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7 inch

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2.5 inch2.5 inchMEDICAL ENTRANCE

PERIODIC TABLE

I N D E X

Topic Page No.

INORGANIC CHEMISTRY

PERIODIC TABLE

01. Introduction 03

02. Development of Periodic Table 03

03. Mendeleev’s Periodic Table 04

04. Modern Periodic Table 06

05. Long form/Present from of Modern Periodic Table 06

06. Classification of Elements 10

07. Classification on the Basis of Sub-shell in which 11

last electron enters

08. Periodicity 13

09. Periodic Property 14

10. Atomic Radius 14

11. Ionization potential or Ionization Energy or Ionization Enthalpy 20

MEDICAL ENTRANCE

PERIODIC TABLE

Topic Page No.

12 ElectronAffinity / Electron Gain Enthalpy 23

13. Electronegativity (EN) 24

14. General Trend of Different Properties in the Period & Groups 31

15. Important Facts to Remember 32

16. Solved Example 34

17. Summary 37

17. Exercise - 1 38

Exercise - 2 46

Exercise - 3 50

Exercise - 4 51

18. Answer Key 53

19. Hints/Solution 56

BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5

ACC_CHEMISTRY_Periodic Table 1

Syllabus :

• Quantum Numbers,Aufbau Principle, Pauli Exclusion Principles and Hund's rule• Modern periodic law and long form of periodic table, periodic trends in properties of

elements- atomic radii, ionic radii, ionization enthalpy, electron gain enthalpy, electronegativity,valence.

Contents :

PERIODIC PROPERTIES

• Introduction

• Development of periodic table

• Modern periodic table (modified mendeleev table)

• Long form/present form of modern periodic table

• Description of groups

• Classificationof elements

• Classification of the basis of subshell in which last electron enters

• Periodicity

• Periodic properties

• Atomicsize

• Ionization potential or ionization energy or ionization enthalpy

• Electron affinity/ electron gain enthalpy

• Electronegativity (EN)

• Important facts to remember

• Solved examples

• Exercise

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PERIODIC TABLE

INTRODUCTION

The arrangement of all the known elements according to their properties in such a way that the elements

of similar properties are grouped together in a tabular form is called periodic table.

DEVELOPMENT OF PERIODIC TABLE

DOBEREINER TRIAD RULE [1817]

(i) He made groups of three elements having similar chemical properties called TRIAD.

(ii) In Dobereiner triad, atomic weight of middle element is equal to the average atomic weight of first and

third element.

e.g.

Ca Sr Ba

40 87.5 137 = 88.540 + 137

2

(iii) Other examples – (K, Rb, Cs), (P, As, Sb), (S, Se, Te), (H, F, Cl), (Sc, Y, La)

NEWLAND OCTAVE RULE [1865]

(i) He arranged the elements in the increasing order of their atomic mass and observe that properties of

every 8th element was similar to the 1st one. like in the case of musical vowels notation.

8

Sa

765432

NeDhaPaMaGaRe

1

Sa

(ii) At that time inert gases were not known.

Li

Na

K

Be

Mg

Ca

B

Al

C

Si

N

P

O

S

F

Cl

H

(iii) The properties of Li are similar to 8th element i.e. Na, Be are similar to Mg and so on.

Drawback or Limitation :

(a) This rule is valid onlyupto Ca. because afterCa due to presence of d-block element there is difference

of 18 elements instead of 8 element.

(b) After the discovery of Inert gas and included in the periodic table it becomes the 8th element from

Alkali metal so this law had to be dropped out.

LOTHER MEYER’S CURVE [1869]

(i) He plotted a curve between atomic weight and atomic volume of different elements.

(ii) The following observation can be made from the curve -

(a) Most electropositive elements i.e. alkali metals (Li, Na, K, Rb, Cs etc.) occupy the peak positions

on the curve.

(b) Less electropositive i.e. alkaline earth metal (Be, Mg, Ca, Sr, Ba) occupy the descending position of

the curve.

(c) Metalloids (B, Se,As, Te,At etc.) and transition metals occupy bottom part of the curve.



Note : Elements having similar properties occupy similar position on the curve.

Ato

mic

Vol

ume

Metalloid and transition metals

Atomic Weight

Li

BeF

Na

Mg

Cl

K

CaBr

Rb

Sr I

Cs

Ba

Conclusion : On the basis of this curve Lother Meyer proposed that the physical properties of the

elements are periodic function of their atomic weight and become the base of Mendeleev’s periodic

table.

MENDELEEV’S PERIODIC TABLE [1869]

(i) Mendeleev’s Periodic Law - The physical and chemical properties of elements are the periodic

function 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.

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

(a) Study of elements - First time all known elements were classified in groups according to their

similar properties. So study of the properties become easier of elements.

(b) Prediction of new elements - It gave encouragement to the discovery of new elements as some

gapswere left in it.Sc(Scandium),Ga(Gallium),Ge(Germanium),Tc(Technetium)were theelements

for whom position and properties were defined by Mendeleev even before their discoveries and he

left 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.

Similarly other elements discovered after mendeleev periodic table were.

Eka aluminium -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.

Atomic Weight =Valency × Equivalent weight.

Initially, it was found that equivalent weight of Be is 4.5 and it is trivalent (V= 3), so the weight of Be

was 13.5 and there is no space in Mendeleev’s table for this element. So, after correction, it was

found that Be is actually divalent (V = 2). So, the weight of Be became 2 × 4.5 = 9 and there was a

space 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) in

properties so Mendeleev could not decide where to place it.

(b) Position of isotopes -As atomic weight of isotopes differs, they should have placed in different

position in Mendeleev’s periodic table. But there were no such places for isotopes in Mendeleev’s

table.

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

increasing order of atomic wts.

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

weights.

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 were

placed in different groups in mendeleev’s table.

Pt Au

VIII 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 they differ in their properties (Only

similar inhavingns1 electronic configuration)



MODERN PERIODIC TABLE (MODIFIED MENDELEEV PERIODIC TABLE)

(i) It was proposed by Moseley.

(ii) Modern periodic table is based on atomic number.

(iii) Moseley did an experiment in which he bombarded high speed electron on different metal surfaces and

obtained X-rays. He found out that Z where = frequency of X-rays

from this experiment, Moseley concluded that the physical and chemical properties of the elements are

periodic function of their atomic number. It means that when the elements are arranged in the increasing

order of their atomic number elements having similar properties gets repeated after a regular interval.

This is also known as ‘Modern Periodic Law’.

(iv) Modern Periodic Law - The physical & chemical properties of elements are a periodic function of the

atomic number.

(v) Characteristics of Modern Periodic Table -

(a) 9 vertical columns called groups.

(b) Ist to VIIth + VIIIth group +0 group of inert gases.

(c) Inert gases were introduced in periodic table by Ramsay.

(d) 7 horizontal series called periods.

LONG FORM/PRESENT FORM OF MODERN PERIODIC TABLE

(i) It consist of 7 horizontal rows (periods) and 18 vertical columns (groups)

(ii) According to I.U.P.A.C. 18 vertical columns are named as 1st to 18th group.

(iii) The co-relation between the groups in long form of periodic table and in modern form of periodic table

are given below.

IA, IIA, IIIB, IVB, VB, VIB, VIIB,

1098

|—————|VIII IB, IIB. IIIA, IVA, VA, VIA, VIIA 0

1 2 3 4 5 6 7 11 12 13 14 15 16 17 18

(iv) Elements belonging to same group having same no. of electrons in the outermost shell so their properties

are similar.

DESCRIPTION OF GROUPS –

1st/IA/Alkali metals :

H = 1s1

Li = 1s2, 2s1

Na = 1s2, 2s2, 2p6, 3s1

K = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1

General electronic configuration = ns1(n = Number of shell)

Number of valence shell electron = 1



2nd/IIA/Alkali earth metals :

Be = 1s2, 2s2

Mg = 1s2, 2s2, 2p6, 3s2

Ca = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2

General electronic configuration = ns2


13th/IIIA/Boron Family :

B = 1s2, 2s2, 2p1

Al = 1s2, 2s2, 2p6, 3s2, 3p1

Ga = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p1

General electronic configuration = ns2 np1


14th/IVA/Carbon Family :

C = 1s2, 2s2, 2p2

Si = 1s2, 2s2, 2p6, 3s2, 3p2

Ge = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p2



15th/VA/Nitrogen family/Pnicogens : (Used in fertilizer as urea)

N = 1s2, 2s2, 2p3

P = 1s2, 2s2, 2p6, 3s2, 3p3

As = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p3



16th/VIA/Oxygen family/Chalcogens : (Ore forming)

O = 1s2, 2s2, 2p4

S = 1s2, 2s2, 2p6, 3s2, 3p4

Se = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p4



17th/VIIA/Halogen family/Halogens : (Salt forming)

F = 1s2, 2s2, 2p5

Cl = 1s2, 2s2, 2p6, 3s2, 3p5

Br = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p5





18th/Zero group/Inert gases/Noble gases :

Ne = 1s2, 2s2, 2p6

Ar = 1s2, 2s2, 2p6, 3s2, 3p6

Kr = 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6

General electronic configuration = ns2 np6 (except. He)


Period Total No. of Starts with Ends with Elements Remark

Elements Elements

1. 2 Hydrogen (1) Helium (2) Very short period

2. 8 Lithium (3) Neon (10) Short period

3. 8 Sodium (11) Argon (18) Short period

4. 18 Potassium (19) Krypton (36) Long period

5. 18 Rubidium (37) Xenon (54) Long period

6. 32 Cesium (55) Radon (86) Very long period

7. 26 Francium (87) (Not named yet)(118) Incomplete period

(v) Important Points :

(a) 2nd periodelements (Li,Be,B)Showsdiagonalrelationshipwith3rdperiodelements (Mg,Al,Si), so(Li,

Be, B) are called Bridge elements. Because of same ionic potential (Ionic potential = Charge/Radius)

value theyshowssimilarityinproperties.

Li Be B

Na Mg Al Si

(b) 3rd period elements (Na, Mg,Al, Si, P, S, Cl) are called typical elements because they represent the

properties of other element of their respective group.

(c) In 6th period all types of elements are included (s, p, d & f)

(d) No inert gas in 7th period.

(e) Normal elements present in all periods.

(f) Atomic No. of last inert gas element is 86.

(g) Long form of modern periodic table can be divided into four portions.

1. Left portion IA & IIA (s-block elements.)

2. Right portion IIIA to VIIA + 0 group (p-block elements.)

3. Middle portion IIIB to VIIB + VIII + IB, IIB (d-block elements.)

4. Bottom portion IIIB (f-block elements)

(h) The group containing most electro positive elements – IA GROUP

(i) The group containing most electro negative elements – VIIA GROUP

(j) The group containing maximum number of gaseous elements – ZERO GROUP (18th GROUP)



(k) The group in which elements have zero valency – ZERO GROUP (18th GROUP)

(l) In the periodic table –

Number of Gaseous elements - 11 (H, N, O, F, Cl + Noble gases)

Number of Liquid elements - 6 (Cs, Fr, Ga, Hg, Br, Uub)

Bromine is the only non-metal which exists in liquid form.

Number of Solid elements - 89 (if discovered elements are 105)

- 95 (if discovered elements are 112)

(m) No p-block elements in 1st and 7th periods.

(n) 1st period has all the elements in gaseous form (H, He)

(o) Zero group (18th group) have all the elements in gaseous form.

(p) 2nd period contains maximum total number of four gases which are : N, O, F, Ne elements

(q) IIIB (3rdgroup) iscalled longest grouphaving32elements including14 Lanthanidesand14Actinides

Sc

Y

La ............... Lanthanides (14)

Ac ...............Actinides (14)

IUPAC nomenclature for the super heavy elements

Atomic Name Symbol Atomic Name Symbol

Number Number

110 un-un-nilium Uun

101 un-nil-unium Unu 111 un-un-unium Uuu

102 un-nil-bium Unb 112 un-un-bium Uub

103 un-nil-trium Unt 113 un-un-trium Uut

104 un-nil-quadium Unq 114 un-un-quadium Uuq

105 un-nil-pentium Unp 115 un-un-pentium Uup

106 un-nil-hexium Unh 116 un-un-hexium Uuh

107 un-nil-septium Uns 117 un-un-septium Uus

108 un-nil-octium Uno 118 un-un-octium Uuo

109 un-nil-ennium Une 119 un-un-ennium Uue

120 un-bi-nilium Ubn

130 un-tri-nilium Utn

140 un-quad-nilium Uqn

150 un-pent-nilium Upn

Note : Hyphens have been put in the name for clarity. They should be omitted.


ACC_CHEMISTRY_Periodic Table 1 0

(vii) Merits of long form of periodic table -

(a) Position of isotopes -Atomic No. of isotopes are similar, so different isotopes can be placed at same

place in periodic table.

(b) Lanthanides and actinides are in IIIB group.

(c) Elements of same group have same general formula of electronic configuration of outer most shell.

(viii) Demerits of long form of periodic table -

(a) Position of hydrogen is still controversial.

(b) ‘He’ is a inert gas but it has different electronic configuration than other inert gas elements.

(c) Lanthanides and actinides are still not placed in main frame.

(d) Isotopes have different physical properties but have same place in periodic table.

CLASSIFICATION OF ELEMENTS

(A) Bohr classification - Elements can be divided in four parts on the basis of electronic configuration.

(i) Inert gas elements -

(a) The elements in which ultimate orbit is completely filled up are called inert gas elements.

(b) General electronic configuration is ns2np6 (Except He = 1s2)

(c) Because of most stable configuration, they are very less reactive. Hence known as noble gas or inert

gas.

