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.

As per KTG, molecules in gaseous state are under continuous motion with no intermolecular forces occurring & the speed of the molecules will keep on changing due to the molecular collections. The distribution pattern remains constant and is given by (dN)/(du) = 4 pi N (M/(2 pi R T))^(3/2) u^2 . e^((-Mu^2)/(2RT)) (symbols have usual meaning). The collisions are assumed to be perfectly elastic. In case there are more than one gas in the container then also presence of one type of gas will not affect movement of other gas than one gas. Based on this information, answer the question that follows. A closed container fitted with a movable piston always operating at fixed pressure is subjected to increase in temperature. Identify the correct option assuming a single gas to be present in the container.

Answer»

NUMBER of collisions made by any one molecule in one second will be directly PROPORTIONAL to root of absolute temperature.
Mean free path will INCREASES linearly with absolute temperature
Total number of BMC in one second will increases.
Average speed of approach will remain unaffected.

Answer :B
Discussion
2.

As per KTG, molecules in gaseous state are under continuous motion with no intermolecular forces occurring & the speed of the molecules will keep on changing due to the molecular collections. The distribution pattern remains constant and is given by (dN)/(du) = 4 pi N (M/(2 pi R T))^(3/2) u^2 . e^((-Mu^2)/(2RT)) (symbols have usual meaning). The collisions are assumed to be perfectly elastic. In case there are more than one gas in the container then also presence of one type of gas will not affect movement of other gas than one gas. Based on this information, answer the question that follows. A container consist of two gases O_2 & SO_2. Assuming them to be non reactive & behaving ideally identify the correct statement.

Answer»

Each molecule of `O_(2)` will cause lesser change in MOMENTUM due to wall COLLISION as compared to `SO_(2(G))`
Average RELATIVE speed of approach of `O_(2)` and `SO_(2)` molecule will be `sqrt((3RT)/(8pi))`
`O_(2(g))` will exert higher pressure irrespective of the AMOUNT taken.
Average translational kinetic energy of mole of `SO_(2)` nad`O_(2` will be different.

Answer :B
Discussion
3.

As per KTG, molecules in gaseous state are under continuous motion with no intermolecular forces occurring & the speed of the molecules will keep on changing due to the molecular collections. The distribution pattern remains constant and is given by (dN)/(du) = 4 pi N (M/(2 pi R T))^(3/2) u^2 . e^((-Mu^2)/(2RT)) (symbols have usual meaning). The collisions are assumed to be perfectly elastic. In case there are more than one gas in the container then also presence of one type of gas will not affect movement of other gas than one gas. Based on this information, answer the question that follows. Which of the following statements regarding distribution pattern is incorrect?

Answer»

The DISTRIBUTION PATTERN is more uniform in case of LIGHTER gas.
FRACTION of molecules with speed equal to `U_(mps)` will be more at lower TEMPERATURE as compared to higher temperature
The ratio of `U_(mps)` to `U_(avg)` is fixed for all gases at all temperature
On increasing temperature, fraction of particles with very less speed increases.

Answer :D
Discussion
4.

As per IUPAC norms, the name of the complex [Co(en)_2 (ONO)Cl]Clis

Answer»

Chloridobis (ETHANE - 1,2 - DIAMINE) nitro - O. cobalt (III) chloride.
Chlorobis (ethylenediamine)) nitro - O - cobalt (III) chloride
Chloridodi (ETHYLENE diamine) nitrocobalt (III) chloride.
Chloroethylenediaminenitro - O cobalt (III) 1 chloride

Answer :a
Discussion
5.

As per IUPAC nomenclature , the name of the complexis :

Answer»

tetraaquadiaminecobalt (III) chloride
tetraaquadimminecobalt (III) chloride
diaminetetraquacobalt (III) chloride
diamminetetraquacobalt (III) chloride

Solution :b) it is the correct IUPAC ANME:
Discussion
6.

As per IUPAC nomenclature, the name of the complex, [Co(H_2O)_4(NH_3)_2]Cl_3 is

Answer»

Tetraaquadiaminecobalt (III) chloride
Tetraaquadiamminecobalt (III) chloride
Diaminetetraaquacobalt (III) chloride
Diamminetetraaquacobalt (III) chloride

Answer :D
Discussion
7.

As per IUPAC guidelines .the name of the complex [ Co(ONO) Cl] Cl is

Answer»

Chlorobisethllyenediaminenitritocobalt (III) chloride
Chloridobis (ethane -1 ,2- DIAMINE ) nitro k-Ocobaltate (III) chlroide
Chloridobis (ethane -1,2-diamine ) nitrito k-Ocobalt (II) chloride
chloridobis (ethane -1,2-diamine )nitro k-Ocobalt (III) chloride

Answer :A
Discussion
8.

