PRINCIPLES AN

OF EXTRACTIO

ND PROCESSES

ON OF METALS

1. The smelting of ironall the processes exa t e p ocesses e

( )(a) combustion(b) reduction(b) reduction (c) slag formation(d) sublimation

n in blast furnace involvxceptcept

• 1. Option (d): It is verygiven processes happg e p ocesses appexcept sublimation.Sublimation is changi• Sublimation is changidirectly without forminnapthalene balls.

y clear by now that all thpen in blast furnace pe b ast u ace

ng solid to vapourng solid to vapourng a liquid. Ex. Campho

2. Concentration methodAl O 2H O iAl2O3.2H2O is

(a) Froth floatation (b) Leaching(b) Leaching(c) Liquation (d) Levigation(d) Levigation

d used for mineral

• 2. Option (b): alumina leaching with NaOH. eac g t aO

is concentrated by

. In the equation, 4M + 8CN- + 2 H2O +O2Identify the metal M.

( )C(a)Copper (b) iron (c)gold(c)gold (d) zinc

2 → 4[ M(CN)2]- + 4OH-2 [ ( )2]

• 3. Option (c):• Noble metals like gold• Noble metals like gold

hydrometallurgy. i.e bNaCN and the compleNaCN and the completreated with preferably

d silver are extracted byd, silver are extracted byy dissolving the ore in

ex of gold obtained isex of gold obtained is y zinc to get gold.

. Vapour phase purificatmetal to ametal to a

(a) Volatile stable comp(b) Volatile unstable co(c) Non-volatile stable (d) Non olatile nstab(d) Non-volatile unstab

tion involves converting

poundompoundcompound

ble compo ndble compound

• 4. Option (a):• As the name suggest• As the name suggest

refining in the vapourformed should be volaformed should be volabe decomposable.

vapour phase meansvapour phase means state, so the compound

atile but stable and canatile but stable and can

5. Higher percentage of zfrom

(a) zincite(a) zincite(b) Zinc blende(c) Calamine(c) Calamine (d) All of these

zinc can be obtained

• 5. Option (b):• Zinc blende contains m• Zinc blende contains m

ore.more zinc and is the chmore zinc and is the ch

6. Identify the reaction tplace in blast furnacplace in blast furnac

(a) CaO + SiO2 → CaS(b) CaCO3 → CaO + CO(b) CaCO3 → CaO + CO(c) CO2 + C→ 2CO(d) FeCO3 → FeO + CO

that does not take cece

SiO3

O2O2

O2

• 6. Option (d):• a) CaO + SiO2 → CaSiO3a) CaO SiO2 CaSiO3

• b) CaCO3 → CaO + CO2

) CO + C 2CO• c) CO2 + C→ 2COThese happen in blast fur

• d ) FeCO3 → FeO + CO2d.) FeCO3 → FeO + CO2

3 is slag formation3 is slag formationis decomposition

h t b tiheat absorptionrnace

happens during roastinhappens during roastin

. Which one of the follow

(a) Horn silver (b) Zincite(c) Bauxite ( )(d) Galena

wing ores is a chloride

• 7. Option (a):• Horn silver is AgCl• Horn silver is AgCl• Zincite is ZnO• Bauxite is Al2O3

• Galena is Pbs• Galena is Pbs

Aluminium is extractedbecausebecause

( ) Al i i i d(a) Aluminium is a good(b) Bauxite cannot be m(c) Aluminium is more (d) Aluminium has high( ) g

d by electrometallurgy

d d t f l t i id conductor of electricimelted easily

electropositiveh affinity for oxygeny yg

• 8. Option (c):• Aluminium lies very m• Aluminium lies very m

diagram. It cannot be agent It is more electragent. It is more electrnegative reduction po

much below in ellinghammuch below in ellinghamreduced by any reducinropositive it has moreropositive, it has more

otential.

