Chem Periodic Properties

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES

Chapter

3

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Classification of Elements and Periodicity in Properties Solutions SECTION - A

Objective Type Questions

1.

In Lothar Meyer’s curve, the element on the peak of curve will be (1) F

(2)

Na

(3)

Mg

(4)

Ne

Sol. Answer (2)

Due to large atomic volume.

2.

According to Mendeleev’s periodic law, physical and chemical properties are function of

(1) Atomic number

(2)

Atomic weight

(3)

Atomic volume

(4)

Number of neutrons

(3)

d

(4)

f

(3)

K

(4)

Rb

Na

(4)

Be

Sol. Answer (2)

Classification is based on Mendeleev’s periodic law.

3.

With which block

30Zn

belongs?

(1) s

(2)

p

Sol. Answer (3)

Last electron enters in ‘d’ sub-shell. Zn belongs with 12th group.

4.

Which of the following has highest electron affinity? (1) Na

(2)

Li

Sol. Answer (2) 5.

Out of following, which has the highest electronegativity? (1) H

(2)

Li

(3)

Sol. Answer (1) H-2.1

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 68 6.

Classification of Elements and Periodicity in Properties

Solution of Assignment

Which of the following has largest size? (1) H –

(2)

He

(3)

Li +

(4)

Be+ 2

Sol. Answer (1) H– these are isoelectronic species. Which of the following orbitals has higher screening power?

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

7.

(1) 4s

(2)

4p

(3)

4d

(4)

4f

(3)

Li

(4)

Mg

(3)

S

(4)

Cl

Ca

(4)

Ar

Sol. Answer (1)

s orbital have more screening power.

8.

The alkali metal which is radioactive is

(1) Fr

(2)

Al

Sol. Answer (1)

Fr which is also liquid at room temperature.

9.

The element which belongs with chalcogen family is

(1) N

(2)

P

Sol. Answer (3)

16th group is chalcogen.

10. The element which has highest 2nd ionisation energy is (1) Na

(2)

Mg

(3)

Sol. Answer (1)

Electron will be removed from completely filled second orbit.

11. An atom with high electronegativity generally has (1) Low electron affinity

(2)

Small atomic number

(3) Large atomic radius

(4)

High ionisation potential

Sol. Answer (4)

Higher electronegativity implies higher ionization potential.

12. Ionic radii are

(1) Greater than the respective atomic radii of elements in general

(2) Greater than the respective atomic radii of electropositive elements (3) Greater than the respective atomic radii of electronegative elements (4) Less than the respective atomic radii of electronegative elements Sol. Answer (3) During formation of Anions, electron is added to neutral atom, hence size of anion is bigger, compared to neutral atom. 13. The first ionisation potentials of Na, Mg, Al and Si are in the order (1) Na < Mg > Al < Si

(2)

Na < Mg < Al > Si

(3) Na > Mg > Al > Si

(4)

Na > Mg > Al < Si

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

69

Sol. Answer (1) Size; Mg < Na Hence, ionization energy of Na < Mg Size; Si < Al Hence, ionization energy of Si > Al Mg = 1s2 2s2 2p6 3s2

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

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

Mg has stable configuration, hence its ionization energy will be higher than Al.  Na < Mg > Al < Si

14. Which of the following has maximum number of unpaired electrons? (1) Mg2+

(2)

Ti3+

(3)

V3+

(4)

Fe2+

Sol. Answer (4)

Fe2+ has four unpaired electrons.

15. Which of the following elements do not belong to the family indicated? (1) Cu – Coinage metal

(2)

Ba – Alkaline earth metal

(3) Zn – Alkaline earth metal

(4)

Xe – Noble gas

Sol. Answer (3)

Zn is not an alkaline earth metal, IIA group elements are called alkaline earth metals.

16. If the atomic number of an element is 33, it will be placed in the periodic table in the (1) 1st group

(2)

3rd group

(3)

15th group

(4)

17th group

Sol. Answer (3)

33 = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 Above is a p-block element

group number = 10 + number of valence electrons = 10 + 5 = 15

17. Which among the following elements has the highest value for third ionisation energy? (1) Mg

(2)

Al

(3)

Na

(4)

Ar

Sol. Answer (1)

Since, Mg belongs to IIA group hence, after removal of 2e–, atom will become stable, and hence, removal of 3rd electron will require high energy.

18. The screening effect of inner electrons on the nucleus causes (1) A decrease in the ionisation energy (2) An increase in the ionisation energy (3) No effect on the ionisation potential (4) An increase in the attraction of the nucleus on the outermost electrons Sol. Answer (1) Due to screening effect, repulsion on electron increases which implies that attraction of nucleus on electrons decreases, hence ionization energy decreases.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 70

Classification of Elements and Periodicity in Properties

Solution of Assignment

19. The diagonal similarities are due to similar polarising powers for the elements The polarising power is directly proportional to (1)

ionic charge ionic radius

(2)

(ionic charge) 2 ionic radius

ionic charge

(3)

(ionic radius)

2

ionic charge

(4)

(ionic radius)1/2

Sol. Answer (3) ionic charge

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Polarising power 

(ionic radius) 2

20. The ionisation potential of isotopes of an element will be (1) Same

(2) Different

(3) Dependent on atomic masses

(4) Dependent on the number of neutrons present in the nucleus

Sol. Answer (2)