As per IUPAC nomenclature, the name of the complex [Co(H_(2)O)_(4)(NH_(3))_(2)]Cl_(3) is

Answer»

tetraaquadiaminecobalt(III) Chloride.
tetraaquadiamminecobalt(III) Chloride.
diaminetetraaquacobalt(III) Chloride.
diamminetetraaquacobalt(III) Chloride.

Answer :D
Discussion
9.

As per IUPAC guidelines, the name of the complex [Co(en)_(2)(ONO)Cl]Cl is……….. .

Answer»

chlorobisethylenediaminenitritocobalt `(III)` chloride
chloridobis(ethane-`1,2-`DIAMINE)nitro k-Ocobaltate`(III)`chloride
chloridobis(ethane-`1,2-`diammine)NITRITO k-Ocobalt`(II)`chloride
chloridobis(ethane-`1,2-`diamine)nitro k-Ocobalt`(III)`chloride

Answer :A::B::C::D
Discussion
10.

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order :

Answer»

`AlCl_(3) gt K_(2)[Fe(CN)_(6)] gt K_(2)CrO_(4) gt KBr = KNO_(3)`
`K_(3)[Fe(CN)_(6)] lt K_(2)CrO_(4) lt AlCl_(3) lt KBrlt KNO_(3)`
`K_(3)[Fe(CN)_(6)]ltK_(2)CrO_(4) lt KBr = KNO_(3) = AlCl_(3)`
`K_(3)[Fe(CN)_(6)]gtAlCl_(3)gtK_(2)CrO_(4) gt KBr gt KNO_(3)`

Solution :`K_(3)[Fe(CN)_(6)]ltK_(2)CrO_(4) lt KBr = KNO_(3) = AlCl_(3)`
The minimum CONCENTRATION of an electrolyte which is required to cause the coagulation or flocculation of a sol is known as flocculation VALUE.
Flocculation value is inversely proportional to coagulation power (coagulation power is directly proportional to the valency of the ions causing coagulation). `Fe(OH)_(3)` sol is a positive sol and THUS we have to consider the valency of the anions in the given electrolytes.
`K_(3)[Fe(CN)_(6)]ltK_(2)CrO_(4) lt KBr = KNO_(3) = AlCl_(3)`
Discussion
11.

As per cooled water freezes spontaneously, its temperature rises to 0^(@)C. DeltaH for the spontaneous process. H_(2)O_((l))(-10^(@)C)toH_(2)O_((s))(0^(@)C) is :

Answer»

zero
`+ve`
`-ve`
EITHER of these

Solution :No energy is TRANSFERRED to or from the system. The energy liberated in the freezing PROCESS warms the system to `0^(@)C`. Thus `DeltaH=0`
Discussion
12.

As per de Broglie's formula a macroscopic patiicle of mass 100 g and moving at a velocity of 100 cm s^(-1) will have a wavelength of

Answer»

`6.6xx10^(-29)cm`
`6.6xx10^(-30)cm`
`6.6xx10^(-31)cm`
`6.6xx10^(-32)cm`

Solution :Mass (m) = 100 G, velocity (v) =100 cm `s^(-1)`
According to de Broglie.s equation,
`lamda=h/(mv)`
`lamda=(6.626xx10^(-27))/(100xx100)=6.6xx10^(-31)` cm
Discussion
13.

As originally developed, cystal field theory was used to describe the electronic structure of metal ion crystals, where they are surrounded by oxide ions or other anions that create an electrostatic filed with symmetry dependent on the crystal structure. Total number of nodal planes of the orbitals involved in hybridisation of central metal ion in complex [MnO_(4)]^(-) is :

Answer»

6
4
2
3

Solution :`d_(XY),d_(YZ),d_(XZ),s impliesd^(3)s "" N.P.=2,2,2`
Discussion
14.

As originally developed, cystal field theory was used to describe the electronic structure of metal ion crystals, where they are surrounded by oxide ions or other anions that create an electrostatic filed with symmetry dependent on the crystal structure. If for the complex K_(4)[FeL_(6)] the value of Delta_(0)=22000" "cm^(-1). Find out the value of crystal pairing energy. Where L is uninegative monodentate ligand.

