Periodic Table & Periodicity - Contents Syllabus

Topic Page No.

Theory 01 - 06

Exercise - 1 07 - 14




Answer Key 28 - 29

Contents

Periodic Table & Periodicity

SyllabusPeriodic Table & Periodicity

Modern Periodic law and present form of the periodic table, s, p, d and f blockelements, periodic trends in properties of elements atomic and ionic radii,ionization enthalpy, electron gain enthalpy, valence, oxidation states andchemical reactivity.

Name : ____________________________ Contact No. __________________

PERIODIC TABLE & PERIODICITYADVANCED # 1

A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)

Physical & Inorganic By

NV SirB.Tech. IIT Delhi

Organic Chemistry By

VKP SirM.Sc. IT-BHU

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PERIODIC TABLE & PERIODICITY

DOBEREINER TRIAD RULE [1817]He made groups of three elements having similar chemical properties called TRIAD.

NEWLAND OCTAVE RULE [1865]He arranged the elements in the increasing order of their atomic mass and observe that properties of every8th element was similar to the 1st one. like in the case of musical vowels notation.

8Sa

765432NeDhaPaMaGaRe

1Sa

LOTHER MEYER’S CURVE [1869]

He plotted a curve between atomic wt. and atomic volume of different elements.

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.

MENDELEEV’S PERIODIC TABLE [1869]Mendeleev’s Periodic’s Law

According to him the physical and chemical properties of the elements are the periodic functions of theiratomic masses.This table was divided into nine vertical columns called groups and seven horizontal rows called periods.The groups were numbered as I, II, III, IV, V, VI, VII, VIII and Zero group

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 andobtained X-rays.

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

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

LONG FORM/PRESENT FORM OF MODERN PERIODIC TABLEIt consist of 7 horizontal periods and 18 vertical columns (groups)

IA, IIIA, 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






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S–Block Elements

1IA

18VIII A

1H

1.007

2II A

13III A

14IV A

15V A

16VI A

17VII A

2He

4.0023Li

6.941

4Be

9.012

5B

10.811

6C

12.011

7N

14.006

8O

15.999

9F

18.998

10Ne

20.17911Na

22.98

12Mg

24.30

3III B

4IV B

5V B

6VI B

7VII B

8VIII

9VIII

10VIII

11I B

12II B

13Al

26.981

14Si

28.085

15P

30.973

16S

32.006

17Cl

35.452

18Ar

39.94819K

39.08

20Ca

40.078

21Sc

44.959

22Ti

47.88

23V

50.9415

24Cr

51.996

25Mn

54.938

26Fe

55.84

27Co

55.933

28Ni

58.693

29Cu

63.546

30Zn

65.39

31Ga

69.723

32Ge

72.61

33As

74.921

34Se

78.96

35Br

79.904

36Kr

83.8037Rb

85.46

38Sr

87.62

39Y

88.905

40Zr

91.224

41Nb

92.906

42Mo

95.94

43Tc98

44Ru

101.07

45Rh

102.905

46Pd

106.42

47Ag

107.868

48Cd

112.411

49In

114.82

50Sn

118.710

51Sb

121.757

52Te

127.60

53I

126.904

54Xe

132.2955Cs

132.90

56Ba

137.27

57La*

138.905

72Hf

178.49

73Ta

180.947

74W

183.85

75Re

186.207

76Os

190.2

77Ir

192.22

78Pt

195.08

79Au

196.666

80Hg

200.59

81Tl

204.383

82Pb

207.2

83Bi

207.980

84Po209

85At

210

86Rn222

87Fr

223

88Ra226

89Ac**227

104Rf

261.11

105Ha

262.114

106Sg

263.118

107Bh

262.12

108Hs265

109Mt266

110Uun269

114Uuq

58Ce

140.115

59Pr

140.907

60Nd

144.24

61Pm145

62Sm

150.36

63Eu

151.965

64Gd

157.25

65Tb

158.925

66Dy

162.50

67Ho

164.930

68Er

167.26

69Tm

168.934

70Yb

173.04

71Lu

174.96790Th

232.038

91Pa231

92U

238.028

93Np237

94Pu244

95Am243

96Cm247

97Bk247

98Cf

251

99Es252

100Fm257

101Md258

102No259

103Lr

260

d –Block Elements

p–Block Elements

Inner - Transition Metals (f-Block elements)

*Lanthanides

**Actinides

Classification of the Elements :s-block elements : When shells upto (n – 1) are completely filled and the last electron enters the s-orbital of theoutermost (nth) shell, the elements of this class are called s-block elements.

p-block elements : When shells upto (n – 1) are completely filled and differentiating electron enters the p-orbital of the nth orbit, elements of this class are called p-block elements.

d-Block elements : When outermost (nth) and penultimate shells (n – 1)th shells are incompletely filled anddifferentiating electron enters the (n – 1) d orbitals (i.e., d-orbital of penultimate shell) then elements of this classare called d-block elements.

(1) Ist transition series i.e. 3d series contains 10 elements and starts from 21Sc – 30Zn. Filling of electronstakes place in 3d sub-shell.

(2) IInd transition series i.e. 4d series contains 10 elements and starts from 39Y – 48Cd. Filling of electronstakes place in 4d sub-shell.

(3) IIIrd transition series i.e. 5d series contains 10 elements and starts from 57La, 72Hf – 80Hg. Filling ofelectrons takes place in 5d sub-shell.

(4) IVth transition series i.e. 6d series contains 10 elements and starts from 89Ac, 104Rf – 112Uub. Filling ofelectrons takes place in 6d sub-shell (incomplete series).

f-Block elements : When n, (n – 1) and (n – 2) shells are incompletely filled and last electron enters into f-orbital of antepenultimate i.e., (n – 2)th shell, elements of this class are called f-block elements. Generalelectronic configuration is (n – 2) f1-14 (n – 1) d0-1 ns2

The elements of f-blocks have been classified into two series.

1. st inner transition or 4 f-series, contains 14 elements 58Ce to 71Lu. Filling of electrons takes place in4f subshell.

2. IInd inner transition or 5 f-series, contains 14 elements 90Th to 103Lr. Filling of electrons takes placein 5f subshell.






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PERIODIC PROPERTIES :VALENCY : It is defined as the combining capacity of the elements. The word valency is derived from anItalian word “Valentia” which menas combining capacity.

DENSITY :(a) In a group volume of an atom increases along with atomic weight but atomic weight increases more than

atomic volue, So density increases in a group.

Density (D) = VolumeMass

= VM

(b) In period - Density first increases till maximum and then decreases. (s-block to d-block increases,d-block to p-block decreases)

(c) In group - From top to bottom in a group density increases regularlye.g. In VIIA gp - F and Cl are gases (Low density)

Br is liquid (density 3.19 gm/cm–3)I is solid (density 4.94 gm/cm–3)

Effective nuclear charge : Between the outer most valence electrons and the nucleus of an atom, there existsnumber of shells containing electrons. Due to the presence of these intervening electrons, the valenceelectrons are unable to experience the attractive pull of the actual number of protons in the nucleus. Theseintervening electrons act as shield between the valence electrons and protons in the nucleus. Thus, thepresence of intervening (shielding) electrons reduces the electrostatic attraction between the protons in thenucleus and the valence electrons because intervening electrons repel the valence electrons. The concept ofeffective nuclear charge helps in understanding the effects of shielding on periodic properties.

The effective nuclear charge (Zeff) is the charge actually felt by the valence electron. Zeff is given byZeff = Z – , (where Z is the actual nuclear charge (atomic number of the element) and is the shielding(screening) constant). The value of i.e. shielding effect can be determined using the Slater’s rules.

Atomic radius :Covalent radius : It is one-half of the distance between the centres of two nuclei (of like atoms) bonded bya single covalent bond as shown in figure.

(a) For homodiatomic molecules dA–A = rA + rA or 2rA so, rA = A Ad2

(b) For heterodiatomic molecules in which electronegativity remains approximately same.dA – B = rA + rB

For heteronuclear diatomic molecule, A–B, where difference between the electronegativity values of atom Aand atom B is relatively larger, dA – B = rA + rB – 9.0 This formula was given by Stevenson & Schomaker. Electronegativity values are given in Pauling units andradius in picometers. = XA – XB where XA and XB are electronegativity values of high electronegative element A and lesselectronegative element B.

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Vander Waal’s radius (Collision radius) : It is one-half of the internuclear distance between two adjacentatoms in two nearest neighbouring molecules of the substance in solid state as shown in figure.

Vander Waal’s radius does not apply to metal and its magnitude depends upon the packing of theatoms when the element is in the solid state.

Metallic radius (Crystal radius) : It is one-half of the distance between the nuclei of two adjacent metalatoms in the metallic crystal lattice as shown in figure.

Thus, the covalent, vander Wall’s and metallic radius magnitude wise follows the order,rcovalent < rcrystal < rvander Walls

IONISATION POTENTIAL OR IONISATION ENERGY OR IONISATION ENTHALPY :

(i) Minimum energy required to remove most loosly held outer most shell e– in ground state from an isolatedgaseous atom is known as Ionisation Potential.

