Chem Periodic Properties

SECTION - A

(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 Sol. Answer (2)

Classification is based on Mendeleev’s periodic law.

3. With which block ₃₀Zn belongs?

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

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 (3) K (4) Rb

Sol. Answer (2)

5. Out of following, which has the highest electronegativity?

(1) H (2) Li (3) Na (4) Be

Sol. Answer (1) H-2.1

Chapter

3

Solutions

Classification of Elements and

Periodicity in Properties

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6. Which of the following has largest size?

(1) H– _{(2) He (3) Li}+ _{(4) Be}+ 2

Sol. Answer (1)

H– _{these are isoelectronic species.}

7. Which of the following orbitals has higher screening power?

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

Sol. Answer (1)

s orbital have more screening power.

8. The alkali metal which is radioactive is

(1) Fr (2) Al (3) Li (4) Mg

Sol. Answer (1)

Fr which is also liquid at room temperature.

9. The element which belongs with chalcogen family is

(1) N (2) P (3) S (4) Cl

Sol. Answer (3)

16th _{group is chalcogen.}

10. The element which has highest 2nd ionisation energy is

(1) Na (2) Mg (3) Ca (4) Ar

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

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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

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) Ti}3+ _{(3) V}3+ _{(4) Fe}2+

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 _{= 1s}2 _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.

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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 radius

charge) ionic ( 2 (3) ₂ radius) (ionic charge ionic (4) _1/2 radius) (ionic charge ionic Sol. Answer (3) 2 ionic charge Polarising power (ionic radius) 

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 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 (3) K (4) B

Sol. Answer (1)

IA group elements have highest IInd ionization energy.

24. The element which has highest electron affinity?

(1) Oxygen (2) Sulphur (3) Nitrogen (4) Phosphorus

Sol. Answer (2)

Resultant of size factor and electronic configuration factor.

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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) E_muliken = 2.8 E_pauling = 2.8 × 1 = 2.8

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 VII

CO₂ Cl₂O₇

SiO₂ 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)

 

– – –2

O e O

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).

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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?

(1) A–2  _A– _+1e– _{(2) A + e}–  _A–

(3) A–  _{A + e}– _{(4) A}+ _{+ e}–  _A

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] 3d5_4s1 _{(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

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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 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_{, 5d}1_{, 6s}2_{. 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_{, 2s}2_p6_{, 3s}1 _{(2) 1s}2_{, 2s}2_p6_{, 3s}2_p1 _{(3) 1s}2_{, 2s}2_p6_{, 3s}2_p2 _{(4) 1s}2_{, 2s}2_p6_{, 3s}2 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 3}rd _{ionization energy (because in 3}rd _{ionization we have to remove electron from a stable} configuration)

37. An element has 56 nucleons in nucleus and if it is isotonic with ₃₀Y60_{. Which group and period does it belong} to?

(1) 8th _{group, 4}th _{period (2) 14}th _{group, 3}rd _period (3) 12th _{group, 3}rd _{period (4) 12}th _{group, 4}th _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

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

I –ClII –IIII –BrIV

(1) III < IV < II < I (2) I < II < IV < III (3) I < III < IV < II (4) III < I < IV < II Sol. Answer (1)

39. Which of the following processes will release energy equal to ionization energy?

(1) M_(g) e M(₍_g)) (2) M_(s)e M( )_(s) (3) M₍_g) e M_(g) (4) M_(g) e 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

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Sol. Answer (1) Number of moles of Li =    –3 –3 70 10 _{10 10} 7 = 10–2 _mole ∵ 1 mole Li requires 520 kJ energy

 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) 1s2_2s2_2p5 _{(2) 1s}2_2s2_2p6_3s1 _{(3) 1s}2_2s2_2p4 _{(4) 1s}2_2s1 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_–2 = 11

2 oxygen atom Hence formula = 1 3 11 2 A A O = 2 11 4O A = A₈O₁₁

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

Sol. Answer (2, 3, 4)

Tl have has lower IE₁ than Ga.

2. Which of the following elements have same Z_effective?

(1) Na (2) Li (3) K (4) Rb

Sol. Answer (1, 3, 4)

s block, except 2nd _{period, all have same electronegativity.}

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3. 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.

4. Choose the correct statement/statements regarding Modern periodic table (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.

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7. Choose the process which is/are endothermic. (1) O–1 _{+ e}–  _O–2 _{(2) Ar}  + r A + e– (3) Ar + e–  _Ar– _{(4) H + e}–  _H– 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

 

1 2

– –

1sH e 1sH , H

– _{is getting noble gas configuration hence it will favors addition of e}–_{. When H is}

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

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Sol. Answer (2, 3, 4) Si 3s2 _3p2 _Si+ _3s2 _3p1 P 3s2 _3p3 _P+ _3s2 _3p2 Cl 3s2 _3p5 _Cl+ _3s2 _3p4 S 3s2 _3p4 _S+ _3s2 _3p3 IV V VI VII Si+ _P+ _S+ _Cl+

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

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 (3) Ra (4) Cr

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)

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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 _5d1 _6s2 _{element belongs to f block} Sol. Answer (1, 4)

[Ar] 3d10 _4s2 _4p6 _{is Kr (36) which is a noble gas element [Ar] 3d}5 _4s1 _{is Cr(24) which is a d block element.} [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 Z}

eff.

