Classification of Elements and Periodic Classification (Package Solutions)

8/10/2019 Classification of Elements and Periodic Classification (Package Solutions)

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Classification of Elements and Periodicity in Properties (Solutions) Solutions of Assignment (Set-2)

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Section-A

Q.No. Solution

1. 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.

3Chapter

Classification of Elements andPeriodicity in Properties



Solutions of Assignment (Set-2) Classification of Elements and Periodicity in Properties (Solutions)


Q.No. Solution

11. 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, comparedto 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 2s

2 2p6 3s

2

Al = 1s2 2s

2 2p6 3s

2 3p1

Mg has stable configuration, hence its ionization energy will be higher than Al.

Na < Mg > Al < Si

14. Answer (4)

Fe2+ has four unpaired electrons.

15. Answer (3)

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

16. Answer (3)33 = 1s

2 2s2 2p

6 3s2 3p

6 4s2 3d

10 4p3 = 1s

2 2s2 2p

6 3s2 3p

6 3d10 4s

2 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 onelectrons decreases, hence ionization energy decreases.

19. Answer (3)

2)radiusionic(

eargchionicpower gsinpolari





Q.No. 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 = 2.8 Epauling

= 2.8 × 1

= 2.8

26. 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 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.





Q.No. Solution

29. Answer (1)

2 – – – OeO

When e – will approach O – , O – will repel the approaching electron, hence, we have to give some energytherefore, 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 2s

2 2p6 3s

2 3p6 4s

1 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 2s

2 2p6 3s

2 3p6 4s

2 3d10 4p

6 5s1 4d

10 (Half filled-full filled rule)

In Ag except 5s1, all subshells are full filled which contains total 46e – out of which 23e – haveanticlockwise 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

33. 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] 4f 7 5d

1 6s2

1s2 2s

2 2p6 3s

2 3p6 4s

2 3d10 4p

6 5s2 4d

10 5p6 6s

2 5d1 4f

7

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





Q.No. Solution

36. Answer (4)

For IIA group elements i.e., elements containing 2e – in outermost shell, there is a sudden jumpbetween values of 2nd and 3rd ionization energy (because in 3rd ionization we have to remove electronfrom 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 2s

2 2p6 3s

2 3p6 4s

2 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 sizeratio 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 = 3 – 3 –

10107

1070

= 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. Answer (4)

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





Q.No. Solution

45. 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 =2 –

111 – =

2

11 oxygen atom

Hence formula =21131 OAA =

2114OA = A8O11

Section-B

Q.No. Solution

1. Answer (2, 3, 4)

Tl have has IE1 than Ga.

2. Answer (1, 3, 4)

s block, except 2nd 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 – 195T – 190

Po – 174

IA IIA

Li Be

Na

K





Q.No. 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.

Be+2 < 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 eAr , 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

– –

1s 1sH e H , 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. 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 3p

2 Si+ 3s2 3p

1

P 3s2 3p

3 P+ 3s2 3p

2

Cl 3s2 3p

5 Cl+ 3s2 3p

4

S 3s2 3p

4 S+ 3s2 3p

3

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 typical elements.





Q.No. Solution

10. 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

* 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

13. 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 4s

2 4p6 is Kr (36) which is a noble gas element [Ar] 3d

5 4s1 is Cr(24) which is a d block

element.

[Xe] 4f 1 5d

1 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 alsoincreases.

Fe forms more +ve ions like Fe+2 and Fe+3 while Cu can form only Cu+ and Cu+2 and Zn can form onlyZn+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 4 th choice. IEof 1st member is greater due to Zeff .





Q.No. Solution

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

Element Radius in Å

Fe 1.17

Co 1.16

Zr 1.45Hf 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 blockelements are called the representative or main group elements

Set 1

blockI53

blockCd48

blockMg12

blockCs55

p

d

s

s

Set 2

blockBi83

blockgasnobleXe54

blockAs33

blockAl13

p

p

p

p

Set 3

blockFr 87

blockI53

blockAs33

blockB3

s

p

p

p

Set 4

blockDy66

blockCs55

blockAs33

blockTi22

f

s

p

d

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

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.





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

2

EAIEEN

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

2. Answer (1)

A B – +

A Bx

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

6.5

)EA()IP(

6.5

)EA()IP(BBAA

3. Answer (1)

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

Section-D

Q.No. Solution

1. 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.





Q.No. Solution

4. 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 electronicrepulsion 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 2s

2

B(5) = 1s2 2s

2 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 thanp electrons





Q.No. Solution

14. Answer (3)

Li, Mg and Be, Al and B, Si show diagonal relationship

Li 1.23 Å

Mg 1.36 Å

Be 0.89 ÅAl 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 2s

2

He and Be both have similar electronic configuration like ns2

Be forms compounds, hence it is not inert.

Section-E

Q.No. Solution

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

Element Electronegativity

C 2.5

N 3.0Al 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

Cu+ 0.96 Å

Ca+ 0.99 Å

Zn+2 0.74 Å

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

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





Section-F

Q.No. Solution

1. 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.





Section-H

Q.No. Solution

1. 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 > AlB = 2.0

Al = 1.5

Ga = 1.6

In = 1.7

Tl = 1.8





Q.No. Solution

11. On Allred Rochow scale,

2eff

r

Z359.0744.0EN

r = radius in Å

82 = 1s2

2s2

2p6

3s2

3p6

3d10

4s2

4p6

4d10

4f 14

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

2)146(

2.5359.0744.0EN

21316

2.5359.0744.0

= 0.744 + 8.75 × 10 –5

= 0.74405

12.Number of moles =

24

1012 3 = 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.

6.5

EAIPEN

03.36.5

413EN

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 andlarger will be IE

Na+ (Z = 11)

Ne (Z = 10)

Hence, Na+ have higher IE.





Q.No. Solution

16. (a) Group II (coinage metals) elements differ from group I elements in that, the penultimate shellcontains 10d electrons. The poor screening by the d electrons makes the atoms of Cu group muchsmaller 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

Be+ 1s2 2s

1

B+ 1s2 2s

2

C+ 1s2 2s

2 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 inatomic size and shielding effect.

* On moving from Al to Ga, the (IE)1 , increases slightly, because on moving from Al to Ga, bothnuclear charge and shielding effect increase, but due to poor shielding d electron (3d

10) in Ga,effective nuclear charge on valence electron increases, resulting in d-block contraction, that is whyionization enthalpies increase.

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

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

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

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Classification of Elements and Periodic Classification (Package Solutions)