(d) These elements are present in ‘0’ group or 18th group and 1st to 6th period of periodic table.

(e) Number of inert gas elements are 6(One in each period upto 6th) He, Ne,Ar, Kr, Xe, Rn

(ii) Normal or Representative elements -

(a) The elements in which ultimate orbit is incomplete, while penultimate orbits are complete are called

normalelements.

(b) Theirgeneral electronic configuration is :

IA IIIA VA VIIAIIA IVA VIA

ns1 ns2np1 ns2np3 ns2np5

ns1–2 ns np2 1–5

ns2 ns2np2 ns2np4

(c) These elements lies in group IA to VIIA and period 1st to 7th

(d) Elements of 2nd period known as bridge elements.

(e) Elements of 3rd period (Na, Mg, Al, Si, P, S and Cl) are called typical elements. These are 7 in

numbers.

(iii) Transition elements -

The elements in which both ultimate (n) as well as penultimate shells (n – 1) are incomplete either in

atomic state or in some oxidation state are called transition elements.

Note:According to this concept, Zn, Cd, Hg and Uub are not transition elements because they do not

have incomplete penultimate shell either in atomic state or in some oxidation state.



Zn = [Ar] 3d10 4s2

Zn+2 = [Ar] 3d10

(a) group – IIIB to VIIB + VIII + IB

Periods – 4th to 7th

(b) Electronic configuration (n – 1)d1–10 ns1 or 2

(c) Total number of d-block elements = 40

Total number of transition elements = 36 (Except Zn, Cd, Hg and Uub)

Note : All transition elements are d-block elements but all d-block elements are not transition elements.

(iv) Inner transition elements -

(a) Theelementsinwhichall the threeshells thatisultimate(n)penultimate(n–1)andpreorantipenultimate

(n – 2) shell are incomplete are called inner transition elements.

58Ce = [Xe] 6s2, 5d1, 4f1

(b) Electronic configuration -

(n – 2)f1 – 14 (n – 1)d0 or 1 ns2

(c) These are 28 in number.

(d) Group – IIIB

(e) Period – 6th & 7th

(f) Inner transition elements are divided into two series.

(A) Lanthanide series/Rare earth elements/Lanthenones {Ce58 – Lu71 (14 elements)}

The first element of this series is Cerium not Lanthanum.

In these elements, last electron enters into 4f subshell.

They are present in IIIB group and 6th period of the periodic table.

Promethium (61Pm) is the only lanthanide which is synthetic and radioactive in nature.

(B) Actinide series/Man made elements/Actinones {Th90 – Lw103, (14 elements)}

The first element of this series isThorium & notActinium.

In these elements, last electron enters into 5f subshell.

they are present in IIIB group and 7th period of the periodic table.

All the actinides are radioactive in nature.

First three elements (Th, Pa, U) are found in nature while others are synthetic in nature.

After U92 i.e. from 93Np onwards elements are called transuranic elements because :

They are heavier than uranium.

They are derived from uranium by nuclear reactions.

CLASSIFICATION ON THE BASIS OF SUB SHELL IN WHICH LAST ELECTRON ENTERS

(i) s-block elements -

(a) In these elements last electron enters in s-subshell.

(b) Groups – IA, IIA + 0 group (He only)

(c) Period – 1st to 7th



(d) Electronic configuration – ns1 – 2

IA group n = 1 to 7

IIA group n = 2 to 7

(e) Total s-block elements are (14) (including H and He)

(f) Total s-block metals are (12) (excluding H), H is non-metal

(ii) p-block elements -

(a) Last electron enters in p-sub shell

(b) Group – IIIA (13) to VIIA(17) + 0 group (18) (except He)

(c) Period – 2nd to 6th

(d) Electronic configuration – ns2 np1 – 6

(e) Total p-block elements – (30)

(iii) d-block elements -

(a) Last electron enters in (n – 1)d subshell

(b) Group – IIIB – VIIB, VIII, IB, IIB or group 3 to 12 (IUPAC)

(c) Period – 4th to 7th

(d) Electronic configuration – (n – 1)d1 – 10 ns1 or 2

(e) Total d-block elements – (40)

Total transition elements – (36). If 112 elements are included in periodic table.

(f) IIB elements (Zn, Cd, Hg, Uub) are d-block elements but not transition elements.

(iv) f-block elements -

(a) Differentiating electron enters in (n – 2)f subshell.

(b) Group – IIIB

(c) Period – 6th and 7th

(d) From atomic number 58 - 71, 6th period ; Lanthanide series 4f1 – 14 5d0 or 1 6s2

90 - 103, 7th period; Actinide series 5f1 – 14 6d0 or 1 7s2

(e) Total number of f-block elements – (28)

(f) All the actinides are radioactive elements.

(g) Transuranic actinides are man made elements (Np93 – Lw103)

(h) Lanthanides are found rarely on earth so these are called rare earth metals.

Determination of period, block and group of an element -

(i) Period No. - The period no. of the element can be predicted from the principal quantum no. (n) of the

valenceshell.

e.g. electronic configuration of iodine is :

1s22s22p6, 3s23p63d10, 4s24p64d10, 5s25p5. Therefore the period number of iodine is 5 as the valence

shell configuration is 5s25p5.

(ii) Block - The type of orbital which receives the last electron known as block.

e.g. An element ‘X’ has its electronic configuration is 1s22s22p63s23p64s23d8. As the last electron

enters in the d-orbital, therefore it is a d-block element.



(iii) Group No. - It is predicted from the number of electrons in the valence shell and penultimate shell as

follows.

(a) For s-block elements, group number is equal to the number of electrons in the valence shell after the

noble gas configuration

e.g. An element ‘Y’ having electronic configuration 1s22s22p63s23p64s2 or [Ar] 4s2 has two electron in

valence shell and it is a s-block element. Therefore it belongs to group 2.

(b) For p-block element, the group number is equal to 10 + number of electrons in valence shell.

An element ‘Z’ with electronic configuration as 1s22s22p63s23p63p104s24p3 has five electrons in its

valence shell and belongs to p-block.

Therefore its group number is 10 + 5 = 15. It belongs to group VAof the Mendeleev’s periodic table.

(c) For d-block elements, group number is equal to the number of electrons in (n – 1)d sub-shell and

valenceshell.

e.g. An elements ‘A’having electronic configuration as.

1s22s22p63s23p63d104s1. So its group number will be 10 + 1 = 11.

PERIODICITY

(i) The regular gradation in properties from top to bottom in a group and from left to right in a period is

called periodicity in properties.

(a) In a period, the ultimate orbit remain same, but the no. of electron gradually increases.

(b) In a group, the no. of electron in the ultimate orbit remains same, but the values of n increases.

(ii) Causes of periodicity -

(a) The cause of periodicity in properties is due to the same outermost shell electronic configuration

coming at regular intervals.

(b) In the periodic table, elements with similar properties occur at intervals of 2, 8, 8, 18, 18 and 32.

These numbers are called as magic numbers.

CALCULATION OF EFFECTIVE NUCLEAR CHARGE (Zeff) (using slater’s rule)

To calculate the shielding constant () for an electron in an np or ns orbital :

1. Write out the electronic configuration of the element in the following order and groupings :

(1s), (2s, 2p) (3s, 3p) (3d) (4s, 4p), (4d) (4f) (5s, 5p), etc.

For s and p electrons :

2. Electrons in any group to the right of the (ns, np) group contribute nothing to the shielding constant.

3. All of the other electrons in the (ns, np) group, shield the valence electron to an extent of 0.35 each.(Except for the 1s orbital for which value is 0.30).

4. All electrons in the (n – 1) shell shield to an extent of 0.85 each.

5. All electrons (n – 2) or lower group shield completely ; that is, their contribution is 1.00 each.



For d and f electrons :

1. Electrons in any group to the right of the nd or nf group contribute nothing to the shielding constant.

2. All of the other electrons in the nd or nf group, shield the valence electron to an extent of 0.35 each.

3. All electrons in groups lying to the left of the nd or nf group contribute 1.00.

PERIODIC PROPERTIES

VALENCY :

It is defined as the combining capacity of the elements. The word valency is derived from an Italian word

“Valentia” whichmeans combiningcapacity.

Old concept : Given by : Grankland

Valency with respect to Hydrogen and Chlorine : 1HofValency

It is defined as the number of hydrogen or Chlorine atoms attached with a particular element.

IA IIA IIIA IVA VA VIA VIIA

NaH MgH2 AlH3 SiH4 PH3 H2S H-Cl

NaCl MgCl2 AlCl3 PCl3 SiCl3 PCl3 Cl-Cl

Valency 1 2 3 4 3 2 1

Note: Valency w.r.t. H across the period increases upto 4 and then again decreases to 1.

Valency with respect to oxygen : 2'O'ofValency

It is defined as twice the number of oxygen atoms attached with a particular atom.

IA IIA IIIA IVA VA VIA VIIA

Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7

Valency 1 2 3 4 5 6 7

Note: Valency with respect to oxygen increases from 1 to 7 across the period. Valency w.r.t. ‘Oxygen’

is equal to the group number.

New concept :This concept is based on the electronic configuration.According to this concept valency

for IAto IVAgroup elements is equal to number of valence shell electron and from VAto zero group, it

is [8– (number of valence electron)].Valency = No. of valence electron Valency = (8 – No. of valence electron)

IA IIA IIIA IVA VA VIA VII 0

Valence shell electron

Valency

1

ns1 ns np2 3

5

1 3

2

ns2 ns np2 4

6

2 2

3

ns np2 1 ns np2 5

7

3 1

4

ns np2 2 ns np2 6

8

4 0

(8 – 5) = 3 (8 – 8) = 0

Note:All the elements of a group have same valencies because they have same number of valence shell

electrons.

ATOMIC SIZE

Atomic radius :

(a) It is distance between outermost electron and nucleus.

(b) Half of the nuclear distance between two atoms is defined as atomic radius.



(c) X-ray diffraction, electron diffraction method and nuclear magnetic resonance (NMR) spectrum

methods are used to determine internuclear distance or bond length.

Atomic radius =2

cetandisnuclearInter

(d) Atomic radius depends on the type of chemical bond between atoms in a molecule. These are :

i) Covalent radius ii) Ionic radius iii)metallic radius iv)Vander waal’s radius

(i) Covalent radius : (SBCR, Single Bonded Covalent Radius)

(a) Covalent bonds are formed by overlapping of atomic orbitals.

(b) Internucleardistance is minimum in this case.

(c) Covalent radius is the half of the internuclear distance between two singly bonded homo atoms e.g.

internuclear distance ofA–A(A2) molecule is (dA–A) and covalent radius is rA then

dA – A = rA + rA or 2rA

rA =2

d AA

e.g. – In Cl2 molecule, internuclear distance is 1.98 Å so rCl =2

98.1= 0.99 Å

(d) SBCR of O, N and C etc. elements can be determined by taking H2O2, N2H4, C2H6 respectively.

Case I - For Heteroatomic molecule with no. electronegativity difference.

ForA– B molecule Electronegativities ofA& B are approximately equal e.g. C – I

E.N. of C & I are approx equal (2.5) & internuclear distance of C – I is 2.13Å

dC–I = rC + rI (rC is 0.77Å)

rI = 2.13 – 0.77 = 1.36Å

Case II - Heteroatomic molecule with electronegativity difference more :

Schoemaker and Stevensos law :

If in a diatomic molecule electronegativities ofA– B have more difference. Then actual bond length

will be reduced.

As per schoemaker & Stevenson – The reduction in bond length depends on the difference in

electronegativities of atoms by following manner -

dA – B = rA + rB – 0.09 (XA – XB)

Here XA is electronegativity ofA& XB is electronegativity of B

e.g. - If bond length of F2 = 1.44Å, Bond length of H2 = 0.74Å.

Find out the bond length of H – F ? (electronegativity of F is 4.0, electronegativity

of H is 2.1)

Solution - dH – F = rF + rH – 0.09 (XF – XH)

dF–F = 1.44/2 = 0.77Å, dH–H = 0.72/2 = 0.37Å

dH–F = 0.77 + 0.37 – 0.09 (4.0 – 2.1)

= 1.09 – (0.09 × 1.9) = 1.09 – 0.171 = 0.919Å



(ii) Ionic Radius :

(A) Cationic radius :

(a) When an neutral atom loses electron it converts into cation (positive charged ion)

Atomic radius > Cationic radius

because after loosing e– no. of electron reduces, but no. of protons remains same, due to its effective

nuclear charge increases, hence electrons pulled towards nucleus and atomic radius decreases,

moreoverafter loosingall the electrons from the outermost shell, penultimate shell becomes ultimate

shell which is nearer to nucleus so size decreases

(b) Size of cationeffZorchargetheofMagnitude

1

e.g. Fe > Fe+2 > Fe+3

Pb+2 > Pb+4

Mn > Mn+2 > Mn+3 > Mn+4 > Mn+5 > Mn+6 > Mn+7

(B) Anionic radius :

(a) When neutral atom gains electron it converts into anion

Anionic radius >Atomic radius

(b) In an anion electron are more than protons so effective nuclear charge reduces, and inter electronic

repulsion increases, which also increases screening effect. So distance between electron and nucleus

increases and size of anion also increases.

e.g. Zof flourine is 9

F F–

Z 9 9

e 9 10

soe

Z=

9

9= 1

10

9= 0.9 As effective nuclear charge of F– is less than F, so size of F– >F

(c) Size of isoelectronic species :

– Thosespecieshavingsame numberof electron but different nuclear charge forms isoelectronic series.