21. Which of the following gradation in the properties is false, as we move from left to right in the periodic table? (1) Metallic to non-metallic character (2) Oxidising to reducing properties

(3) Metallic solids through network solids to molecular solids (4) Base forming to acid forming character

Sol. Answer (2) Factual

22. The first electron affinity values of ‘O’, S & Se are given correctly as (1) O > S > Se

(2)

S > Se > O

(3)

Se > O > S

(4)

Se > S > O

(3)

K

(4)

B

(3)

Nitrogen

(4)

Phosphorus

Sol. Answer (2) Element

Electron gain enthalpy

Electron affinity

O

–144

144

S

–200

200

Se

–195

195

 S > Se > O

23. The element which has highest IInd I.E.? (1) Li

(2)

Be

Sol. Answer (1) IA group elements have highest IInd ionization energy. 24. The element which has highest electron affinity? (1) Oxygen

(2)

Sulphur

Sol. Answer (2) Resultant of size factor and electronic configuration factor.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

71

25. Electronegativity of an element is 1.0 on the Pauling scale. Its value on Mulliken scale will be (1) 2.8

(2)

1

(3)

2.0

(4)

1.5

Sol. Answer (1) Emuliken =

2.8 Epauling 2.8 × 1

=

2.8

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

=

26. Which of the following element has highest shielding constant? (1) Mg

(2)

Al

(3)

K

(4)

Ga

Sol. Answer (4) Mg

12e–

Al

13e–

K

19e–

Ga

31e–

Ga has highest atomic number (i.e. number of electrons), hence, it will have highest shielding constant.

27. Which one of the following order is correct? (1) I > I+ > I–

(radii)

(2) I– > I > I+

(radii)

(3) I– > I > I+

(Ionisation energy)

(4) I+5 < I+ < I+7

(Ionisation energy)

Sol. Answer (2)

Anions are larger than the neutral atom, while cations are smaller than the neutral atom.

28. Which one of the following oxides has highest acidic character? (1) CO2

(2)

Cl2O7

(3)

SiO2

(4)

SO2

Sol. Answer (2) IV

VI

CO2

SiO2

VII

Cl2O7

SO2

As we move from left to right across a period, acidic character of oxides of elements increases. As we move from top to bottom in a group, acidic character of oxides of element decreases.

29. Electron gain enthalpy will be positive in (1) O–2 is formed from O–1

(2)

O–1 is formed from O

(3) S–1 is formed from S

(4)

Na– is formed from Na

Sol. Answer (1) O –  e –  O –2

When e– will approach O–, O– will repel the approaching electron, hence, we have to give some energy therefore, electron gain enthalpy will be +ve (endothermic).

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 72

Classification of Elements and Periodicity in Properties

Solution of Assignment

30. Enthalpy change in the following process is A + e–  A– H = –X kJ/mole Which of the following processes has enthalpy change = X kJ/mole? (2)

(3) A–  A + e–

(4)

A + e–  A– A+ + e–  A

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

(1) A–2  A– +1e–

Sol. Answer (3)

A + e–  A– ; H = –x

A–  A + e– ; H = +x

As direction of reaction is reversed, sign of heat exchanged is also reversed.

31. According to Mulliken method, electronegativity is a function of (1) Ionization energy and Electron affinity

(2)

Ionization energy and size

(3) Effective nuclear charge and size

(4)

Effective nuclear charge and Electron affinity

Sol. Answer (1) Fact

32. Which of the following statement(s) is incorrect?

(1) The electronic configuration of Cr is [Ar] 3d 54s1 (Atomic No. of Cr = 24) (2) The magnetic quantum number may have a negative value

(3) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type (Atomic number of Ag = 47)

(4) Vander Waal radius of Cl is less than its covalent radius

Sol. Answer (4)

Cr (24) = 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (Half filled-full filled rule)

Magnetic quantum number varies from +l to –l through zero for a given value of l. Ag(47) = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d10 (Half filled-full filled rule)

In Ag except 5s1, all subshells are full filled which contains total 46e– out of which 23e– have anticlockwise spin, hence total number of e– having clockwise spin = 23 + 1(5s1) = 24 Total no. of e– having anticlockwise spin = 23 In HN3 0.5 of N is non zero N

N

N

H

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

73

33. Uub is the symbol for the element with atomic number (1) 102

(2)

108

(3)

110

(4)

112

Sol. Answer (4) Uub U = un = 1

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

u = un = 1 b = bium = 2

Hence element has atomic number 112.

34. Zr and Hf have nearly same size because (1) They belong to same group

(2)

They belong to same period

(3) Of lanthanoid contraction

(4)

Of poor screening of d orbitals

Sol. Answer (3)

Zr and Hf have nearly same size due to lanthanide contraction

35. An element has electronic configuration [Xe] 4f7, 5d1, 6s2. It belongs to ..... block of the periodic table. (1) s

(2)

p

(3)

d

(4)

f

Sol. Answer (4)

[Xe] 4f7 5d1 6s2

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d1 4f7

Last electron enters the f sub shell, hence, it belongs to f block.