Answer»

`-46200" "cm^(-1)`
`-52800" "cm^(-1)`
`+2200" "cm^(-1)`
`-22200" "cm^(-1)`

SOLUTION :`K_(4)[overset(+2)FeL_(6)] d^(6) to t_(2g)^(2,2,2)EG^(0,0)`
CFSE`=-0.4xx6Delta_(0)=-2.4Delta_(0)`
`implies "CFSE"=-2.4xx22000=-52800 cm^(-1)`
Discussion
15.

As O_(2) is cooled at 1 atm pressure, it freezes to form solid I at 54.5 K. At a lower temperature, solid I rearranges to solid II, which has differentcrystal structure. Thermal measurements show that for the phase transition solid I to solid II, DeltaH=-743.1 J mol^(-1) and DeltaS=-17.0 JK^(-1) mol^(-1). At what temperature are solids I and II in equilibrium

Answer»

2.06 K
31.5 K
43.7 K
53.4 K

Answer :C
Discussion
16.

As O_2 (I) is cooled at 1 atm pressure , it freezes to form solid I at 54.5 K. At a lower temperature , solid rearrange to solid II,which has a different crystal that for the phase transition solid to slid II , DeltaH=-743.1 Jmol^(-1) and DeltaS=-17.0JK^(-1) mol^(-1) . At what temperature are solids I and II in equilibrium ?

Answer»

2.06 K
31.6 K
43.7 K
53.4 K

ANSWER :C
Discussion
17.

As lead storage battery is charged :

Answer»

lead dioxide DISSOLVES
sulphuric ACID is REGENERATED
lead metal gets COATED with lead sulphate
The concentration of sulphuric acid DECREASES.

Solution :(b) is the correct answer.
Discussion
18.

As ............ is lowered in the presence of catalyst, more molecules take part in the reaction and hence the rate of the reaction increases.

Answer»

SOLUTION :ACTIVATION ENERGY
Discussion
19.

As increase in both atomic and ionic radii with atomic number occurs in any group of the periodic table and in accordance with this, the ionic radii of Ti^(4+) and Zr ^(4+) ion are 68 Pm and 74 Pm respectivelt. But for Hf ^(4+) ion, the ionci radius is 75 Pm. Which is most same as that for Zr ^(4+) ion. This is due to

Answer»

greater degree of COVALENCY in compound of `Hf ^(4+)`
LANTHANIDE contraction
difference in co-ordination number of `Zr ^(4+) and Hf ^(4+)` in their compound
actinide contraction

Solution :In the same group atomic size increases ad atomic number increaes. But due to the lanthanide contraction atomic size of the elements just following Hf, Ta, W, Re, Os, etc.
ONWARDS are close to the element just above them in the respective group i.e. Zr, Nb, Mo, Tc, Ru etc.
Due to the same size of Hf and Zr, Ta and Nb, W and Mo, Re and Tc, Os and Ru etc., the two properties and they occurs together in nture and are difficult to separate. This pair of elements is known as CHEMICAL twins.
As a result of lanthanide contraction the elements of `2 ^(nd)and 3^(rd)` row transition series (d-block) resembles each other much more than the elements of `1^(st) and 2^(nd)` row trsnsition series. So they do not shwo close similarities.
Discussion
20.

As ideal gas undergoingcyclie process ABC .A consistingof isothermal expansionAB, isobariccompression BC & adiabatic copmressionCA. Findthe %(approximate) efficiencyof cycle. [Given: T_(A)= T_(B)= 400 K , gamma =1.5, In2= 0.7,2^(1//3)=0.8]

Answer»

Solution :`T_(C) = T_(A) [(P_(A))/(P_(C))]^((1-gamma)/(1.5)) = 400 [(2)/(1)]^((1-1.5)/(1.5)) = (400)/(2^(1//3))= 400 XX 0.8= 320`
`eta= - (W_("Total") )/(q_("ABSORBED"))`
`W_(AB) = -n 400R lno2 =-280 n R`
`W_(BC) = - nR (T_(C) - T_(B)) = - n R(320-400) = 80 n R`
`W_(CA) = (nR)/(gamma-1)(T_(A) -T_(C))`
Discussion
21.

As dillution takes place

Answer»

oxidation potential decreases
reduction potential increases
oxidation potential increases
cell potential may increases

Solution :From Nemst EQUATION
`E_(M^(2+)//M)=E_(M^(1+//M)^(@)- (2.303)/(NF) log (1)/([M^(n+)])`
`rArr E_(M^(1+)//M)=E_(M^(1+)//M)^(@)+(2.303)/(nF) log [M^(n+)]`
`E^(M//M^(1+))=E_(M//M^(2+))^(@)-(2.303)/(nF) log [M^(n+)]`
Thus on increasing the concentralion keeping olher factors conslant.
Reduction potenial `(E_(M^(1+)//M))` increases oxidation potential `(E_(M//M^(+))` decreases while on dilution reducion polential decreases, oxidation potential increases. Hence choices (C) and (D) are correct while (A) and (B) are INCORRECT.
Discussion
22.