– for isoelectronic species the atomic radius increases with decrease in nuclear charge

K+ Ca+2 Ar S–2 Cl–

Z 19 20 18 16 17

e 18 18 18 18 18

e

z

18

19

18

20

18

18

18

16

18

17

Order of radius: (S–2 > Cl– > Ar > K+ > Ca+2), (N3– > O2– > F– > Ne > Na+2 > Mg+2 > Al+3)

(iii)Metallic radius :

(a) Half of the nuclear distance between two adjacent metallic atoms in crystalline lattice structure.

(b) There is no overlapping of atomic orbitals

(c) so, metallic radius > Covalent radius



(d) Metallic radiusstrengthbondMetallic

1

(e) More metallic radius loose crystal packing less bond strength, (body centred packing)

(f) Less metallic radiusTight crystal packing FCCHigh bond strength, (Hexagonal close packing)

(iv)Vander Waal’s radius :

(a) Those atoms (like noble gases) which are not bonded with each other, experiences a weak attractive

force to come nearer.

(b) the half of the distance between the nuclei of adjacently placed atoms in solid state of a noble gas isVanderWaal’s radius.

(c) Vander Waal radius > Covalent radius.

(d) Inert gas have only Vander Waal radius.

(e) In molecules of non–metals both covalent and VanderWaal radius exists.

Cl2

Cl2

Covalent radius

Vander Waal distance

Cl2 momecules

Covalent radius = 0.99ÅVander Waal radius = 1.80Å

radiusalentcov2radiusWaalVander

(f) Vander Waal radius > Metallic radius > Covalent radius

(g) Vander Waal force of attractionMolecular weight or atomic weight (in inert gases)

(h) In a period from left to right Vander Waal radius decreases.

(i) In a group from top to bottom its values increases.

(v) Factors affecting atomic size are following :

(a) eargchnuclearEffective

1radiusAtomic Li > Be > B > C > N > O > F

(b) shellsof.NoradiusAtomic Li < Na < K < Rb < Cs

(c) effectScreeningradiusAtomic

(d) eargchpositiveofMagnitude

1sizeAtomic Mn > Mn+2 > Mn+3 > Mn+4

eargchnegativeofMagnitudesizeAtomic O < O– < O–2

(e) orderBond

1radiusAtomic > N – N < > – N = N – > – N N–



(vi) Periodic variation of atomic size :

i- Across a period – It decreases from left to right in a period as nuclear charge increases

e.g. Li > Be > B > C > N > O > F < Ne

ii- In a group – It increases from top to bottom in a group as number of shell increases

e.g. Li < Na < K < Rb < Cs

ATOMIC RADIUS

ALKALI METALS ALKALINE EARTH METALS

Li, 1.23

Na, 1.57

K, 2.03Rb, 2.16

Cs, 2.35

0

0.5

1

1.5

2

2.5

Li Na K Rb Cs

inÅ

Be, 0.89

Mg, 1.6

Ca, 1.97Sr, 2.15 Ba, 2.22

0

0.5

1

1.5

2

2.5

Be Mg Ca Sr Ba

(in

Å)

BORON FAMILY

B, 0.8

Al, 1.24 Ga, 1.24

In, 1.5 Tl, 1.55

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

B Al Ga In Tl

(in

Å)

CARBON FAMILY NITROGEN FAMILY (PNICOGENS)

C, 0.77

Si, 1.11Ge, 1.22

Sn, 1.41 Pb, 1.44

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

C Si Ge Sn Pb

(in

Å)

N, 0.74

Pb, 1.1As, 1.21

Sb, 1.41Bi, 1.52

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

N Pb As Sb Bi

(in

Å)



CHALCOGENS HALOGENS

O, 0.74

S, 1.04Se, 1.14

Te, 1.37

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

O S Se Te

(in

Å)

F, 0.72

Cl, 0.99

Br, 1.14

I, 1.33

0

0.2

0.4

0.6

0.8

1

1.2

1.4

F Cl Br I

(in

Å)

Exceptions : Transition elements

(a) From Sc21 to Zn30 in first transition series effective nuclear charge continuously increases, but atomic

size does not decreases continuously.

(b) From Sc21 to Mn25 size decreases

(c) From Fe26 to Ni28 size almost constant

Nuclear charge ~ Screening effect

(d) From Cu29 to Zn30 size increases. nuclear charge is less effective than screening effect.

(vii) Lanthanide Contraction :

(a) Outermost electronic configuration of inner transition elements is

(n – 2)f1 – 14, (n – 1)s2p6d0–1, ns2 (n = 6 to 7)

(b) Electron enters in (n – 2) f orbitals

(c) Mutual screening effect of electron is very less, because of complicated structure of f-orbital

(d) Nuclear charge increases by one unit (+1) in lanthanides and actinides, so atomic size of these

elements slightly decreases. it is known as lanthanide contraction.

Here, Nuclear charge > Screening effect.

(e) In transition series 1st, 2nd and 3rd

III B IV B

size Sc Ti size increases

IncreasesLa

Y

Hf

ZrEqual due to lanthanide contraction

Radii 3d < 4d 5d, except III B.



(viii) Orders of atomic and ionic radii are following :

H+ < H < H–

I+ < I < I–

Ti < Zr Hf

Sc < Y < La

Na+ < Ne < F–

Be < Li < Na

Ni < Cu < Zn

Sc > Ti > V > Cr

Ar > Na > Cl

Cu < Au < Ag

IONIZATION POTENTIAL OR IONIZATION ENERGY OR IONIZATION ENTHALPY

(i) Minimum energy required to remove most loosly held outer most shell electron in ground state from an

isolated gaseous atom is known as ionization Potential.

(ii) Successive I.E.

(a) For an atom M, successive ionization energies are as follows -

M + E1 M+ + e– E1 = IP1

M+ + E2 M+2 + e– E2 = IP2

M+2 + E3 M+3 + e– E3 = IP3

IP3 > IP2 > IP1

(b) electron cannot be removed from solid state of an atom, it has to convert in gaseous form, Energy

required for conversion from solid state to gaseous state is called Sublimation energy.

(c) IP is always an endothermic process (H = +ve)

(d) It is measured in eV/atom (electron volt/atom) or K.cal/mole or KJ/mole

(iii) Factors affecting ionization potential :

(A) Atomic size : Larger the atomic size, smaller is the I.P. It is due to that the size of atom increases theoutermost electrons farther away from the nucleus and nucleus loses the attraction on that electronsand hence can be easily removed.

IP sizeAtomic

1

(B) Effective nuclearcharge (Zeff) : I.P. increases with the increase in nuclear charge between outermostelectrons and nucleus.

(C) Screening effect : Higher is the screening effect on the outer most electrons causes less attractionfrom the nucleus and can be easily removed, which is leading to the lower value of I.P.

IP effectScreening

1

(D)Penetration power of sub shells :

(a) Order of attraction of subshell towards nucleus. (Penetration power) is - s > p > d > f

(b) As s-subshell is more closer to nucleus so more energy will be required to remove electron incomparison to p, d & f.



(c) Order of decreasing IP is s > p > d > f

Example IP1 of Be > IP1 of B

(1s22s2) (1s22s22p1)

After loosing one electron, B attains electronic configuration of Be, so 2nd IP of B is more than Be.

IP2 of B > Be

(E) Stability of half filled and fully filled orbitals :

(a) Half filled p3, d5, f7 or fully filled s2, p6, d10, f14 are more stable than others so it requires more

energy.

e.g. In C, N, O, F

IP1 order is C < O < N < F

(b) Because of half filled p-orbitals in N, its ionization energy (stability) is higher than O.

IP1 order Na < Al < Mg

(c) Because s-orbital in Mg is completely filled and its penetration power is also higher than p-orbital

(Al).

IP2 order Mg+ < Al+ < Na+

(2, 8, 1) (2, 8, 2) (2, 8)

(F) Oxidation state :

I.P. oxidation state of an atom. Ion with high oxidation state will have ionization potential e.g.

Fe+3 > Fe+2 > Fe

(iv) IP & Periodic table :

In a group : Size increase so IE decrease.

decreasesIE

increasesSize

CsRbKNaLi

Exception :

IP ofAl = I.P. of Ga (While I.P. decreases down the group it is due to Transition contraction)

d4d5

ZrofIPHfofIP(While I.P. should decreases down the group)

It is due to lanthanide contraction



In a period : In a period atomic size decreases and effective nuclear charge increases so removal of

electron become difficult and ionization energy increases.

.increasesenergyionization

increaseseargchnucleareffective

,decreasessizeatomic

NeFONCBBeLi

Ionization energy Ne > F > N > O > C > Be > B > Li

(v) Application of ionization potential :

(A) Metallic and non metallic character :

Metallic I.P. Low (Na, K, Rb etc.)

nonmetallic I.P. High (F, Cl, Br etc.)

I.P. propertyIonisation

1

(B) Reactivity :

Reducing character characterIonisation

1

(a) IAgroup has minimum I.P. so they are strong reducing agents in gaseous state

(Li < Na < K < Rb < Cs)

(b) VIIAgroup has maximum I.P. so they are strong oxidising agents (F > Cl > Br > I)

(C) Stability of oxidation states

(a) If the difference between two successive I.P. 16eV, then lower oxidation state is stable.

e.g.eV7.42

IPaNNa

IPaNNa

22

1

So Na+ is stable.

(b) If the difference between two successive I.P. 11eV, higher oxidation state is stable.

e.g.eV4.7

IPMgMg

IPMgMg

22

1

So Mg+2 is stable.

Al Al+

Al Al+2+

Al Al+3+2

Al is stable only in gaseous state+

Al is stable in liquid and solid state.+3

12.8 eV So Al is stable+

6.0 eV So Al is stable+3

(D)Basic nature – It is property of elements with loosely held electrons

I.P. opertyPrBasic

1

M2O > MO, Na2O > MgO, Cs2O > Rb2O > K2O > Na2O > Li2O, NaOH > Mg(OH)2



ELECTRON AFFINITY / ELECTRON GAIN ENTHALPY

(i) The amount of energy released or absorbed when electron is added to the valence shell of an isolated

gaseous atom.

X + e– X– + E.A.

known as Electron affinity.

(ii) Mostly energy is released in the process of first E.A.

X + e– X– + EA1

(iii) If EA1 is exothermic then (eg H (electron gain enthalpy) = – ve)

–EA1 is mostly positive

(iv) EAof neutral atom is equal to I.P. of its anion.

{Energy evolved in E.A. to convert neutral atom to anion} = {Energy absorbed to remove electron from

anion}

i.e. X + e–EA

IPX– + Energy

On adding 2nd electron in anion X–

X– + e– X–2 – EA2

or X– + e– + EA2 X–2

(v) eg is endothermic process, i.e. energy has to given to introduce 2nd electron in an anion, because

repulsion between negative charge on anion (X–) and 2nd electron.

(vi) Factors affecting electron affinity :

(a) Atomic size – EAsizeAtomic

1

(b) Screening effect – EAeffectScreening

1

(c) Effective nuclear charge (Zeff) – EA Zeff

(d) Stability of completely filled or half filled orbitals -

Electron affinity of filled or half filled orbital is very less or zero or energy is given to introduce any

electron. It is because of its stability.

(vii) In period

(a) Electron affinity increases along the period due to increase in effective nuclear charge and decrease

in atomic size.

(b) Halogensposses maximumelectron affinitydue to small sizeand maximumZeff and aftergainingone

electron. They attain a stable inert gas configuration.

(c) Electron affinity of IIA and noble gases is zero.

(d) In 2nd period from 6C to 7N value of electron affinity decreases, because of half filled orbitals

present in 7N which is more stable.

(e) In 3rd period value of electron affinity of Si is higher than P, because of half filled orbitals present in P.

(f) The value of electron affinity of inert gases is zero due to fully p-orbitals.



zeroLessZero

Ar

Ne

ClS

FO

P

N

SiAl

CB

Mg

Be

Na

Li

F > O > C > B > N > Be > Ne

Cl > S > Si > Al > P > Mg > Ar

(viii) In Group :

(a) Electron affinity decreases down the group.As the size increase and attraction on additional electron

is less.

(b) Electron affinity of 3rd period element is greater than electron affinity of 2nd period elements of the

respective group.

F 2s22p5

Cl 3s23p5

Due to small size of fluorine, electron density around the nucleus increases. The incoming electron

suffers more repulsion. In case of chlorine electron density decreases due to large size, decreasing order

of electron affinity of halogen. Cl > F > Br > I

S > O > P > N

Si > C > P > N

N & P have low electron affinity due to stable half filled configuration.

1 18

H He+ 0.754 2 13 14 15 16 17 –0.5

Li Be B C N O F Ne+ 0.618 0

0

+ 0.277 +1.263 – 0.07 +1.461 +3.399 –1.2

Na Mg Al Si P S Cl Ar+ 0.548 + 0.441 +1.385 +0.747 +2.077 +3.617 –1.0

K Ca Ga Ge As Se Br Kr+ 0.502 + 0.02 + 0.30 +1.2 + 0.81 +2.021 +3.365 –1.0

Rb Sr In Sn Sb Te I Xe+ 0.486 + 0.05 + 0.3 +1.2 +1.07 +1.971 +3.059 –0.8

Electron affinity of the main-group elements (in electronvolts)

ELECTRONEGATIVITY(EN)

(i) The tendency of an atom to attract shared electrons towards itself is called electronegativity.

(ii) EN and EA both have tendency to attract electrons but electron affinity is for isolated atoms. Where aselectronegativity is for bonded atoms.

(iii) Apolar covalent or ionic bond ofA– B may be broken

as (a) A – B : AA– + B+ (EN A > EN B)

or (b) A – B AA+ + :B– (EN A < EN B)

depending on their tendency to attract bonded electron.

(iv) There is no unit of electronegativity as EN is tendency of a bonded atom not an energy

(v) Paulingexplained it first time.