36. A sudden large jump between the values of second and third ionisation energies of an element would be associated with the electronic configuration (1) 1s2, 2s2p6, 3s1

(2)

1s2, 2s2p6, 3s2p1

(3)

1s2, 2s2p6, 3s2p2

(4)

1s2, 2s2p6, 3s2

Sol. Answer (4)

For IIA group elements i.e., elements containing 2e– in outermost shell, there is a sudden jump between values of 2nd and 3rd ionization energy (because in 3rd ionization we have to remove electron from a stable configuration)

37. An element has 56 nucleons in nucleus and if it is isotonic with to?

60 30Y .

Which group and period does it belong

(1) 8th group, 4th period

(2)

14th group, 3rd period

(3) 12th group, 3rd period

(4)

12th group, 4th period

Sol. Answer (1) n + p = 56 for y, n = 60 – 30 = 30 hence for given element n = 30 p = 26 P(26) = 1s2 2s2 2p6 3s2 3p6 4s2 3d6 period = 4, group number = 2 + 6 = 8

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 74

Classification of Elements and Periodicity in Properties

Solution of Assignment

38. Arrange the given species in the increasing order of electronegativity –F I

–Cl II

–I III

(1) III < IV < II < I

–Br IV

(2)

I < II < IV < III

(3)

I < III < IV < II

(4)

III < I < IV < II

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Sol. Answer (1) 39. Which of the following processes will release energy equal to ionization energy? 

e (1) M( g)   M(( g))

(2)

e () M(s)  M(s)

(3)



e M(g)   M( g)

(4)



e M( g)   M(( g))

Sol. Answer (3)

The process is opposite of Ionization energy.

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

(1) The properties of elements are the periodic function of their atomic numbers (2) Non-metallic elements are lesser in number than metallic elements

(3) The first ionisation energy of elements along a period do not vary in a regular manner with increase in atomic number

(4) For transition elements the d-subshells are filled with electrons with increase in atomic numbers

Sol. Answer (3)

Generally IE increases along period with few exceptions.

41. The chemistry of Be is very similar to that of aluminium, because (1) They belong to same group

(2) They belong to same period

(3) Both have nearly the same ionic size

(4) The ratio of their charge to size is nearly the same

Sol. Answer (4)

Be and Al show diagonal relationship hence we can say that, for these two elements charge to size ratio is nearly same. 42. Hypothetically if one orbital bear three electrons then how many groups are present in 2nd period? (1) Twelve

(2)

Eight

(3)

Six

(4)

Nine

Sol. Answer (1) 43. The ionization potential of lithium is 520 kJ/mole. The energy required to convert 70 mg of lithium atoms in gaseous state into Li+ ions is (1) 5.2 kJ

(2)

52 kJ

(3)

520 kJ

(4)

52 J

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

75

Sol. Answer (1) 70  10 –3  10  10 –3 7

Number of moles of Li =

= 10–2 mole

∵ 1 mole Li requires 520 kJ energy

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

 10–2 mole Li will require = 10–2 × 520 kJ

= 5.2 kJ energy

44. Which of the following configuration is associated with the biggest jump between first and second ionization energy? (1) 1s22s22p5

(2)

1s22s22p63s1

(3)

1s22s22p4

(4)

1s22s1

Sol. Answer (4)

IA group element (ns1) have biggest jump between Ist and IInd ionization energy.

45. If an element A shows two oxidation states +2 and +3 and form oxides in such a way that ratio of the element showing +2 and +3 state is 1 : 3 in a compound. Formula of the compound will be (1) A8O11

(2)

A4O11

(3)

A9O11

(4)

A5O11

Sol. Answer (1)

Total +ve charge (A1A3) =

1 × (+2) + 3 × (+3)

=

2 + 9 = 11

For compound to be neutral, O must contain total 11 unit negative charge ∵ –2 unit charge possessed by one oxygen atom

 –11 unit charge possessed by =

–11  1 11 = oxygen atom –2 2

Hence formula = A1A3O 11 = A4O 11 = A8O11 2

2

SECTION - B

Objective Type Questions (More than one options are correct)

1.

Choose the correct pair regarding ionisation energy (IE1) (1) B > Be

(2)

Tl > Ga

(3)

N>O

(4)

Li > Na

(3)

K

(4)

Rb

Sol. Answer (2, 3, 4) Tl have has lower IE1 than Ga. 2.

Which of the following elements have same Zeffective? (1) Na

(2)

Li

Sol. Answer (1, 3, 4) s block, except 2nd period, all have same electronegativity.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 76 3.

Classification of Elements and Periodicity in Properties

Solution of Assignment

The elements which belong to p-block is/are (1) Fe

(2)

Ga

(3)

Na

(4)

I

Sol. Answer (2, 4) Ga and iodine belong to p block. Choose the correct statement/statements regarding Modern periodic table

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

4.

(1) Actinoids are placed in main body of periodic table

(2) Chemical properties are periodic function of atomic number (3) In periodic table, 18 groups are present (4) 7th period is incomplete

Sol. Answer (2, 3, 4)

Actinoids are placed separately.

5.

Which of the following is/are correct pair regarding size?

(1) Zn > Cu

(2)

N>O

(3)

Al > Ca

(4)

B>C

Sol. Answer (1, 4)

Zn has larger size than Cu.

6.

