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.

The intermediate formed in the acid catalysed dehydration of n-propyl alcohol is

Answer»

`CH_(3)-CH_(2)-CH_(2)^(+)`
`CH_(3)-CH^(+)-CH_(3)`
`CH_(3)-CH_(2)-OVERSET(.)(C)H_(2)`
`CH_(3)-overset(.)(C)H_(2)-CH_(3)`

SOLUTION :
Discussion
2.

The intermediate during the addition of HCl to propene in the presence of peroxide is

Answer»

`CH_(3)overset(.)(C)HCH_(2)CL`
`CH_(3)overset(+)(C)HCH_(3)`
`CH_(3)CH_(2)overset(.)(C)H_(2)`
`CH_(3)CH_(2)overset(+)(C)H_(2)`

Answer :B
Discussion
3.

The intermediate compound formed when acetylene is hydrated in presence of dil. H_(2)SO_(4) and HgSO_(4) is

Answer»

ACETALDEHYDE
Ethenol
Vinyl chloride
Ethenal

ANSWER :B
Discussion
4.

The intermediate carbocation formed in the reaction of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^(-1), 363.7kJ mol^(-1) and 296.8 kJ mol^(-1) respectively. What will be the order of reactivity of these halogen acids ?

Answer»

Solution :Hydrogen halides (HCl, HBr, HI) ADD upto a alkanes to form alkyl halides. The order of reactivity of the hydrogen halides is `hI gt HBr gt HCl`. Reactivity of HX depends on bond dissociation enthalpy of HX. Bond enthalpy of HI `(296.kJ mol^(-1)) lt HBr (363.7 kJ mol^(-1)) lt HCl (430.5 kJ mol^(-1))` is in this order hence there activity is in reverse order : `HCl lt HBr lt HI`
`CH_(3)-CH=CH_(2)+H^(+)` slow step `RARR CH_(3-) ""^(+)CH-CH_(3)+X^(-)` fast `rarr CH_(3)-CHX-CH_(3)`
Slow step is RATE determining step and it depends on the availability of `H^(+)`. Availability of `H^(+)` depends on bond dissociation enthalpy of HX.
Discussion
5.

The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 "kJ mol"^(-1), 363.7 "kJ mol"^(-1) and 296.8 "kJ mol"^(-1) respectively. What will be the order of reactivity of these halogen acids ?

Answer»

Solution :In the reaction of propene with HCl , HBr or HI, first a `H^+` adds to give the same carbocation intermediate (I) which then undergoes nucleophilic attack by the HALIDE ion `(X^-)` to give the addition product. Here, first STEP is slow and hence is the rate-determining step of the reaction while the second step is first and hence does not affect the rate of the reaction.

The rate of the first step depends upon how easily is the proton available. The availability of the proton, in turn, depends upon the bond dissociation energy of the H-X MOLECULE. Since the bond dissociation energy decreases in the order : HCl `(430.5 "kJ mol"^(-1)) GT HBr(36. "kJ mol"^(-1)) gt HI (296.8 "kJ mol"^(-1))`, therefore , the REACTIVITY of the halogen acids decreases in the reverse order : HI gt HBr gt HCl
Discussion
6.

The intermediate as well as final products formed in the reaction .

Answer»




Solution :nucleophillic addition with HYDROXYL followed ellectrophllic attack of NITROGEN to benzene RING.
Discussion
7.

The intermediate as well as final products feomed in the reaction .

Answer»




Solution :HEMIKETAL ACID catalyses to hydroxy ketone. HYDROXYL group attacks aldehyde SINCE is more REACTIVE then ketone
Discussion
8.

The interionic distance for cesium chloride crystal will be

Answer»

a
a/2
`sqrt3 a//2`
`2a//sqrt3`

Solution :CSCL have SIMPLE cubic structure. In this structure body idagenal of simple cube `sqrt3 a = 2 XX (r_(CS^(+)) + r_(Cl^(-)))` So interionic distance `= 2 xx (r_(CS^(+)) + r_(Cl^(-)))`
Discussion
9.

