+7 (size)

(2) Pb+2 > Pb+4

The size of an anion is greater than the size of its corresponding atom, because the number of electrons

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

decrease in the effective nuclear charge.

Size of anion 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

Size in an isoelectronic series

1

Nuclear charge

No. of protons

Illustration

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

H–1 H+1 H

1p 1p 1p

2e 0e 1e

Sol. H+1 < H < H–1



Exercise

1. Which of the following has the largest size

(A) N–3 (B) O–2 (C) K+1 (D) Ca+2

Ans. (A)

2. Which one of the following 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 most looselybound electron from the outermost orbit of one mole of

isolated gaseous atoms ofan element, is called ionisationpotential (IP). This ionisation is an endergonic

or energy-absorbing process.

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

converted to gaseous state and the energy required for this is called sublimation energy.

A stI IPA+1 ndII IP AA+2 rdIII IPA+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). The energyrequired to convert a monopositive cation to

a diapositive cation is called second ionisation potential (IInd IP) of an atom

Ist IP < IInd IP < IIIrd IP because as the electrons go out of the atom, the ionic size goes on decreasing

and the amount of positive charge goes on increasing.

FactorsAffecting Ionisation Potential

(i) Atomic size : When the size of an atom is very large the electron of the outermost orbit bound to the

nucleus byweaker attractive forces. Such an electron will be readilyremoved from the atom. Therefore.

The value of ionisation potential will be low.

1Ionisation potential

atomic size

(ii) Effective Nuclear Charge : Atomic size decreases with increase in effective nuclear charge because,

higher the effective nuclear charge stronger will be the attraction of the nucleus towards the electron of

the outermost orbit and higher will be the ionisation potential

Ionisation potential Effective nuclear charge

(iii) Shielding Effect : The electrons of internal orbits repel the electrons of the electron of the outermost

orbit due to which the attraction of the nucleus towards the electron of the outermost orbit decreases and

thus atomic size increases and the value of ionisation potential decreases.

1Ionisation potential

shielding effect



(iv) Stability of half filled and fully filled orbitals : The atoms whose orbitals are half-filled (p3, d5, f7) or

fully-filled (s2 , p6 , d10 , f14) have greater stability than the others. Therefore, they required greater

energyto for removing out electron. However stabilityof fullyfilled orbitals is greater than that of the

half filledorbitals

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, greater energy is 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

Ion isa tion poten tia l penetration pow er

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 increases and 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 ofAl due to high stabilityof fullyfilled orbitals.

In second period ionisation potential of N is greater than O and in the third period ionisation potential of

P is greater than that of S, due to stability of half filled orbitals.

The increasing order of the values of ionisation potential of the second period elements is

Li < B < Be < C < O < N < F < Ne

The increasing order of the values of ionisation potential of the third period elements is

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

InnerTransition Elements : Thesizeof inner transitionelements is greater than that ofdblockelements.

Therefore the value of ionisation potential of f block elements is smaller than that of d block elements

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

potential remains more constant than that of d block elements.

In a Group

The value of ionisation potential normallydecreases on going from top to bottom in a group because

both atomic size and shielding effect increase.

Exception :

The value of ionisation potential remains almost constant fromAl to Ga in the IIIAgroup.

(B>Al , Ga > In)

IP1(Sn) < IP1(Pb)

IP1(In) < IP1(Tl)

In the periodic table the element havinghighest value of ionisation potential is He.

The values of ionisation potential of noble gases are extremelyhigh, because the orbitals of outermost

orbit are fully-filled (ns2 , np6) and provide great stability.

In a period, the element having least value of ionisation potential is an alkali metal (group IA) and that

having highest value is inert gas (Group 0)



Applications of Ionisation Potential

The elements havinghigh values of ionisation potential have low reactivity, e.g. inert gases.

The value of ionisation potential decreases more on going from top to bottom in a group in comparison

to a period. Therefore, reactivity of metal increases and the atom forms a cation by loss of electron.

Ionisation potentialmetalofactivityRe

1

The elements having low value of ionisation potential readily lose electron and thus behave as strong

reducing agents.

Ionisation potentialpropertyducingRe

1

The elements having low value of ionization potential readilylose electron and thus exhibit greater

metallic property.

Ionisation potential1

Metallic property

The elements having low value of ionisation potential readily lose electron and thus have basic property.