As compared with the Boiling point of straight chain isomers the Boiling point of branched chain alkanes is-

Answer»

Lower
Equal
Higher
INDEPENDENT of branchong

Answer :A
Discussion
23.

As compared to pure atomic orbitals, hybrid orbitals have

Answer»

Low ENERGY
Same energy
HIGH energy
NONE of these

Solution :As COMPARE to pure atomic orbitals, HYBRID orbitals have low energy.
Discussion
24.

As compared to ethylene, the bond distance between the two carbon atoms in acetylene is :

Answer»

Longer
Shorter
Same
None

ANSWER :B
Discussion
25.

As compared to potassium, sodium has

Answer»

Lower electronegativity
Higher ionization potential
Greater ATOMIC radius
Lower mp

SOLUTION :`{:("Element","Na","K"),(IE_(1),496,419),(IE_(2),4562,3051):}`
Sodium has higher I.E. because of SMALLER atomic size.
Discussion
26.

As compared to covalent compounds, electrovalent compounds generally possess

Answer»

HIGH MELTING POINT and high BOILING point
LOW melting point and low boiling point
low melting point and high boiling point
high melting point and low boiling point

Answer :A
Discussion
27.

As compared to B.P. of straight chain isomers, the B.P. of branched chain alkane is :

Answer»

lower
higher
equal
independent UPON branching

Answer :B
Discussion
28.

As atomic number increases, the ionic radii of Ln ^(3+)ion

Answer»

increases
first increases and then decreases
decreases
first decreases and then increases

Solution :We know that from` CE ^(3+)` to `LU^(3+)` ion nucleus charge increase. The 4f orbitals are not able to SHIELD perfectly the attraction between nucleus and outer most ELECTRONIC orbit. Thus increased nuclear charge is more than that of shielding EFFECT of 4f electrons. Hence from `Ce ^(3+)` to `Lu^(3+)` ionic radii decreases.
Discussion
29.

As a result of the isobaric change by 80 K, one mole of a certain ideal gas obtain an amount of heat equal to 1.60 kJ. What is change in enthalpy of gas

Answer»

0
20 KJ
`-1.60` kJ
`+1.60` kJ

Answer :D
Discussion
30.

As a result osmosis the volume of solution

Answer»

INCREASES
DECREASES
REMAINS contant
Increases or decreases

ANSWER :A
Discussion
31.

Asa result of Frenkel defect ,

Answer»

there is no effect on the density
there in no effect on theconductivity
there is no effect on the DIELECTRIC constant
there is no effect on all the THREE abov.

Solution :ASA result of Frenkel defect,there is NOEFFECT on density .
Discussion
32.

As a result of osmosis the volume of solution:

Answer»

REMAINS constant
Increases
Decreases
Increases or decreases

Answer :D
Discussion
33.

As a result of osmosis, the volume of the concentrated solution:

Answer»

GRADUALLY decreases
Gradually increases
Suddenly increases
None

Answer :B
Discussion
34.

Aryl halides undergo nucleophilic substitution reaction only is extreme condition. Except to certain industrial processes where very severe conditions are feasible, one does not ordinarily prepare phenols (ArOH), ethers (ArOR), amines (ArNH_(2)) or nitriles (ArCN) by nucleophilic attact on aryl halides. The presence of electron withdrawing groups like -NO_(2), -CF_(3) at ortho or para positioin to the halogen atom makes the aryl halides more susceptible to nucleophilic attack. The reaction of unactivated ordeactivated aryl halide with strong bases or at highh temperature proceed via the benzyne intermediate. underset(LiqNH_(3))overset(KNH_(2))to(A) Identity A in the above reaction

Answer»




NONE of these

Solution :
Discussion
35.

Arylamines develop colour due to __________.

Answer»

hydroscopic nature
atmosphericoxidation
moisture
atmospheric REDUCTION

ANSWER :B
Discussion
36.

Aryl halids are less reactive rtowards nucleophilic substitution reactions than alkyl halides due to :

Answer»

FORMATION of less STABLE carbocation
resonance stabilisation
longer carbon-halogen bond
`sp^(2)` hybridised carbon ATTACHED to halogen.

Solution :both are CORRECT answer.
Discussion
37.