(vi) In Pauling’s scale, elements having almost same electronegativity are –

N = Cl = 3.0

C = S = I = 2.5

P = H = 2.1

Cs = Fr = 0.7

Be = Al = 1.5

(vii) EN of some other elements are as follows –

H

2.1

Li Be B C N O F

1.0 1.5 2.0 2.5 3.0 3.5 4.0

Na Mg Al Si P S Cl

0.9 1.2 1.5 1.8 2.1 2.5 3.0

K Br

78.0

Rb

8.0

5.2

I

8.2

7.0

Fr

72.0

Cs

– Small atoms are normally having more EN than larger atoms.

(viii) Factors Affecting electronegativity :

(a) Atomic size -

electronegativitysizeAtomic

1

(b) Effective nuclear charge (Zeff) -

ElectronegativityZeff

(c) Hybridisation state of an atom -

Electronegativity% of s character in hybridised atom

sp > sp2 > sp3

% s character 50% 33% 25%

Electronegativity 3.25 2.75 2.5

Because s-orbital is nearer to nucleus so by increasing s-character in hybridisation state, EN also

increases.

(d) Oxidation state -

Electronegativityoxidation state

Mn+2 < Mn+4 < Mn+7

O– < O < O+

Fe < Fe+2 < Fe+3



– As atomic radius decreases by increasing oxidation state of cation species, EN increases.

– In anionic species, the order of electronegativity is O–2 < O– < O

(e) Electronegativity does not depends on filled or half filled orbitals, because it is a tendency to attract

bonded electron, not to gain electron from out side.

(ix) Periodic table & Electronegativity :

(a) Electronegativity decreases down the group.

(b) In period on moving from left to right EN increases.

Exceptions -

(Zero group)– Electronegativityof ‘0’group is always zero, because inert gas do not form molecule.

(c) Electronegativity of Cs and Fr in 6th and 7th period (IA) are equal, it is because from 55Cs to 87Fronly one shell increases but nuclear charge (No. of proton) increases by +32. So effect of nuclearcharge balanced the effect of increase in number of shell.

(d) Electronegativity of F > Cl but electron affinity of Cl > F

So Fluorine is called Black sheep element.

(e) In group of IIB elements (Zn, Cd, Hg) value of electronegativity increases down the group, becauseof lanthanide contraction

(f) In IIIA group, value of electronegativity increases down the group, because of transition contraction(+ 18 charge)

EN of Ga > EN of Al

ELECTRONEGATIVITY

ALKALI METALS ALKALINE EARTH METALS

Li, 1Na, 0.9

K, 0.8 Rb, 0.78Cs, 0.7

0

0.2

0.4

0.6

0.8

1

1.2

Li Na K Rb Cs

E.N

.

Be, 1.47

Mg, 1.23

Ca, 1.04 Sr, 1 Ba, 0.97

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Be Mg Ca Sr Ba

E.N

.

BORON FAMILY

B, 2

Al, 1.5Ga, 1.6 In, 1.7 Tl, 1.8

0

0.5

1

1.5

2

2.5

B Al Ga In Tl

E.N

.



CARBON FAMILY NITROGEN FAMILY (PNICOGENS)

C, 2.5

Si, 1.8Ge, 2

Sn, 1.7 Pb, 1.6

0

0.5

1

1.5

2

2.5

3

C Si Ge Sn Pb

E.N

.

N, 3

Pb, 2.1 As, 2 Sb, 1.9 Bi, 1.9

0

0.5

1

1.5

2

2.5

3

3.5

N Pb As Sb Bi

Ele

ctro

neg

ativ

ity

CHALCOGENS HALOGENS

O, 3.5

S, 2.58Se, 2.4

Te, 2.1 Po, 2

0

0.5

1

1.5

2

2.5

3

3.5

4

O S Se Te Po

E.N

.

F, 4

Cl, 3Br, 2.8

I, 2.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

F Cl Br I

E.N

.

(x) Application of electronegativity :

(A) Metallic and non metallic nature -

Low electronegativity Metals

High electronegativity Non metals

Metallic character increases down the group but decreases along a period.

(B) Bond length -

EN lengthBond

1

Here EN = difference in electronegativities of bonded atoms

HF < HCl < HBr < HI

– HF has minimum bond length because of much difference in the electronegativities of H and F.

(C) Bond energy - By increasing EN bond length decreases and hence bond energy increases.

Bond energyElectronegativity difference

HF > HCl > HBr > HI

(D) Acidic strength of hydrides -

Bond energy (Strength) stability of molecule.

– Order of stability of hydrohalides is

HF > HCl > HBr > HI and so order of their acidic strength will be -

HF < HCl < HBr < HI



– In VA group -NH3 *Thermal stability decreases

H3P *Basic character decreases

AsH3 *Acidic character increases

In PH3 andAsH3 there is less difference in the electronegativities of XA and XB, so their bond energydecreases and hence acidic character (losing H+ ion) increases.

(E) Reactivity -

Bond energyStabilityactivityRe

1

As bond energy difference of electronegativities

So, Electronegativity differenceStabilityactivityRe

1

activityRe

HIHBrHClHF

– HI is most reactive hydrohalide or strongest acid among all hydrohalides.

(F) Nature of bonds -

(a) It can be determined by Hanny & Smith formula -

Ionic % = 16(XA – XB) + 3.5 (XA – XB)2

Here XA = ElectronegativityofA

XB = Electronegativityof B

If XA – XB 2.1 Ionic % > 50% i.e. Ionic bond

If XA – XB 2.1 Ionic % < 50% i.e. covalent bond

(b) Gallis experimental values are -

XA – XB 1.7 Ionic

XA – XB 1.7 Covalent

If XA = XB; then A – B will be non polar, e.g. H – H, F – F

If XA > XB and difference of EN is small, then

A– —— B+ bond will be polar covalent

e.g. H2O (H+ —— O– —— H+)

If XA >> XB and XA – XB, EN is high then, A– B+ bond will be polar or ionic

e.g. Na+Cl–

In HF, XA – XB = 1.9, which is more than 1.7, even then it is covalent compound.

(G) Nature of hydroxides -

(a) As per Gallis, InAOH if electronegativity ofAis more than 1.7 (Non metal) then it is acidic in nature.

(b) If electronegativity of ‘A’is less than 1.7 (metal) thenAOH will be basic in nature



e.g. NaOH ClOH

XA 0.9 3.0

Nature Basic Acidic

(c) If XA – XO XO – XH

thenAO bond will be more polar and will break up as

A OH A + OH+ –

It show basic nature

(d) If XA – XO XO – XH

A O H H + AO— + –

It show acidic nature

e.g. In NaOH

XO – XNa (2.6) > XO – XH (1.4)

So hydroxide is basic

In ClOH –

XO – XCl (0.5) < XO – XH (1.4)

So hydroxide is acidic

(H) Nature of oxides - Consider an oxide AO

If XA – XO > 2.3 Basic oxide

If XA – XO = 2.3 Amphoteric oxide

If XA – XO < 2.3 Acidic oxide

(a) Along a period acidic nature increases.

(b) Down the group basic nature increases

Li Be B C N O F

Na Mg Al Si P S Cl

X – X > 2.3A O

X – X = 2.3A O

X – X < 2.3A O

Basic

Amphoteric

Acidic

i.e. when in periodic table the distance between the element and oxygen increases, basic character

increases.

NO2 > ZnO > K2O

acidic character decreases

– BeO, Al2O3, ZnO, SnO, PbO, SnO2, PbO2, Sb2O3 etc. are amphoteric oxides.

– CO, H2O, NO, N2O are neutral oxides.

Acidic strength of oxide and oxyacidEN



increasenatureacidic,increaseEN

ONCOOB 52232

H3BO3 < H2CO3 < HNO3

SO3 H SO42 EN

SeO3 H SeO2 4 Acidic

TeO3 H TeO2 4

H3PO4 > H3AsO4 > H3SbO4

H2SO3 > H2SeO3 > H2TeSO3

HOF > HOCl > HOBr > HOI

HClO4 > HBrO4 > HIO4

N2O5 > P2O5 > As2O5

N2O3 > P2O3 > As2O3 > Sb2O3

Acidic nature oxidation state

Acidic properties increases with increasing oxidation state of an element

HClO4 > HClO3 > HClO2 > HClO

HNO3 > HNO2

H2SO4 > H2SO3

SO3 > N2O3

Sb2O3 < Sb2O5

N2O < NO < NO2 < N2O5



IMPORTANT FACTS TO REMEMBER

1. Lowest electronegativity : Cs

2. Highestelectronegativity : F

3. Highest ionisation potential : He

4. Lowest ionisation potential : Cs

5. Lowest electron affinity : Noble gases

6. Highestelectronaffinity : Chlorine

7. Least electropositive element : F

8. Lowest m. pt. metal : Hg

9. Highest m. pt. and b. pt. metal : W(Tungsten)

10. Lowest m. pt. and b. pt. non metal : He

11. Notorious element : Hydrogen

12. Lightestelement : Hydrogen

13. Smallest atomic size : H

14. Largest atomic size : Cs

15. Largest anionic size : I–

16. Smallest cation : H+

17. Most electropositive element : Cs

18. Elementwithelectronegativity

next to Fluorine : Oxygen

19. Groupcontainingmaximum no.

of gaseous elements in periodic table : Zero group

20. Total number of gaseous elements

in periodic table : 11 (H, N, O, F, Cl, He, Ne, Ar, Kr, Xe, Rn)

21. Total number of liquid elements in

periodic table : 6 (Ga, Br, Cs, Hg, Fr, Uub)

22. Smallestanion : F–

23. Liquid element of radioactive nature : Fr

24. Total number of radioactive elements in

periodic table : 25

25. Volatile d-block elements : Zn, Cd, Hg, Uub

26. Element containingno neutron : H

27. Most aboundant element on earth : Oxygen

28. Rarest element on earth : At (astatine)

29. Most aboundant metal on earth : Al

30. Elementhavingmaximumtendency

for catenation : Carbon



31. Non metal having highest m. pt., b. pt. : Carbon (diamond)

32. Metals showing highest oxidation number : Os (+8)

33. Most electrovalent compound : CsF

34. Most stable carbonate : Cs2CO3

35. Strongest alkali : CsOH

36. Strongest basic oxide : Cs2O

37. Best electricity conductor among metals : Ag

38. Best electricity conductor among non metals: graphite

39. Most poisonous element : Pu(Plutonium)

40. Liquid non metals : Br

41. Element kept in water : Phosphorous

42. Elements kept in kerosene : IA group element (except Li)

43. Elements sublimeonheating : I2

44. Noble metals : Au, Pt. etc.

45. Amphoteric metal : Be, Zn, Al, Sn, Pb

46. Amphoteric non metal : Si

47. Metalloids elements : B, Si, As, Te, At, Ge, Sb

48. Non metals havingmetallic lusture : Graphite, Iodine

49. Heaviest naturally occuringelement : Uranium

50. Poorest conductor of electricity : Diamond

51. Hardest naturally occuring element : Diamond

52. Lightest solid metal : Li

53. Amphoteric oxides : BeO, Al2O3, ZnO, PbO, PbO2, SnO, SnO2,

Sb2O3, As2O3 etc.

54. Neutral oxides of non metals : NO, CO, H2O, N2O

55. Dry bleacher : H2O2

56. Dry ice : Solid CO2

57. First man made element : Tc43 (Technicium)

58. Smallest period : Ist (2 elements)

59. Largest period in periodic table : 6th (32 elements)

60. Largest group in periodic table : IIIB (32 elements)

61. Most abundant d-block metal : Fe

62. Most abundant s-block metal : Ca

63. Highest density (Metals) : Os, Ir

64. Highest density (Non Metals) : Boron

65. Most abundant gas in atmosphere : Nitrogen

66. Most abundant element in the universe : Hydrogen



SOLVED EXAMPLES

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

(1) Crystal, vander Waal and covalent radii

(2) Covalent, crystal and vander Waal radii

(3) Vander Waal, ionic and covalent

(4) None is correct

Ans. (2)

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

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

(1) NO2 (2) NO (3) N2O (4) N2O3

Ans. (3)

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.

(1) TFTF (2) TFFT (3) FFTF (4) TTFF

Ans. (1)

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 -

(1) 36 (2) 34 (3) 33 (4) 32

Ans. (2)

Sol. At. No. = 16(S)

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



Q.5 Which of the following relation is correct -(1) 2 I.P. – E.A. – E.N. = 0 (2) 2 I.P. – E.N. + E.A. = 0(3) 2 E.N. – I.P. – E.A. = 0 (4) E.N. – I.P. – E.A. = 0

Ans. (3)Sol. Where E.N. stands for electronegativity E.A. stands for electron affinity and I.P. stands for Ionization

potential.It is observed that for an element, E.A. and E.N. and I.P. usually vary in the same direction. Hencewhen E.A. and E.N. increase the I.P. also increases E.N. has the mean valued of I.P. and E.A.

E.N. =2

.A.E.P.I

2E.N. = I.P. + E.A. or 2E.N. – I.P. – E.A. = 0

Q.6 Pd has exceptional valence shell electronic configuration of 4d105s0. It is a member of -(1) 5th Period, Group 10 (2) 4th Period, Group 12(3) 6th Period, Group 10 (4) 5th Period, Group 14

Ans. (1)Sol. 4d105s0 is the exceptional configuration of Pd. Its electronic configuration should be [Kr]364d8, 5s2.

Thus itsPeriod = 5thGroup = ns + (n – 1)d electrons

= 2 + 8 = 10

Q.7 The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2eV respectively. Theelement is likely to be -(1) Na (2) Si (3) F (4) Ca

Ans. (2)Sol. The jump in IP values exist in IP5 and thus removal of fifth electron occurs from inner shell. Thus

element contains four electrons in its valency shell.

