Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

A cubic unit cell is characterised by a=b=c and alpha=beta=gamma=90^@ . Calculate the number of atoms in one unit cell in the above three cases.

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Solution :BCC, NUMBER of atoms = `8xxfrac{1}{8}+1=2`
2.

Cubic unit cells are divided into primitive, bcc and fcc. Calculate the number of atoms in a unit cell of each of the following: ** bcc ** fcc

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SOLUTION :BCC-`2(8xx1/8+1=2)`
FCC-`4(8xx1/8+6xx1/2=4)
3.

Cubic close-packed (ccp) latticealso called bcc is it true or false?

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SOLUTION :face centred cubic (FEE) lattice
4.

Cubic close-packed (ccp) lattice is also called_____

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SOLUTION :face centred CUBIC (fcc) LATTICE
5.

Cu_(2)S+2C_(2)O to 6Cu+SO_(2) In which process of metallurgy of copper, above equation is involved ?

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Roasting
SELF reduction
Refining
Purification

Solution :This process is also caiied autoreduction process or air reduction process. The sulphide ores of less ELECTROPOSITIVE metals are heated in air to convert part of the ore into OXIDE or SULPHATE which then react with the remaining sulphide ore to give the metal and SULPHUR dioxide.
`2Cu_(2)S+3O_(2)to 2Cu_(2)O+2SO_(2)`
`Cu_(2)S+2Cu_(2)O to 6Cu+SO_(2)`
6.

Cu^(64)(" half life" =12.8 "hours" ) decay by beta^(c-)- emission (38%), beta^(o+)- emission(19%), and electron capture (43%). Write the decay products and calculate partial half lives for each of the decay processes.

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SOLUTION :The nuclear reactions are `:`
`._(29)Cu^(64) rarr ._(30)Zn^(64)+._(-1)E^(0)(._(-1)beta^(0))`
`._(29)Cu^(64) rarr ._(28)Ni^(64)+._(+1)e^(0)(._(+1)beta^(0))`
`._(29)Cu^(64)+._(-1)e^(0) rarr ._(28)Ni^(64)`
Given, `lambda_(avg)=(0.693)/(12.8)hr^(-1)`
`:. lambda_(1)+lambda_(2)+lambda_(3)=lambda_(avg)=(0.693)/(12.8)=5.41xx10^(-2)hr^(-1)......(i)`
Also for parallel PATH decay
`lambda_(1)=` Fractional yield of`._(30)Zn^(64)xxlambda_(avg)`
`lambda_(2)=` Fractional yield of `._(28)Ni^(64)xxlambda_(avg)`
`lambda_(3)=` Fractional yield of `_(28)Ni^(64^(**))xxlambda_(avg)`
`(lambda_(1))/(lambda_(2))=(38)/(19)....(ii)`
and `(lambda_(1))/(lambda_(3))=(38)/(43) ....(iii)`
From Eqs. `(i), (ii),` and `(iii), lambda_(1)=2.056xx10^(-2)hr^(-1)`
`lambda_(2)=1.028xx10^(-2)hr^(-1),lambda_(3)=2.327xx10^(-2)hr^(-1)`
`t_(1//2)` for `beta^(c-)-` emission `=(0.693)/(2.056xx10^(-2))=33.70hr`
`t_(1//2)` for `beta^(o+)-` emission`=(0.693)/(1.028xx10^(-2)0=67.41hr`
`t_(1//2)` for electron capture `=(0.693)/(2.327xx10^(-2))=29.78hr`
7.

Cu_2O is :

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Black oxide of copper
Copper (II)oxide
RED oxide of copper
CUPRIC oxide

Answer :C
8.

Cu_2Cl_2 absorbs:

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`CO_2`
`SO_2`
`H_2SO_4`
`CO`

ANSWER :D
9.

Cu^(2+) ions would be reducd to cuprous ion if their solution are mixedwith an aqueous :

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KL SOLUTION
KCL solution
`KCO_3` solution
`K_2SO_4` solution

Answer :A
10.

Cu^(2+) ions will be reduced to Cu^(+) ions by the addition of an aqueous solution of

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KF
KCl
Kl
KOH

Solution :`UNDERSET("CUPROUS IODIDE")(2CuSO_(4)+4KI to 2CuI+2K_(2)SO_(4)+I_(2))`
11.