Choose the correct option(s)

(1) Oxygen has highest electron gain enthalpy among chalcogens (2) Be+2 < Li+ < Na+ < K+ (radii)

(3) F is 2nd most electronegative element

(4) Cl has highest negative electron gain enthalpy

Sol. Answer (2, 4) Element

Electron gain enthalpy

O

– 144

S

– 200

Se

– 195

T

– 190

Po

– 174

IA

IIA

Li

Be

Na K As we move from left to right across a period, size decreases. As we move from top to bottom in a group, size increases.

 Be+2 < Li+ < Na+ < K+ F is most electronegative element known. Cl has highest negative electron gain enthalpy.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

7.

Classification of Elements and Periodicity in Properties

77

Choose the process which is/are endothermic. (1) O–1 + e–  O –2

(2)

Ar  Ar+ + e –

(3) Ar + e–  Ar –

(4)

H + e–  H –

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Sol. Answer (1, 2, 3) O– will repel electron, hence, addition of electron in O– is endothermic Ar  Ar   e – , process is ionization, hence, endothermic

Ar  e –   Ar – , Ar has noble gas configuration, hence, addition of e– is not favored hence we have to give

some energy, therefore, process is endothermic

H  e–   H2 – , H– is getting noble gas configuration hence it will favors addition of e–. When H is

1s1

1s

converted to H–, it gains stability hence loses energy.

8.

Which of the following is / are correct order regarding radius? (1) Be+2 < B+3 < Li+

(2)

B+3 < Be+2 < Li+

(3) F– < O–2 < N–3

(4)

B+3 < Ga+3 < Al+3

Sol. Answer (2, 3, 4) B+3 = 0.02 Å

Be+2 = 0.31 Å Li+ = 0.76 Å

B+3 < Be+2 < Li+

F–, O–2, N–3 are isoelectroelectronic, hence higher the charge on nucleus lesser will be radius F– < O–2 < N–3

B+3 < Ga+3 < Al+3

Size increases due to poor shielding of d electrons.

9.

Choose the correct order (1) Cl < S < P < Si (2nd Ionisation energy) (2) Si < P < Cl < S (2nd Ionisation energy) (3) Be+2 < Li+ < Na+ < K+ (Inverse of charge density) (4) Na & Mg are electropositive elements

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 78

Classification of Elements and Periodicity in Properties

Solution of Assignment

Sol. Answer (2, 3, 4) Si 3s2 3p2

Si+ 3s2 3p1

3s2 3p3

P+ 3s2 3p2

Cl 3s2 3p5

Cl+ 3s2 3p4

3s2 3p4

S+ 3s2 3p3

P

S

V VI VII

Si+

P+ S+ Cl+

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

IV

S+ has stable configuration so second ionization energy of S will be more than Cl+ Si < P < Cl < S Size

Li+ < Na+ < K+

Size of

Be+2 < Li+ < Na+ < K+

Lesser cation is more hydrated, hence has lower ionic mobility So ionic mobility  size

So correct order of ionic mobility is Be+2 < Li+ < Na+ < K+ Na and Mg are electropositive elements.

10. Which of the following pairs contain metalloids? (1) Si, Ge

(2)

As, Te

(3)

I, Sb

(4)

In, Tl

(3)

Ra

(4)

Cr

Sol. Answer (1, 2)

Si, Ge, As, Sb, Te are metalloids

11. Which of the following elements belongs to 16th group? (1) Se

(2)

Te

Sol. Answer (1, 2)

12. The correct statement about d block element is/are (1) They are all metals

(2) They show variable valency

(3) All these elements have full (n – 1) d subshell (4) They have strong tendency to gain electron

Sol. Answer (1, 2) 13. Choose the correct statement(s) (1) Cu, Ag, Au are known as coinage metals (2) Ce, Gd, U are Lanthanoids (3) H is placed in 1st group (4) N has lower first ionization energy, than F Sol. Answer (1, 4)

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

79

14. Choose the correct statement(s). (1) [Ar] 3d10 4s2 4p6 element is a noble gas

(2)

[Ar] 3d 5 4s1 element belongs to s block

(3) 1s2 element belongs to f block

(4)

[Xe] 4f 1 5d 1 6s2 element belongs to f block

Sol. Answer (1, 4) [Ar] 3d10 4s2 4p6 is Kr (36) which is a noble gas element [Ar] 3d5 4s1 is Cr(24) which is a d block element.

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

[Xe] 4f1 5d1 6s2 has atomic number 58 which is Ce (58) a lanthanoid belonging to f block. 15. Which of the following is/are correct order in accordance of electropositive nature of metal? (1) Fe < Mg < Cu

(2)

Na > Mg > Al

(3)

Mg < Ca < Sr

(4)

Fe > Cu > Zn

Sol. Answer (2, 3, 4) IA IIA IIIA

Na Mg Al

In a period electropositive (metallic) character decreases. IIA

Mg Ca Sr

Going top to bottom in a group, size increases, hence electropositive (metallic) character also increases.

Fe forms more +ve ions like Fe+2 and Fe+3 while Cu can form only Cu+ and Cu+2 and Zn can form only Zn+2

16. Choose the pair in which IE1 of first element is greater than IE1 of second element but in case of IE2 order is/are reversed (1) N, O

(2)

P, S

(3)

Be, B

(4)

F, O

Sol. Answer (1, 2, 3, 4)

In I, II and III stable electronic configuration of the first element is the reason while for the 4th choice. IE of 1st member is greater due to Zeff.