(ii) Successive I.E.(a) For an atom M, successive ionisation energies are as follows -

M + E1 M+ + e– E1 = IP1M+ + E2 M+2 + e– E2 = IP2M+2 + E3 M+3 + e– E3 = IP3

IP3 > IP2 > IP1(b) e– can not 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)

In a period : In a period atomic size decreases and zeff increases so removal of electron become difficultand IE increases.

.increasesIEincreaseszeff

,decreasessizeatomicNeFONCBBeLi

IE = Ne > F > N > O > C > Be > B > Li

ELECTRON AFFINITY / ELECTRON GAIN ENTHALPYThe amount of energy released or absored when electron is added to the valence shell of an isolated gaseousatom.X + e– X– + E.A.known as Electron affinity.

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Mostly energy is released in the process of first E.A.X + e– X– + EA1

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

In period : Electron affinity increases along the period due to increase in Zeff and decrease in atomic size.

zeroLessZeroArNe

ClSFO

PN

SiAlCB

MgBe

NaLi

F > O > C > B > N > Be > NeCl > S > Si > Al > P > Mg > Ar

In Group : Electron affinity of 3rd period element is greater than electron affinity of 2nd period elements of therespective group.

F 2s22p5

Cl 3s23p5

Due to small size of fluorine, electron density around the nucleus increses. The incoming electronsuffers more repulsion. In case of chlorine electron density decreases due to large size, decreasing order ofelectron affinity of halogen. Cl > F > Br > I

S > O > P > NSi > C > P > N

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

Difference between EN and EAElectronegativity Electron Affinity

– Tendency of an atom in a molecule – Energy released when an electron is added toto attract the bonded electrons neutral isolated gaseous atom

– Relative value of an atom – Absolute value of an atom

– It regularly changed in a period – It does not changes regularlyor group

– It has no unit – It is measured in eV/atom or KJ mol–1

or K.cal mole–1

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 as

electronegativity is for bonded atoms.(iii) A polar covalent or ionic bond of A – 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

H2.1

Li Be B C N O F1.0 1.5 2.0 2.5 3.0 3.5 4.0Na Mg Al Si P S Cl0.9 1.2 1.5 1.8 2.1 2.5 3.0K Br

8.0Rb

8.0

5.2I

8.2

7.0Fr

7.0Cs

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

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PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) : Atomic and Ionic RadiusA-1. The correct order of atomic size of C, N, P, S follows the order -

(A) N < C < S < P (B) N < C < P < S (C) C < N < S < P (D) C < N < P < S

A-2. Match list-I with list-II and select the correct answer using the codes given below - List -I List-II Ion Radius (in pm)(I) Li+ (a) 216(II) Na+ (b) 195(III) Br– (c) 60(IV) I– (d) 95Codes :

I II III IV(A) a b d c(B) b c a d(C) c d b a(D) d c b a

A-3. The ionic radii of N3–, O2– and F– are respectively given by -(A) 1.36, 1.40, 1.71 (B) 1.36, 1.71, 1.40 (C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40

A-4. The screening effect of d-electrons is -(A) Equal to the p-electrons (B) Much more than p-electrons(C) Same as f-electrons (D) Less than p-electrons

A-5. In which of the following compounds manganese shows maximum radius -(A) MnO2 (B) KMnO4 (C) MnO (D) K3[Mn(CN)6]

A-6. Arrange in the increasing order of atomic radii of the following elements O, C, F, Cl, Br -(A) F < O < C < Cl Ni > Pd (D) Pd > Pt > Ni

A-8. Which group of atoms have nearly same atomic radius -(A) Na, K, Rb, Cs (B) Li, Be, B, C (C) Fe, Co, Ni (D) F, Cl, Br, I

A-9. In the ions P3–, S2– and Cl– the increasing order of size is -(A) Cl– < S2– < P3– (B) P3– < S2– < Cl– (C) S2– < Cl– < P3– (D) S2– < P3– < Cl–

A-10. Atomic radii of Fluorine and Neon in Angstrom units are given by -(A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) None of these

A-11. Which of the following has largest radius -(A) 1s2, 2s2, 2p6, 3s2 (B) 1s2, 2s2, 2p6, 3s2, 3p1

(C) 1s2, 2s2, 2p6, 3s2, 3p3 (D) 1s2, 2s2, 2p6, 3s2, 3p5

A-12. Arrange the elements in increasing order of atomic radius Na, Rb, K, Mg -(A) Na < K < Mg < Rb (B) K < Na < Mg < Rb(C) Mg < Na < K < Rb (D) Rb < K < Mg < Na

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A-13. Arrange the following in order of increasing atomic radii Na, Si, Al, Ar -(A) Na < Si < Al < Ar (B) Si < Al < Na < Ar(C) Ar < Al < Si < Na (D) Na < Al < Si < Ar

A-14. Consider the isoelectronic series :K+, S2–, Cl– and Ca2+, the radii of the ions decrease as -(A) Ca2+ > K+ > Cl– > S2– (B) Cl– > S2– > K+ > Ca2+

(C) S2– > Cl– > K+ > Ca2+ (D) K+ > Ca2+ > S2– > Cl–

A-15. Which of the following is not isoelectronic series-(A) Cl–, P3–, Ar (B) N3–, Ne, Mg+2 (C) B+3, He, Li+ (D) N3–, S2–, Cl–

A-16. In the isoelectronic species the ionic radii (Å) of N3–, Ne and Al+3 are respectively given by -(A) 1.36, 1.40, 1.71 (B) 1.36, 1.71, 1.40 (C) 1.71, 1.40, 1.36 (D) 1.71, 1.36, 1.40

A-17. The size of the following species increases in the order :(A) Mg2+ < Na+ < F– (B) F– < Na+ < Mg2+ (C) Mg < F– < Na+ (D) Na+ < F– < Mg2+

A-18. Highest size will be of-(A) Br– (B) I (C) I– (D) I+

A-19. Element Hg has two oxidation staters Hg+1 & Hg+2, the right order of radii of these ions.(A) Hg+1 > Hg+2 (B) Hg+2 > Hg+1 (C) Hg+1 = Hg+2 (D) None of these

A-20. The correct order of increasing atomic size of element N,F, Si & P.(A) N < F < Si N < P < Si (C) F < N < P < Si (D) F < N < Si < P

A-21. The correct order of atomic / ionic size(A) N < Li < B (B) Cl < Mg < Ca (C) Ca+2 < S–2 < Cl¯ (D) Na+ < Mg+2 < Cl¯

Section (B) : Ionization Energy or Potential

B-1. Correct orders of Ist I.P. are -(i) Li < B < Be < C(ii) O < N < F(iii) Be < N < Ne(A) (i), (ii) (B) (ii), (iii) (C) (i), (iii) (D) (i), (ii), (iii)

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

(C) 1s2, 2s22p6, 3s23p2 (D) 1s2, 2s22p6, 3s23p3

B-3. The second ionisation potentials in electron volts of oxygen and fluorine atoms are respectively givenby :(A) 35.1., 38.3 (B) 38.3, 38.3 (C) 38.3, 35.1 (D) 35.1, 35.1

B-4. A sudden large jump between the values of 2nd and 3rd IP of an element would be associated withthe electronic configuration-(A) 1s2, 2s2 2p6, 3s1 (B) 1s2, 2s2 2p6, 3s2 3p5

(C) 1s2, 2s2 2p6, 3s2 3p2 (D) 1s2, 2s2 2p6 3s2

B-5. The ionization energy of sodium is 495 kJ mol–1. How much energy is needed to convert atoms persentin 2.3 mg of sodium into sodium ions -(A) 4.95 J (B) 49.5 J (C) 495 J (D) 0.495 J

B-6. Ionisation energy increases in the order -(A) Be < B < C < N (B) B < Be < C < N (C) C < N < Be < B (D) N < C < Be < B

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B-7. IP1 and IP2 of Mg are 178 and 348 K.cal mol–1. The enthalpy required for the reqctionMg Mg2+ + 2e– is -(A) + 170 K.cal (B) + 526 K.cal (C) – 170 K.cal (D) – 526 K.cal

B-8. Highest ionisation potential in a period is shown by -(A) Alkali metals (B) Noble gases (C) Halogens (D) Representative elements

B-9. In which of the following electronic configuration ionisation energy will be maximum(A) [Ne] 3s2 3p1 (B) [Ne] 3s2 3p2 (C) [Ne] 3s2 3p3 (D) [Ar] 3d10 4s2 4p3

B-10. The ionization energy will be maximum for the process.(A) Ba Ba++ (B) Be Be++ (C) Cs Cs+ (D) Li Li+

B-11. The correct order of second I.P.(A) Na < Mg > Al < Si (B) Na > Mg < Al > Si (C) Na > Mg > Al < Si (D) Na > Mg > Al > Si

B-12. Amongst the following, the incorrect statement is(A) IE1 (Al) < IE1 (Mg) (B) IE1 (Na) < IE1(Mg) (C) IE2 (Mg) > IE2 (Na) (D) IE3 (Mg) > IE3 (Al)

Section (C) : Electron affinity or Electron Gain Enthalpy

C-1. In which case the energy released is minimum -(A) Cl Cl– (B) P P– (C) N N– (D) C C–