9. Aluminium is most abit is obtained from bauit is obtained from bau

a) Bauxite is available ina) Bauxite is available inb) Of easy extraction of

) B it t ic) Bauxite contains maxd) Bauxite is less impur

undant in earth crust yeuxite becauseuxite because

n larger quantityn larger quantity f aluminium from it

i l i iximum aluminiumre

• 9. Option (c):• Bauxite is the chief or• Bauxite is the chief or

percentage of aluminire with maximumre with maximum um

10. Copper pyrites are co

(a) Electromagnetic m(b) Gravity method (c) Froth floatation pro( ) p(d) All the above meth

oncentrated by

method

ocess hods

• 10. Option (c):• Copper pyrites is CuF• Copper pyrites is CuF

is concentrated by froFeS2 a sulphide ore whiFeS2, a sulphide ore whioth floatation process

1 I th l t l i f1. In the electrolysis of aadded to

(1) lower the melting point (1) lower the melting point (2) increase the electrical c(3) remove impurities from(3) remove impurities from(4) to dissolve alumina

(a) both 1 & 2(c) both 2 & 3(c) both 2 & 3

l i lit ialumina, cryolite is

of alumina of alumina conductivity alumina alumina

(b) both 1 & 3 (d) both 2 & 4(d) both 2 & 4

• 11. Option (a):• Alumina does not mel• Alumina does not mel

conduct current, hencincrease conductivityincrease conductivity added.

lt and also does notlt and also does not ce to lower M.P and to cryolite(Na AlF ) iscryolite(Na3AlF6) is

2. In the zone refining m

(a) contains impurities(b) contains purified m(b) contains purified m(c) contains more imp

original metaloriginal metal(d) moves to either sid

method, the molten zone

s metal onlymetal onlypurities than the

de

• 12. Option (c):• As the metal melts th• As the metal melts, th

soluble in molten mete impurities are moree impurities are more al.

3. Ores like magnetite oti ttin ores are concentr

(a) Froth-floatation (b) Magnetic separation( ) g p(c) Gravity separation (d) Electrostatic separa(d) Electrostatic separa

r tungstates present in t d brated by

n

ationation

• 13. Option (b):• Magentite and tugstat• Magentite and tugstat

nature hence, magnettes in tin are magnetic intes in tin are magnetic intic separation

4. The metal used to reccomplex [Ag(CN) ]- iscomplex [Ag(CN)2]- isbecause

(a) Zinc is lighter than (b) Zinc is more electro(b) Zinc is more electro(c) Zinc melts at a lowe(d) None of the abo e(d) None of the above

cover silver from the s zinc and not coppers zinc and not copper

copperopositive than copperopositive than copperer temperature

• 14. Option (b):• Zn and copper both ar• Zn and copper both ar

compared to silver, soreplacing silver from treplacing silver from tbeing more electroposcopper, zinc is used.

• Eo of Zn = -0.76V, Cu =E of Zn 0.76V, Cu

re more electropositivere more electropositive o both are capable of the complex but zincthe complex, but zinc sitive and cheaper than

= +0.34V, Ag = + 0.8V 0.34V, Ag 0.8V

5. ZnS can be separatedd i f th fl t tiduring froth floatation

(a) Froth stabilisers(b) depressants ( ) p(c) Frothing agents(d) Washing with wate(d) Washing with wate

d from PbS in an ore in using

erer

• 15. Option (b):In case of an ore contaIn case of an ore contadepressant used is NaCZnS from coming to theZnS from coming to thecome with the forth.

aining Zns and PbS theaining Zns and PbS, the CN. it selectively preventse forth but allow PbS toe forth but allow PbS to

6. Bauxite is concentrat

(a) Serpeck’s process (b) B ’(b) Baeyer’s process (c) Thermite process (d) Down’s process

ted by

• 16. Option (b):• Concentration of baux• Concentration of baux

NaOH is Baeyer’s procxite by leaching withxite by leaching with cess

7. Which of the followinpurified by van Arkel

(a) Ni and Fe (b) Ga and In(b) Ga and In (c) Zr and Ti (d) Ag and Au(d) Ag and Au

ng pairs of metals is method?