The interionic distance for cesium chloride crystals will be

Answer»

a
`a//2`
`sqrt3a//2`
`2a//sqrt3`

SOLUTION :As CSCL is BODY CENTRED, `d=sqrt3 a//2`.
Discussion
10.

The intercepts and slope of graph of logK_(p)^(@) "Vs" (1)/(T) are :

Answer»

`DeltaG^@, DELTAH^@`
`(DeltaG^@)/(2.303 \ R), (DeltaH^@)/(2.303 \ R)`
`(DELTAS^@)/(2.303R), (-DeltaH^@)/(2.303R)`
`(DeltaH^@)/(2.303 \ R),DeltaS^@`

Solution :In `K=(-Delta H^(@))/(RT)+(DeltaS^(@))/(R)`
Discussion
11.

The interactions that are results of temporary dipoles induced in the ordinarily non polar molecules are

Answer»

Dispersion FORCES 
Dipole – Dipole 
Dipole INDUCED dipole 
HYDROGEN bonding 

ANSWER :A
Discussion
12.

The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles but their magnitude depends upon

Answer»

charge of interacting particles.
mass of interacting particles.
polarisability of interacting particles.
strength of permanent dipoles in the particles.

Solution :LONDON dispersion forces exist only over very short distance. The ENERGY of INTERACTION varies as `(1)/("(distance between two interactingparticles)"^(6))`
More complex or large are the molecules, higher is the magnitude of London forces. This is due to the fact that the large ELECTRON clouds are easily polarised.
HENCE, higher the polarisability of interacting particles, higher is the magnitude of the interaction energy.
Discussion
13.

The interaction energy of London force is inversely proportional to sixth of the distance between two interacting particles but their magnitude depends upon

Answer»

charge of intercting PARTICLES
MASS of interacting particles
polarisability of interacting particles
strength of permanent dipoles in the particles

Solution :Greater the polarizability of the interacting particles, greater is the MAGNITUDE of the interaction energy.
Discussion
14.

The inter molecular forces present in inert gases are

Answer»

Ion-ion
Ion-dipole
Dipole - dipole
DISPERSION

Solution :INERT GASES have only LONDON dispersion FORCES.
Discussion
15.

The intensive property among the quantitiesis _____

Answer»

SOLUTION :MASS /VOLUME
Discussion
16.

………….is an intensive property .

Answer»

MASS
VOLUME
ENTHALPY
`("mass")/("volume")`

ANSWER :D
Discussion
17.

The intensive property among these quantities is

Answer»

Weight
Density
Entropy
Volume

Answer :B
Discussion
18.

Theintensityof linesin spectrumdependson whichfactor ?

Answer»

SOLUTION :Emitteda ABSORB NUMBEROF PHOTON.
Discussion
19.

The instrument measuring heat changes at constant pressure ..........

Answer»

PH meter
calorie meter
galvano meter
hydro meter

Answer :B
Discussion
20.

The inorganic compound obtained by the auto oxidation of 2-alkylanthraquinol is

Answer»

`H_(2)O`
`H_(2)O_(2)`
`H_(2)`
`O_(2)`

ANSWER :B
Discussion
21.

The inner shell to which the electrons jump from the outer shells in the formation of Pfund series in the hydrogen spectrum is

Answer»


Solution :For Pfund SERIES, `n_(1) = 5, n_(2) = 6,7`...
Discussion
22.

The initial temperature of a gas 1s 27 to temperature 80 cm^(3) of the gas should be heated to increases the volume by 25%.

Answer»

Solution :`v_(1)80 cm^(3), v_(2)+80+(25)/(100)xx80=100 cm^(3)`
`T_(1)=27+273=300 K, T_(2)=?`
`(v_(1))/(T_(1))=(v_(2))/(T_(2)) orT_(2)=(T_(1)V_(2))/(V_(1))=(300xx100)/(80)`
`:. T_(2) =375 K=375-273=102^(@)C`
`:. T_(2)=102^(@)C`
`:. T_(2)=102^(@)C`
Discussion
23.