C-2. In the formation of a chloride ion, from an isolated gaseous chlorine atom, 3.8 eV energy is released,which would be equal to -(A) Electron affinity of Cl– (B) Ionisation potential of Cl(C) Electronegativity of Cl (D) Ionisation potential of Cl–

C-3. The correct order of electron affinity is -(A) Be < B < C < N (B) Be < N < B < C (C) N < Be < C < B (D) N < C < B < Be

C-4. Electron addition would be easier in -(A) O (B) O+ (C) O– (D) O+2

C-5. In the process Cl(g) + e– H Cl–(g), H is -(A) Positive (B) Negative (C) Zero (D) None

C-6. O(g) + 2e– 2)g(O Heg = 744.7 KJ/mole. The positive value of Heg is due to -

(A) Energy is released to add to 1 e– to O–1 (B) Energy is required to add to 1 e– to O–1

(C) Energy is needed to add on 1 e– to O (D) None of the above is correct

C-7. Which of the following process energy is liberated -(A) Cl Cl+ + e– (B) HCl H+ + Cl– (C) Cl + e– Cl– (D) O– + e– O–2

C-8. Second electron affinity of an element is -(A) Always exothermic (B) Endothermic for few elements(C) Exothermic for few elements (D) Always endothermic

C-9. The element having very high ionization enthalpy but zero electron gain enthalpy is-(A) H (B) F (C) He (D) Be

C-10. The electron affinity values for the halogens shown the following trend -(A) F < Cl > Br > I (B) F < Cl Cl > Br > I (D) F < Cl > Br < I

C-11. The process requiring the absorption of energy is -(A) F F– (B) Cl Cl– (C) O O2– (D) H H–

C-12. Which of the following configuration will have least electron affinity -(A) ns2np5 (B) ns2np2 (C) ns2np3 (D) ns2np4

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C-13. The electron affinity of the members of oxygen family of the periodic table, follows the sequence(A) O > S > Se (B) S > O < Se (C) O < S > Se (D) Se > O > S

C-14. Of the following elements, which possesses the highest electron affinity?(A) As (B) O (C) S (D) Se

C-15. Electron affinities of O,F,S and Cl are in the order.(A) O < S < Cl < F (B) O < S < F < Cl (C) S < O < Cl < F (D) S < O < F < Cl

C-16. Which of the following statement is not true?(A) F atom can hold additional electron more tightly than Cl atom(B) Cl atom can hold additional electron more tightly than F atom(C) The incoming electron encounters greater repulsion for F atom than for Cl atom(D) It is easier to remove an electron from F¯ than Cl¯

C-17. Increasing order of 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

(A) a < d < b < c (B) d < a < c < b (C) a < b < c < d (D) a < b < d < c

C-18. Highest electron affinity is shown by(A) F¯ (B) Cl¯ (C) Li+ (D) Na+

Section (D) : ElectronegativityD-1. In the following which configuration of element has maximum electronegativity.

(A) 1s2, 2s2 2p5 (B) 1s2, 2s2 2p6 (C) 1s2, 2s2 2p4 (D) 1s2, 2s2 2p6, 3s2 3p3

D-2. On the Pauling’s electronegativity scale, which element is next to F .(A) Cl (B) O (C) Br (D) Ne

D-3. Bond distance C–F in (CF4) & Si–F in (SiF4) are respective 1.33Å & 1.54 Å. C–Si bond is 1.87 Å. Calculationthe covalent radius of F atom ignoring the electronegativity differences.

(A) 0.64 Å (B) 1.33 1.54 1.8

3

Å (C) 0.5 Å (D) 1.54

2Å

D-4. Two elements A & B are such that B. E. of A–A, B–B & A–B are respectively 81 Kcal / mole,64 Kcal / mole, 76 Kcal / mole & if electronegativity of B is 2.4 then the electronegativity of 'A' may beapproximately(A) 2.81 (B) 1.8 (C) 1.99 (D) 3.0

D-5. The lowest electronegativity of the element from the following atomic number is.(A) 37 (B) 55 (C) 9 (D) 35

D-6. Which one is not correct order of electronegativity.(A) F > Cl > Br > I (B) Si > Al > Mg > Na (C) Cl > S > P > Si (D) None of these

D-7. 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 Å andelectronegativities of C and X are 2.0 and 3.0 respectively.(A) 2.74 Å (B) 1.37 Å (C) 1.46 Å (D) 1.28 Å

D-8. Outermost electronic configuration of the most electronegative element is -(A) ns2np3 (B) ns2np6 (C) ns2 (D) ns2np5

D-9. Electronegativity of the following elements increases in the order -(A) O < N < S < P (B) P < S < N < O (C) P < N < S < O (D) S < P < N < O

D-10. Which one of the following is incorrect ?(A) An element which has high electronegativity always has high electron gain enthalpy(B) Electron gain enthalpy is the property of an isolated atom(C) Electronegativity is the property of a bonded atom(D) Both electronegativity and electron gain enthalpy are usually directly related to nuclear charge andinversely related to atomic size






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PART - II : MISCELLANEOUS QUESTIONS

Comprehensions Type

Comprehension # 1Read the following passage carefully and answer the questions.The minimum amount of energy required to remove the most loosely bound electron from an isolatedatom in the gaseous state is known as ionisation energy or first ionisation energy or ionisationenthalpy (IE1) of the element. The energy required to remove the second electron from the monvalentcation is called second ionisation enthalpy (IE2). Similarly, we have third, fourth ..... ionistion enthalpies.The values of ionisation energy depends on a number of factors such as (i) size of the atom (ii)screening effect (iii) nuclear charge (iv) half filled and fully filled orbitals.In a group, the ionisation energy decreases from top to bottom. In a period, the value of ionisationenergy increases from left to right with breaks where atoms have somewhat stable configurations.

1. Compared to the second ionisation energy (IE2) of an atom, the third ionisation energy (IE3) is -(A) The same (B) Greater (C) Smaller (D) Half

2. In a period, the ionisation energy is lowest for the -(A) Noble gases (B) Halogens(C) Alkaline earth metals (D) Alkali metals

3. The electronic configurations of some neutral elements are given below -(a) 1s2, 2s2 (b) 1s2, 2s22p1

(c) 1s2, 2s1 (d) 1s2, 2s22p3

Which of these electronic confugarations would be expected to have highest second ionisation energy(IE2)(A) (a) (B) (b) (C) (c) (D) (d)

4. The first (IE1) and second (IE2) ionisation energies (kJ mol–1) of a few elements are shown below -IE1 IE2

(a) 2372 5251(b) 520 7300(c) 900 1760(d) 1680 3380Which of the above elements is likely to be a noble gas -(A) (a) (B) (b) (C) (c) (D) (d)

Comprehension # 2

Pauling gave method to calculate univalent ion radii by assuming that

(i) In ionic crystal (let M+X¯ ) cations and anions are is contact of each other and sum of their radii is equal tointerionic distance, i.e.

¯)X—M(d = XM

rr

(ii) The radius of an ion having noble gas configuration is inversely proportional to the effective nuclear charge

felt at the periphery of the ion, i.e. )M(r =

)M(.effZC

and ¯)X(r =

¯)X(.effZC

Here C is constant of proportionality whose value depends on electronic configuration of ion. Thus,

¯)XM(d

= )M(.effZ

C

+ ¯)X(.effZ

C pm

Zeff. is the effective nuclear charge whose value can be calculated by the formula : Zeff.= Z – . Here isshielding constant and for neon, the value of when calculate by Slater’s rule, is found to be 4.5.

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5. The value of constant C for NaF crystals is [given that interionic distance of NaF = 231 pm] :(A) 231 (B) 115.5 (C) 614.5 (D) 307.25

6. The value of univalent radii for F¯ as calculated by Pauling method is (given that interionic distance ofNaF = 231 pm) :(A) 94.5 pm (B) 136.5 pm (C) 111.68 pm (D) 115.5 pm

7. The value of ‘C’ for Na+, Mg2+ and Al3+ will be in the order :(A) Al3+ > Mg2+ > Na+ (B) Al3+ < Mg2+ < Na+ (C) Al3+ = Mg2+ = Na+ (D) Can’t be compared.

Comprehension # 3The amount of energy required to remove, an electron from the last orbit of an isolated (free) atom ingaseous state is known as ionisation energy or first ionisation energy of the element. Similarly the energyrequired for the removal of the electron from the unipositive ion (M+ produced above) is referred to assecond ionization energy and thus the third, fourth etc.The ionisation energy depends on various factors like nuclear charge, size of atom, type of configurations,screening effect and penetration power of the electrons.

8. Which of the following statement is correct ?(A) Ionisation energies of elements decrease along the period.(B) Ionisation energies of the IIA group elements are less than that of the corresponding III A group elements.(C) Ionisation energies of group 15 elements are less than that of the corresponding group 16 elements.(D) Ionisation energy of Ga is greater than Al.