17. Which of the following pairs have nearly same size? (1) Fe, Co

(2)

Zr, Hf

(3)

Ru, Rh

(4)

Nb, Ta

Sol. Answer (1, 2, 3, 4) Element

Radius in Å

Fe

1.17

Co

1.16

Zr

1.45

Hf

1.44

Ru

1.24

Rh

1.25

Nb

1.34

Ta

1.34

1, 2, 3, 4 all have nearly same size

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 80

Classification of Elements and Periodicity in Properties

Solution of Assignment

18. Which of the following pair of oxides are acidic? (1) CO2 and NO2

(2)

SO3 and SO2

(3)

MgO and Na2O

(4)

Li2O and K2O

Sol. Answer (1, 2) 19. Which of the following sequence contains atomic number of only representative elements? (2)

13, 33, 54, 83

(3)

3, 33, 53, 87

(4)

22, 33, 55, 66

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

(1) 55, 12, 48, 53 Sol. Answer (2, 3)

The p block elements comprise those belonging to group 13 to 18 and these together with s block elements are called the representative or main group elements

Set 1

Set 2

Set 3

Set 4

⎡55 Cs s block ⎢12 Mg s block ⎢ ⎢ 48 Cd d block ⎢ p block ⎣53 I ⎡13 ⎢33 ⎢ ⎢54 ⎢ ⎣83

Al

p block

As p block Xe noble gas  p block

Bi

p block

⎡3 ⎢33 ⎢ ⎢53 ⎢ ⎣87

B p block As p block I p block

⎡22 ⎢33 ⎢ ⎢55 ⎢ ⎣66

Ti d block As p block Cs s block

Fr

s block

Dy f block

All elements of Set 2, 3 only belongs to either s or p block, hence these sets belong to representative elements.

20. Choose the correct order

(1) Mo(II) > Mo(III) > Mo(IV) (Electronegativity order) (2) Fe(I) < Fe(II) < Fe(III) (Electronegativity order) (3) Fe(III) < Fe(II) < Fe(I) (Ionic radius)

(4) Mo (IV) > Mo (III) > Mo(II) (Electronegativity order) Sol. Answer (2, 3, 4)

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

81

SECTION - C Linked Comprehension Type Questions Comprehension I Ionisation energy is the amount of energy required to remove the outermost e– from a gaseous atom. It’s unit is kJ/mole or kcal/mole.

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Successive ionisation energy – It is the amount of energy required to remove electron successively from a gaseous ion. These are termed as IE2, IE3, IE4 etc. The difference in the values of IE1, IE2 and IE3 helps to determine electronic configuration of the element. Element

IE1

IE2

IE3

EA

A

150

350

1920

–50

B

52

729

1181

60

C

418

1091

1652

349

D

550

1025

1500

–495

All data is reported in Kcal/mol 1.

Which element forms stable unipositive ion?

(1) A

(2)

B

(3)

C

(4)

D

Sol. Answer (2)

It is an alkali metal because it shows greatest jump between 1st and 2nd IE.

2.

Which of the following is a non-metal?

(1) A

(2)

B

(3)

C

(4)

D

B

(3)

C

(4)

D

Sol. Answer (3, 4)

Both (3) & (4) are possible.

3.

Which element is a noble gas?

(1) A

(2)

Sol. Answer (4)

Noble gas have very high ionization energy.

Comprehension II

Mulliken defined the electronegativity of an atom as the arithmetic mean of its ionisation energy and electron affinity. 1 (I.P.  E.A.) 2 One more relationship given by him, if the values are given in eV is A 

A 

Ionisation potential  Electron affinity 5.6

When there is pure covalent bond between A – B (IP) A  (EA) A (IP)B  (EA)B  5.6 5.6

 A  B

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 82 1.

Classification of Elements and Periodicity in Properties

Solution of Assignment

According to Mulliken, electronegativity depends on (1) Ionisation potential

(2)

Electron gain enthalpy

(3) Electron affinity

(4)

Both (1) & (3)

Sol. Answer (4) According Mulliken IE  EA 2

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

EN 

Hence EN depends on both ionization energy as well as electron affinity 

2.



When there is formation of A  B bond then condition will be

(1)

(IP ) A  (EA ) A (IP )B  (EA )B  5 .6 5 .6

(2)

(IP )B  (EA )B (IP ) A  (EA )B  5 .6 5 .6

(3)

(IP ) A  (EA )B (IP ) A  (EA )B  2 .8 2 .8

(4)

(IP) A  (EA )B (IP) A  (EA ) A  5 .6 5 .6

(3)

E.A. data

Sol. Answer (1) –

+

A B

A

x

B

which implies that (EN)A > (EN)B

(IP)A  (EA)A (IP)B  (EA)B  5.6 5.6

3.

Pauling’s electronegativity scale is based on

(1) Thermochemical data (2)

I.E. data

(4)

Both (2) & (3)

Sol. Answer (1)

Pauling’s electronegativity is based on bond energy data i.e. thermo chemical data.

SECTION - D

Assertion-Reason Type Questions

1.