The initial concentrations of A and B are same for A +2B harr 3C and when equilibrium concentrations of B and C are same then K_(c) is

Answer»

0.25
0.75
0.5
1

Solution :
[B]=[C] `x=a // 5`
`K_(C)=([C]^(3))/([A][B]^(2))=(((3A)/(5))^(3))/((4a // 5)(3a // 5)^(2))=(3)/(4)`
Discussion
24.

The inflow of nutrients from fertilizers into lakes resulting in the increase of phosphate in lake water and depletion of dissolved oxygen causing the fish to die is called _____

Answer»


ANSWER :EUTROPHICATION
Discussion
25.

Theinert gases are ________atomic.

Answer»

MONO
DI
TRI
ZERO

ANSWER :a
Discussion
26.

In which of the following the inert pair effect is most prominent?

Answer»

C
Si
Ge
PB

Solution :Pb
Discussion
27.

The industrial dihydrogen is produced ……. from coal.

Answer»

`77%`
`18%`
`4%`
`50%`

Answer :B
Discussion
28.

The indicator used in iodometric titrations is :

Answer»

phenolphthalein
litmus
potassium iodide
starch

Solution :N//A
Discussion
29.

Theincreasingorder orreactivity amonggroup 1 element is Li lt Na lt K lt Rb lt Cs whereasthe of group17isF gt C1 gt Br gt I. Explain.

Answer»

Solution :Theelements of group1 haveonly oneelectronin theirrespectivevalenceshellsand thusa strong tendencyto losethiselectron. Thetendencyto loseelectrons in turn DEPENDS upon theionizationenthalpy. Sincethe ionizationenthalpydecreasesdownthe groupthereforethe reactivityof group1 elements increasesin the sameorder : `LI lt Na lt K lt Rb ltCs.` In contrastthe elements of group17 haveseven electrons in theirrespectivevalenceshellsand thushave astrong tendencyto accept one moreelectron
Thetendency to accept electronsin turn dependsupontheirelectrode potentials.Sincetheelectrode potentialsof group17 elements decreasesin the order : `F (+ 2.87 V ) GT C1 ( + 1.36 V ) , Br ( 1.08 V )` and `I (+ 0.53 V)` thereforetheirreactivitiesalso decreases in the sameorder `: F gt C1 gt Br gt I`
Alternativelytendency toaccept electrons can belinked to electrongain enthalpy. sinceelectrongainenthalpybecomesless and lessnegativeas we move fromC1 to I thereforereactivityincreasesfromto I, F ,howeveris themostreactive due toits lowbonddissociationenergy .
Discussion
30.

Theincreasingorderof theionicradiiof the givenisoelectronicspecies is

Answer»

`S^(2-) , C1^(-),CA^(2+) ,K^(+)`
`Ca^(2+) , K^(+) , C1^(-) ,S^(2-)`
`K^(+), S^(2-),Ca^(2+) ,C1^(-)`
`C1^(-) ,Ca^(2+) ,K^(+) S^(2-)`

SOLUTION :Referto Ans . 27 Option( b) is correct
Discussion
31.

The increasing order of the ionic radii of the given isoelectronic species is :-

Answer»

`K^(+),S^(2-),CA^(2+),CL^(-)`
`Cl^(-),Ca^(2+),K^(+),S^(2-)`
`S^(2-),Cl^(-),Ca^(2+),K^(+)`
`Ca^(2+),K^(+),Cl^(-),S^(2-)`

ANSWER :D
Discussion
32.

The increasing order of the ionic radii of the given isoelectronic species is :

Answer»

`CL^(-),Ca^(2+), K^(+), S^(2-)`
`S^(2-), Cl^(-), Ca^(2+), K^(+)`
`Ca^(2+), K^(+), Cl^(-), S^(2-)`
`K^(+), S^(2-), Ca^(2+), Cl^(-)`

Solution :
From `Ca^(2+) to K^(+)to Cl^(-) to S^(2-) ` nuclear chaege Z is DECREASING in them are EQUAL 18.
THEREFORE the ionic radii of them is increasing.
Discussion
33.