• 17. Option (c):• Van Arkel method is v• Van Arkel method is v• Zr and Ti forms volatil

and then the compoungives the metal.

vapour phase refiningvapour phase refining.le compound with iodinnd on decomposition

. The extraction of nickeinvolves

(a) The formation of Ni(b) The decomposition (c) The formation and th

Ni(CO)4(d) The formation and c( )

Ni(CO)4

el by Mond’s process

i(CO)44of Ni(CO)4hermal decomposition o

catalytic decomposition y p

• 18. Option (C):• Mond’s process is als• Mond s process is alsso vapour phase refiningso vapour phase refining

9. Silica is added to roalti i d tsmelting in order to r

(a) Cuprous sulphide(b) Cuprous oxide ( ) p(c) Ferrous oxide (d) Ferrous sulphide(d) Ferrous sulphide

asted copper ore duringremove

• 19. Option (c):• FeO is formed after ro• FeO is formed after ro

(CuFeS2), this is remoferrous silicateferrous silicate.

oasting copper oreoasting copper ore oved by adding silica as

0. When a metal is to beif the gangue associag gsilica, then

(a) An acidic flux is ne(b) A basic flux is need(b) A basic flux is need(c) Both acidic and ba(d) N ith f th i(d) Neither of them is n

e extracted from its ore,ated with the ore is

eeded deddedsic flux are needed

d dneeded

• 20. Option (b):• If the gangue is acidic• If the gangue is acidic

used. For basic ganguusedused.

• Note: silica is used asextraction, to remove added.added.

c like silica a basic fluxc like silica, a basic flux ue like CaO, acidic flux i

s acidic flux. Ex. copper basic FeO silica is

21.The impurities associi t ll llin metallurgy are coll

(a) Slag (b) Flux ( )(c) Gangue (d) Ore(d) Ore

iated with minerals usedl ti l ll dlectively called

• 21. Option (c):• Earthy impurities and• Earthy impurities and others are gangueothers are gangue.

22. Roasting of copper p

(a) To remove moistur(b) T bt i id f(b) To obtain oxide fre(c) To decompose pyr(d) All of the above

pyrites is done

re and volatile impuritief l hee of sulphur

rites into Cu2S and FeS

• 22. Option (d):• Roasting is mainly do• Roasting is mainly do

metal so that reductioduring the process mduring the process, mimpurities also are rem

one to get the oxide of tone to get the oxide of ton becomes easy. But moisture volatilemoisture, volatile moved.

23. Go Vs T plot in Ellingdownwards for the r

(a) CO + ½ O2 → CO(b) C ½ O CO(b) C + ½ O2 → CO(c) 2Ag + ½ O2 → 2A2(d) Mg + ½ O2 → Mg

gham diagram slope reaction

O2

AgOgO

• 23. Option (b):• Generally lines in EllinghaGenerally lines in Ellinghaexcept for one reaction i.ereaction becomes +ve Threaction becomes +ve. Th

• ∆G = ∆H‐T∆S will become temperature increases he

m diagram slope upwardsm diagram slope upwards e 2C + O2 → 2CO.  ∆S for thihen ∆G as per the equationhen ∆G as per the equation 

more and more –ve as ence it slopes down.

4. Identify the reaction tduring the smelting pduring the smelting pextraction

(a) 2FeS + 3O2 → 2 FeO(b) C O F S C S(b) Cu2O + FeS → Cu2S(c) 2Cu2S + 3O2 → 2Cu2 2(d) FeO + SiO2 → FeSi

that doesn’t take place rocess of copperrocess of copper

O + 2SO2 ↑ S F OS + FeOu2O + 2SO2 ↑ 2 2O3

• 24. Option (c):• All the reactions given• All the reactions given

but the reactionC S O C O• 2Cu2S + 3O2 → 2Cu2O

• Occurs during roastinOccurs during roastin

n occur during smeltingn occur during smelting

O SOO + 2SO2 ↑ ngng

25. In smelting of iron, wreactions takes placereactions takes place400oC - 600oC

(a) CaO + SiO2 → CaSi(b) 2FeS + 3O2 → 2Fe +(c) FeO + SiO2 → FeSi( ) 2(d) Fe2O3 + 3CO → 2Fe

which of the following e in Blast furnace ate in Blast furnace at

iO3+ SO2iO33e + 3CO2

• 25. Option (d):• As per Ellingham diag• As per Ellingham diag

agent for iron oxide at1083K so option d is t1083K. so option d is t

gram CO is the reducinggram, CO is the reducingt temperature below the reactionthe reaction.