Q.8 The electronic configurations for some neutral atoms are given below :(i) 1s2, 2s2 2p6, 3s2 (ii) 1s2, 2s2 2p6, 3s1

(iii) 1s2, 2s2 2p6, 3s23p2 (iv) 1s2, 2s2 2p6,3s2 3p3

Which of these is expected to have the highest second ionization enthalpy ?(1) (i) (2) (ii) (3) (iii) (4) (iv)

Ans. (2)Sol. B atom after losing outermost electron acquires noble gas configuration (stable configuration).

It is difficult to remove the next electron from B+ (1s2, 2s2 2p6) ion.

Q.9 What should be the atomic number of the next inert gas if discovered in future ?(1) 115 (2) 119 (3) 118 (4) 121

Ans. (3)Sol. The nexthalogen willhave 7s27p5 outerconfiguration. Since, the fillingof7p-orbitalswill begin after5f

and 6d-orbitals, thus the atomic number of the new halogen will be 112 (up to the filling of 6d-orbitals)plus 5, i.e., 117.

Q.10 Arrange Ce3+, La3+, Pm3 and Yb3+ in increasing order of their size -(1) Yb3+ < Pm3+ < Ce3+ < La3+ (2) Ce3+ < Yb3+ < Pm3+ < La3+

(3) Yb3+ < Pm3+ < La3+ < Ce3+ (4) Pm3+ < La3+ < Ce3+ < Yb3+

Ans. (1)Sol. Lanthanide contraction isobserved in these ions, i.e., ionic radius decreases asatomic number increases.



Q.11 The paramagnetic species among the following is -Na+, Zn2+, Cu+, Fe3+

(1) Na+ (2) Zn2+ (3) Cu+ (4) Fe3+

Ans. (4)Sol. Paramagnetic species haveat least one unpaired electron.Write theelectronicconfigurationandobserve

the unpaired orbital.Na+ : 1s2, 2s2, 2p6 – all paired :(11)Zn2+; 1s2, 2s22p6, 3s23p63d10 – All paired(30)Cu+ : 1s2, 2s22s6, 3s23p63d10 – All paired(29)Fe3+ : 1s2, 2s2, 2p6, 3s23p63d5 – Unpaired orbitals(26)

Q.12 Calculate the energy needed to convert three moles of sodium atoms in the gaseous state to sodiumions. The ionization energy of sodium is 495 kJ mol–1.(1) 1485 kJ (2) 495 kJ (3) 148.5 kJ (4) None

Ans. (1)Sol. Energy needed to convert 1 mole of sodium (g) to sodium ions = 495 kJ

energy needed to convert 3 mole of Na (g) to Na+ ions = 495 × 3 = 1485 kJ

Q.13 The ionic radius of Cr is minimum in which of the following compounds ?(1) CrF3 (2) CrCl3 (3) Cr2O3 (4) K2CrO4

Ans. (4)Sol. Cr has maximum oxidation number (+6) in K2CrO4 and thus, has minimum ionic radius.

Q.14 The correct decreasing order of atomic size among the following species is : K+, Cl–, S2–, Ca2+

(1) Ca2+ > K+ > Cl– > S2– (2) K+ > Ca2+ > Cl– > S2–

(3) S2– > Cl– > K+ > Ca2+ (4) S2– > Cl– > Ca2+ > K+

Ans. (3)Sol. In isoelectronic ions, the atomic size decreases as z/e ratio increases.

S2–e

z=

18

16; Cl–

e

z=

18

17; K+

e

z=

18

19; Ca2+

e

z=

18

20

Q.15 The E.N. of H, X, O are 2.1, 0.8 and 3.5 respectively comment on the nature of the compoundH–O–X, that is :(1) Basic (2)Acidic (3)Amphoteric (4) Can't be predicted

Ans. (1)

Sol. X—O—H

O–H bond will break basic

4.1

5.31.27.2

8.0

More electronegativitydifference it would break first it is basic in nature



SUMMARY

In this Unit, you have studied the development of the Periodic Law and the PeriodicTable. Mendeleevís

Periodic Table was based on atomic masses. Modern Periodic Table arranges the elements in the

order of their atomic numbers in seven horizontal rows (periods) and eighteen vertical columns (groups

or families).Atomic numbers in a period are consecutive, whereas in a group they increase in a pattern.

Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit

similar chemical properties. However, the elements of the same period have incrementally increasing

number of electrons from left to right, and, therefore, have different valencies. Four types of elements

can be recognized in the periodic table on the basis of their electronic configurations. These are s-block,

p-block, d-block and f-block elements. Hydrogen with one electron in the 1s orbital occupies a unique

position in the periodic table. Metals comprise more than seventy eight per cent of the known elements.

Nonmetals, which are located at the top of the periodic table, are less than twenty in number. Elements

which lie at the border line between metals and non-metals (e.g., Si, Ge,As) are called metalloids or

semi-metals. Metallic character increases with increasing atomic number in a group whereas decreases

from left to right in a period. The physical and chemical properties of elements varyperiodicallywith their

atomic numbers.

Periodic trends are observed in atomic sizes, ionization enthalpies, electron gain enthalpies,

electronegativity and valence. The atomic radii decrease while going from left to right in a period and

increase with atomic number in a group. Ionization enthalpies generally increase across a period and

decrease down a group. Electronegativityalso shows a similar trend. Electron gain enthalpies, in general,

become more negative across a period and less negative down a group. There is some periodicity in

valence, forexample, amongrepresentative elements, the valence is eitherequal to thenumberofelectrons

in the outermost orbitals or eight minus this number. Chemical reactivity is hightest at the two extremes

of a period and is lowest in the centre. The reactivity on the left extreme of a period is because of the

ease of electron loss (or low ionization enthalpy). Highly reactive elements do not occur in nature in free

state; they usually occur in the combined form. Oxides formed of the elements on the left are basic and

of the elements on the right are acidic in nature. Oxides of elements in the centre are amphoteric or

neutral.



EXERCISE-1

[SINGLE CORRECT CHOICE TYPE]

Q.1 Three element X, Y, Z are following Doberiner's Triad rule. If the atomic weight of X andY are 10 and

26 respectively, then atomic weight of Z will be

(1) 34 (2) 40 (3) 42 (4) 18

Q.2 IfAufbau rule is not followed, K – 19 will be placed in ________ block

(1) s (2) p (3) d (4) f

Q.3 Which of the following set of elements obeys Newland’s octave rule -

(1) Na, K, Rb (2) F, Cl, Br (3) Be, Mg, Ca (4) B,Al, Ga

Q.4 Elements which occupied position in the Lother mayer curve, on the peaks, were -

(1)Alkali metals (2) Highly electro positive elements

(3)Elements having large atomic volume (4)All of these

Q.5 The places that were left empty by Mendeleef were for which elements

(1)Aluminium&Silicon (2)Galliumandgerminium

(3)Arsenic and antimony (4) Molybdenum and tungsten

Q.6 Which of the following pairs of elements do not follow octave rule -

(1) Na, K (2) Ca, Sr (3) F, Cl (4) O, S

Q.7 In the general electronic configuration -

(n – 2)f1–14 (n – 1)d0–1 ns2, if value of n = 7 the configuration will be -

(1) Lanthanoides (2)Actinoides

(3)Transition elements (4) None

Q.8 Which is not anomalous pair of elements in the Medeleeves periodic table -

(1) Ar and K (2) Co and Ni (3) Te and I (4)Al and Si

Q.9 Which are correct match -(i) Eka silicon – Be (ii)Eka aluminium – Ga(iii) Eka mangenese – Tc (iv) Eka scandium – B(1) (ii) & (iii) (2) (i), (ii) & (iv) (3) (i) & (iv) (4)All

Q.10 Z/e ratio for N3–, O2– and F– respectively will be -(1) 0.7, 0.8 and 0.9 (2) 0.9, 0.8 and 0.7(3) 7, 8 and 9 (4) 9, 8 and 7

Q.11 If each orbital can hold a maximum of three electrons, the number of elements in 9th period of periodictable (long form) are(1) 48 (2) 162 (3) 50 (4) 75



Q.12 The number of elements in 5th and 6th period of periodic table are respectively -(1) 8, 18 (2) 18, 18 (3) 18, 32 (4) 18, 28

Q.13 Given atomic number 15, 33, 51 represents the following family -(1) Carbon family (2)Nitrogen family (3)Oxygen family (4) None

Q.14 The atom having the valence shell electronic configuration 4s2 4p2 would be in -(1) Group II A and period 3 (2) Group II B and period 4(3) Group IVA and period 4 (4) Group IVA and period 3

Q.15 Atomic number is the base of -(i) Lother mayer curve (ii) Newland octave rule(iii) Modern periodic table (iv) Doeberiener triad rule(v) Long form of periodic table(1) (i), (ii), (iv) (2) (iii), (v) (3) (i), (iv) (4) (i), (iii), (v)

Q.16 What is the atomic number of Ununhexium ?

(1) 106 (2) 96 (3) 116 (4) 118

Q.17 Which of the following is known as the bridge element of 2nd group in Mendeleev’s table?

(1) Zn (2) Sr (3) Mg (4) Hg

Q.18 The statement that is not correct for the periodic classification of element is

(1) The properties of the elements are the periodic function of their atomic numbers

(2) Non-metallic elements are lesser in number than the metallic elements

(3)The first ionization energies of elements along the period do not vary in a regularmanner with increasein atomic number

(4) d–subshell is filled by final electron with increasing atomic number of inner transition elements.

Q.19 Which of the following elements never has an oxidation state of +1 ?

(1) F (2) Cl (3) Br (4) I

Q.20 Which pair is different from the others?

(1) Li – Mg (2) B – Si (3) Be – Al (4) Li – Na

Q.21 The ionic radii of N3–, O2– and F– are respectively given by -

(1) 1.36, 1.40, 1.71 (2) 1.36, 1.71, 1.40

(3) 1.71, 1.40, 1.36 (4) 1.71, 1.36, 1.40

Q.22 In which of the following compounds manganese show maximum radius -

(1) MnO2 (2) KMnO4 (3) MnO (4) K3[Mn(CN)6]

Q.23 The correct order of size would be -

(1) Ni < Pd ~ Pt (2) Pd < Pt < Ni (3) Pt > Ni > Pd (4) Pd > Pt > Ni

Q.24 Atomic radii of Fluorine and Neon inAngstrom units are given by -

(1) 0.72, 1.60 (2) 1.60, 1.60

(3) 0.72, 0.72 (4) None of these



Q.25 Which of the following has largest radius -(1) 1s2, 2s2, 2p6, 3s2 (2) 1s2, 2s2, 2p6, 3s2, 3p1

(3) 1s2, 2s2, 2p6, 3s2, 3p3 (4) 1s2, 2s2, 2p6, 3s2, 3p5

Q.26 Which of the following sequences is correct for decreasing order of ionic radius -(1) Se–2 > I– > Br– > O–2 > F– (2) I– > Se–2 > O–2 > Br– > F–

(3) Se–2 > I– > Br– > F– > O–2 (4) I– > Se–2 > Br– > O–2 > F–

Q.27 The correct order of atomic size of C, N, P, S follows the order -(1) N < C < S < P (2) N < C < P < S (3) C < N < S < P (4) C < N < P < S

Q.28 Which of the following is arranged in decreasing order of size ?(1) Mg2+ > Al3+ > O2– (2) O2– > Mg2+>Al3+ (3) Al3+ > Mg2+ > O2– (4) Mg2+ = Al3+ > O2–

Q.29 Which has maximum ionic size ?(1) Li+ (2) Mg2+ (3)Al3+ (4) O2–

Q.30 The maximum tendency to form unipositive ion is for the element with the electronic configuration-(1) 1s2, 2s2, 2p6, 3s2 (2) 1s2, 2s22p6, 3s23p1

(3) 1s2, 2s22p6, 3s23p2 (4) 1s2, 2s22p6, 3s23p3

Q.31 The second ionisation potentials in electron volts of oxygen and fluorine atoms are respectivelygiven by-(1) 35.1., 38.3 (2) 38.3, 38.3 (3) 38.3, 35.1 (4) 35.1, 35.1

Q.32 The correct order of stability ofAl+,Al+2,Al+3 is-(1)Al+3 > Al+2 > Al+ (2)Al+2 > Al+3 > Al+ (3)Al+2 < Al+ > Al+3 (4)Al+3 > Al+ > Al+2

Q.33 Amongst the following, the incorrect order is(1) IE1 (Al) < IE1 (Mg) (2) IE1 (Na) < IE1(Mg)(3) IE2 (Mg) > IE2 (Na) (4) IE3 (Mg) > IE3 (Al)

Q.34 IP1 and IP2 of Mg are 178 and 348 K.cal mol–1. The enthalpy required for the reactionMg Mg2+ + 2e– is -(1) + 170 K.cal (2) + 526 K.cal(3) – 170 K.cal (4) – 526 K.cal

Q.35 Which of the following has maximum value of effective nuclear charge ?

(1) C (2) O (3) N (4) Ne

Q.36 Of the followingelements, which possess the highest electron affinity?

(1)As (2) O (3) S (4) Se

Q.37 Increasingorderof Electron affinity for following configuration.