Cu^(2+) ions are more stable than Cu^(+) ions because Cu^(2+) ions___________ as compared to Cu^(+) ions.

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have HIGHER charge
have SMALLER IONIC size
undergo EXTENSIVE hydration
All of these

Answer :D
12.

Cu^(2+) ions are coloured but Zn^(2+) ions are colourless. Give reason.

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Solution :`CU^(2+)` ions CONTAIN one unpaired electron in its d-subshell and can undergo d-d TRANSITION. Therefore it is COLOURED.
In `ZN^(2+)`, there is no unpaired electrons, hence it does not undergo d-d transition. Hence it is colourless
13.

Cu^(2+) gives white ppt. of (a)____ with (b) ____ and deep blue colour of _____(c ) ______with (d) ____.

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Answer :a.`Cu_(2)I_(2)` B. `KJ`, C. `[Cu(NH_(3))_94)]^(2+)`
14.

Cu^(2+)(aq) + Zn(s) rightarrow Cu(s) + Zn^(2+)(aq), E^(@) = 1.1V Zn^(2+)(aq) + 2e^(-) rightarrow Zn(s), E^(@) = -0.76V Cu^(2+)(aq) + 4NH_(3)(aq) rightarrow [Cu(NH_(3))_(4)]^(2+)(aq) + 2e^(-) is:

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0.34 V
0.76 V
0.26 V
0.14 V

Answer :C
15.

Cu^(2+)+2e^(-)toCu,E^(@)=+0.34V,Ag^(+)+1e^(-)toAg,E^(@)=+0.80V (i) Construct a galvanic cell using the above data. (ii) For what concentration of Ag^(+) ions will the emf of the cell be zero at 25^(@)C, if the concentration of Cu^(2+) is 0.01 M? (log 3.919=0.593).

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Solution :(i) For `E_(cell)^(@)` to be +ve, oxidation will occur at copper ELECTRODE and reduction at SILVER electrode. Hence, the cell will be represented as : `Cu|Cu^(2+)||Ag^(+)|Ag`
(ii) `Cu+2Ag^(+)toCu^(2+)+2Ag,n=2`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))`. hence, `0.46V-(0.0591)/(2)"log"(0.1)/([Ag^(+)]^(2))`. CALCULATE `[Ag^(+)]`
16.

Cu^(2+)+2e^(-) rarr Cu. For this, graph between E_(red) versus ln[Cu^(2+)] is a straight line of intercept 0.34V, then the electrode oxidation potential of the half cell Cu|Cu^(2+)(0.1M) will be

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`-0.34+(0.0591)/(2)V`
`0.34+0.0591" V"`
`0.34" V"`
NONE of these.

SOLUTION :(a) `E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)-(0.0591)/(2)LOG[Cu^(2+)]`
It `log[Cu^(2+)]=0`, then `Cu^(2+)=1`
`E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)`
`:. OA=E_(Cu//Cu^(2+))^(@)=-E_(Cu//Cu^(2+))^(@)=-0.34" V"`
Hence `E_(Cu//Cu^(2+))=-0.34-(0.0591)/(2)log 0.1=-0.34+(0.0591)/(2)`
17.

Cu^(2+) + 2e rarr Cu(s). For the graph between log[Cu^(2+)] versus E_("red") a straight line of intercept 0.34 V is obtained. Then electrode potential of the half-cell Cu// Cu^(2+)(0.1 M) will be :

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`[0.34+(0.0591)/2]V`
`[-0.34-(0.0591)/2]V`
`0.34 V`
`[-0.34+(0.0591)/2]V`

ANSWER :D
18.

Cu^(2+) +2e rarr Cu, log [Cu^(2+)] vs E_(red).graph is of the type as shown in figure where OA =0.34 V then electrode otential of the half cell of Cu^(2+)(0.1M) will be :

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`-0.34+0.091/2 V`
`0.34 + 0.0591V`
`0.34V`
None

Answer :A
19.

Cu^(+), Zn^(2+), Sc^(3+), Ti^(4+) are colourless. Prove this statement.

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Solution :`(i) Cu^(+)`, `ZN^(2+)` have `d^(10)` CONFIGURATION and `SC^(3+)`, `Ti^(4+)` have `d^(1)` configuration .
`(ii)d-d` transition is not possible in the above COMPLEXES. So they are colourless.
20.