26. In the commercial elefor aluminium extracfor aluminium extracused is

(a) Al(OH)3 in NaOH so(b) An aqueous solutio(b) An aqueous solutio(c) A molten mixture o(d) A molten mixture o(d) A molten mixture o

ectrochemical process ction the electrolytection, the electrolyte

olution on of Al (SO4 )on of Al2(SO4 )3of Al2O3 and Na3AlF6of AlO(OH) and Al(OH)of AlO(OH) and Al(OH)3

• 26. Option (c):• In the extraction of alu• In the extraction of alu

to alumina and electrouminium cryolite is addeuminium cryolite is addeolysis is carried out.

27. In froth floatation prpurification of ores,purification of ores, because

(a) Their surface is no(b) They are light( ) Th i l bl(c) They are insoluble(d) All the above

ocess for the the particles of ore floathe particles of ore floa

ot easily wetted by wate

e

• 27. Option (d):• Particles float when th• Particles float when th

and also not wetted byhey are light insolublehey are light, insoluble y water.

8. Which of the followinroasting during metag ggetting the metal oxid

(a) Horn silver (b) Zinc blende(b) Zinc blende(c) Malachite (d) Bauxite

ng ores is subjected to llurgical operations for g pde

• 28. Option (b):• Zinc blende is ZnS he• Zinc blende is ZnS, he

ZnO will then be reducence roasted to get ZnOence roasted to get ZnOced to zinc.

29. Which of the followinplace in blast furnaceplace in blast furnace

(a) Fe2O3 + 3CO →Fe + (b) C + O2 → CO2(b) C + O2 → CO2

(c) 2C + O2 → 2CO2

(d) CaCO3 → CaO + CO

ng reactions taking e in endothermice in endothermic

3CO2

O2

• 29. Option (d):The reactions in blast furnace arC + O2 → CO2 &  2 Fe2O3 + 3CO →Combustion and reduction reacCO2 + C →2CO  &   CaCO3 → CaOWhile formation of CO and decoendothermic. 

In the options formation of CO Option b is same as c and decom

re→Fe + 3CO2

ctions are exothermicO + CO2

omposition of CaCO3 are p 3

is not given, mposition of CaCO3 is option d

0. Pb and Sn are extractbyby

(a) Carbon reduction a(a) Carbon reduction a(b) Self reduction and (c) Electrolysis and se(d) Self reduction and ( )

ted from their chief ore

and self reductionand self reduction carbon reduction

elf reduction electrolysisy

• 30. Option (b):• We can us carbon to r• We can us carbon to r

also undergoes self reS O S• PbS + 2PbO→3Pb + S

reduce lead oxide or itreduce lead oxide or it eduction like copperOO2

31. The slag formed in this

(a)CuFeS2(a)Cu eS2(b)Cu2O +FeS(c) Cu2S +FeO(c) Cu2S FeO(d)FeSiO3

he extraction of copper

• 31. Option (d):• FeO + SiO2→ FeSiO3• FeO + SiO2→ FeSiO3

• Gangue + flux slag

32. The gas liberated at taluminium extractioaluminium extractio

a) Oxygena) Oxygenb) Carbon dioxide

) W tc) Water d) Hydrogen

the anode in the n from bauxite isn from bauxite is

• 32. Option (a):• In aluminium extraction• In aluminium extraction

anode.oxygen is liberated atoxygen is liberated at

33. Limestone is used in

(a) Oxidation of Fe ore(b) Reduction of Fe or( )(c) Formation of slag (d) Purification of Fe f( )

n Fe extraction for

e re

formed

• 33. Option (C):• Lime stone (CaCO3) g• Lime stone (CaCO3) g

with silica to form calcive CaO which combineive CaO which combinecium silicate slag.

34. A metal obtained diresulphide ore issulphide ore is

(a) Cu(a) Cu (b) Pb(c) Hg(c) Hg(d) Zn

ectly by roasting of its

• 34. Option (c):• 2HgS + 3O2→ 2HgO +• 2HgS + 3O2→ 2HgO + • 2HgO → 2Hg + O2

SO2SO2

35. Gangue is removed

(a) Gangue is infusibl(b) Gangue is fusible(b) Gangue is fusible (c) Gangue floats on f(d) Sl f t(d) Slag fuses at very

as slag because

le while slag is fusiblewhile slag is infusiblewhile slag is infusiblefused metall t tlow temperature

• 35. Option (a):• Gangue i e mud sand• Gangue i.e mud, sand

melt(fuse), so it is mixwhich melts and floatswhich melts and floats

d etc impurities will notd etc impurities will not xed with flux to form slas on molten metals on molten metal.