9. The dominating factor responsible for the decreasing ionisation energies of the elements on moving downthe group is :(A) atomic radius (B) type of electron to be removed(C) the valence shell electron configuration (D) all of these

10. Which of the following order is not correct ?(A) E(I) of Be > E(I) of B but E(II) of Be < E(II) of B(B) E(I) of Be < E(I) of B but E(II) of Be < E(II) of B(C) E(II) of O > E(II) of N(D) E(I) of Mg > E(I) of Al

Match the column

11. Match list-I (atomic number of element) with list-II (position of elements in periodic table) and selectthe corect anwer using the codes given gelow the lists :List-I List-II(A) 19 (p) p-block(B) 22 (q) f-block(C) 32 (r) d-block(D) 64 (s) s-block

12. Match the electronic configurations of the elements given in column-(I) with their correct characteristic(s)given in column-(II).Column-I Column-II(A) 1s2 (p) Show highest negative oxidation state.(B) 1s2 2s2 2p5 (q) Show highest first ionisation energy.(C) 1s2 2s1 (r) Show highest reducing power in aqueous solution.(D) 1s2 2s2 2p3 (s) Show highest electron affinity.

(t) Show highest electronegativity.

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13. Column-I contains some increasing orders of various species and column-II has the properties of theelements / ions. Accordingly match the column-I and column-II.Column – I Column – II(A) Na+ < F– < O2– < N3 – (p) Electronegativity(B) Li+ < Na+ < K+ < Rb+ < Cs+ (q) Mobility of hydrated ions(C) O < S < F < Cl (r) Ionisation energy(D) Cl– < K+ < Ca2+ < Sc3+ (s) Electron affinity

(t) Ionic radius.14. In Column-I, there are given electronic configurations of some elements. Match these with the correct

metals given in Column-II :Column-I Column-II(A) ns2, np5 (p) Chromium(B) (n – 1) d10 , ns1 (q) Copper(C) (n – 1) d5 , ns1 (r) Krypton(D) (n – 1) d10 , ns2 , np6 (s) Bromine

15. Match the metals given in Column-II with their type given in Column-I :Column-I Column-II(A) Metalloid (p) Sulphur(B) Radioactive (q) Gold(C) Transition metal (r) Arsenic(D) Chalcogen (s) Uranium

16. Match the metals given in Column-II with their type given in Column-I :Column-I Column-II(A) Representative element (p) Cerium(B) Lanthanide (q) Aluminium(C) Coinage metal (r) Thorium(D) Actinide (s) Gold

17. Match the particulars given in Column-I with the process/metal / species given in Column-II.Column-I Column-II

(A) Isoelectronic species (p) A+(g) + energy A ++(g) + e– (g)

(B) Half filled orbital (q) Ar, K+, Ca++

(C) Second ionisation energy (r) Lutetium(D) Inner transition element (s) Antimony

18. Match the type of elements / characteristic of the elements listed in Column-I with the correct elementlisted in Column-II.Column-I Column-II(A) Highest 1st ionisation energy (p) Technitium(B) Highest electronegativity (q) Lithium(C) Synthetic element (r) Helium(D) Strongest reducing agent (s) Fluorine

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19. The Column-I has certain details about the elements of s-, p- and d-block elements. Match those withthe group number of the elements listed in Column-II.Column-I Column-II

(element / elements) (group number)(A) An element whose fourth shell contains two p-electrons (p) 8th group(B) An element whose valence shell contains one unpaired p-electron (q) 12th group

(C) An element which receives last electron in (n – 1) d-subshell (r) 14th group

(D) An element with the ground-state electron configuration [Ar]4s23d10 (s) 17th group

Assertion / Reasoning

DIRECTIONS :Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.

20. Statmemt-1 : F atom has less electron affinity than Cl atom.Statmemt-2 : Additional electrons are repelled more strongly by 3p electrons in Cl atom than by 2pelectrons in F atom.

21. Statmemt-1 : Noble gases have highest ionization enthalpies in their respective periods.Statmemt-2 : Noble gases have stable closed shell electronic configuration.

22. Statmemt-1 : Electron gain enthalpy of oxygen is less than that of fluorine but greater than that ofnitrogen.Statmemt-2 : Ionization enthalpy is as follows : N > O > F

23. Statmemt-1 : Cs and F combines violently to form CsF.Statmemt-2 : Cs is most electropositive and F is most electronegative.

24. Statmemt-1 : Nitrogen has higher IE than that of oxygen.Statmemt-2 : Nitrogen atom has smaller atomic size than that of oxygen.

25. Statement-1 : Electron affinity values of the 3rd period elements on extreme right of the periodic tableexcept noble gases are generally more than the 2nd period element of the same group.Statement-2 : Due to smaller atomic size of the 2nd period element, its electron density increaseswhich eases the addition of electron.

26. Statement-1 : In a period noble gas has largest atomic radius.Statement-2 : In case of noble gases vander Waal's radius is defined and there is much inter electronicrepulsions.

27. Statement-1 : The 5th period of periodic table contains 18 elements not 32.Statement-2 : n = 5 , = 0, 1, 2, 3. The order in which the energy of available orbitals 4d , 5s and 5pincreases is 5s < 4d < 5p and the total number of orbitals available are 9 and thus 18 electrons can beaccommodated.

True / False28. Among K+, Mg2+ and Al3+ ions, Al3+ is the smallest one.

29. The negative value of electron gain enthalpy of Cl > F because there is weak electron-electronrepulsion in the bigger 3-p sub-shell of Cl as compared to compact 2p-subshell of F.

30. Formation of S2– and Ar–, both require the absorption of energy.

31. The following set of elements does not represent the correct order of electron affinity valuesS > Se > Te > O.

32. The size of the isoelectronic species is effected by electron-electron interaction in the outer orbitals.

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PART - I : MIXED OBJECTIVE

Single choice type1. Which of the following element has highest metallic character .

Element IP(A) P 17 eV(B) Q 2 eV(C) R 10 eV(D) S 13 eV

2. The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p4. The atomic number and the groupnumber of the element ‘X’ which is just below the above element in the periodic table are respectively.(A) 24 & 6 (B) 24 & 15 (C) 34 & 16 (D) 34 & 8

3. Choose the s-block element from the following:(A) 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 (B) 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s1

(C) 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 (D) all of the above

4. False statement for periodic classification of elements is(A) The properties of the elements are periodic function of their atomic numbers.(B) No. of non-metallic elements is less than the no. of metallic elements.(C) First ionization energy of elements does not increase regularly with the increasing of atomic number in

a period.(D) d-subshell is filled by final electron with increasing atomic number of inner transition elements.

5. Pick out the isoelectronic species from the following:

I. +CH3 II. H3O+ III. NH3 IV.

3CH(A) I and II (B) III and IV (C) I and III (D) II, III and IV

6. If there were 10 periods in the periodic table then how many elements would this period can maximumcomprise of.(A) 50 (B) 72 (C) 32 (D) 98

7. Among the following which species is/are paramagnetic(i) Sr2+ (ii) Fe3+ (iii) Co2+ (iv) S2– (v) Pb2+

(A) i, iv, v (B) i, ii, iii (C) ii, iii (D) iv, v

8. If each orbital can hold a maximum of three electrons, the number of elements in 9th period of periodic table(long form) are(A) 48 (B) 162 (C) 50 (D) 75

9. Atomic radii of F & Ne in Angstrom are respectively given by :(A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) 1.60, 0.72.

10. Which one of the following is the smallest in size ?(A) N 3 (B) O 2 (C) F (D) Na +

11. Which one of the following is correct order of increase in size ?(A) Mg < Na+ < O2– < Al (B) Na+ < Al < Mg < O2– (C) Na+ < O2– < Al < Mg (D) Na+ < O2– < Mg < Al

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12. The first Ionization energy of Na, Mg, Al & Si are in the order :(A) Na < Mg > Al < Si (B) Na > Mg > Al > Si (C) Na < Mg < Al > Si (D) Na > Mg > Al < Si

13. The first ionisation energy in eV of N & O are respectively given by :(A) 14.6, 13.6 (B) 13.6, 14.6 (C) 13.6, 13.6 (D) 14.6, 14.6

14. As one move down the group from top to bottom then which one among the following will not be observed?(A) Ionisation energy increases (B) Electron affinity decreases(C) Electronegativity decreases (D) Atomic radii increase.

15. Which of the following relation is correct with respect to first (I) and second (II) ionization energies ofsodium and magnesium ?(A) IMg = IINa (B) INa > IMg (C) IIMg > IINa (D) IINa > IIMg

16. The ionization energy will be higher when the electron is removed from .............. if other factors beingequal.(A) s-orbital (B) p-orbital (C) d-orbital (D) f-orbital

17. Which of the following isoelectronic ion has the lowest 1st ionization energy ?(A) Na+ (B) F– (C) Mg2+ (D) O2–.

18. A large difference between the third and fourth ionization energies indicates the presence of :(A) 4 valence electrons in an atom. (B) 5 valence electrons in an atom.(C) 3 valence electrons in an atom. (D) 8 valence electrons in an atom.

19. The second ionization energies of elements are always higher than their first ionization energies because:(A) the cation is smaller than its parent atom.(B) it is easier to remove electron from cation.(C) ionization is an endothermic process.(D) cation formed always have stable half filled or completely filled valence shell electron configuration.

20. Which is a true statement ?(A) Larger is the value of ionisation energy easier is the formation of cation.(B) Larger is the value of electron affinity easier is the formation of anion.(C) Larger is the value of ionisation energy as well as electron affinity the smaller is the electronegativity of atom.(D) Larger is the Zeff larger is the size of atom.