The increasing order of the density of alkali metals is

Answer»

`Li lt K lt NA lt Rb lt CS `
`Li lt Na lt K lt Rb lt Cs `
`Cs lt Rb lt Na lt K lt Li `
`Cs lt Rb lt K lt Na lt Li`

SOLUTION :DENSITY of K is LOWER than that of Na .
Discussion
34.

The increasing order of the first ionisation enthalpies of the element B, P, S and F is-

Answer»

<P>`F LT S lt P lt B`
`P lt S lt B lt F`
`B lt P lt S lt F`
`B lt S lt P lt F`

ANSWER :D
Discussion
35.

The increasingorder of thefirstionizationenthalpyfo the elementB, P, S andF (lowestfirst )is

Answer»

`B lt Plt S lt F`
`b lt S lt P lt F`
`F lt S lt P lt B`
`P lt S lt Blt F`

SOLUTION :B andF liein the 2ndperiodwhileP andS LIE in charge `Delta_(i) H_(1) `ofP andS arehigherthan thatof Bbut lowerthan thatof F. AmongP andS, P hashigher `Delta_(i) H_(1) ` due toexactly HALF- filledconfiguration.Thusthe overall INCREASINGORDER of `Delta_(i) H_(i)` is: `B lt S lt P lt F` , i.e.,option ( b)is correct.
Discussion
36.

The increasing order of the acid strength is correct in

Answer»

PHENOL, ACETIC acid, Ethanol, Chloroacetic acid
acetic acid, Ethanol, Phenol, Chloro acetic acid
Ethanol, Phenol, Acetic acid, Chloro acetic acid
chloroacetic acid, Acetic acid, Ethanol, Phenol

Solution :`C_(2)H_(5)OH gt C_(6)H_(5)OH gt CH_(3)COOH gt ClCH_(2)COOH`
Discussion
37.

The increasing order of stability of the following free radicals is……….

Answer»

`(CH_(3))_(2) overset(.)(C )H lt (CH_(3))_(3) overset(.)( C) lt (C_(6)H_(5))_(2) overset(.)(C )H lt (C_(6)H_(5)) overset(.)(C )`
`(C_(6)H_(5))_(3) overset(.)(C ) lt (C_(6)H_(5))_(2) overset(.)(C )H lt (CH_(3))_(3) overset(.)(C )lt (CH_(3))_(2) overset(.)(C )H`
`(C_(6)H_(5))_(2) overset(.)(C )H lt (C_(6)H_(5))_(3) overset(.)( C) lt (CH_(3))_(3) overset(.)(C )lt (CH_(3))_(2) overset(.)(C )H`
`(CH_(3))_(2) overset(.)(C )H lt (CH_(3))_(3) overset(.)(C )lt (C_(6)H_(5))_(3) overset(.)(C ) lt (C_(6)H_(5))_(2) overset(.)(C )H`

Solution :`(CH_(3))_(2) CH lt (CH_(3))_(3) overset(.)(C )lt (C_(6)H_(5))_(2) overset(.)(C )H lt (C_(6)H_(5))_(2) overset(.)(C )`
`2^(@) lt 3^(@) lt "benzylic " 2^(@) lt "benzylic " 3^(@)`
Stability of free readicals `ul(1^(@)lt 2^(@)lt 3^(@) 4^(@) lt "benzylic"1^(@))lt` benzylic `2^(@)`
alkylic `lt` benzylic `3^(@)`
Discussion
38.