Cu^(+), Zn^(2+) are diamagnetic. Prove it.

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Solution :`Cu^(+), Zn^(2+)` electronic configuration `[AR] 3d^(10)`
The number of unpaired electron is 0.
`MU= sqrt( 0(0+2)) = 0 mu_(B)`. So `Cu^(+), Zn^(2+)` are DIAMAGNETIC.
21.

[Cu ( NH _ 3 )_4 ]^ (2+)show thefollowinghybridization

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` dsp ^ 2`
`sp^ 3 d`
`dsp^3 `
`sp ^ 3`

Solution : ` "" _ (29 ) Cu= [Ar]3d^( 10) 4S ^ 1`
` Cu^ ( 2 +)= [Ar] 3 d^( 9) `
22.

Cu^(+) ion is not stable in aqueous solution because of disproportionation reaction. E^(@) value for disproportionation of Cu^(+) is (Given E_(Cu^(2+)//Cu^(+))^(@)=0.15,E_(Cr^(2+)//Cu)^(@)=0.34V)

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`-0.49V`
`0.49V`
`-0.38V`
`0.38V`

Solution :The REQUIRED REACTION `(Cu^(++)+Cuto2Cu^(+))` can be obtained by using the following reactions.
`Cu^(++)+e^(-)toCu^(+),E_(Cu^(++)//Cu^(+))^(o)=0.15V` . . .(i)
`Cu^(++)+2e^(-)toCu,E_(Cu^(++)//Cu)^(o)=0.34V` . .(ii)
Multiplying eq. (i) by 2 we get
`2CU^(++)+2e^(-)to2Cu^(+)`. . . (iii)
`DeltaG_(1)=-nFE=-2xxFxx0.15`
`Cu^(++)+2e^(-)toCu`. . . (iv)
Subtract the eq. (iv) from (iii)
`Cu^(++)+Cuto2Cu^(+)`
`DeltaG_(3)=-nFE=-1xxFxxE^(o)`
Also `DeltaG_(3)=DeltaG_(1)-DeltaG_(2)`
`-1FE^(o)=(-2Fxx0.15)-(-2Fxx0.34)`
`E^(o)=-0.38`
this is the value of the reaction
`Cu^(++)+Cuto2Cu^(+)`
but the given reaction is just REVERSE of it
`thereforeE_(cell)` for given reaction =+0.38V.
23.

Cu^(+) ion is not stable in aqueous solution because because of dispropotionation reaction. E^(@) value of disproportionation of Cu^(+) is [E_(Cu^(2+)//Cu^(+))^(@)=+ 0.15 V, E_(Cu^(2+)//Cu)^(@)=0.34 V]

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SECOND ionisation entyhalpy of copper is less than the first ionisation enthalpy
LARGE value of second ionisation enthalpy of copper is compensated by MUCH more negative hydration energy of `Cu_((aq))^(2+)`
hydration energy of `Cu_((aq))^(2+)` is much more negative than that of `Cu_((aq))^(2+)`
MANY copper (I) COMPOUNDS are unstable in aqueous solution and undergo disproportionation rection.

Answer :B
24.

Cu^+ ion is not stable in aqueous solution because of diproportionation reaction . E^(@)value for the disproportionation of Cu^(+) is : (E_(cu^(2+)// cu^+)^(@) = + 0.15 V , E_(cu^(+) // cu)^(@)= + 0.34 V )

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`Cu_((aq))^(+2)` is more STABLE than `Cu_((aq))^(+)`
`Cur_((aq))^(+2)`can form INTERSTITIAL compounds in aqueous solution
The equilibrium CONSTANT for `Cu_((aq))^(+) hArr Cu_((aq))^(+2) + Cu` is very high
The equilibrium constant for `Cu_((aq))^(+2) hArr Cu_((aq))^(+2) = Cu` is very low

Solution :`Cu^(+2)` is more stable is aqueous MEDIUM due to hydration energy
25.

Cu (II) sulphate solution is treated separately with KCl and KI In which case, Cu^(2+) be reduced to Cu^+:

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With KCI
With KI
With both (a) and (B)
None

Answer :B
26.

Cu+H_(2)SO_(4)

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SOLUTION :`Cu+2H_(2)SO_(4) to SO_(4)+SO_(2)+2H_(2)O`
27.