6. Consider the isoelectroniF- and O2-. The correct orlength of their radii is

(a) F- < Mg2+ < Na+< O2–

(b) Mg2+ < Na+ < F– < O2( ) g(c) O2– < F– < Na+ < Mg2

(d) O2– < F– < Mg2+ < Na(d) O < F < Mg < Na

ic species, Na+, Mg2+, rder of increasing

–

2–

2+

a+a

• 36. Option (b):• Isoeletronic species a• Isoeletronic species a

number of electrons.• As the nuclear charge

decreases.

are the ones having samare the ones having sam

e increases, radii

7 Whi h f th f ll i7. Which of the followincorrect?

a) Sodium has the highesb) Chl i h l lb) Chlorine has less value

than fluorine.c) Mercury and bromine ac) Mercury and bromine a

temperature.d) In any period atomic rd) In any period, atomic r

least.

t t t ing statements is

st first ionisation enthalpy f l t i th le of electron gain enthalpy

are liquids at roomare liquids at room

radius of alkali metal is theradius of alkali metal is the

• 37. Option (c):

38 Th l i h38. The element with atbelongs to

(a) s-block(a) s block(b) p-block(c) d-block(c) d-block(d) f-block

i b 39tomic number 39

• 38. Option (c):• 39 is 36 + 3• 39 is 36 + 3 • 36 is inert gas krypton

blockn + 3 then it will be in d-

39. The order of screeniof s p d and f orbitaof s, p, d and f orbitaatom on its outer she

(a) s > p > d > f(b) f d(b) f > d > p > s(c) p < d < s > f(d) f > p > s > d

ing effect of electrons als of a given shell of anals of a given shell of anell electrons is:

• 39. Option (a):• Shape of the orbital lo• Shape of the orbital, lo

orbital, more time in thThese factors decide tThese factors decide t

• Based on this the orde

ower energy of theower energy of the he vicinity of the nucleuthe screening effectthe screening effect. er is s > p > d > f

. In a period, atomic rada) outer electrons repela) outer electrons repel b) closer packing amon

particles is achievedc) the number of neutroc) the number of neutrod) increasing nuclear ch

attractive force

dii decreases becauseinner electronsinner electrons

ng the nuclear

ons increasesons increasesharge exerts a greater

• 40. Option (d):• Along a period as the• Along a period, as the

increases effective nuthis pulls the electronthis pulls the electrondecreases.

e atomic numbere atomic number uclear charge increases,

closer and the radii closer and the radii

1. The number of radial

(a) 3(b) 4(c) 2(c) 2(d) 1

nodes for 3p orbital is

• 41. Option (d):• Nodes are points of ze• Nodes are points of ze• Radial nodes given by• For 3p, n=3, n-2 = 3-2

ero electron densityero electron density.y n-2= 1

2. Total number of orbitthird shell will bethird shell will be

(a) 2(b) 4(b) 4(c) 9(d) 3

tals associated with

• 42. Option (c):• When n = 3• When n = 3• Orbitals l = 0 s -1 orbit• l = 1 p – 3 orb• l = 2 d- 5 orbi• l = 2 d- 5 orbi• Total 9

talbitalstalstals

3. The conclusion not derexperiment?experiment?

a) Most of the space in theb) The radius of the atom

is 10-15 m.c) Electrons move in circud) Electrons and the n cled) Electrons and the nucle

electrostatic forces of a

rived from Rutherford’s

e atom is empty.10is 10-10 m and nucleus

ular orbits of fixed energye s are held together beus are held together by attraction.