21. Which of the following is the correct order of ionisation energy ?(A) O2– < F– < Na+ < Mg2+ (B) F– < O2 – < Na+ < Mg2+

(C) O2 – < Na+ < F– < Mg2+ (D) Mg2+ < Na+ < F– < O2 –

22. Values of E1, E2 and E3 of an element are 9.32, 18.21 and 553.83 eV. What informations do thesedata convey ?(A) The element has two electrons in the valence shell.(B) The element has two p–electrons in the valence shell.(C) (A) and (B) both.(D) None of the two.

23. The electron gain enthalpies of halogens are as given below.F = – 332, Cl = – 349, Br = – 324, = – 295 kJ mol–1.The less negative value for F as compared to that of Cl is due to :(A) strong electron-electron repulsions in the compact 2–p sub shell of F.(B) strong electron -electron repulsions in the bigger 3–p sub shell of Cl(C) higher electronegativity value of Cl.(D) higher effective nuclear charge of F.

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24. The formation of the oxide ion O2– (g) requires first an exothermic and then an endothermic step as shownbelow

O(g) + e– = O– (g) ; egH = – 142 kJmol–1

O– (g) + e– = O2– (g) ; egH = 844 kJmol–1

This is because :(A) O– ion will tend to resist the addition of another electron on account of same charge.(B) oxygen has high electron affinity.(C) oxygen is more electronegative.(D) O– ion has comparatively larger size than oxygen atom.

25. For electron affinity of halogens which of the following is correct ?(A) Br > F (B) F > Cl (C) Br < Cl (D) F > I

26. Which of the following is the correct order of ionisation energy ?(1) Be+ > Be (2) Be > Be+ (3) C > Be (4) B > Be(A) 2, 3 (B) 3, 4 (C) 1, 3 (D) None of these

27. The correct order of electronegativity is :(A) F > Cl I (B) F > O > N > C (C) S < O < Se < Te (D) All

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

29. Consider the following statements.1. E(I) of nitrogen atom is more than E(I) of oxygen atom.2. Electron affinity of oxygen is less than sulphur atom3. Electronegativity of an atom has no relation with its ionization enthalpy and electron gain enthalpy.Which of the above statements are correct ?(A) 1, 2 (B) 1, 3 (C) 2, 4 (D) 1, 2 & 3

30. Which of the following statement is wrong ?(A) Vander Waal’s radius of iodine is more than its covalent radius.(B) All isoelectronic ions belong to same period of the periodic table.(C) I.E(I) of N is higher than that of O while I.E(II) of O is higher than that of N.(D) The electron affinity of N is less than that of P.

31. 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 -(A) Na (B) Si (C) F (D) Ca

32. With reference to ionisation potential which one of the following sets is correct -(A) Li > K > B (B) B > Li > K (C) Cs > Li > K (D) Cs < Li < K

33. Successive ionisation energies of an element ‘X’ are given below (in K.Cal) :IP1 IP2 IP3 IP4165 195 556 595Electronic configuration of the element ‘X’ is -(A) 1s2, 2s2 2p6, 3s2 3p2 (B) 1s2, 2s1

(C) 1s2, 2s2 2p2 (D) 1s2, 2s2 2p6, 3s2

34. The ionistion energy of B and Al as compared to Be and Mg are -(A) Lower (B) Higher (C) Equal (D) None of these

35. Which of the following has 2nd IP < Ist IP(A) Mg (B) Ne (C) C (D) None

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36. Among the following elements (Whose electronic configuration is give below) the one having thehighest ionisation energy is -(A) [Ne] 3s2 3p3 (B) [Ne] 3s2 3p4 (C) [Ne] 3s2 3p5 (D) [Ne] 3s2

37. The correct order of decreasing first ionization energy is -(A) Si > Al > Mg > Na (B) Si > Mg > Al > Na(C) Al > Si > Mg > Na (D) None of these

38. Which of the following transitions involves maximum amount of energy -(A) M–

(g) M(g) (B) M(g) M+(g) (C) M+

(g) M2+(g) (D) M2+

(g) M3+(g)

39. Element X, Y and Z have atomic numbers 19, 37 and 55 respectively. Which of the followingstatements is true -(A) Their ionisation potential would increase with the increasing atomic number(B) ‘Y’ would have an ionisation potential in between those of ‘X’ and ‘Z’(C) ‘Z’ would have the highest ionisation potential(D) ‘Y’ would have the highest ionisation potential

40. 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(A) I 2372 5251(B) II 520 7300(C) III 900 1760(D) IV 1680 3380

41. The decreasing order of the ionization potential of the following elements is -(A) Ne > Cl > P > S > Al > Mg (B) Ne > Cl > P > S > Mg > Al(C) Ne > Cl > S > P > Mg > Al (D) Ne > Cl > S > P > Al > Mg

42. For the processes K+(g) I K(g) II K(s)-(A) Energy is released in (I) and abosrbed in (II)(B) Energy is absorbed in (I) and released in (II)(C) Energy is absorbed in both the processes(D) Energy is released in both the processes

43. The order of first ionisation enthalpies of the elments Li, Be, B, Na is -(A) Li > Be > B > Na (B) Be > B > Li > Na(C) Na > Li > B > Be (D) Be > Li > B > Na

44. Arrange the elements S, P, As in order of increasing ionization enthalpy -(A) S N > O > F (B) O > N > F > C (C) O > F > N > C (D) F > O > N > C

46. First, second and third IP values are 100eV, 150eV and 1500eV. Element can be -(A) Be (B) B (C) F (D) Na

47. M(g) M+(g) + e–, H = 100 eVM(g) M2+(g) + 2e–, H = 250 eVWhich is incorrect statement ?(A) IE1 of M(g) is 100 eV (B) IE1 of M+(g) is 150 eV(C) IE2 of M(g) is 250 eV (D) IE2 of M(g) is 150 eV

48. I.P. of sodium is 5.14 eV. then I.P. of potassium will be -(A) Equal to sodium (B) 5.68 eV (C) 4.34 eV (D) 10.28 eV

49. The correct values of ionization enthalpies (in kJ mol–1) of Si, P, Cl and S respectively are -(A) 786, 1012, 999, 1256 (B) 1012, 786, 999, 1256(C) 786, 1012, 1256, 999 (D) 786, 999, 1012, 1256

50. Consider the following changes -A A+ + e– : E1 and A+ A2 + e– : E2The energy required to pull out the two electrons are E1 and E2 respectively. The correct relationshipbetween two energies would be -(A) E1 < E2 (B) E1 = E2 (C) E1 > E2 (D) E1 E2

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51. Electron affinity is a -(A) Relative strength to attract the shared electron pair(B) Necessary energy required to remove the electron from the ultimate orbit(C) Energy released when an electron is added to the outermost shell(D) Energy released when an electron is added to the inner shell

52. The electron affinityes of , O, S and Cl are such that -(A) N < O < S < Cl (B) O < N < Cl < S (C) O Cl < N S (D) O < S < Cl < N

53. The correct order of electron affinity of B, C, N, O is -(A) O > C > N > B (B) B > N > C > O (C) O > C > B > N (D) O > B > C > N

54. The correct order of electron affinity for the different families is -(A) Halogen > carbon > nitrogen > oxygen (B) Halogen > oxygen > nitrogen > carbon(C) Halogen > nitrogen > carbon > oxygen (D) Halogen > oxygen > carbon > nitrogen

55. Highest electron-affinity is associated with the configuration -(A) 2s2, 2p0 (B) 2s2, 2p2 (C) 2s2, 2p3 (D) 2s2, 2p1

56. Adding electron a neutral gaseous atom usually leads to -(i) Liberation of energy (ii) Formation of anion(iii) Proton/electron ratio decrease (iv) Proton/electron ratio increase(A) (i), (iv) (B) (ii), (iii), (iv) (C) (i), (ii), (iv) (D) (i), (ii), (iii)

57. Which statement is correct -(A) The E.A. of carbon is greater than oxygen (B) The E.A. of sulphur is less than oxygen(C) The E.A. of iodine is greater than bromine (D) The E.A. of bromine is less than chlorine

58. Fluorine has low electron affinity than chlorine because of -(A) Smaller radius of fluorine, high density (B) Smaller radius of chlorine, high density(C) Bigger radius of fluorine, less density (D) Smaller radius of chlorine, less density

59. Which of the following element is expected to have highest electron gain enthalpy -(A) 1s22s22p63s23p5 (B) 1s22s22p3 (C) 1s22s22p4 (D) 1s22s22p5

60. Arrange N, O and S in order of decreasing electron affinity -(A) S > O > N (B) O > S > N (C) N > O > S (D) S > N > O

61. Which of the following set of elements exhibits positive and negative oxidation states-(A) O, Cl, H (B) F, Li, Be (C) Na, Mg, Al (D) H, Ba, Ne

62. A, B and C are hydroxy-compounds of the elements X, Y and Z respectively. X, Y and Z are in thesame period of the periodic table. A gives an aqueous solution of pH less than seven. B reacts withboth strong acids and strong alkalis. C gives an aqueous solution which is strongly alkalineWhich of the following statements is/are true:I : The three elements are metalsII : The electronegativities decrease from X to Y to Z.III : The atomic radius decreases in the order X, Y and Z.IV : X, Y and Z could be phosphorus, aluminium and sodium respectively :(A) I, II, III only correct (B) I, III only correct(C) II, IV only correct (D) II, III, IV only correct

63. The correct set of decreasing order of electronegativity is -(A) Li, H, Na (B) Na, H, Li (C) H, Li, Na (D) Li, Na, H

64. Increasing order of electronegativity is -(A) Bi F > N > C (B) F > N > C > O (C) C > F > N > O (D) F > O > N > C






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More than one choice type66. M(g) M+ (g) + e– ; H = 100 eV.