Cu can reduce Ag^(+)to metallic Ag but Ag cancan't reduce CU^(2+)to metallicCu. Explain

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ANSWER :N//A
28.

Cu^(+) (aq) is unstablein solutionand undergoessimultaneous oxidationand reduction according to the reaction 2cu^(+)(Aq)rarrCu^(2+)(aq)+cu(s) choose correct E^(2)for above reactionif E_(Cu^(2+)//Cu)^(@) =0.34 V and E_+(Cu^(2+)//Cu^(+))^(2) =0.15 V

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`-0.38 V`
`+0.49 V`
`+0.38 V`
`-0.19 V`

Solution :From given data (suign `TRIANGLE G^(2)=- nFE_(cell)^(@)`
`(i) Cu (s) rarr Cu^(2+) (aq)+2E^(-)`
`triangle G_(1)^(2)=-2xx(-0.34)xxxF`
`(ii) Cu^(2+) +(aq)+e^(-) rarr Cu^(+) (aq).triangle G_(2)^(@)=- 1xx(0.15)"F on addition"`
`Cu(s) rarr Cu^(+) (aq) +e^(-) triangleG_(3)^(@) =- 1xxE^(@)xxF`
`triangle G_(3)^(2)=triangleG_(1)^(2)+triangle_(2)^(@)=(-2xx0.34 xxF)+(-1xx0.15 xxF)`
`=+0.68 F-0.15 F` or `E^(@)=- 0.53 V`
29.

Cu, Ag and Au are called white metals. Is it true or false?

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SOLUTION :COINAGE METALS
30.

Cu^(+2) and and 'en gives a chelate complex. It is ... Membered

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SOLUTION :`H_(2)OVERSET(**)N-CH_(2)-CH_(2)-overset(**)NH_(2)`
31.

Cu^(+2) can oxidise the halide

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`I^-`<BR>`Br^-`
`CL^-`
`F^-`

ANSWER :A
32.

CsCl has bec arrangement and its unit cell edge length is 400 pm. Calculate the inter-ionic distance in CsCl.

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Solution :Edge LENGTH, a = 400 PM. Inter-ionic distance, `d = (a SQRT3)//2 = (400 ×x 1.732)//2` = 346.4 pm
33.

CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is

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400 pm
800 pm
`sqrt(3)XX100` pm
`(sqrt(3)/2)xx400` pm

Solution :`sqrt(3)a=r_(Cs^(+))+2r_(CL^(-))+r_(Cs^(+))`
`(sqrt(3)/2)a=(r_(Cs^(+))+r_(Cl^(-)))" " (sqrt(3)/(2))=400=` inter ionic distance
34.

CsCl has bcc arrangement. Its unit cell edge length is 400 pm. Its interionic distance is …….. .

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400 PM 
800 pm 
73 x 100 pm 
`sqrt(3)/(2) xx 400` pm 

Solution :In body-centred UNIT cell : `4r = sqrt(3)a`
`r = (SQRT3)/(4) a`
`thereforer = (sqrt3)/(4) xx 400`
`implies r = sqrt(3) xx 100 "pm"`.
35.

CsCl has bcc arrangement, its unit cell edge length is 400 pm, its inter atomic distance is ……………… .

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400 pm
800 pm
`sqrt3xx"100 pm"`
`((SQRT3)/(2))xx"400 pm"`

Solution :Hint : `sqrt3a=r_(Cs^(+))+2r_(CL^(-))+r_(Cs^(+))`
`((sqrt3)/(2))a=(r_(Cs^(+))+r_(Cl^(-)))((sqrt3)/(2))400=" inter ionic distance"`
36.

CsCl crystallises in body centred cubic lattice. If 'a' is its edge length, then which of the following expressions is correct?

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`r_(CS) + r_(CL^(-)) = 3a`
`r_(Cs^(+)) + r_(Cl^(-)) = (3a)/2`
`r_(Cs^(+)) + r_(Cl^(-)) = (sqrt(3)a)/2`
`r_(Cs^(+)) + r_(Cl^(-)) = sqrt(3)a`

ANSWER :C
37.

CsCl crystallises in body-centred cubic lattice. If 'a' is its edge length then which of the following expressions is correct ?

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`r_(CS^+) + r_(CL^-) = sqrt(3)a`
`r_(Cs^+) + r_(Cl^-) = 3a`
`r_(Cs^+) + r_(Cl^-) = (3a)/2`
`r_(Cs^+) + r_(Cl^-) = (sqrt(3))/2 a`

ANSWER :D
38.