(a) 1s2, 2s2 2p3 (b) 1s2, 2s2 2p4

(c) 1s2, 2s2 2p6 3s2 3p4 (d) 1s2, 2s2 2p6, 3s2 3p3

(1) a < d < b < c (2) d < a < c < b (3) a < b < c < d (4) a < b < d < c



Q.38 Which of the following case is endothermic -

(1) Cl Cl– (2) P P– (3) N N– (4) C C–

Q.39 In the formation of a chloride ion, from an isolated gaseous chlorine atom, 3.8 eV energy is released,which would be equal to -

(1) Electron affinity of Cl– (2) Ionisation potential of Cl

(3) Electronegativity of Cl (4) Ionisation potential of Cl–

Q.40 The correct order of electron affinity is -

(1) Be < B < C < N (2) Be < N < B < C (3) N < Be < C < B (4) N < C < B < Be

Q.41 Which of the following process energy is liberated -

(1) Cl Cl+ + e– (2) HCl H+ + Cl– (3) Cl + e– Cl– (4) O– + e– O–2

Q.42 Which of the followingconfiguration will have least electron affinity -

(1) ns2np5 (2) ns2np2 (3) ns2np3 (4) ns2np4

Q.43 Second electron affinity of an element is -

(1)Always exothermic (2) Endothermic for few elements

(3) Exothermic for few elements (4)Always endothermic

Q.44 Electronegativity of the following elements increases in the order -

(1) O < N < S < P (2) P < S < N < O(3) P < N < S < O (4) S < P < N < O

Q.45 On the Pauling’s electronegativity scale, which element is next to F .(1) Cl (2) O (3) Br (4) Ne

Q.46 3

54321

HCHCHCCCH Which carbon atom will show minimum electronegativity -

(1) Fifth (B) Third (3) First (4) Second

Q.47 Which one of the following is incorrect ?(1)Anelementwhichhashighelectronegativityalways has high electron gain enthalpy(2) Electron gain enthalpy is the property of an isolated atom(3) Electronegativity is the property of a bonded atom(4) Both electronegativity and electron gain enthalpy are usually directly related to nuclear charge and

inversely related to atomic size

Q.48 Which of the following has least density ?(1) Na (2) Li (3) Mg (4) K

Q.49 The order of first ionization enthalpies of the elments Li, Be, B, Na is -

(1) Li > Be > B > Na (2) Be > B > Li > Na

(3) Na > Li > B > Be (4) Be > Li > B > Na



Q.50 Which of the following statement is not true?

(1) F atom can hold additional electron more tightly than Cl atom

(2) Cl atom can hold additional electron more tightly than F atom

(3) The incoming electron encounters greater repulsion for F atom than for Cl atom

(4) It is easier to remove an electron from F¯ than Cl¯.

Q.51 Successive ionization energies of an element ‘X’ are given below (in K.Cal) :

IP1 IP2 IP3 IP4

165 195 556 595

Electronic configuration of the element ‘X’ is -

(1) 1s2, 2s2 2p6, 3s2 3p2 (2) 1s2, 2s1

(3) 1s2, 2s2 2p2 (4) 1s2, 2s2 2p6, 3s2

Q.52 Amongthe followingelements (Whose electronic configuration is give below) the one having the highestionization energy is -

(1) [Ne] 3s2 3p3 (2) [Ne] 3s2 3p4 (3) [Ne] 3s2 3p5 (4) [Ne] 3s2

Q.53 Electron affinities of O,F,S and Cl are in the order.

(1) O < S < Cl < F (2) O < S < F < Cl (3) S < O < Cl < F (4) S < O < F < Cl

Q.54 Element X,Yand Zhave atomic numbers 19, 37 and 55 respectively.Which of the following statementsis true -

(1) Their ionization potential would increase with the increasing atomic number

(2) ‘Y’ would have an ionization potential in between those of ‘X’ and ‘Z’

(3) ‘Z’ would have the highest ionization potential

(4) ‘Y’ would have the highest ionization potential

Q.55 The first (IE1) and second (IE2) ionization energies (kJ/mol) of a few elements designated by Romannumerals are given below. Which of these would be an alkali metal ?

IE1 IE2

(1) I 2372 5251

(2) II 520 7300

(3) III 900 1760

(4) IV 1680 3380

Q.56 Select the correct order (s).

(1) IE1 of F < IE1 of Cl (2) E A of O > E A of S

(3) Ionic radius of Cl¯ > Ionic radius of K+ (4) None of these

Q.57 The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. Theelement is likely to be -

(1) Na (2) Si (3) F (4) Ca



Q.58 The correct order of second ionization potential of C, N, O and F is -

(1) C > N > O > F (2) O > N > F > C

(3) O > F > N > C (4) F > O > N > C

Q.59 First, second and third IP values are 100eV, 150eV and 1500eV. Element can be -

(1) Be (2) B (3) F (4) Na

Q.60 M(g) M+(g) + e–, H = 100 eV

M(g) M2+(g) + 2e–, H = 250 eV

Which is incorrect statement ?

(1) IE1 of M(g) is 100 eV (2) IE1 of M+(g) is 150 eV

(3) IE2 of M(g) is 250 eV (4) IE2 of M(g) is 150 eV

Q.61 The correct values of ionization enthalpies (in kJ mol–1) of Si, P, Cl and S respectively are -

(1) 786, 1012, 999, 1256 (2) 1012, 786, 999, 1256

(3) 786, 1012, 1256, 999 (4) 786, 999, 1012, 1256

Q.62 Consider the following changes -

AA+ + e– : E1 and A+ A2+ + e– : E2

The energy required to pull out the two electrons are E1 and E2 respectively. The correct relationshipbetween two energies would be -

(1) E1 < E2 (2) E1 = E2 (3) E1 > E2 (4) E1 E2

Q.63 Electron affinity is a -

(1) Relative strength to attract the shared electron pair

(2) Necessary energy required to remove the electron from the ultimate orbit

(3) Energy released when an electron is added to the outermost shell

(4) Energy released when an electron is added to the inner shell

Q.64 The correct order of electron affinity of B, C, N, O is -

(1) O > C > N > B (2) B > N > C > O

(3) O > C > B > N (4) O > B > C > N

Q.65 If internuclear distance between Cl atoms in Cl2 is 10 Å & between H atoms in H2 is 2 Å, then calculateinternuclear distance between H & Cl (Electronegativity of H = 2.1 & Cl = 3.0).

(1) 5.919 Å (2) 6.9 Å (3) 7.919 Å (4) 4.91 Å

Q.66 Which of the following element is expected to have highest electron gain enthalpy -

(1) 1s22s22p63s23p5 (2) 1s22s22p3 (3) 1s22s22p4 (4) 1s22s22p5

Q.67 Calculate the bond length of C–X bond if C – C bond length is 1.54 Å and X–X bond length is 1.2 Åand electronegativities of C and X are 2.0 and 3.0 respectively.

(1) 2.74 Å (2) 1.37 Å (3) 1.46 Å (4) 1.28 Å



Q.68 The correct set of decreasing order of electronegativity is -(1) Li, H, Na (2) Na, H, Li (3) H, Li, Na (4) Li, Na, H

Q.69 Arrange F, C, O, N in the decreasing order of electronegativity -(1) O > F > N > C (2) F > N > C > O(3) C > F > N > O (4) F > O > N > C

Q.70 The increasing order of acidic nature of Li2O, BeO, B2O3(1) Li2O > BeO < B2O3 (2) Li2O < BeO < B2O3(3) Li2O < BeO > B2O3 (4) Li2O > BeO > B2O3

Q.71 Choose the correct Ionization potential order :(1) O¯ > O (2) N¯ > N (3) H¯ > H (4) None of these

Q.72 For an element ‘A’.

A 1IE A+ 2IE A2+ 3IE AA3+ ...................The IE1 and IE3 values are 27 kJ/mole and 51 kJ/mole respectively. Then the value of IE2 is _____kJ/mole.(1) 21 (2) 33 (3) 59 (4) 63

Q.73 Sodium forms Na+ ion but it does not form Na+2 because -

(1) Very low value of IE1 and IE2 (2) Very high value of IE1 and IE2

(3) High value of IE1 and low value of IE2 (4) Low value of IE1 and high value of IE2

Q.74 Consider the M(OH)3 formed by all the group 13 elements. The correct sequence of acidic strength ofhydroxides [M(OH)3] is -

(1) B(OH)3 < Al(OH)3 > Ga(OH)3 > In(OH)3 > Tl(OH)3

(2) B(OH)3 > Tl(OH)3 > Al(OH)3 > In(OH)3 > Ga(OH)3

(3) Al(OH)3 > Ga(OH)3 > B(OH)3 > In(OH)3 > Tl(OH)3

(4) B(OH)3 > Al(OH)3 > Ga(OH)3 > In(OH)3 > Tl(OH)3

Q.75 Arrange the following hydrides in their increasing acid strength [CH4, H2S, PH3 and SiH4] -

(1) H2S < PH3 < SiH4 < CH4 (2) CH4 < SiH4 < PH3 < H2S

(3) SiH4 < CH4 < PH3 < H2S (4) CH4 < H2S < PH3 < SiH4

Q.76 Arrange in the order of increasing basicity (NO2, K2O, ZnO) -

(1) NO2 < ZnO < K2O (2) K2O < ZnO < NO2

(3) NO2 < K2O < ZnO (4) K2O < NO2 < ZnO

Q.77 The order in which the following oxides are arranged according to decreasing basic nature is-

(1) Na2O > MgO > Al2O3 > SiO2 (2) SiO2 > Al2O3 > MgO > Na2O

(3) Al2O3 > SiO2 > MgO > Na2O (4) SiO2 > MgO > Na2O > Al2O3



Q.78 Set containing isoelectronic species is -

(1) C22+, NO+, CN–, O2

2+ (2) CO, NO, O2, CN

(3) CO2, NO2, O2, N2O5 (4) CO, CO2, NO, NO2

Q.79 The IE values of Al(g) = Al+ + e is 577.5 kJ mol–1 and H for Al(g) = Al3+ + 3e is 5140 kJ mol–1. If

second and third IE values are in the ratio 2 : 3. Calculate IE2 and IE3.

(1) IE2 = 1825 kJ/mole, IE3 = 2737.5 kJ/mol]




Q.80 Which of the following in increasing order of paramagnetism ?

(1) Al < Mg < O < N (2) Mg < Al < N < O

(3) Mg < Al < O < N (4) N < O < Al < Mg

Q.81 The correct order of increasing size is -

(1) Mg2+ < Na+ < F– < Al (2) F– < Al < Na+ < Mg2+

(3) Al < Mg2+ < F– < Na+ (4) Na+ < Al < F– < Mg2+

Q.82 Which of the following oder is wrong -

(1) NH3 < PH3 < AsH3 Acidic character

(2) Li < Be < B < C IE1

(3) Al2O3 < MgO < Na2O < K2O Basic character

(4) Li+ < Na+ < K+ < Cs+ Ionic radius

Q.83 Strongest reducing agent among the following is -

(1) F– (2) Cl– (3) Br– (4) I–

Q.84 Which one of the following represents the electronic configuration of the most electropositive element

(1) [He] 2s1 (2) [Xe] 6s1 (3) [He] 2s2 (4) [Xe] 6s2

Q.85 Alkali metals in each period have

(1) smallest size (2) Lowest ionization potential

(3) Highest ionization potential (4) Highest electronegativity

Q.86 In II period most acidic oxide is formed by ?(1) F (2) N (3) O (4) Li

Q.87 Decreasing order of atomic weight is correct of the elements given below ?(1) Fe > Co > Ni (2) Ni > Co > Fe (3) Co > Ni > Fe (4) Co > Fe > Ni

Q.88 Lanthanide contraction can explain ?(1)Atomic number of the series (2) Number of extra nuclear electrons(3) Density of the series (4) Ionic radius of series



EXERCISE-2

[PREVIOUS YEAR'S QUESTIONS]

Q.1 Hydrogen can be put in halogen group because [RPMT 2000]

(1) It has deuterium and tritium as isotopes (2) It forms hydrides like chlorides

(3) It contains one electron only (4) It is light

Q.2 Correct order of 1st IP among following elements Be, B, C, N,O is [CPMT 2001]

(1) B < Be < C < O < N (2) B < Be < C < N < O

(3) Be < B < C < N < O (4) Be < B < C < O < N

Q.3 Outer electronic configuration of an element is 4s1 3d10. The element is expected to be

[RPMT 2001]

(1)Ametal (2)Anonmetal

(3) Element of 10th group (4) Liquid at 2981°C

Q.4 Electron affinityof oxygen, sulphur and selenium follows the order ` [RPMT 2001]

(1) O > S > Se (2) S > O > Se (3) Se > O > S (4) Se > S > O

Q.5 Element ‘X’ having electronic configuration 1s2 2s2 2p6 3s2 3p3 forms compound with Ca. The com-

pound is [RPMT 2001]

(1) Ca2 X3 (2) Ca3X (3) Ca3X2 (4) CaX

Q.6 Triad of transuranic element is [RPMT 2001]

(1) Th, NP , Pu (2) Bk, Cf, Fm (3) Tm, Nd , Pm (4) Pa, Fm, Md

Q.7 Most covalent halide of aluminium is [RPMT 2001]

(1)AlI3 (2)AlBr3 (3)AlCl3 (4)AlF3

Q.8 Which of the following order is wrong [CBSE 2002]

(1) NH3 < PH3 < AsH3 –Acidic

(2) Li < Be < B < C – IE1

(3) Al2O3 < MgO < Na2O < K2O – Basic

(4) Li+ < Na+ < K+ < Cs+ – Ionic radius

Q.9 General electronic configuration of lanthanides is [CBSE 2002]

(1) (n–2) f 1–14 , (n–1) s2 p6 d0–1, ns2 (2) (n–2) f 10–14 , (n–1)d10–1 , ns2

(3) (n–2) f 0–14 , (n–1) d10 , ns2 (4) (n–2) d0–1, (n–1) f1–14 , ns2

Q.10 Which of the following orders is correct for the first ionisation potential of B, C and N

[MP-PMT 2002]

(1) B > C > N (2) N > C > B (3) N > C < B (4) N < C < B



Q.11 Which of the following does not have valence electron in 3d-subshell [AIIMS 2002]

(1) Fe (III) (2) Cr (I) (3) Mn (II) (4) P (0)

Q.12 Which of the followingelements is most metallic [MP-PMT 2002]

(1) P (2)As (3) Sb (4) Bi

Q.13 The ions O2– , F– , Na+ , Mg2+ andAl3+ are isoelectronic. Their ionic radii show [CBSE,PMT 2003]

(1)Asignificant decrease from O2– toAl3+

(2)An increase from O2– to F– and then decrease from Na+ to Al3+

(3)A decrease from O2– to F– and then increase from Na+ to Al3+

(4)Asignificant increase from O2– toAl3+

Q.14 The ions O2–, F–, Na+, Mg2+ andAl3+ are isoelectronic. Their ionic radii show [CBSEAIPMT 2003](1)An increase from O2– to F– and then decrease from Na+ to Al3+

(2)A decrease from O2– to F– and then increase from Na+ to Al3+

(3)Asingnificant increase from O2– toAl3+

(4)Asignificant decrease from O2– toAl3+.