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

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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?

(1) 55, 12, 48, 53 (2) 13, 33, 54, 83 (3) 3, 33, 53, 87 (4) 22, 33, 55, 66 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 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 55 Cs block 12 Mg block 48 Cd block 53 I block s s d p Set 2 ⎡ ⎢ ⎢ ⎢  ⎢ ⎣ 13 Al block 33 As block

54 Xe noble gas block

83 Bi block p p p p Set 3 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 3 B block 33 As block 53 I block 87 Fr block p p p s Set 4 ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 22 Ti block 33 As block 55 Cs block 66 Dy block d p s f

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)

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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.

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 2}nd _IE.

2. Which of the following is a non-metal?

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

Sol. Answer (3, 4)

Both (3) & (4) are possible.

3. Which element is a noble gas?

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

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.

E.A.) (I.P. 2 1 A   

One more relationship given by him, if the values are given in eV is

A Ionisation potential Electron affinity_5.6

 

When there is pure covalent bond between A – B

5.6 (EA) (IP) 5.6 (EA) (IP)_A _{A  B} _B B A  

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1. 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 EN 2

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) 6 . 5 ) EA ( ) IP ( 6 . 5 ) EA ( ) IP ( _A _{A  B} _B (2) 6 . 5 ) EA ( ) IP ( 6 . 5 ) EA ( ) IP ( _B _{B  A} _B (3) 8 . 2 ) EA ( ) IP ( 8 . 2 ) EA ( ) IP ( _A _{B  A} _B (4) 6 . 5 ) EA ( ) IP ( 6 . 5 ) EA ( ) IP ( _A _{B  A} _A Sol. Answer (1) A B– + _{A B} x

which implies that (EN)_A > (EN)_B

 _ 

A A B B

(IP) (EA) (IP) (EA)

5.6 5.6

3. Pauling’s electronegativity scale is based on

(1) Thermochemical data (2) I.E. data (3) E.A. 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.

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3. STATEMENT-1 : Noble gases have high ionisation energy and

STATEMENT-2 : Noble gases belong to 18th _group. Sol. Answer (2)

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 _{& 17}th _{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.

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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+_{, Mg}2+_{, Al}3+ _{are isoelectronic.} and

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 Z_eff 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.

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Sol. Answer (3)

Li, Mg and Be, Al and B, Si show diagonal relationship Li 1.23 Å Mg 1.36 Å Be 0.89 Å Al 1.25 Å B 0.80 Å Si 1.17 Å

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

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

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2. 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}

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

Zn++ _{< Cu}+ _{< Ca}+ _{< K}+ K+ _{1.33 Å} Cu+ _{0.96 Å} Ca+ _{0.99 Å} Zn+2 _{0.74 Å}

3. 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.

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2. 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) 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) T T T (3) F T T (4) F F T

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) F F T (3) T F T (4) TFF

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) T T T (2) T F T (3) F F T (4) F F F

Sol. Answer (2)

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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}+

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

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

Sol. Answer (1)

Due to low ionization energy.

5. In which compound Mn has highest electronegativity?

(1) MnO (2) Mn₃O₄ (3) MnO₂ (4) Mn₂O₅

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.

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7. The oxide which has highest acidic character is

(1) MnO (2) MnO₂ (3) Mn₂O₃ (4) Equal in all of these

Sol. Answer (2) MnO₂

8. Choose the correct statement

(1) Isotopes have nearly same chemical properties (2) Isoelectronic species may be neutral

(3) Na and K have nearly same Z_effective (4) All of these

Sol. Answer (4) Fact

9. Maximum number of electrons in nth _{shell is}

(1) n2 _{(2) 2(l + 1) (3) 2n}2 _{(4) (l + 1)}2

Sol. Answer (3)

10. Choose the correct regarding E.N.

(1) B > Al > Ga > In (2) B > Al = Ga = In (3) B > In > Ga = Al (4) B > In > Ga > Al 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,

2 0.359 0.744 Zeff EN r   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

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Z* = S –  = 82 – 76.8 Z* _{= 5.2} 2 0.359 5.2 0.744 (146) EN   0.359 5.2 0.744 21316    = 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 =

3 12 10 24   = 0.0005 = 5 × 10–4

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IP EA_5.6

13 4 3.03 5.6

EN  

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.}

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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)

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.

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19. Why Al₂O₃ is amphoteric while B₂O₃ is acidic?

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.

  

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