• 43. Option (c):• This was the drawbac• This was the drawbac• This was given by Boh

ck of Rutherford modelck of Rutherford modelhr

44. The first ionisation pand S follows the ordand S follows the ord

( ) S(a)Mg < Al < P < S(b)Al < Mg < P < S (c)Al < Mg < S < P(d)Mg < Al < S < P(d)Mg Al S P

potential of Mg, Al, P derder

• 44. Option (C):• Al is 13 [Ne] 3s2 3p1--- ea• Mg - 12 [Ne] 3s2 –compl• S- 16 [Ne] 3s2 3p4 –easyS 16 [Ne] 3s 3p easy

configuration• P- 15 [Ne] 3s2 3p3 half filP 15 [Ne] 3s 3p half fil

• therefore the order Al <therefore the order Al

asy to remove last e-yete orbital so stable

y because form half filledy because form half filled

lled so stablelled so stable

< Mg < S < P Mg S P

45. The lowest energy welectrons identified belectrons identified b

(a) n=4 l=0(a) n=4, l=0(b) n=4, l=1(c) n=3 l=2(c) n=3, l=2(d) n=3, l=1

will be associated with by quantum numbersby quantum numbers

• 45. Option (d):(a)n=4, l=0 --4s(a) , 0 s(b) n=4, l=1 ----4p(c) n=3 l=2 ----3d(c) n=3, l=2 ----3d(d) n=3, l=1-----3p

Lowest energy is 3pp

46. An example of a met

(a)Mg(b) S(b) S(c) Si(d) Ar(d) Ar

alloid is

• 46. Option (c):• Silicon is both a meta• Silicon is both a metal and a non-metall and a non-metal

47. The diagonal relationis due tois due to

(a)Similarity of ionisa(a)Similarity of ionisa(b) Similarity of electr(c) Similarity of ionic(c) Similarity of ionic (d) Dissimilar atomic

nship between Li and M

ation potentialation potentialronegativityradiiradiiradii

• 47. Option (c):

48. Electronic configuratthe atomic number othe atomic number o

(a) 24(a) 24(b) 25(c) 26(c) 26(d) 27

tion of X3+ is [Ar] 3d3 , of X is of X is

• 48. Option (a):

• X3+ is [Ar] 3d3 = 18(Ar)felectrons so for X we

• therefore X = 21 + 3 =therefore X 21 3 • X = [Ar] 4s1 3d5

) + (d)3 = 21 after losingshould add 3 e-.2424

49. The atom having vale4s2 4p2 would be in4s 4p would be in

(a)Group 13 and peri(a)Group 13 and peri(b) Group 14 and per(c) Group 13 and per(c) Group 13 and per(d) Group 14 and per

ence shell configuration

od 3od 3riod 4iod 4iod 4

riod 3

• 49. Option (b):• 4p2 that means n=4 p• 4p that means n=4, p• p2 means group 14 (ca

eriod 4eriod 4arbon gp)

50. Which of the followinatomic size.atomic size.

(a) Cs(a) Cs(b) F(c) Na(c) Na(d) Mg

ng has the smallest

• 50. Option (b):• Alkali metals have larg• Alkali metals have larg

are more. S• Size decreases along So F is small

ger size so Na Cs Mgger size so Na, Cs, Mg

a period, F is in the end

51. The ionisation energthan oxygen becausethan oxygen because

(a)The greater attract(a)The greater attractnucleus

(b) The extra stability(b) The extra stability(c) The smaller size o(d) More penetration e

gy of nitrogen is more e of,e of,

ion of electrons byion of electrons by

y of half filled orbitalsy of half filled orbitalsof nitrogeneffect

• 51. Option (b):• Nitrogen =7 1s2 2s2 2p3• Nitrogen =7 1s 2s 2p3

52. Out of four halogenselectron gain enthalpelectron gain enthalp

(a) Fluorine(a) Fluorine (b) Chlorine (c) Bromine(c) Bromine (d) Iodine

s, the one with highest py is .py is .