CsCl crystalises in body centred cubic lattice. If 'a' is its edge length then which of the following expression is correct ?

Answer»

`r_(Cs^(+))+r_(Cl^(-)) = 3a`
`r_(Cs^(+))+r_(Cl^(-)) = (3a)/(2)`
`r_(Cs^(+))+r_(Cl^(-)) = (sqrt(3))/(2)a`
`r_(Cs^(+))+r_(Cl^(-)) = sqrt(3)a`

Solution :EDGE LENGTH =a
Face diagonal = `sqrt( 2) a `
Body diagonal `= sqrt( 3) a `
Since `Cl^(-)` ions are PRESENT at the corners of the CUBE and `Cs^(+)`ion is at the body centre.
`2 ( R _(Cs^(+)) + r_(Cl^(-))) = sqrt( 3). a `
`:. r_(Cs^(+)) + r_(Cl^(-))= ( sqrt( 3) a)/( 2)`
39.

CSCI possess bcc arrangement, its edge length in unit cell is 400 pm, then inter ionic distance in them are = .......... pm

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`(SQRT3)/2 xx 400`
`sqrt(3) xx 100`
`400`
`800`

Answer :B
40.

CsBr has BCC structure with edge length 4.3 pm. The shortest interionic distance in between Cs^(+) and Br^(-) is:

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3.72 PM
1.86 pm
7.44 pm
4.3 pm

ANSWER :A
41.

CsBr has bcc structure with edge length 4.3. The shortest interionic distance in between Cs^+ and Br^- is :

Answer»

3.72
1.86
7.44
4.3

Answer :A
42.

CsBr crystallise in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02xx1023 mol^(-1), the density of CsBr is

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`8.25 g//cm^(3)`
`4.25 g//cm^(3)`
`42.5 g//cm^(3)`
`0.425 g//cm^(3)`

ANSWER :A
43.

CsBr crystal has bcc structure . It has an edge length of 4.3 overset(@)A. The shortest interionic distance between Cs^(+) and Br^(-) ions is

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`1.86 OVERSET(@)A`
`3.72 overset(@)A`
`4.3 overset(@)A`
`7.44 overset(@)A`

SOLUTION :Closest approach in bcc lattice
`=(1)/(2)` of body diagonal `=(1)/(2)xxsqrt(3)a =SQRT(3)/(2)xx4.3=3.72overset(@)A`.
44.

C(s)+O_(2)(g)rarrCO_(2), DeltaH=-94 kcalCO(g)+(1)/(2)O_(2)(g)rarrCO_(2), DeltaH=+135.2 kcal Then heat of formation of CO(g) is

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`-26.4` KCAL
`41.2` kcal
26.4 kcal
`-229.2` kcal

Solution :By (i) - (II) :`C+(1)/(2)O_(2)rarrCO, DeltaH=-229.2`
45.

Cs^+ ion imparts violet colour to the flame. This is due to the fact that the emitted radiations have:

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HIGH ENERGY
LOW energy
Longer wavelength
None of three

Answer :B
46.

Crystals which are good conductor of heat and electricity are -

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IONIC CRYSTALS
COVALENT crystals
Metallic crystals
Molecular crystals

Answer :C
47.

Crystals which are good conductor of electricity and heat are known as -

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IONIC CRYSTALS
Covalence crystals
Metallic crystals
Molecular crystals

ANSWER :C
48.

Crystals where dipoles may align themselves in an ordered mannerso that there is a net dipole moment, exhibit

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pyro-electricity
para,agmetosm
dimagnetism
antiferro-electricity

Solution :CRYSTALS having a NET dipole MOMENT SHOW paramagnetism.
49.

Crystals of which pair are isomorphous

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`ZnSO_(4),SnSO_(4)`
`MgSO_(4),CaSO_(4)`
`ZnSO_(4),MgSO_(4)`
`PbSO_(4),NiSO_(4)`

Solution :`ZnSO_(4),MgSO_(4)` are isomorphous i.e. having same structure.
50.

Crystals of KCl appears violet due to ...........

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SCHOTTKY defects 
Metal excess defects 
Metal deficiency DEFECT 
IMPURITY of violet colour 

ANSWER :B