The increasing order of stability of the following free radicals is

Answer»

`(CH_(3))_(2)overset(**)(C)H lt (CH_(3))_(3)overset(**)(C) lt (C_(6)H_(5))_(2)overset(**)(C)Hlt(C_(6)H_(5))_(3)overset(**)(C)`
`(C_(6)H_(5))_(3)overset(**)(C) lt (C_(6)H_(5))_(2)overset(**)(C)H lt (CH_(3))_(3)overset(**)(C)lt(CH_(3))_(2)overset(**)(C)H`
`(C_(6)H_(5))_(2)overset(**)(C)H lt (C_(6)H_(5))_(3)overset(**)(C) lt(CH_(3))_(3)overset(**)(C)lt(CH_(3))_(2)overset(**)(C)H`
`(CH_(3))_(2)overset(**)(C)H lt (CH_(3))_(3)overset(**)(C)lt(C_(6)H_(5))_(3)overset(**)(C)lt(C_(6)H_(5))_(2)overset(**)(C)H`

Solution :On the basis of hyperconjugation, the order of stability of free radicals is as follows `3^(@) gt 2^(@) gt 1^(@)`. BENZYL free radicals are STABILISED by resonance and hence are more STABLE than alkyl free radicals. More the number of phenyl groups attached to the carbon atom, more is the stability of free radical.
Discussion
39.

The increasing order of stability of the following free radicals is :

Answer»

`(CH_(3))_(2)overset(*)(CH) lt (CH_(3))_(3) overset(*) lt (C_(6)H_(5))_(2)overset(*)(CH) lt (C_(6)H_(5))_(3)overset(*)C`
`(C_(6)H_(5))_(2)overset(*)C lt (C_(6)H_(5))_(2)overset(*)CH lt (CH_(3))_(3) overset(*)C lt (CH_(3))_(2)overset(*)(CH)`
`(C_(6)H_(5))_(2)Hoverset(*)C lt (C_(6)H_(5))_(3) lt (CH_(3))_(3) overset(*)C lt (CH_(3))_(2) overset(*)(CH)`
`(CH_(3))_(2)H overset(*)C lt (CH_(3))_(3) overset(*)C lt (C_(6)H_(5))_(3) overset(*)C lt (C_(6)H_(5))_(2) overset(*)(CH)`

Answer :A
Discussion
40.

The increasing order of reduction of alkyl halides with zinc and dilute HCl is …..

Answer»

`R-CL LT R-I lt R-Br`
`R-Cl lt R-Br lt R-I`
`R-I lt R-Br lt R-Cl`
`R-Br lt R-I lt R-Cl`

Solution :The REACTIVITY with alkane is `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. Hence reduction of alkyl halide with Zn/HCl follow reverse ORDER. The reactivity of reduction increases as the bond strength of reduction increases as the bond strength of C-X decrease. It depends on the size of X. As the size of X increases, bond strength decreases ad reactivity increases.
Discussion
41.

The increasing order of reduction of alkyl halides with zinc and dilute HCl is

Answer»

R-Cl lt R-I lt R-Br
R-Cl lt R-Br lt R-I
R-I lt R-Br lt R-Cl
R-Br lt R-I lt R-Cl

Solution :The reactivity of REDUCTION of alkyl halides with Zn/HCl increases as the strength of the C-X BOND DECREASES , i.e., R-Cl lt R-Br lt R-I
Discussion
42.

The increasing order of reactivity among group 1 elements isLi< Na< K< Rb < Cs whereas in group 17 F> Cl> Br> I Explain.

Answer»

Solution :The ELEMENTS of group I have only one electron in their respective VALENCE shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization ENTHALPY. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the same ORDER LiltNaltKltRbltCs. In contrast. the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to ACCEPT one more electron to make stable configuration. It is linked with electron gain enthalpy and electronegativity. Since both of them decreases down the group, the reactivity also decreases.
Discussion
43.

The increasing order of pK_(a) values of the following compounds is :

Answer»

`C LT B LTA lt D`
`B lt C lt D lt A`
`B lt C lt A lt D`
`D lt A lt C lt B`

ANSWER :C
Discussion
44.

The increasing order of reactivity among group-1 elements is Li lt Na lt K lt Rb lt Cs whereas that among group 17 elements is F gt CI gt Br gt I.