Ionisation potential1

Basic property

Illustration

1. Which of the following should be the order of increasing values 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. The second ionisation potential means removal of electron from a cation

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

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

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

F+1 (8e) = 1s2, 2s2 , 2p4

Therefore C < N < F < O

2. Which of the following should be correct for Z1 and Z2 in the following two processes

M+ + Z1 M+2 + e–

M+2 + Z2 M+3 + e–

(A)1

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

1

2Z2 (D) Z1 < Z2

Ans. (D)

Sol. Z1 = second ionisation potential and Z2 = Third ionisation potential.

Second ionisation potential is always less than the third ionisation potential.



Exercise

1. One mole ofmagnesium in the vapour state absorbed 1200 kJ ofenergy. If the first and second ionization

enthalpies of magnesium are 750 and 1450 kJ mol–1 respectively, the final composition of the mixture is

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

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

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

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

Ans. (A)

2. The incorrect statement in the following is

(A)The third ionisation potential of Mg is greater than the third ionisation potential ofAl

(B) The first ionisation potential of Na is less than first I.P of Mg

(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 energy released on adding up one mole of electron to one mole of neutral atom (A) in its gaseous

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

associated with an 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 the system. Therefore, the value of the second electron affinity is positive.

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

FactorsAffecting ElectronAffinity

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

EA increases.

Electron affinity Effective nuclear charge

Atomic Size orAtomic Radius : When the size or radius of an atom increases, the electron entering the

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

1Electron affinity

Atomic size



Shielding Effect : Shielding effect is directly proportional to atomic size and atomic size is inversely

proportional to electron affinity.

1Electron affinity

Shielding effect

Stability of Fully-Filled and Half-Filled Orbitals :

The stabilityof theconfiguration having fully-filled orbitals (p6, d10, f14) andhalf-filled orbital (p3 , d5 , f7)

is relativelyhigher than that of other configurations.

Periodic Table and ElectronAffinity :

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 are half-filled (2p3) orbitals in the outermost orbit of N, which are more

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

In the third period, the value of electron affinityof Si is greater than that of P. This is because electronic

configuration of the outermost orbit in P atom is 3p3 , which being half-filled, is relativelymore stable

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

orbitals.

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

value of electron affinityof the elements of group IIA is zero because ns orbitals are fully-filled and such

orbitals have no tendency to accept electrons.

In a Group

The values of electron affinitynormally decrease on going from top to bottom in a group because the

atomic size increases which decreases the actual force of attraction by the nucleus.

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

The value of electron affinityof F is lower than thatof Cl, because the size of F is verysmall and compact

and thecharge densityishighon thesurface.Therefore, the incomingelectron experiencesmore repulsion

in comparison to Cl . That is why the value of electron affinityof Cl is highest in the periodic table.

The values of electron affinityof alkali metals and alkaline earth metals can be regarded as zero, because

they do not have tendency to form anions by accepting electron.



Illustration

1. O(g) + 2e–1 O–2 (g) –E = + 744.7

The reason for the positive value of E is

(A) endothermic reaction (B) exothermic reaction

(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 the energy

will be positive i.e there will be absorption of energyduring the process.

2. The increasing order of electron affinityvalues of O,S and Se is

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

Ans. (C)

Sol. Atomic size of Se is very large.

Exercise

1. The least electronaffinityis found in

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

Ans. (A)

2. Of the followingelement of which electronicconfiguration will have the highest 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 capacityor tendencyof an atom to attract the shared pair of electrons of the covalent

bond towards itself is called electronegativity of that atom.

Electronegativity is a relative value that indicates the tendency of an atom to attract shared electrons

more than the other atom bonded to it. Therefore it does not have anyunit. Pauling was the first scientist

to put forward the concept of electronegativity.

The numerical value of electronegativityof an atom depends on its ionisation potential and electron

affinityvalues.

FactorsAffecting Electronegativity

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

1Electronegativity

Atomic size



SOLVED EXAMPLES

Q.1 For the element X, student Surbhi measured its radius as 102 nm, Mr.Gupta as 113 nm and Mr.Agarwalas 100 nm using same apparatus. Their teacher explained that measurements were correct bysaying thatrecorded values by three students were

(A) Crystal, vander Waal and covalent radii (B) Covalent, crystal and vander Waal radii

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

Ans. (A)

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

Q.2 Which oxide of N is isoelectronic with CO2 -

(A) NO2 (B) NO (C) N2O (D) N2O3

Ans. (C)

Sol. N2O

Total electrons 14 + 8 = 22

CO2

Total electrons 6 + 16 = 22

Q.3 Give the correct order of initials T (true) or F (false) for following statements.