M(g) M2+ (g) + 2e– ; H = 250 eV.Which is/are correct statement(s) ?(A) IE1 of M(g) is 100 eV (B) IE1 of M+ (g) is 150 eV.(C) IE2 of M(g) is 250 eV. (D) IE2 of M(g) is 150 eV.

67. In halogens, which of the following increases from iodine to fluorine?(A) Bond length (B) Electronegativity(C) The ionization energy of the element (D) Oxidizing power

68. Which of the following have isoelectronic structures?(i) CH3

+ (ii) H3O+ (iii) CH–

3 (iv) NH3(A) (i) and (ii) (B) (i) and (iii) (C) (iii) and (iv) (D) (ii), (iii) and (iv)

69. Which among the following are isostructural pairs?(A) NF3 and H3O

+ (B) NO–3 and BF3 (C) NF3 and NO–

3 (D) NF3 and NH3

70. Which of the following statements are correct?(A) F is the most electronegative and Cs the most electropositive element(B) The electronegativity of halogens decreases from F to I.(C) The electron affinity of Cl is higher than that of F though their electronegativities are in the reverse

order.(D) The electron affinity of noble gases in almost zero

71. Which of the following species are isoelectronic with Ne?(A) N3– (B) Mg2+ (C) Al3+ (D) Ca2+

72. In which of the following are the orders of electron affinity of the elements or ions shown correctly?(A) S > O– (B) O > S– (C) O– > S– (D) N– > P

73. Which of the following salts does not undergo hydrolysis ?(A) NaCl (B) KCN (C) KClO4 (D) Na2CO3

74. Which of the following halides are soluble in water?(A) AlF3 (B) AgI (C) AgCl (D) AgF

75. Which of the following pairs have approximately the same atomic radii?(A) Pd and Pt (B) Al and Mg (C) Al and Ga (D) Na and Ne

76. The ionic bonds X+ Y– are formed when the :(A) electron affinity of Y is high (B) ionization energy of X is low(C) lattice energy of XY is high (D) lattice energy of XY is low

77. Which of the following pairs do not show the inert-pair effect?(A) Cu and Au (B) Si and Ge (C) TI and Pb (D) Bi and Sn

PART - II : SUBJECTIVE QUESTIONS

1. Elements A, B, C, D and E have the following electronic configurations :A : 1s2 2s2 2p1 B : 1s2 2s2 2p6 3s2 3p1

C : 1s2 2s2 2p6 3s2 3p3 D : 1s2 2s2 2p6 3s2 3p5

E : 1s2 2s2 2p6 3s2 3p6 4s2

Which among these will belong to the same group in the periodic table ?

2. An element X with Z = 112 has been recently discovered. What is the electronic configuration of theelement ? To which group and period will it belong ?






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3. What is the effective nuclear charge at the periphery of nitrogen atom when a extra electron is addedduring the formation of an anion. Compare the value of Zeff when the atom is ionized to N+.

4. X – X bond length is 1.00 Å and C–C bond length is 1.54 Å. If electronegativities of X and C are 3.0 and 2.0respectively, then C–X bond length is likely to be ? (using Stevension & Schomaker formula).

5. Atomic radius of Li is 1.23 Å and ionic radius of Li+ is 0.76Å. Calculate the percentage of volumeoccupied by single valence electron in Li.

6. Select from each group the species which has the smallest radius stating appropriate reason.(a) O, O–, O2– (b) P3+ , P4+ , P5+

7. Mg2+ is smaller than O2– in size, though both have same electronic configuration. Explain ?

8. From each set, choose the atom which has the largest ionization enthalpy and explain your answerwith suitable reasons. (a) F, O, N (b) Mg, P, Ar

9. First and second ionisation energies of magnesium are 7.646 eV and 15.035 eV respectively. The amountof energy in kJ needed to convert all the atoms of magnesium into Mg2+ ions present in 12 mg of magnesiumvapour will be ? [Given 1 eV = 96.5 kJ ml–1].

10. Consider the elements N, P, O and S and arrange them in order of increasing negative electron gainenthalpy.

11. Why do halogens have high electron gain enthalpies (i.e. –egH) ?

12. The amount of energy when million atoms of iodine are completely converted into I– ions in the vapour stateaccording to the equation, I (g) + e– (g) I– (g) is 5.0 × 10–13 J.Calculate the electron gain enthalpy of iodine in terms of kJ mol–1 and eV per atom.

13. Account for the large decrease in electron affinity between Li and Be despite the increase in nuclearcharge.

14. Calculate the electronegativity of carbon from the following data :EH–H = 104.2 kcal mol–1 , EC–C = 83.1 kcal mol–1

EC–H = 98.8 kcal mol–1| , XH = 2.1

15. Arrange the following in decreasing basic nature LiOH, NaOH, KOH, CsOH.LiOH, NaOH, KOH, CsOH

16. Arrange the following compounds in increasing order of acidic strength Al2O3, SiO2, P2O3 and SO2.

Al2O3, SiO2, P2O3 SO2.

17. A M2+ ion derived from a metal in the first transition metal series has four electrons in 3d subshell. Whatelement might M be ?

18. Following are the valence shell electronic configurations of some elements.(i) 3s2 3p5 (ii) 3d10 4s2 (iii) 2s2 3p6 4s1 (iv) 1s2 2s2

Find out the blocks to which they belong in the periodic table ?

19. Find out the group of the element having the electronic configuration, 1s2 2s2 2p6 3s2 3p6 3d6 4s2 .






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20. Arrange the following ions in the increasing order of their size : Be2+ , Cl–, S2– , Na+ , Mg2+ , Br– ?

21. The (IE1) and the (IE2) in kJ mol–1 of a few elements designated by Roman numerals are shownbelow:

I II IIIIE1 403 549 1142IE2 2640 1060 2080Which of the above elements is likely to be a(a) non-metal (b) alkali metal (c) alkaline earth metal ?

22. Ionisation energy and electron affinity of fluorine are respectively 17.42 and 3.45 eV. Calculateelectronegativity of fluorine atom.

23. Select neutral, acidic,basic and amphoteric oxides from the following.CO, BeO, Na2O , N2O5

24. Why the electron gain enthalpy values of alkaline earth metals are lower (i.e. less negative) or positive ?

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

1. Which one of the following has largest size ? [IIT-1993](A) Na (B) Na+ (C) Na– (D) None of these

2. The decreasing order of the second ionization energy of K, Ca and Ba is (At. Nos. K = 19, Ca = 20, Ba = 56)[IIT-1993]

(A) K > Ca > Ba (B) Ca > Ba > K (C) Ba > K > Ca (D) K > Ba > Ca

3. Element Hg has two oxidation states Hg+1 & Hg+2. The right order of radii of these ions - [IIT-1994](A) Hg+1 > Hg+2 (B) Hg+2 > Hg+1 (C) Hg+1 = Hg+2 (D) Hg+2 Hg+1

4. Which has most stable + 2 oxidation state : [IIT-1995](A) Sn (B) Pb (C) Fe (D) Ag

5. Moving from right to left in a periodic table, the atomic size is- [IIT-1995](A) Increased (B) Decreased (C) Remains constant (D) None of these

6. The increasing order of electronegativity in the following elements- [IIT-1995](A) C, N, Si, P (B) N, Si, C, P (C) Si, P, C, N (D) P, Si, N, C

7. One element has atomic weight 39. Its electronic configuration is 1s2, 2s2 2p6, 3s2 3p6 4s1. The true statementfor that element is- [IIT-1995](A) More (IE)1 (B) Transition element (C) Isotone with 18Ar38 (D) Stable oxide M2O

8. The number of paired electrons in oxygen is - [IIT-1995](A) 6 (B) 16 (C) 8 (D) 32

9. Which of the following oxide is neutral ? [IIT-1996](A) CO (B) SnO2 (C) ZnO (D) SiO2

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10. Which of the following has the maximum number of unpaired electrons - [IIT-1996](A) Mg2+ (B) Ti3+ (C) V3+ (D) Fe2+

11. The incorrect statement among the following is - [IIT-1997](A) the first ionisation potential of Al is less than the first ionisation potential of Mg(B) the second ionisation potential of Mg is greater that the second ionisation potential of Na(C) the first ionisation potential of Na is less than the first ionisation potential of Mg(D) the third ionisation potential of Mg is greater than the third ionisation potential of Al

12. The correct order of radii is - [IIT-2000](A) N < Be Na > Li (B) Be >Mg > Ca (C) B > C > N (D) Ge > Si > C

14. The correct order of acidic strength is - [IIT-2000](A) Cl2O7 > SO3 > P4O10 (B) CO2 > N2O5 < SO3(C) Na2O > MgO > Al2O3 (D) K2O > CaO > MgO