STATEMENT-1 : Nitrogen and oxygen have nearly same size. and

STATEMENT-2 : Electron-electron repulsions tend to increase the size. Sol. Answer (2) Due to electron-electron repulsion attraction of nucleus decreases. 2.

STATEMENT-1 : Nitrogen can form maximum four bonds with hydrogen. and STATEMENT-2 : Valency is the combining capacity of element and it is always constant.

Sol. Answer (3) Valency is available quantity.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

3.

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

Classification of Elements and Periodicity in Properties

83

STATEMENT-1 : Noble gases have high ionisation energy and STATEMENT-2 : Noble gases belong to 18th group.

Sol. Answer (2)

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Due to inert configuration, I.E. is very high. 4.

STATEMENT-1 : Actinoides belongs to f block.

and

STATEMENT-2 : Lanthanoid belongs to 3rd group.

Sol. Answer (2) 5.

STATEMENT-1 : Fluorine has less electron affinity than chlorine.

and

STATEMENT-2 : Due to small size, more electron-electron repulsions are observed in F.

Sol. Answer (1)

It is due to very high shielding effect in F.

6.

STATEMENT-1 : Long form of periodic table exactly explain the position of hydrogen. and

STATEMENT-2 : Hydrogen is most abundent element of the universe.

Sol. Answer (4)

Properties of hydrogen match with Ist & 17th group both.

7.

STATEMENT-1 : He and H– have same size

and

STATEMENT-2 : He and H– have same number of electrons in valence shell.

Sol. Answer (4)

In H– and He, effective nucleus charges are different.

8.

STATEMENT-1 : B2O3 is more acidic than BeO.

and STATEMENT-2 : Ionisation energy of B is more than Be. Sol. Answer (3) Ionisation energy of Be is greater than B. 9.

STATEMENT-1 : Bond dissociation energy of F2 is more than that of Cl2. and STATEMENT-2 : Due to smaller size of fluorine there are greater electron repulsions between the F atoms than Cl atoms.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES 84

Classification of Elements and Periodicity in Properties

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Solution of Assignment

Sol. Answer (4) Bond dissociation energy of F2 is less than that of Cl2 due to smaller size of fluorine there is greater electronic repulsion between F atoms than Cl atoms. 10. STATEMENT-1 : Ions Na+, Mg2+, Al3+ are isoelectronic. and

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

STATEMENT-2 : In each ion, the total number of electrons are 10.

Sol. Answer (1)

Na+ = 11 – 1 = 10

Mg+2 = 12 – 2 = 10 Al+3 = 13 – 3 = 10

Na+, Mg+2, Al+3 all have 10e–, hence, isoelectronic

11. STATEMENT-1 : First Ionisation energy of beryllium is more than that of boron. and

STATEMENT-2 : In boron, 2p orbital is fully filled, whereas, in beryllium, it is not fully filled.

Sol. Answer (3)

Be has stable outermost electronic configuration, hence has high ionisation energy Be(4) = 1s2 2s2

B(5) = 1s2 2s2 2p1

In boron 2p orbital is not full-filled In Be 2p orbital is empty

12. STATEMENT-1 : Fluorine is more electronegative than chlorine. and

STATEMENT-2 : Fluorine is smaller in size than chlorine.

Sol. Answer (2)

F is highest electronegative element known F is smaller in size than chlorine.

Both size and Zeff explains the equation.

13. STATEMENT-1 : Ionization energy of s-electrons are more than the p-electrons for the same shell. and STATEMENT-2 : s-electrons are closer to the nucleus than p-electrons, hence, more tightly attached. Sol. Answer (1) s electrons are closer to nucleus than p electrons, hence ionisation energy of s electrons is higher than p electrons 14. STATEMENT-1 : Li and Mg show diagonal relationship. and STATEMENT-2 : Li and Mg have nearly same atomic radius.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

Classification of Elements and Periodicity in Properties

85

Sol. Answer (3) Li, Mg and Be, Al and B, Si show diagonal relationship 1.23 Å

Mg

1.36 Å

Be

0.89 Å

Al

1.25 Å

B

0.80 Å

Si

1.17 Å

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Li

Li and Mg do not have same atomic size

15. STATEMENT-1 : He and Be both have the same outer electronic configuration like ns2 type. and

STATEMENT-2 : Both are chemically inert.

Sol. Answer (3) He 1s2

Be 1s2 2s2

He and Be both have similar electronic configuration like ns2 Be forms compounds, hence it is not inert.

SECTION - E

Matrix-Match Type Questions

1.

Match the following Column I

Column II

(Element)

(Electronegativity on Pauling scale)

(A) Carbon

(p)

0.8

(B) Nitrogen

(q)

1.6

(C) Aluminium

(r)

2.5

(D) Cesium

(s)

3.0

Sol. Answer A(r), B(s), C(q), D(p) Element

Electronegativity

C

2.5

N

3.0

Al

1.6

Cs

0.8

Increasing order of EN Cs < Al < C < N3 Cs is strongly electropositive hence least electronegative

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES 86

Solution of Assignment

Match the following Column I

Column II

(Element)

(Size (r) in Å)

(A) K+

(p)

0.74

(B) Cu+

(q)

0.99

(C) Ca+

(r)

0.96

(D) Zn+2

(s)

1.33

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

2.