(I) Top positions of Lother-Mayer’s atomic volume 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 take maximum of six 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+ attained inert gas configuration

IE2 of Na is greater than Mg.

(IV) p-orbital can have 2 electrons but p-subshell can have 6 electrons.

Q.4 The electronic configuration of an element is 1s2, 2s22p6,3s13p33d2. The atomic number of the elementpresent just below this element in the periodic table -

(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



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 The Zeff for3d electron of Cr4s electron of Cr3d electron of Cr3+

3s electron of Cr3+ are in 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 Periodic Law of elements, the Variation in properties of elements is related to their ?

(A) Nuclear masses (B)Atomic numbers

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

Q.4 Atomic radius decreases in a period, but after halogens, the atomic radius suddenly increases. Thus,inert gases has almost highest radius in a period. The explanation for such an increase is-(A) Inert gases has most stable configuration(B) Inert gases do not take part in bonding(C) Vander Wall’s radius is reported in case of inert gases(D) None of these

[2010110748]

Q.5 Which one of the following 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 the followingstatements:

I. The radius of an anion is larger than that of parent atom

II. The I.E. increases from left to right in a period generally

III. The electronegativity of an element is the tendency 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 The atomic numbers of vanadium (V). Chromium (Cr), manganese (Mn) and iron (Fe) respectively23,24, 25 and 26. Which one of these may be expected to have the higher second ionization enthalpy ?

(A) Cr (B) Mn (C) Fe (D) V

[2010111695]



EXERCISE-2 (Exercise for JEE Advanced)

[PARAGRAPH TYPE]

Paragraph for Question Nos. 1 to 2

Four elements P, 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 which of the following option represent the correct order of true (T) & false (F) statement

I size of P < size of Q II size of R < size of S

III size of P < size of R (appreciable difference) IV size of Q < size of S (appreciable difference)

(A) TTTT (B) TTTF (C) FFTT (D) TTFF

Q.2 Order of IE1 values among the following is

(A) P > R > S Q (B) P < R < S < Q (C) R > S > P > Q (D) P > S > R > Q


The electron affinity is a inherent property of the atom and it depends upon several factors.

[2010111642]

Q.3 The correct electron affinityorder 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 affinity ofA

(B) 2nd electron affinity is always greater than 1st electron affinity.

(C) OO2– process is endothermic

(D) LiLi+ process is endothermic.


Atoms of metals have only a few electrons in their valence shells while atoms of non-metals generallyhave more electrons in their valence shells. Metallic character is closely related to atomic radius andionization enthalpy. Metallic character increases from top to bottom in a group and decreases from left toright in a period. Metallic character is inverselyrelated to electronegativity.

[2010111495]

Q.5 Which of the following groups contains metals, 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 elements

Q.7 Considering the 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



EXERCISE-3 (Miscellaneous Exercise)

Subjective Type

Q.1 Use thefollowingsystem of namingelements inwhich first alphabets of the digits arewritten collectively,0 1 2 3 4 5 6 7 8 9nil uni bi tri quad pent hex sept oct ennto write three-letter symbols for the elements with atomic number 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 Why Ca2+ 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 Whythe first ionisation energyof carbon atom is greater than that of boron atom whereas, the reverse istrue for the second ionisation energy. [2010110601]

Q.6 The IE do not follow a regular trend in II & III periods with increasing atomic number. Why?

[2010110943]

Q.7 Explain whyafew elements suchas Be (+0.6), N(+0.3)&He(+0.6) havepositiveelectron gain enthalpieswhile majorityof elements do have negative values. [2010110599]

Q.8 Which bond in 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)The element with highest electron gain enthalpy.(ii) The elementwith lowest ionisation potential.(iii) The element whose oxide is amphoteric.(iv) The elementwhich has smallest radii.(v) The element whose atom has 8 electrons in the outermost shell. [2010110597]

Q.10 In the ionic compound KF, the K+ and F– 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 The basic nature of hydroxides of group 13 (III-A) decreases progressivelydown the group. Comment.