15. Identify the least stable ion amongst the following- [IIT-2002](A) Li¯ (B) Be¯ (C) B¯ (D) C¯

16. Identify the correct order of acidic strengths of CO2, CO, CuO, CaO, H2O – [IIT-2002](A) CaO < CuO < H2O < CO2 (B) H2O < CuO < CaO < CO2(C) CaO < H2O < CuO < CO2 (D) H2O < CO2 < CaO < CuO

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. The correct order of ionic radius is - [AIEEE-2002](1) Ce > Sm > Tb > Lu (2) Lu > Tb > Sm > Ce(3) Tb > Lu > Sm > Ce (4) Sm > Tb > Lu > Ce

2. Ce3+, La3+, Pm3+ and Yb3+ have ionic radii in the increasing order as – [AIEEE-2002](1) La3+ < Ce3+ < Pm3+ < Yb3+ (2) Yb3+ < Pm3+ < Ce3+ < La3+

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

3. According to the Periodic Law of elements, the Variation in properties of elements is related to their ?[AIEEE-2003]

(1) Nuclear masses (2) Atomic numbers(3) Nuclear neutron-proton number ratio (4) Atomic masses

4. The reduction in atomic size with increase in atomic number is a characteristic of elements of -[AIEEE-2003]

(1) d-block (2) f-block (3) Radioactive series (4) High atomic masses

5. Which one of the following groups represent a collection of isoelectronic species ?(At. no. Cs = 55, Br = 35) [AIEEE-2003](1) N3–, F–, Na+ (2) Be, Al3+, Cl– (3) Ca2+, Cs+, Br (4) Na+, Ca2+, Mg2+

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

[AIEEE-2003](1) Cr (2) Mn (3) Fe (4) V

7. Which one of the following sets of ions represents the collection of isoelectronic species ? [AIEEE-2004](1) K+, Cl–, Mg2+, Sc3+ (2) Na+, Ca2+, Sc3+, F– (3) K+, Ca2+, Sc3+, Cl– (4) Na+, Mg2+, Al3+, Cl–

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8. Which one of the following ions has the highest value of ionic radius ? [AIEEE-2004](1) O2– (2) B3+ (3) Li+ (4) F–

9. Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is : [AIEEE-2004](1) Al2O3 < SiO2 < SO2 < P2O3 (2) SiO2 < SO2 < Al2O3 < P2O3(3) SO2 < P2O3 < SiO2 < Al2O3 (4) Al2O3 < SiO2 < P2O3 < SO2

10. The formation of the oxide ion 2gO requires first an exothermic and then an endothermic step as shown

below : [AIEEE-2004]O(g) + e– = O–

(g) H° = – 142 kJ mol–1

O–(g) + e– = O2–

(g) H° = 844 kJ mol–1

This is because of :(1) O– ion will tend to resist the addition of another electron(2) Oxygen has high electron affinity(3) Oxygen is more electronegative(4) O– ion has comparatively larger size than oxygen atom

11. In which of the following arrangements the order is NOT according to the property indicated against it ?[AIEEE-2005]

(1) Al3+ < Mg2+ < Na < F– – increasing ionic size(2) B < C < N < O – increasing first ionization enthalpy(3) I < Br < F < Cl – increasing electron gain enthalpy (with negative sign)(4) Li < Na < K < Rb – increasing metallic radius

12. Which of the following oxides is amphoteric in character ? [AIEEE-2005](1) SnO2 (2) SiO2 (3) CO2 (4) CaO

13. Pick out the isoelectronic structure from the following : [AIEEE-2005]I. +CH3 II. H3O

+ III. NH3 IV. CH3–

(1) I and II (2) III and IV (3) I and III (4) II, III and IV

14. The lanthanide contraction is responsible for the fact that [AIEEE-2005](1) Zr and Y have about the same radius (2) Zr and Nb have similar oxidation state(3) Zr and Hf have about the same radius (4) Zr and Zn have the same oxidation state

15. Which of the following factors may be regarded as the main cause the lanthanide contraction ?[AIEEE-2005]

(1) Poor shielding of one of 4f electron by another in the subshell(2) Effective shielding of one of 4f electrons by another in the subshell(3) Poorer shielding of 5d electrons by 4f electrons(4) greater shielding of 5d electrons by 4f electrons

16. The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is–[AIEEE-2006]

(1) F < S < P < B (2) P < S < B < F (3) B < P < S < F (4) B < S < P < F

17. Which one of the following sets of ions represents a collection of isoelectronic species ? [AIEEE-2006](1) N3–, O2–, F–, S2– (2) Li+, Na+, Mg2+, Ca2+ (3) K+, Cl–, Ca2+, Sc3+ (4) Ba2+, Sr2+, K+, Ca2+

18. lanthanoid contraction is caused due to - [AIEEE-2006](1) The same effective nuclear charge from Ce to Lu(2) The imperfect shielding on outer electrons by 4f electrons from the nuclear charge(3) The appreciable shielding on outer electrons by 4f electrons from the nuclear charge(4) The appreciable shielding on outer electrons by 5d electrons from the nuclear charge

19. Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Seand Ar ? [AIEEE-2013](1) Ca < S < Ba < Se < Ar (2) S < Se < Ca < Ba < Ar(3) Ba < Ca < Se < S < Ar (4) Ca < Ba < S < Se < Ar

20. The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be :[AIEEE-2013](1) – 2.55 eV (2) – 5.1 eV (3) – 10.2 eV (4) +2.55 eV

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NCERT QUESTIONS

1. What is the basic theme of organisation in the periodic table?

2. Which important property did Mendeleev use to classify the elements in his periodic table and did he stickto that?

3. What is the basic difference in approach between the Mendeleevís Periodic Law and the Modern PeriodicLaw?

4. On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

5. In terms of period and group where would you locate the element with Z = 114?

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

7. Which element do you think would have been named by(i) Lawrence Berkeley Laboratory (ii) Seaborgis group?

8. Why do elements in the same group have similar physical and chemical properties?

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

10. How do atomic radius vary in a period and in a group? How do you explain the variation?

11. What do you understand by isoelectronic species? Name a species that will be isoelectronic with each ofthe following atoms or ions.(i) F– (ii) Ar (iii) Mg2+ (iv) Rb+

12. Consider the following species :N3–, O2–, F–, Na+, Mg2+ and Al3+

(a) What is common in them?(b) Arrange them in the order of increasing ionic radii.

13. Explain why cation are smaller and anions larger in radii than their parent atoms?

14. What is the significance of the terms - 'isolated gaseous atom' and 'ground state' while defining the ionizationenthalpy and electron gain enthalpy?Hint : Requirements for comparison purposes.

15. Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10–18 J. Calculate the ionizationenthalpy of atomic hydrogen in terms of J mol–1.Hint: Apply the idea of mole concept to derive the answer.

16. Among the second period elements the actual ionization enthalpies are in the orderLi < B < Be < C < O < N < F < Ne.Explain why(i) Be has higher iH than B (ii) O has lower iH than N and F?

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17. How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesiumbut its second ionization enthalpy is higher than that of magnesium?

18. What are the various factors due to which the ionization enthalpy of the main group elements tends todecrease down a group?

19. The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :B Al Ga In Tl801 577 579 558 589How would you explain this deviation from the general trend ?

20. Which of the following pairs of elements would have a more negative electron gain enthalpy?(i) O or F (ii) F or Cl

21. Would you expect the second electron gain enthalpy of O as positive, more negative or less negative thanthe first? Justify your answer.

22. What is the basic difference between the terms electron gain enthalpy and electronegativity?

23. How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all thenitrogen compounds?

24. Describe the theory associated with the radius of an atom as it(a) gains an electron (b) loses an electron

25. Would you expect the first ionization enthalpies for two isotopes of the same element to be the same ordifferent? Justify your answer.

26. What are the major differences between metals and non-metals?

27. Use the periodic table to answer the following questions.(a) Identify an element with five electrons in the outer subshell.(b) Identify an element that would tend to lose two electrons.(c) Identify an element that would tend to gain two electrons.(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

28. The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that amonggroup 17 elements is F > CI > Br > I. Explain.

29. Write the general outer electronic configuration of s-, p-, d- and f- block elements.

30. Assign the position of the element having outer electronic configuration(i) ns2np4 for n = 3(ii) (n - 1)d2ns2 for n = 4, and(iii) (n - 2) f 7 (n - 1)d1ns2 for n = 6, in the periodic table.

31. The first (iH1) and the second (iH2) ionization enthalpies (in kJ mol–1) and the (egH) electron gain enthalpy(in kJ mol–1) of a few elements are given below :Elements H1 H2 eg1HI 520 7300 – 60II 419 3051 – 48III 1681 3374 – 328IV 1008 1846 – 295V 2372 5251 + 48VI 738 1451 – 40Which of the above elements is likely to be :(a) the least reactive element. (b) the most reactive metal.(c) the most reactive non-metal. (d) the least reactive non-metal.(e) the metal which can form a stable binary halide of the formula MX2(X = halogen).(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?






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32. Predict the formulas of the stable binary compounds that would be formed by the combination of thefollowing pairs of elements.(a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine(d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine

33. In the modern periodic table, the period indicates the value of :(a) atomic number (b) atomic mass(c) principal quantum number (d) azimuthal quantum number.