• 52. Option (b):• Electron gain enthalpy• Electron gain enthalpy

the group therefore Clsmall it cannot accomsmall it cannot accomhence its EGE is less

y(EGE) decreases downy(EGE) decreases downl > Br > I. Since fluorine

mmodate an electronmmodate an electron than Cl

53. The number of electrcarbon dioxide iscarbon dioxide is

(a) 22(a) 22(b) 44(c) 66(c) 66(d) 88

rons in a molecule of

• 53. Option (a):• C = 6e- and 2O = 8+8• C = 6e- and 2O = 8+8• CO2 is 6 + 16 = 22 e-

8=16e-8=16e-

54. The element with atoresembles the most behaviour with the enumber

(a) 8( )(b) 10(c) 17( )(d) 19

omic number 9 in its chemical

element having atomic

• 54. Option (c):• Element with atomic n• Element with atomic n

halogen. 17 is chlorinenumber 9 is flourine anumber 9 is flourine a e

55. Which of the followinwavelength providedwavelength provided

(a) CO molecule(a) CO2 molecule(b) Electron (c) NH molecule(c) NH3 molecule (d) proton

ng has largest de Broglid all have equal velocityd all have equal velocity

• 55. Option (b):• De Broglie equation λ• De Broglie equation λ• Lower the mass more

velocity is same.• Among these, electronAmong these, electron

= h/mv= h/mvis the wavelength when

n is of lowest mass.n is of lowest mass.

56. Which of the followin

(a)λ = υ/c(b) E = mc2(b) E = mc(c) E/υ = h(d) λ = h/p(d) λ = h/p

ng is not correct.

• 56. Option (a):• Correct equation is• Correct equation is • C = λυ   velocity = waveleength x frequency

57. Bohr atomic model c

(a) The spectrum of h(b) Spectrum of an at(b) Spectrum of an at

one electron only(c) The spectrum of h(c) The spectrum of h(d) The solar spectru

can explain

hydrogen atom onlytom or ion containingtom or ion containing

hydrogen molecule onlyhydrogen molecule onlyum

• 57. Option (b):• It is the limitation of B• It is the limitation of BBohr’s modelBohr s model

58. Which electronic leveto absorb a photon bto absorb a photon b

(a) 3s(a) 3s(b) 2p(c) 1s(c) 1s(d) 3d

el would allow hydrogebut not emit a photon but not emit a photon

• 58. Option (c):• That is the lowest ene• That is the lowest eneergy levelergy level.

59. The configuration of because it violatesbecause it violates

(a) Hund’s rule( )(b) Pauli’s rule(c) Aufbau principle(c) Aufbau principle(d) (n+l) rule

nitrogen cannot be 1s7

• 59. Option (b):• Pauli’s rule says two e• Pauli s rule says two e

cannot have same valquantum numbersquantum numbers.

• If it is 1s7, more numbsame values for all qu

electrons in an atomelectrons in an atom ues for all the four

ber of electrons will haveuantum numbers.

60 Pi k th t t60. Pick the wrong statemephotoelectric effect

(a) electron emission depradiation

(b) Kinetic energy of ejecthe intensity of light

(c) Emission of electrons(c) Emission of electronsmetal

(d) There is no time lag b(d) e e s o t e ag bsurface and ejection of

t ith t tent with respect to

pends on frequency of

cted electrons depends on

s depends on the nature os depends on the nature o

between light striking the bet ee g t st g t ef electron

• 60. Option (b):• Kinetic energy of ejec• Kinetic energy of ejec

the frequency of light cted electrons depends octed electrons depends o

and not intensity.

61. Which of the followinnumbers is not allownumbers is not allow

(a) 2 1 0 +1/2(a) 2, 1, 0, +1/2 (b) 2, 2, 1, +1/2(c) 2 1 +1 1/2(c) 2, 1, +1, -1/2(d) 3, 2, 0, -1/2

ng set of quantum wed(n, l, m, s)wed(n, l, m, s)

• 61. Option (b):• When n=2 l can be = 0• When n=2, l can be = 00 1 only and not 20,1 only and not 2

62. The correct set of quunpaired electron of unpaired electron of

(a) 2 1 0(a) 2, 1, 0(b) 2, 1, 1(c) 3 1 1(c) 3, 1, 1(d) 3, 0, 0

uantum numbers for thechlorine atom( n, l, m)chlorine atom( n, l, m)

• 62. Option (c):• Cl = 17 [Ne] 3s2 3p5• Cl = 17 [Ne] 3s 3p• One unpaired electron• Assume 3px = -1, 3py =

n, so n =3, l= 1, m=10, 3pz= +1

63. The de Broglie wavemass 60g moving wmass 60g moving wis approximately (h =

(a) 10-33 m(b) 10-31 m(b) 10 m(c) 10-16 m(d) 10-25 m(d) 10 25 m

length of a tennis ball owith a velocity of 10m/sewith a velocity of 10m/se

= 6.63 x 10-34Js)

