Classification of Elements and Periodic Classification (Package Solutions)

Classification of Elements and Periodicity in Properties (Solutions)

3 Chapter

Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties Section-A

Q.No. 1.

Solution Answer (2) Due to large atomic volume.

2.

Answer (2) Classification is based on Mendeleev’s periodic law.

3.

Answer (3) Last electron enters in ‘d’ sub-shell. Zn belongs with 12th group.

4.

Answer (2) Pm is radioactive.

5.

Answer (1) H-2.1

6.

Answer (1) H– these are electronic species.

7.

Answer (1) s orbital have more screening power.

8.

Answer (1) Fr which is also liquid at room temperature.

9.

Answer (3) 16th group is chalcogen.

10.

Answer (1) Electron will be removed from completely filled second orbit.

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 11.

Solution Answer (4) Higher electronegativity implies higher ionization potential.

12.

Answer (3) During formation of Anions, electron is added to neutral atom, hence size of anion is bigger, compared to neutral atom.

13.

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.

Answer (4) 2+

Fe 15.

has four unpaired electrons.

Answer (3) Zn is not an alkaline earth metal, IIA group elements are called as alkaline earth metals.

16.

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.

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.

Answer (1) Due to screening effect, repulsion on electron increases which implies that attraction of nucleus on electrons decreases, hence ionization energy decreases.

19.

Answer (3) polari sing power 

ionic ch arg e (ionic radius) 2

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

Solutions of Assignment (Set-2)

Solution

20.

Answer (2)

21.

Answer (2) Factual.

22.

Answer (2) Element

Electron gain enthalpy

Electron affinity

O

–144

144

S

–200

200

Se

–195

195

 S > Se > O 23.

Answer (1) IA group elements have highest IInd ionization energy.

24.

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

25.

Answer (1) Emuliken

26.

=

2.8 Epauling

=

2.8 × 1

=

2.8

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.

Answer (2) Anions are larger than the neutral atom, while cations are smaller than the neutral atom.

28.

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. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 29.

Solution 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). 30.

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.

Answer (3) Fact.

32.

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

33.

N

N

H

Answer (4) Uub U = un = 1 u = un = 1 b = bium = 2 Hence element has atomic number 112.

34.

Answer (3) Zr and Hf have nearly same size due to lanthanide contraction

35.

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.

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

36.

Solutions of Assignment (Set-2)

Solution

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.

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

38.

Answer (1)

39.

Answer (3) The process is opposite of Ionization energy.

40.

Answer (3) Generally IE increases along period with few exceptions.

41.

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.

Answer (1) Mercury is the only metal which is liquid at 0°C this is due to weak metallic bonding.

43.

Answer (1) Number of moles of Li =

70  10 –3  10  10 – 3 7

= 10–2 mole

 1 mole Li requires 520 kJ energy  10

–2

–2

mole Li will require = 10

× 520 kJ

= 5.2 kJ energy 44.

Answer (4) IA group element (ns1) have biggest jump between Ist and IInd ionization energy.

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

45.

Solution

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 = A 1A 3 O 11 = A 4 O 11 = A8O11 2

2

Section-B Q.No. 1.

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

2.

Answer (1, 3, 4) nd

s block, except 2 period, all have same electronegativity. 3.

Answer (2, 4) Ga and iodine belong with p block.

4.

Answer (2, 3, 4) Actinoids are placed separate.

5.

Answer (1, 4) Zn has larger size than Cu.

6.

Answer (2, 4) Element

Electron gain enthalpy

O

– 144

S

– 200

Se

– 195

T

– 190

Po

– 174

IA

IIA

Li

Be

Na K

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

Solutions of Assignment (Set-2)

Solution As we move from left to right across a period, size decreases. As we move from top to bottom in a group, size increases. +2 + + +  Be < Li < Na < K

F is most electronegative element known. Cl has highest negative electron gain enthalpy. 7.

Answer (1, 2, 3) O– will repel e–, hence, addition of e– in O– is endothermic A   A   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

1s1

–

–

–

, H is getting noble gas configuration hence it will favors addition of e . When H is

1s

converted to H–, it gains stability hence loses energy. 8.

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.

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

Si+ 3s2 3p1

P 3s2 3p3

P+ 3s2 3p2

2

5

Cl 3s 3p

3s2 3p4

S+ 3s2 3p3

Cl 3s 3p S IV +

Si

+

2

4

V VI VII 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 typical elements. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 10.

Solution Answer (1, 2) Si, Ge, As, Sb, Te are metalloids

11.

Answer (1, 2) Elements Technetium (Tc) (Z = 43) Promethium (Pm) (Z = 61), Astatine (At) (Z = 85), Francium, (Fr) (Z = 87) and all elements with Z > 92 are artificial elements.

12.

Answer (1, 2) For d-block element

13.

*

All are metals

*

d block elements show variable valency.

*

Some transition metals are colorless also.