34. Which of the following statements related to the modern periodic table is incorrect?(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in ad-subshell.(c) Each block contains a number of columns equal to the number of electrons that can occupy thatsubshell.(d) The block indicates value of azimuthal quantum number () for the last subshell that received electronsin building up the electronic configuration.

35. Anything that influences the valence electrons will affect the chemistry of the element. Which one of thefollowing factors does not affect the valence shell?(a) Valence principal quantum number (n)(b) Nuclear charge (Z )(c) Nuclear mass(d) Number of core electrons.

36. The size of isoelectronic species :F–, Ne and Na+ is affected by :(a) nuclear charge (Z)(b) valence principal quantum number (n)(c) electron-electron interaction in the outer orbitals(d) none of the factors because their size is the same.

37. Which one of the following statements is incorrect in relation to ionization enthalpy?(a) Ionization enthalpy increases for each successive electron.(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gasconfiguration.(c) End of valence electrons is marked by a big jump in ionization enthalpy.(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having highern value.

38. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :(a) B > Al > Mg > K (b) Al > Mg > B > K(c) Mg > Al > K > B (d) K > Mg > Al > B

39. Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is :(a) B > C > Si > N > F (b) Si > C > B > N > F(c) F > N > C > B > Si (d) F > N > C > Si > B

40. Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizingproperty is :(a) F > Cl > O > N (b) F > O > Cl > N(c) Cl > F > O > N (d) O > F > N > Cl

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Exercise-1PART - I

A-1. (A) A-2. (C) A-3. (C) A-4. (D) A-5. (C) A-6. (A) A-7. (A)*A-8. (C) A-9. (A) A-10. (A) A-11. (A) A-12. (C) A-13. (B) A-14. (C)A-15. (D) A-16. (C) A-17. (A) A-18. (C) A-19. (A) A-20. (C) A-21. (B)B-1. (D) B-2. (B) B-3. (C) B-4. (D) B-5. (D) B-6. (B) B-7. (B)B-8. (B) B-9. (C) B-10. (B) B-11. (B) B-12. (C) C-1. (C) C-2. (D)C-3. (B) C-4. (D) C-5. (B) C-6. (B) C-7. (C) C-8. (D) C-9. (C)C-10. (A) C-11. (C) C-12. (C) C-13. (C) C-14. (C) C-15. (B) C-16. (A)C-17. (A) C-18. (C) D-1. (A) D-2. (B) D-3. (C) D-4. (A) D-5. (B)D-6. (D) D-7. (D) D-8. (D) D-9. (B) D-10. (A)

PART - II1. (B) 2. (D) 3. (C) 4. (A) 5. (C) 6. (B) 7. (C)8. (D) 9. (A) 10. (B) 11. (A) s ; (B) r; (C) p (D) q12. (A) q ; (B) s, t ; (C) r ; (D) p 13. (A) t ; (B) q, t ; (C) s ; (D) p, r, s14. (A) s ; (B) q ; (C) p ; (D) r 15. (A) r ; (B) s ; (C) q ; (D) p16. (A) q ; (B) p ; (C) s ; (D) r 17. (A) q ; (B) s ; (C) p ; (D) r18. (A) r ; (B) s ; (C) p ; (D) q 19. (A) r ; (B) (s) ; (C) p, q ; (D) q20. (C) 21. (A) 22. (C) 23. (A) 24. (C) 25. (C) 26. (A)27. (A) 28. T 29. T 30. T 31. F 32. F

Exercise-2PART - I

1. (B) 2. (D) 3. (C) 4. (D) 5. (D) 6. (B) 7. (C)8. (D) 9. (A) 10. (D) 11. (B) 12. (A) 13. (A) 14. (A)15. (D) 16. (A) 17. (D) 18. (C) 19. (A) 20. (B) 21. (A)22. (A) 23. (A) 24. (A) 25. (C) 26. (C) 27. (B) 28. (C)29. (A) 30. (B) 31. (B) 32. (B) 33. (D) 34. (A) 35. (D)36. (C) 37. (B) 38. (D) 39. (B) 40. (B) 41. (B) 42. (D)43. (B) 44. (C) 45. (A) 46. (A) 47. (C) 48. (B) 49. (C)50. (A) 51. (C) 52. (A) 53. (C) 54. (D) 55. (B) 56. (D)57. (D) 58. (A) 59. (A) 60. (A) 61. (A) 62. (C) 63. (C)64. (A) 65. (D) 66. (ABD) 67. (BCD) 68. (CD) 69. (AB)70. (ABCD)71. (AC) 72. (AB) 73. (AC) 74. (AD) 75. (AC)76. (AB) 77. (AB)

PART - II1. Out of these, elements A and B will belong to the same group of the periodic table because they have

same outer electronic configuration, ns2 np1.2. (a) The electronic configuration of element X is [Rn]86 5f14 6d107s2

(b) It belongs to d-block as last electron enters in d subshell.(c) As number of electrons in (n – 1)d subshell and valence shell is equal to twelve i.e. 10 + 2. So itbelongs to group 12.(d) It belongs to period 7 of the periodic table as principal quantum number of valence shell is 7 (i.e.,7s2).

3. 4.25 4. C–X bond length = 1. 18 Å. 5. 76.4%6. (a) O is having smallest radius. Anion is larger than its parent atom. Also the anion of the same

atom with higher negative charge is bigger in size as compared to anion with smaller negative charge asproton to electron ratio decreases thus attraction between valence shell electrons and nucleus decreases.Hence electron cloud expands.(b) The ionic radius decreases as more electrons are ionized off that is as the valency increases. So thecorrect order is P5+ < P4+ < P3+.

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7. Mg2+ and O2– both are isoelectronic i.e., have same number of electrons. But Mg2+ having 12 protonsin its nucleus exerts higher effective nuclear charge than O2– having 8 protons and thus valence shellas well as inner shells electrons are more strongly attracted by the nucleus in Mg2+ resulting smallersize than O2–.

8. (a) Fluorine (F) has the largest ionization enthalpy because in moving from left to right in a period, atomicsize decreases and electrons are held more tightly. Since F has the smallest size and maximum nuclearcharge. It has the largest ionization enthalpy among these elements.(b) Argon (Ar) has the largest ionization enthalpy as argon has extra stable fully filled configuration.

9. 1.0943 kJ10. Order of increasing negative electron gain enthalpy is N < P < O < S. For detail refer text.11. The valence shell electronic configuration of halogens is ns2np5 and thus they require one electron to

acquire the stable noble gas configuration ns2np6. Because of this they have strong tendency to accept anadditional electron and hence have high electron gain enthalpies.

12. –3.12.13. The electron configurations of Li and Be are [He]2s1 and [He]2s2, respectively. The additional electron

enters the 2s orbital of Li but the 2p orbital of Be and hence is much less tightly bound. In fact, the nuclearcharge is so well shielded in Be that electron gain is endothermic.

14. 2.515. The basic nature of hydroxides of elements of group 1st increases on descending the group with increase

in size of cation as CsOH > RbOH > KOH > NaOH > LiOH.16. Al2O3 < SiO2 < P2O3 < SO2.17. Electron configuration of M2+ is : [Ar]18 4s0 3d4

Electron configuration of M is : [Ar]18 4s1 3d5 (and not 4s2 3d4)So total number of electrons = 24.Hence, metal M is chromium (Cr).

18. (i) p-block (ii) d-block (iii) s-block (iv) s-block19. As last electron enters in d-subshell, therefore this belongs to d-block. For d-block element the group

number is equal to the number of valence shell electrons + number of electrons in (n-1) d-subshell.So, group number = 6 + 2 = 8.

20. Be2+ < Mg2+ < Na+ < Cl– < S2– < Br–

21. (a) non-metal(III) – Due to highest ionisation energy, (IE1) and (IE2).(b) alkali metal (I) – Due to lowest ionisation energy, (IE1) and there is quite high jump in (IE2) due to inertgas configuration.(c) alkaline earth metal (II) – There is little difference in (IE1) and (IE2) and the value of (IE1) is slightlygreater than(I) due to stable configuration(ns2).

22. P = 3.72623. BeO is amphoteric and reacts with acids and bases forming salts.

CO is neutral as it is neutral towards litmus.Na2O is basic and dissolves in water forming base (NaOH).N2O5 is acidic and dissolves in water forming acid (HNO3)

24. The general valence shell electron configuration of alkaline earth metals is ns2 (stable configuration). Theextra electron must enter np subshell, which is effectively shielding by the two ns electrons and the innerelectrons. Consequently, the alkaline earth metals have little or no tendency to pick up an extra electron.

Exercise-3PART - I

1. (C) 2. (A) 3. (A) 4. (B) 5. (A) 6. (C) 7. (C)8. (A) 9. (A) 10. (D) 11. (B) 12. (B) 13. (B) 14. (A)15. (B) 16. (A)

PART - II1. (1) 2. (2) 3. (2) 4. (2) 5. (1) 6. (1) 7. (3)8. (1) 9. (4) 10. (1) 11. (2) 12. (1) 13. (4) 14. (3)15. (3) 16. (4) 17. (3) 18. (2) 19. (3) 20. (2)

Periodic Table & Periodicity - Contents Syllabus

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