• 63. Option (a):• λ= h/mv = (6 6 x 10-34)/• λ= h/mv = (6.6 x 10 )/• = 1.1 x 10-33m

/(60 x 10-3kg) x 100m/s/(60 x 10 kg) x 100m/sm

64. The angular momentatom depend on atom depend on

(a) Principal quantum(a) Principal quantum(b) Azimuthal quantu(c) Both a and b(c) Both a and b(d) Neither a nor b

tum of an electron in an

m numberm numberum number

• 64. Option (a):• Angular momentum =• Angular momentum =• n =principal quantum

mvr = nh/2π mvr = nh/2πnumber

65. Arrange the followiincreasing order ofincreasing order of

(b) microwaves, (c)

(a) a < b < c < d( )(b) d < c < b < a(c) a < b < c < d(c) a b c d(d) b < a < c < d

ing radiations in f frequency (a) IR raysf frequency (a) IR raysUV rays, (d) X-rays

• 65. Option (d):

66. Which one of the follboth as a stream of pboth as a stream of p

(a)Diffraction(a)Diffraction (b) Interference (c) λ = h/p(c) λ = h/p(d) Photoelectric effe

lowing explains light particles and as waveparticles and as wave

ect

• 66. Option (c):

67. In hydrogen spectrumLyman series are obsLyman series are obs

(a)UV(a)UV(b) IR(c) Visible(c) Visible(d) Far IR

m spectral lines of served in the region of served in the region of

• 67. Option (a):

68. The azimuthal quantshape of the orbital ishape of the orbital i

(a) Spherical(a) Spherical (b) Dumb-bell (c) Double dumble be(c) Double dumble-be(d) Circular

um number, l =1, the s s

ellell

• 68. Option (b):• l = 1 means p-orbital=• l = 1 means p-orbital= dumb-bell shapeddumb-bell shaped

69. Hydrogen loses one resembles in this proresembles in this pro

(a)Alkali metals(a)Alkali metals(b) halogens(c) Alkaline earth met(c) Alkaline earth met(d) Noble gases

electron and forms H+, operty with.operty with.

talstals

• 69. Option (a):• Alkali metals lose one• Alkali metals lose one• Similarly hydrogen.•

e electron to form M+e electron to form M

70. The nucleus of heliu

(a) Four protons(b) Four neutrons(b) Four neutrons(c) Two neutrons and(d) Four protons and(d) Four protons and

m atom contains

d two protonstwo electrons two electrons

• 70. Option (c):• Helium atomic number• Helium atomic number

neutrons.is 2 i e 2 protons and hasis 2 i.e 2 protons and has

71. The number of unpai1s2 2s2 2p4 is 1s 2s 2p is

(a) 4(a) 4(b) 2(c) 0(c) 0(d) 1

ired electrons in

• 71. Option (b):• 2p 2 2p 1 2p 1• 2px 2py 2pz

• Two unpaired electron

72. 2p orbitals have

(a) n=1, l=2(b) n=1 l=0(b) n=1, l=0(c) n=2, l=1(d) n=2 l=0(d) n=2, l=0

• 72. Option (c):

73. Who discovered neu

(a) James Chadwick(b) William Crooks(b) William Crooks(c) J. J. Thomson(d) Rutherford(d) Rutherford

tron

• 73. Option (a):

74. The number of electrelement is 18 and 20element is 18 and 20is

(a) 17(b) 37(b) 37(c) 2(d) 38

rons and neutrons of an respy. Its mass numbe respy. Its mass numbe

• 74. Option (d):• Proton number = elect• Proton number = elect• Therefore mass numb•

trons = 18trons = 18ber = proton + neutron

= 18 + 20 = 38

75. Bohr’s model of an a

(a) Pauli’s exclusion (b) Planck quantum t(b) Planck quantum t(c) Heisenberg uncer(d) All of these(d) All of these

atom is contradicted by

principletheorytheoryrtainty principle

• 75. Option (c):