[2010110596]

Q.13 Based on location in P.T., which of the following would you expect to be acidic & which basic.

(a) CsOH (b) IOH (c) Sr(OH)2 (d) Se(OH)2 (e) FrOH (f) BrOH

[2010110868]



EXERCISE-4

SECTION-A

(IIT JEE Previous Year's Questions)

Q.1 Moving from right to left in a periodic table, the atomic size is: [JEE 1995](A) increased (B) decreased (C) remains constant (D) none of these

[2010112001]

Q.2 The increasingorder of electronegativity in the following 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 truestatement for that element is:(A) Highest value of IE (B)Transition element(C) Isotone with

18Ar38 (D) None [JEE 1995]

[2010111632]

Q.4 The number of paired electrons in oxygen atom is: [JEE 1995](A) 6 (B) 16 (C) 8 (D) 32

[2010112648]

Q.5 The decreasing size of K+, Ca2+, Cl– & S2– follows the 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 Which of the following oxide is neutral? [JEE 1996](A) CO (B) SnO

2(C) ZnO (D) SiO

2

[2010111977]

Q.7 Which of the following has the maximum numberof unpaired electrons [JEE 1996](A) Mg2+ (B)Ti3+ (C) V3+ (D) Fe2+

[2010111613]

Q.8 The following acids have been arranged in the order of decreasing acid strength. Identify the correctorder [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 The incorrect statement among the following is: [JEE 1997](A) the first ionisation potential ofAl is less than the first ionisation potential of Mg(B) the second ionisation 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 potential of Mg is greater than the third ionisation potential ofAl

[2010112041]



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 following arrangements, the sequence is not strictly according to the property writtenagainst it ? [AIEEE-2009]

(A) H2O < H2S < H2Se < H2Te : Increasing acidic strength

(B) HF < HCl < HBr < HI : Increasing acidic strength

(C*) NH3 > PH3 <AsH3 < SbH3 : Increasing basic strength

(D) B < C < O < N : increasing first ionization enthalpy

[2010113518]

Q.22 The set representing the correct order of ionic radius is – [AIEEE-2009]

(A) Li+ > Be2+ > Na+ > Mg2+ (B) Na+ > Li+ > Mg2+ > Be2+

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

[2010113569]

Q.23 The correct sequence which shows decreasing order of the ionic 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 one of the following ordered presents the correct sequence of the increasing basic nature of thegiven oxides ? [AIEEE-2011]

(A) Al2O3 < MgO < Na2O < K2O (B) MgO < K2O < Al2O3 < Na2O

(C) Na2O < K2O < MgO < Al2O3 (D) K2O < Na2O < Al2O3 < MgO

[2010113671]

Q.25 The increasing order of the ionic radii of the given isoelectronic species 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 ionisation potential of Na 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 of the following represents the correct order of increasing first ionization enthalpyfor Ca, Ba, S,Se andAr?(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]



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 109Unu 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 filled config.]

Q.6 [half filled&fullyfilledorbitals] Q.7 [half filledandfullyfilledorbitals]

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 [rk > 1.34Å > rF] 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]



EXERCISE-1

Q.1 [Due to lanthanide contraction; left to right atomic size decreases. ]Q.2 [Cr = 1s2 2s2 2p6 3s2 3p6 4s1 3d5

Zeff =Atomic nos for 4s electron (valance shell e¯ –1) × 0.35 +

Penultimate shell e¯ × 0.85 + remaining e¯= (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 electrons are same ]

Q.6 [After the removal of one electron Cr acquired half filled stable d-orbital configuration ]

Q.7 [Basic characters electropositive / metallic character ]

Q.8 [Metals oxide are basic as well as they are electropositively in nature ]

Q.10 [(Heg)2 is an endothermic process ]

Q.11 [Due to thehalf filled stable configuration of N its ionisation potential 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 [rc + rF = 1.33 Årsi + rF = 1.54 Å----------------------rc + rsi + 2rF = 2.87 Å

rc + rsi = 1.87 Å

1.87Å + 2 rF = 2.87 Å2 rF = 1.00 Å

rF = .5 Å]

Q.23 [Electronegativity sizeatomic

1]

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Inorganic Chemistry - Periodic Properties