Q.15 Which of these have no unit ? [AFMC 2004](1)Electronegativity (2)Electronaffinity (3) Ionisation energy (4) Excitation potential

Q.16 The Correct sequence of increasing covalent character is represented by [CBSE PMT 2005](1) LiCl < NaCl K BeCl2 (2) BeCl2 < NaCl < LiCl(3) NaCl < LiCl < BeCl2 (4) BeCl2 < LiCl < NaCl

Q.17 Ionic compounds are formed most easily with [DPMT 2005](1)Low electron affinity, high ionisation energy (2)High electron affinity, low ionisation energy(3) Low electron affinity, low ionisation energy (4)High electron affinity, high ionisation energy

Q.18 Which one of the following arrangements represents the correct order of electron gain enthalpy (withnegative sign) of the given atomic species. [CBSE PMT 2005](1) Cl < F < S < O (2) O < S < F < Cl (3) S < O < Cl < F (4) F < Cl < O < S

Q.19 The pair of amphoteric hydroxides is - [AIIMS 2005](1)Al(OH)3, LiOH (2) Be(OH)2, Mg(OH)2(3) B(OH)3, Be(OH)2 (4) Be(OH)2, Zn(OH)2

Q.20 Variable valency is a general feature of [MP PMT 2006](1) s-block elements (2) p-block elements(3) d-block elements (4)All of these

Q.21 Identify the correct order of the size of the following [CBSE AIPMT 2007](1) Ca2+ < K+ < Ar < S2– < Cl– (2) Ca2+ < K+ < Ar < Cl– < S2–

(3) Ar < Ca2+ < K+ < Cl– < S2– (4) Ca2+ < Ar < K+ < Cl– < S2–

Q.22 The long form of Periodic Table based on [BHU 2007](1)Atomic number (2)Atomic mass(3)Electronic configuration (4) Effective nuclear charge



Q.23 Total number of rare earth elements is [BHU 2008](1) 8 (2) 32 (3) 14 (4) 10

Q.24 Element with atomic number 38, belongs to [MP PMT 2008](1) IIA group and 5th period. (2) IIA group and 2nd period(3) VA group and 2nd period. (4) IIIA group and 5th period.

Q.25 Match List-I with List-II and select the correct answer using the codes given below.

IE1 IE2 IE3 (kJ mol–1

)

1 1312 — — A H

2 520 7297 11810 B Li

3 900 1758 14810 C Be

4 800 2428 3660 D B

List I (Successive IE) List II (Elements)

Codes [Punjab PMET 2008]

(1) A-2, B-1, C-4, D-3 (2) A-3, B-4, C-2, D-1

(3) A-4, B-3, C-1, D-2 (4) A-1, B-2, C-3, D-4

Q.26 Amongst the elements with following electronic configurations which one of them may have the highest

ionisation energy? [CBSE AIPMT 2009]

(1) [Ne]3s23p3 (2) [Ne]3s23p2 (3) [Ar]3d104s24p3 (4) [Ne]3s23p1

Q.27 The correct order of electron affinities of N, O, S and Cl is [AIIMS 2009]

(1) N < O < S < Cl (2) O < N < Cl < S

(3) O Cl < N S (4) O < S < Cl < N

Q.28 Which one of the following ions has electronic configuration [Ar]3d6 ? (At nos. Mn = 25, Fe = 26,Co = 27, Ni = 28) [CBSE AIPMT 2010]

(1) Co3+ (2) Ni3+ (3) Mn3+ (4) Fe3+

Q.29 Electronic configuration of an element 'X' having atomic number 24 is [CPMT, OJEE 2010]

(1) [Ar]3d54s1 (2) [Ar]3d44s2 (3) [Ne]2p53s1 (4) [Ar]3d64s2

Q.30 The electronic configuration of chromium (Z = 24) is [MP PMT 2010]

(1) 2, 8, 14 (2) 2, 8, 8, 6 (3) 2, 8, 12, 2 (4) 2, 8, 13, 1

Q.31 The electronic configuration of neon is [MP PMT 2010]

(1) 1s2, 2s1, 2p7 (2) 1s2, 2s2, 2p4 (3) 1s2, 2s2, 2p6 (4) 1s2, 2s2, 2p8

Q.32 The correct order of the decreasing ionic radii among the following isoelectronic species is

[CBSE AIPMT, BVP2010](1) K+ > Ca2+ > Cl– > S2– (2) Ca2+ > K+ > S2– > Cl–

(3) Cl– > S2– > Ca2+ > K+ (4) S2– > Cl– > K+ > Ca2+



Q.33 Which of the following represents the correct order of increasing electron gain enthalpy with negative

sign for the elements O, S, F and Cl ? [CBSE AIPMT 2010]

(1) S < O < Cl < F (2) Cl < F < O < S (3) O < S < F < Cl (4) F < S < O < Cl

Q.34 Which of the followinghas the highest electron affinity ? [AFMC 2010]

(1) F– (2) O– (3) O (4) Na

Q.35 The electronegativity of the following elements increases in the order [AFMC 2010]

(1) C, N, Si, P (2) N, Si, C, P (3) Si, P, C, N (4) P, Si, N, C

Q.36 Ease of formation of the cation is favoured by [RPMT 2010]

(1) lower value of ionisation potential (2) lower value of electron affinity

(3) highervalue of electron affinity (4) lower value of electronegativity

Q.37 Which of the following has the highest ionisation enthalpy ? [RPMT 2010]

(1) P (2) N (3)As (4) Sb

Q.38 Aqueous solutions of lithium salts are not good conductors of electricity, because of : [RPMT 2011]

(1) High hydration energy of Li ion (2) High ionization energy of Li ion

(3) Small size of Li ion (4) Non-metallic character of Li ion

Q.39 The modern periodic table has been divided into : [RPMT 2011]

(1) Seven periods : three short and four long (2) Nine periods : six short and three long

(3) Eighteen periods : ten short and eight long (4) Nine periods from zero to VIII.

Q.40 Among K, K+, Sr2+,Ar, which atom/ion will have the smallest radius ? [RPMT 2011]

(1) K+ (2) Sr2+ (3)Ar (4) K

Q.41 The melting point of NaCl is high because [RPMT 2011]

(1) The distance between the ions is large. (2) Repulsion exists within NaCl lattice.

(3) Lattice energy is high. (4) Size of sodium as well as chlorine is big

Q.42 What is the value of Electron gain enthalpy of Na+ if IE1 of Na= 5.1 eV? [CBSE 2011]

(1) +2.55 eV (2) +10.2eV (3) – 5.1 eV (4) – 10.2 eV

Q.43 Identify the wrongstatement in the following : [CBSE 2012]

(1)Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.

(2)Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius.

(3)Atomic radius of the elements increases as one moves down the first group of the periodic table.

(4)Atomic radius of the elements decreases as one moves across from left to right in the 2nd period ofthe periodic table.



EXERCISE-3

[ASSERTION & REASON TYPE QUESTIONs]

Each of the questions given below consist of Assertion and Reason. Use the following Key to choose

the appropriate answer.

(A) If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.

(B) If both Assertion and Reason are correct, and Reason is not the correct explanation of Assertion.

(C) If Assertion is correct and Reason is incorrect.

(D) IfAssertion is incorrect but Reason is correct.

(E) If bothAssertion and Reason are incorrect.

Q.1 Assertion : First ionisation energy for nitrogen is lower than oxygen.

Reason : Across a period effective nuclear charge decreases. [AIIMS 2008](1)A (2) B (3) C (4) D (5) E

Q.2 Assertion : The ionic size of Mg2+ > Al3+. [AIIMS 2008]

Reason : In isoelectronic species, greater the nuclear charge, less is the size.(1)A (2) B (3) C (4) D (5) E

Q.3 Assertion : F atom has less electron affinity than Cl atom.

Reason :Additional electrons are repelled more strongly by 3p electrons in Cl atom than by 2pelectrons in F atom.(1)A (2) B (3) C (4) D (5) E

Q.4 Assertion: OO2– is associated with energy absorbed but yet several oxides are stable enough

Reason : The lattice energy involved to form its oxides overcomes the other energy absorbed.(1)A (2) B (3) C (4) D (5) E

Q.5 Assertion :Electron gain enthalpyof oxygen is less than that of fluorine but greater than that of nitrogen.

Reason : Ionization enthalpy is as follows : N > O > F.

(1)A (2) B (3) C (4) D (5) E

Q.6 Assertion : Cs and F combines violently to form CsF.

Reason : Cs is most electropositive and F is most electronegative.(1)A (2) B (3) C (4) D (5) E

Q.7 Assertion : Lithium is a better reducing agent than Cs.

Reason : Sublimation energy and Ionization energy of lithium is less than that of Cs.

(1)A (2) B (3) C (4) D (5) E

Q.8 Assertion : F atom has a less negative electron gain enthalpy than Cl atom.

Reason :Additional electron is repelled more efficiently by 3p electron in Cl atom than by 2p electronin F atom.(1)A (2) B (3) C (4) D (5) E

Q.9 Assertion : Al(OH)3is amphoteric in nature.

Reason : Al –O and O – H bonds can be broken with equal ease in Al(OH)3.

(1)A (2) B (3) C (4) D (5) E



EXERCISE-4

[SUBJECTIVE BOARD QUESTIONS]

Very Short Answer Type Question (1 marks)

Q.1 Defineelectronegativity.

Q.2 Why does the first ionisation energy increase as we go from left to right along a given period of periodictable ?

Q.3 State the modern periodic law.

Q.4 On the basis of quantum numbers, justify that the sixth period of the periodic law and the modernperiodic law ?

Q.5 Write the atomic number of the element present in the third period and seventeenth group of the periodictable.

Q.6 Mg2+, O2–, Na+, F–, N3– (Arrange in decreasing order of ionic size)

Q.7 Why Ca2+ has a smaller ionic radius than K+.

Q.8 Why the first ionisation energy of carbon atom is greater than that of boron atom whereas, the reverse istrue for the second ionisation energy.

Q.9 The IE do not follow a regular trend in II & III periods with increasing atomic number. Why?

Q.10 Compare the following giving reasonsAcidic nature of oxides: CaO, CO, CO2, N2O5, SO3

Short Answer Type Question (2 marks)

Q.11 Why are group 1 elements called alkali metals and group 17 are called halogens ?

Q.12 Give any two features of Mendeleev’s periodic table.

Q.13 What does atomic radius and ionic radius really mean to you ?

Q.14 What is meant by ‘diagonal relationship’ in the periodic table ? What is it due to ?

Q.15 How does electronegativitydiffer from electron affinity?

Q.16 If internuclear distance between Cl atoms in Cl2 is 10 Å & between H atoms in H2 is 2 Å, then calculateinternuclear distance between H & Cl (Electronegativity of H = 2.1 & Cl = 3.0).

Q.17 TheAs-Cl bond distance inAsCl3 is 2.20 Å. Estimate the SBCR (single bond covalent radius) ofAs.

(Assume EN of both to be same and radius of Cl = 0.99 Å.)



Q.18 The ionisation potentials of atoms A and B are 400 and 300 kcal mol–1 respectively. The electron

affinities of these atoms are 80.0 and 85.0 k cal mol–1 respectively. Prove that which of the atoms has

higherelectronegativity.

Short Answer Type Question (3 marks)

Q.19 ElementsAand B have atomic number 11 and 24 respectively. Write their electronic configuration andpredict (i) group, (ii) period, (iii) block to which they belong. Which of them is representative element ?

Q.20 How does electronegativity vary down in the group 17 and why ? How does it vary from left to right inperiod ? Name an element having highest electronegativity.

Q.21 Among the elements of second period Li to Ne, pick out element :

(i) with the highest first ionisation energy (ii)with highest electronegativity

(iii)with largest atomic radius (iv) that is most reactive nonmetal

(v) that is most reactive metal (vi) with valency equal to 4

Q.22 Give the reasons of the following :

First ionisation enthalpy of ‘Mg’ is more than that of ‘Na’ but second ionisation enthalpy of ‘Mg’ is lessthan that of ‘Na’

Q.23 Whyhas magnesium higher first ionisation potential than aluminium atom ?

Q.24 Use the followingsystem of namingelements in which first alphabets of the digits are written 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....]

Q.25 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 element with lowest ionisation potential.(iii) The element whose oxide is amphoteric.(iv) The element which has smallest radii.(v) The element whose atom has 8 electrons in the outermost shell.