Classification of Elements and Periodicity in Properties

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

Sol. Answer A(s), B(r), C(q), D(p) Zn++ < Cu+ < Ca+ < K+

3.

K+

1.33 Å

Cu+

0.96 Å

Ca+

0.99 Å

Zn+2

0.74 Å

Match the following Column I

Column II

(A) Zirconium (Atomic no. 40)

(p)

Group no. 15; Period no. 4

(B) Thalium (Atomic no. 81)

(q)

Group no. 4; Period no. 5

(C) Arsenic (Atomic no. 33)

(r)

Group no. 14; Period no. 5

(D) Tin (Atomic no. 50)

(s)

Group no. 13; Period no. 6

Sol. Answer A(q), B(s), C(p), D(r) 4.

Match the following Column I

Column II

(Element)

(Effective nuclear charge for the outermost electron)

(A) Magnesium

(p)

2.20

(B) Potassium

(q)

2.60

(C) Aluminium

(r)

2.85

(D) Boron

(s)

3.50

Sol. Answer A(r), B(p), C(s), D(q)

SECTION - F Integer Answer Type Questions 1.

What is the group number of element Une?

Sol. Answer (9) Atomic number is 109.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

2.

Classification of Elements and Periodicity in Properties

87

What is screening constant for outer electron of H?

Sol. Answer (0) In H only one electron is present so, no other electron which can screen it. 3.

Which group will show lowest second ionization energy? (Non radioactive element)

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Sol. Answer (2) Fact

4.

The element which has highest electron affinity will belong with group number x. Then x – 10 will be.

Sol. Answer (7)

Cl(17th group) have maximum affinity. 17 – 10 = 7

5.

To which group Ce belongs?

Sol. Answer (3)

All f-block elements belongs to IIIrd group.

6.

The largest group in Modern periodic table is _______.

Sol. Answer (3)

All f-block elements & other five elements belongs to IIIrd group.

SECTION - G

Multiple True-False Type Questions

1.

STATEMENT-1 : Electronegativity of an element depends on ionisation energy of that element. STATEMENT-2 : Electronegativity regularly decreases along the group. STATEMENT-3 : Fluorine is the most electronegative element. (1) T F T

(2)

TTT

(3)

FTT

(4)

FFT

Sol. Answer (1) 2.

STATEMENT-1 : In Mendeleev’s periodic table arrangement of elements depend on their atomic weight.

STATEMENT-2 : Some vacant sites were present in this periodic table. STATEMENT-3 : Position of isotope was well explained by Mendeleev (1) T T F

(2)

FFT

(3)

TFT

(4)

TFF

FFT

(4)

FFF

Sol. Answer (1) 3.

STATEMENT-1 : In Tl Lanthanoide contraction is observed STATEMENT-2 : In Ga actinoide contraction is observed STATEMENT-3 : Ionisation energy of Pb is more than Sn. (1) TTT

(2)

TFT

(3)

Sol. Answer (2)

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 88

Classification of Elements and Periodicity in Properties

Solution of Assignment

SECTION - H Aakash Challengers Questions 1.

Which of the following has the largest size in aqueous medium? (1) Li+

(2)

Na+

(3)

Cs+

(4)

K+

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Sol. Answer (1) Li+ due to high hydration energy.

2.

Which of the following has the largest ionisation energy?

(1) Zn

(2)

Sc

(3)

Cd

(4)

Hg

(4)

Mn2O5

Sol. Answer (4)

Hg ; 1007 kJ/mole

3.

s-orbital is more penetrating because

(1) Probability of finding electron is maximum on the surface of nucleus (2) s-orbital is non-directional

(3) s-orbital is spherical in shape (4) All of these

Sol. Answer (1) Fact

4.

Cs metal imparts colour to the flame because

(1) Cs has low ionisation energy

(2)

Cs is a soft metal

(3) Cs has low density

(4)

Cs has large size

(3)

MnO2

Sol. Answer (1)

Due to low ionization energy.

5.

In which compound Mn has highest electronegativity? (1) MnO

(2)

Mn3O4

Sol. Answer(4)

Higher will be oxidation state higher will be electronegative. 6.

The correct pair regarding property given in bracket is (1) F2 > Cl2

(Oxidising character)

(2) F > Cl

(Electron affinity)

(3) F < Cl

(Electronegativity)

(4) O > N

(Ionisation energy)

Sol. Answer (1) F2 is stronger oxidizing agent.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

7.

Classification of Elements and Periodicity in Properties

89

The oxide which has highest acidic character is (1) MnO

(2)

MnO2

(3)

Mn2O3

(4)

Equal in all of these

Sol. Answer (2) MnO2 Choose the correct statement

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

8.

(1) Isotopes have nearly same chemical properties (2) Isoelectronic species may be neutral (3) Na and K have nearly same Zeffective

(4) All of these

Sol. Answer (4) Fact

9.