*

In 3rd series ionic radius decreases but for 4d and 5d it is not true

Answer (1, 4) Cu, Ag, Au are coinage metals. U is not a lanthanoid. Reducing character increases down the group. Li is strongest reducing agent due to very high enthalpy of hydration

14.

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. [Xe] 4f1 5d1 6s2 has atomic number 58 which is Ce (58) a lanthanoid belonging to f block.

15.

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.

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 st of 1 member is greater due to Zeff.

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 17.

Solutions of Assignment (Set-2)

Solution 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 18.

Answer (1, 2) CO, H2O, NO, N2O all are neutral

19.

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 s block  12 Mg s block 48 Cd d block  p block 53 I

Set 2

13  33 54  83

Set 3

3  33 53  87

B p block As p block I p block Fr

s block

22  33 55  66

Ti

d block

Set 4

Al

p block

As p block Xe noble gas  p block Bi p block

As p block Cs 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.

Answer (2, 3, 4) As oxidation number (i.e., charge) of metal ion increases, electronegativity also increases So Mo(II) < Mo(III) < Mo(IV) and Fe(I) < Fe(II) < Fe(III) are correct increasing order of electronegativity. Lesser the ionisation energy greater will be reducing strength.

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Section-C Q.No.

Solution Comprehension-I

1.

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

2.

Answer (3, 4) Both (3) & (4) are possible.

3.

Answer (4) Noble gas have very high ionization energy.

Comprehension-II 1.

Answer (4) According Mulliken EN 

IE  EA 2

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

2.

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.

Answer (1) Pauling’s electronegativity is based on bond energy data i.e. thermo chemical data.

Section-D Q.No. 1.

Solution Answer (2) Due to electron-electron repulsion attraction of nucleus decreases.

2.

Answer (3) Valency is available quantity.

3.

Answer (2) Due to inert configuration, I.E. is very high.

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 4.

Solutions of Assignment (Set-2)

Solution Answer (2) In actinoids + 2 to +6 states are possible.

5.

Answer (1) It is due to very high shielding effect in F.

6.

Answer (4) Properties of hydrogen match with Ist & 17th group both.

7.

Answer (4) In H– and He, effective nucleus charges are different.

8.

Answer (3) Ionisation energy of Be is greater than B.

9.

Answer (4) Bond dissociation energy of F2 is less than Cl2 due to smaller size of fluorine there is greater electronic repulsion between F atoms than Cl atoms.

10.

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.

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.

Answer (2) F is highest electronegative element known F is smaller in size than chlorine. Both size and Zeff explains the equation.

13.

Answer (1)

s electrons are closer to nucleus than p electrons, hence ionisation energy of s electrons is higher than p electrons Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No. 14.

Solution 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.

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 Q.No. 1.

Solution 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 < N Cs is strongly electropositive hence least electronegative 2.

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

1.33 Å +

0.96 Å

+

0.99 Å

+2

0.74 Å

Cu Ca Zn 3.

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

4.

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

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Classification of Elements and Periodicity in Properties (Solutions)

Solutions of Assignment (Set-2)

Section-F Q.No. 1.

Solution Answer (9) Atomic number is 109.

2.

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

3.

Answer (2) Fact.

4.

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

5.

Answer (3) All f-block elements belongs to IIIrd group.

6.

Answer (3) All f-block elements & other five elements belongs to IIIrd group.

Section-G Q.No.

Solution

1.

Answer (1)

2.

Answer (1)

3.

Answer (2)

4.

Answer (2) Fact.

5.

Answer (4) Fact.

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Section-H Q.No. 1.

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

2.

Answer (4) Hg ; 1007 kJ/mole

3.

Answer (1) Fact

4.

Answer (1) Due to low ionization energy.

5.

Answer (4) Higher will be oxidation state higher will be electronegative.

6.

Answer (1) F2 is stronger oxidizing agent.

7.

Answer (2) MnO2

8.

Answer (4) Fact

9.

Answer (3)

10.

Answer (4) B > Tl > In > Ga > Al B = 2.0 Al = 1.5 Ga = 1.6 In = 1.7 Tl = 1.8

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Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

Solutions of Assignment (Set-2)

Solution

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

0.359  5.2 (146 )2

 0.744 

0.359  5.2 21316

= 0.744 + 8.75 × 10–5 = 0.74405 12. 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 –4 –4  5 × 10 mole  5 × 10 × 2188.5 kJ

 1.09425 kJ 13.

EN 

IP  EA 5 .6

EN 

13  4  3.03 5 .6

14. 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. +

–

15. 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.

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Solutions of Assignment (Set-2)

Classification of Elements and Periodicity in Properties (Solutions)

Q.No.

Solution

16. (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. The initial separation of lanthanides and actinides are based on slight differences in solubility. 18. (a) Order of acidic strength is Na2O < P2O5 < Cl2O < N2O < NO2 (b) Li+ 1s2 +

2

1

Be 1s 2s

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.

19. 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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