Q.26 The IE values of Al(g) = Al+ +e is 577.5 kJ mol–1 and H for Al(g) = Al3+ +3e is 5140 kJ

mol–1. If second and third IE values are in the ratio 2 : 3. Calculate IE2 and IE3.

Q.27 Calculate the electronegativity of fluorine from the following data :

EH – H = 104.2 kcal mol–1 EF – F = 36.2 kcal mol–1

EH – F = 134.6 kcal mol–1 XH = 2.1

Q.28 For the gaseous reaction,

K + FK+ F–,H was calculated to be 19 kcal/mol under conditions where the cations and anions

were prevented by electrostatic separation from combining with each other.The ionisation potential of K

is 4.3 eV / atom. What is the electron affinity of F?



EXERCISE-1

Q.1 3 Q.2 3 Q.3 3 Q.4 4

Q.5 2 Q.6 2 Q.7 2 Q.8 4Q.9 1 Q.10 1 Q.11 4 Q.12 3Q.13 2 Q.14 3 Q.15 2 Q.16 3

Q.17 3 Q.18 4 Q.19 1 Q.20 4

Q.21 3 Q.22 3 Q.23 1 Q.24 1

Q.25 1 Q.26 4 Q.27 1 Q.28 2Q.29 4 Q.30 2 Q.31 3 Q.32 3Q.33 3 Q.34 2 Q.35 4 Q.36 3

Q.37 1 Q.38 3 Q.39 4 Q.40 2

Q.41 3 Q.42 3 Q.43 4 Q.44 2

Q.45 2 Q.46 1 Q.47 1 Q.48 2

Q.49 2 Q.50 1 Q.51 4 Q.52 3

Q.53 2 Q.54 2 Q.55 2 Q.56 3

Q.57 2 Q.58 3 Q.59 1 Q.60 3

Q.61 3 Q.62 1 Q.63 3 Q.64 3

Q.65 1 Q.66 1 Q.67 4 Q.68 3

Q.69 4 Q.70 2 Q.71 4 Q.72 2

Q.73 4 Q.74 4 Q.75 2 Q.76 1

Q.77 1 Q.78 1 Q.79 1 Q.80 3

Q.81 1 Q.82 2 Q.83 4 Q.84 2

Q.85 2 Q.86 2 Q.87 3 Q.88 4

EXERCISE-2

Q.1 2 Q.2 1 Q.3 1 Q.4 2

Q.5 3 Q.6 2 Q.7 1 Q.8 2

Q.9 1 Q.10 2 Q.11 4 Q.12 4

Q.13 2 Q.14 4 Q.15 1 Q.16 3Q.17 2 Q.18 2 Q.19 4 Q.20 3Q.21 2 Q.22 3 Q.23 3 Q.24 1Q.25 4 Q.26 1 Q.27 1 Q.28 1Q.29 1 Q.30 4 Q.31 3 Q.32 4

Q.33 3 Q.34 3 Q.35 3 Q.36 1

Q.37 2 Q.38 1 Q.39 1 Q.40 2

Q.41 3 Q.42 3 Q.43 1



EXERCISE-3

Q.1 5 Q.2 1 Q.3 3 Q.4 1

Q.5 3 Q.6 1 Q.7 3 Q.8 3

Q.9 3

EXERCISE-4

Q.1 It is the measure of tendency of a bonded atom to attract a shared pair of electron towards it self.

Q.2 Left to right Zeff increases in a period so I.E. increases.

Q.3 Chemical and physical properties of elements are periodic function of their atomic number.

Q.4 For 6th period type of subshells : 6s, 5d, 4f, 6p

Total electrons = 32

so elements = 32

Q.5 Cl - 35

Q.6 N3– > O2– > F– > Na+ > Mg2+

Q.7 Isoelectronic Ca+2(Value of Zeff is higher)

Q.8 Zeff&half filledconfig.

Q.9 half filled&fullyfilledorbitals

Q.10 CaO < CO < CO2 < N2O5 < SO3

Q.11 Alkali metal due to basic nature (NaOH)

Halogen due to salt formation (NaCl)

Q.12 (i) Same property elements are placed in same group (Li, Na, K, Rb, Cs)

(ii) Place in periodic table are left for elements which are not discovered yet (Ga, Se, Ge, Tc)

Q.13 Atomic radius : Radius of neutral atom

Ionic radius : Radius of cation /Anion

Q.14 (Li, Mg), (Be,Al), (B, Si) have almost same property, they are diagonally present in periodic table.

Q.15 E.A. is energy release to gain an electron while EN is tendency to attract shared pair of electron.

Q.16 5.919 Å

Q.17 1.21 Å

Q.18 EN1 > EN2

Q.19 11 1s2 2s2 2p6 3s1 24 1s2 2s2 2p6 3s2 3p6 3d5 4s1

Group = Ist Group = 6th

Period = 3rd Period = 4th

Block = s Block = d

Atomic number 11 is representative element

Q.20 Top to bottom in a group size increases, so electronegativity decreases. From left to right in a period sizedecreases, Zeff increases so EN increases.



Q.21 (i) Ne; (ii) F; (iii) Ne; (iv) F; (v) Li ; (vi) C

Q.22 Na, Mg

Ist IE [Ne] 3s1 < [Ne] 3s2

Due to penetration power

IInd IE [Ne] < [Ne] 3s1

Due to noble gas configuration.

Q.23 Due to penetration power

Q.24 101 102 103 104 105 106 107 108 109Unu Unb Unt Unq Unp Unh Uns Uno Une

Q.25 (i) Cl (ii) Cs (iii)Al (iv) F (v) Xe

Q.26 IE2 = 1825 kJ/mole, IE3 = 2737.5 kJ/mol

Q.27 3.8752

Q.28 3.476 eV



EXERCISE-1

1. According to triad rule at wt. of Y =2

ZX

26 =2

Z10

Z = 42

2. According toAufbau '4s' is lower in energy then 3d

7. If n = 7 then it will be 5f or actinide series.

10. N3– , O2–, F¯

e

z=

10

7

10

8

10

9

11. 9th period will have total orbital

=

2

2

1n

=

2

2

19

= 25

so element = 25 × 3 = 75

12. Oribtals in 5th period

=

2

2

15

= 9

so elements = 9 × 2 = 18

orbitals in 6th period =

2

2

26

= 16

elements = 16 × 2 = 32

13. At n. 15 = 2, 8, 533 = 2, 8, 18, 551 = 2, 8, 18, 18, 5

So all belongs to nitrogen family as outermost configuration is ns2 np3 or belong to 15th group.

14. n represent period & total electrons in outermost shell represent group.

16. Un – 1

Un – 1

Hex – 6 so at no. = 116



18. In inner transition elements last electron goes into f orbital.

19. F is most electronegative.

20. Except Li–Na, all have diagonal relationship.

21. Ionic radii order N3– > O2– > F¯

22. Radius of ion stateoxidation

1

23. Due to lanthamide contraction size of 4d5d elements.

24. In noble gas vander waals radii is calculated which is normally then double of other type of radii.

25. As we go from left to right in a period radius decreases.

26. Down the group radius increases.

27. Down the group radius increases.

28. Among isoelectronic ions lesser ise

z, more is radius.

O2– Mg2+ Al+3

e

z=

10

8

10

12

10

13

30. Lower is ionisation energy more easily cation is formed.

31. Second 1E of O > F

32. Al+ is more stable as second electron is removed from fully filled orbital.

33. 1E2 (Na) > 1E2 (Mg)because second electron is Na is removed from noble gas configuration.

34. Mg Mg++ e¯ 1E1 = 178Mg+ Mg2++ e¯ 1E2 = 348

-----------------------------------------------Mg Mg2++ e¯ so 1E = 1E1 + 1E2

= 178 + 348 = 526

35. Zeff increases from left to right in the period.

37. EA of third period element is greater than second period in the group.



38. Since N is having half filled stability of orbital.

39. 1P = – EA

40. EA increases from left to right in period & nitrogen family is having endothermic EA.

41. EA of chlorine is exothermic.

42. In half filled orbital it is difficult to give electron.

43. Due to strong repulsion second EA is endothermic.

46. EN% s character.

47. EA of F is less than Cl.

49. 1E increases from left to right along the period.

50. Due to compact size EA of F is less than Cl.

51. From data it seems that it can loose two electrons easily.

52. 1E increases from left to right in the period.

54. All belong to s block & 1E decreases down the group.

55. Since after loosing one e¯ next e¯ in alkali metal goes from noble gas core.

56. In isoelectronic ife

zis high, size is less.

Cl¯ K+

e

z=

18

17

18

19

57. From data it can loose 4 electron easily.

58. Due to half filled & full filled stability of orbital 1E2 order is

O > F > N > C

2p3 2p4 2p2 2p1

59. It can loose 2 electron easily

60. 1E2 is 150 eV



61. 1E increases along the period.

62. It is difficult to remove e¯ from positive ion.

65. Bl HCl = rH + rCl – .09 |XH– XCl|

= 5 + 1 – .09 (0.9) = 5.919 Å

66. Cl has highest E.A.

67. C – X = rC + rX – .09 |XC – XA|

=2

2.1

2

54.1 – .09 (1)

1.28 Å

68. EN decreases down the group.

69. EN increases along the period.

70. Acidic nature increases along the period from left to right.

71. 1Eof anion < 1 E of atom.

72. 1E2 will be > 1E1 < 1E3.

74. Acidic nature of hydroxide decreases down the group.

75. Acid nature of hydride increases in period.

76. Alkali metals are basic.

77. Basic nature of oxide decreases along the period.

78. C22– , NO+, CN¯, O2

2+

14e¯ 14e¯ 14e¯ 14e¯

79. H = 1E1 + 1E2 + 1E2

5140 = 577.5 + 2X + 3X

5X = 4562.5

X = 912.5

So 1E2 = 2 × 912.5 = 1825 kJ/mol

1E3 = 3 × 912.5 = 2737.5 kJ/mol



80. Element no. of unpaired electron

Mg O

Al 1

O 2

N 3

81. In isoelectronic anion is bigger in size.

82. 1E1 of Be > B because of fully filled stability of orbital.

83. I¯ can be easily oxidised.

84. Alkali metals are most electropositive & among alkali metals Cs is most.

86. Nitrogen in + 3 & + 5 state forms acidic N2O3 & N2O5.

EXERCISE-2

2. 1P increases along period.

3. It belongs to d block.

5. Electrovalency of X = 3 & Ca is 2.

7. According to Fajan's rule bigger is size of anion more is polarisation and more covalent character.

8. 1E1 of Be > 1E1 of B

10. 1E increases along period.

12. Metallic character increases down the group.

13. More ise

z, lesser is size.

O2– F¯ Na+ Mg2+ Al3+

e

z= 10

8

10

9

10

11

10

12

10

13

16. According to Fajan's rule more is charge/size ratio of cation more is covalency.

21. Isoelectronic & more ise

zlesser is size.

24. Atomic no. 38 = 2, 8, 8, 18, 2so fifth period & second group.



25. 1E1of H > Li > B > Be

26. 1E increases along the period and decreases down the group.

28. Co = [Ar] 3d7 4s2

so Co+3 = [Ar] 3d6

34. Second EA is endothermic.

35. EN increases along the period and decreases down the group.

37. 1E decreases down the group.

38. Due to high hydration of Li+ ion its mobility decreases.

40. Size increases down the group.

41. m.pt. depend upon lattice energy.

42. EA1 = – 1E1 = – 5.1 eV

43. More is charge on cation, lesser is size of cation.

EXERCISE-3

1. 1E1 of N > 1E1 of O

3. Due to compact size of F.

4. Although formation of O¯ is endothermic but lattice energy is high as LE¯r

¯z

5. EA increases along the period. Whereas 1E1 of F > N > O.

6. High lattice energy is released.

7. Due to high hydration energy Li+ ion formed.

8. Due to compact size of F electron is repelled more in F.

EXERCISE-4

6. In a isoelectronic species, if the positive charge increases atomic size decreases. If the negative chargeincreases, atomic size increase.

7.e

z atomic size



8. C = 1s2 2s2 2p2

B = 1s2 2s2 2p1

Left to right atomic size decreases; I.E1After the removal of 1e¯, B acquire fullyfilled stable s-orbital configuration ; therefore IE2 of B > C.

10. CONeutral oxideCaOBasic oxideAcidic nature

16. rCl =2

10= 5 Å

rH =2

2= 1 Å

rH–Cl = rH + rCl – .09 (3–2.1)= 5 + 1 – .081 = 5.919 Å

17. dAs-Cl = rAs + rCl

2.20 = rAs + 0.99

rAs = 1.21 Å

18. E.N of A =x

80400

E.N of B =x

85300; E. N. of A > B

26. Al(g) = Al+ + e¯; I.E1 = 577.5

I.E1 + IE2 + IE3 = 5140

3

2

IE

IE

3

2 ; I.E2 =3

2; IE2 =

3

2IE3

27. xF – 2.1 = 0.208 2.362.1046.134

xF = 3.87

28. 1 eV = 23 Kcal/mol

H = IE + EA; E.A = 3476eV

Bansal Quick Review Table

Instruction to fill(A) Write down the Question Number you are unable to solve in columnAbelow, by Pen.(B) After discussing the Questions written in columnAwith faculties, strikes off them in the manner

so that you can see at the time of Revision also, to solve these questions again.(C) Write down the Question Number you feel are important or good in the column B.

COLUMN : A COLUMN : B

Exercise I

Exercise II

Exercise III

EXERCISE NO. Question I am unable tosolve in first attempt

Good / Importantquestions

Exercise IV

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Shiksha Test Series · 2019. 12. 24. · BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road...

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