Maximum number of electrons in nth shell is

(1) n2

(2)

2(l + 1)

(3)

2n2

(4)

(l + 1)2

(3)

B > In > Ga = Al

(4)

B > In > Ga > Al

Sol. Answer (3)

10. Choose the correct regarding E.N. (1) B > Al > Ga > In

(2)

B > Al = Ga = In

Sol. Answer (4)

B > Tl > In > Ga > Al B = 2.0

Al = 1.5

Ga = 1.6 In = 1.7 Tl = 1.8

11. Covalent radius of an element having 82 electrons in extranuclear part and 82 protons in the nucleus is 146 Å. Calculate the electronegativity on Allred Rochow scale of that element. Sol. On Allred Rochow scale, EN  0.744 

0.359 Z eff r2

r = radius in Å 82 = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p2  = 3 × 0.35 + 18 ×0.85 + 60 × 1.0 = 1.05 + 15.3 + 60  = 76.8

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES 90

Classification of Elements and Periodicity in Properties

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX Solution of Assignment

Z* = S –  = 82 – 76.8 Z* = 5.2 EN  0.744 

0.359  5.2 (146) 2 0.359  5.2 21316

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

 0.744 

= 0.744 + 8.75 × 10–5 = 0.74405

12. Calculate the energy required to convert all the atoms of M (atomic number : 12) to M2+ ions present in 12 mg of metal vapours. First and second ionization enthalpies of M are 737.77 and 1450.73 kJ mol–1 respectively. Sol. Number of moles =

12  10 3 = 0.0005 = 5 × 10–4 24

Total energy required for ionization of 1 mole Mg from M to M+2 = 737.77 + 1450.73 = 2188.5

∵ 1 mole  2188.5 kJ

 5 × 10–4 mole  5 × 10–4 × 2188.5 kJ  1.09425 kJ

13. Calculate the electronegativity value of an atom on Pauling scale is, IP = 13.0 eV and EA = 4.0 eV.

Sol. EN 

EN 

IP  EA 5.6

13  4  3.03 5.6

14. There are three elements I, II, III and their ionisation energies are given below. IE1

IE2

I

2372

5251

II

900

1760

III

520

7300

Now identify (a) Which element is most reactive? (b) Which element is noble gas? (c) Which can form a stable dihalide? Sol. Element III has least (IE)1 hence, most reactive Element I has highest (IE)1 hence, it is a noble gas Element II has least (IE)2 hence, it can form easily M+2 ion, hence, it can form stable dihalide.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

R. K. MALIK’S NEWTON CLASSES Solution of Assignment

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

Classification of Elements and Periodicity in Properties

91

15. Out of Ne and Na+, which has higher ionisation energy and why? Sol. Na+ & Ne both have 10e– i.e. isoelectronic, hence, greater the nuclear charge, lesser will be radius and larger will be IE Na+ (Z = 11) Ne (Z = 10)

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Hence, Na+ have higher IE.

16. Alkali metals and coinage metals both have ns1 electron in outermost orbitals.

(a) Out of these two groups, which group elements have higher stability. Explain. (b) Compare the polarising power of their cation.

Sol. (a) Group II (coinage metals) elements differ from group I elements in that, the penultimate shell contains 10d electrons. The poor screening by the d electrons makes the atoms of Cu group much smaller in size. This results in higher stability of coinage metals. (b) Coinage metals having smaller size have high polarizing power.

17. Why lanthanides and actinoides have been placed separately in the periodic table?

Sol. The initial separation of lanthanides and actinides are based on slight differences in solubility. 18. Arrange the following in increasing order as directed (a) Na2O, Cl2O, NO2, N2O, P2O5 (Acidic strength)

(b) Li+, Be+, B+, C+ (Stability)

(c) B, Al, Ga, In, Tl (First Ionisation energy)

Sol. (a) Order of acidic strength is Na2O < P2O5 < Cl2O < N2O < NO2 (b) Li+

1s2

Be+

1s2 2s1

B+

1s2 2s2

C+

1s2 2s2 2p1

Hence, order of stability is Li+ > B+ > C+ > Be+ (c) B, Tl, Ga, Al, In (Ist ionization energy) *

On moving from B to Al (IE)1 decreases as expected and this decrease is due to an increase in atomic size and shielding effect.

*

On moving from Al to Ga, the (IE)1 , increases slightly, because on moving from Al to Ga, both nuclear charge and shielding effect increase, but due to poor shielding d electron (3d10) in Ga, effective nuclear charge on valence electron increases, resulting in d-block contraction, that is why ionization enthalpies increase.

*

On moving from Ga to In again there is slight decrease in ionization enthalpies due to increased shielding effect of additional ten 4d electrons, which outweighs the effect of increased nuclear charge.

*

On moving from In to Tl, ionization enthalpies show an increase again because 14, 4f electrons shield valence electron poorly (order of shielding effect (s > p > d > f) and so effective nuclear charge increases, consequently ionization enthalpies increase.

Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968

JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, X & IX

R. K. MALIK’S NEWTON CLASSES 92

Classification of Elements and Periodicity in Properties

Solution of Assignment

19. Why Al2O3 is amphoteric while B2O3 is acidic?

R. NE K . WT M ON A LI RA C NC L K’ A HI S S SE S

Sol. Acidic nature of boron oxide (or hydroxide) may be explained on the basis of small size of boron. Due to small size, there is high positive charge density on atom. This pulls off electron pair from water so O–H bond is weakened facilitating the release of H+ giving acidic solution. As Al+3 and Ga+3 ions are relatively larger hence their tendency to rupture the O–H bond becomes somewhat less. The result of this is, that acidic nature decreases and oxides become amphoteric.







Office.: 606 , 6th Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2562523, 9835508812, 8507613968