Answer»

SOLUTION :The elements of GROUP-1 have only one electron in their outermost shells and therefore, have strong tendency to lose this electron. The tendency of these elements to lose the valence electron depends upon the ionization enthalpy.
Since ionization enthalpy decreases down the group, therefore the reactivity of group-1 elements increases in the same order `Li lt Na lt K lt KB lt Cs`.
On the other hand, the elements of group-17 have seven ELECTRONS in their respective valence shells and therefore, they, have strong tendency to accept one more electron .
The tendency to accept ADDITIONAL electrons depends upon the electrode potentials of group17 elements.
The electrode potential of group 17 elements decreases fromF to I `[F = +2.86V, Cl = + 1.36V, " Br " = + 1.08 V " and " I = +0.53V]` and therefore , their reactivities also decrease in the same order as `F gt Cl gt Br gt I`.
Discussion
45.

The increasing order of ionic character of CsF, LiI, NaBr and KCl is

Answer»

`NaBr lt KCL lt LII lt CsF`
`CsF lt KCl lt NaBr lt LiI`
`LiI lt NaBr lt KCl lt CsF`
`LiF lt KCl lt CsF lt NaBr`.

ANSWER :C
Discussion
46.

The increasing order of first ionization enthalpies of the elements B, P, S and F (lowest first) is .........

Answer»

`P lt S lt B lt F`
`F lt S lt P lt B`
`B lt S lt P lt F`
`B lt P lt S lt F`

SOLUTION :"Generally in a period as atomic number is higher its first IONISATION enthalpy is greater",
ACCORDINGLY first ionisation enthalpy of `B lt F`
(both of second period) and of `P lt S` (both of third period)
The `I^(ST)` ionisation enthalpy decreases from top to botton in a GROUP .
` :. I^(st)`ionisation potential order should be: `B lt F " and " P lt S`
`P [Ne] 3s^(2) 3px^(1) 3py^(1) 3p_(2)^(1)`
and `S [Ne] 3s^(2) 3px^(2) 3py^(1) 3p_(z)^(1)`
P has stable half filled (p)
Electronic configuration, hence`Delta_(i)H(P) gt Delta_(i)H(S)`
` :. ` The correct order is `B lt F " and " S lt P `
F is of smallest size among these all.
`:. F`has highest `Delta_(i)H`
` :. `Orderis `B lt S lt P lt F`
Discussion
47.

The increasing order of hyperconjugative effect ofCH_(3)-,C_(2)H_(5)-,(CH_3)_(2)CH-" and "(CH_3)_(3)C- when one of these is attached to benzene ring is

Answer»

`CH_(3),C_(2)H_(5),(CH_3)_(2)CH" and "(CH_3)_(3)C`
`(CH_3)_3C,(CH_2)CH,CH_3,C_2H_5`
`(CH_2)_2CH,(CH_3),CH_3C_2H_5`
NONE of the above

ANSWER :A
Discussion
48.

The increasing order of electron donating inductive effect of alkyl groups is

Answer»

`-H lt -CH_(3) lt -C_(2)H_(5) lt -C_(3)H_(7)`
`-Hgt-CH_(3)gt-C_(2)H_(5)gt-C_(3)H_(7)`
`-H lt -C_(2)H_(5) lt -CH_(3) lt -C_(3)H_(7)`
`-H gt -C_(2)H_(5) gt -CH_(3) gt -C_(3)H_(7)`

Solution :The ORDER of +I effect of alkyl group is given as `-H lt -CH_(3) lt -CH_(2)-CH_(3) lt CH_(3)-overset(CH_(3))overset(|)(C)H-`
Discussion
49.

The increasing order of effusion among the gases H_2O_2, NH_3 and CO_2 is

Answer»

`H_2 , CO_2 ,NH_3 ,O_2`
`H_2 , NH_3 ,O_2 ,CO_2`
`H_2 , O_2 ,NH_3 ,CO_2`
`CO_2 ,O_2 ,NH_3 ,H_2`

ANSWER :D
Discussion
50.

The increasing order of diazotization of the following compound is:

Answer»

`(a) LT (B) lt ( C) lt (d)`
`lt (d) lt (b) lt (c )`
`(a) lt (d) lt ( c) lt (b)`
`(d) lt ( c) lt (b) lt (a)`

SOLUTION :N//A
Discussion