Aryl halides undergo nucleophilic substitution reaction only is extreme condition. Except to certain industrial processes where very severe conditions are feasible, one does not ordinarily prepare phenols (ArOH), ethers (ArOR), amines (ArNH_(2)) or nitriles (ArCN) by nucleophilic attact on aryl halides. The presence of electron withdrawing groups like -NO_(2), -CF_(3) at ortho or para positioin to the halogen atom makes the aryl halides more susceptible to nucleophilic attack. The reaction of unactivated ordeactivated aryl halide with strong bases or at highh temperature proceed via the benzyne intermediate. underset(NH_(3))overset(NaNH_(2))to Product. The predominant product is

Answer»




SOLUTION :
Discussion
38.

Aryl halides do not undergo nucleophilic substitution reactions under ordinary conditions because: (1). Approach of nucleophile is retarded (2). Carbon carrying halogen atom is sp^(3) hybridised. (3). The substrate molecule is destabilised due to resonance (4). of partial double bond character between carbon and halogen.

Answer»

2 and 3 only
1 and 4 only
2 and 3 only
2,3 and 4 only.

Solution :Both these explanation are correct.
Discussion
39.

Aryl halides do not undergo uncleophilic substitution reactions under ordinary conditions because (1) Approach of nucleophile is retarded (2) Carbon carrying halogen atom is sp^(3) hybridised (3) The substrate molecule is destabilised due to resonance (4) Partial double bond character between carbon and halogen

Answer»

2 and 3 only
2, 3 and 4 only
1 and 4 only
1 and 3 only

Answer :C
Discussion
40.

Aryl halides can be prepared by

Answer»

SAND MAYER's method
Friedel - CRAFT reaction
Gattermann's reaction
1 and 3

Answer :D
Discussion
41.

Aryl halides are less reactive towards nucleophilic substitution reaction. Give four reasons.

Answer»

Solution :i. Reasonance EFFECT
II. Difference in hybridisation of CARBON atom in `c-x` BOND.
Discussion
42.

Aryl halides are less reactive towards nucleophilic substitution because

Answer»

LESS stable carbonium ion
Due to large` C-CI` bond energy
INDUCTIVE effect
Resonance stabilization and `sp^(2)`-hybridisation of `C` ATTACHED to halide .

SOLUTION :Resonance stablization and the hybridisation of `C` attached to halides cis `p^(2)` .
Discussion
43.

Aryl halides are less reactive towards electrophiles than alkyl halides due to :

Answer»

Resonance
stability of CARBONIUM ions
high BOILING point
None of these

Answer :A
Discussion
44.

Aryl halides are less reactive than alkyl halides due to the presence partial double character of C - X bond in aryl halides. Arly halides undergo nucleophilic substitution reactions, if electron withdrawing groups are introduced in ortho & para - positions in aryl halides. The reaction mechanism involves two steps. Keeping these points in view answer the following questions.The number of resonance structure possible for chlorobenzene is

Answer»

three
four
FIVE
two

Solution :
Discussion
45.

Aryl halides are less reactive than alkyl halides towards nucleophilic substitution reactions because of :

Answer»

formation of LESS stable CARBONIUM ION
inductive effect
larger C-halogen bond
RESONANCE stabilization

Answer :D
Discussion
46.

Aryl halides are less reactive towards nucleophiles than alkyl halides due to :

Answer»

Resonance
stability of CARBONIUM ions
high BOILING point
None of these

Answer :A
Discussion
47.

Aryl halides are less reactive than alkyl halides due to the presence partial double character of C - X bond in aryl halides. Arly halides undergo nucleophilic substitution reactions, if electron withdrawing groups are introduced in ortho & para - positions in aryl halides. The reaction mechanism involves two steps. Keeping these points in view answer the following questions.Which compound undergo nucleophilic substitution under mild conditions

Answer»




SOLUTION :More the E.W.G an CHLOROBENZENE more will be the REACTIVITY
Discussion
48.

Aryl halides are less reactive than alkyl halides because of

Answer»

Longer bond LENGTH of C-X than alkyl halides
SHORTER bond length of C-X than alkyl halides
Equal bond length of C - X as that of alkyl halides
None of these

Solution :Aryl halides has shorter bond length of C-X than alkyl halide.
Discussion
49.

Aryl halides are less reactive than alkyl halide in mucleophilic substitution reaction which is due to The formation of less stable coarboniumion 2. Reasonance stabilization 3. Longer C-X bond 4. the indicutive effect 5. sp^(2)- hybdrizied carbon atom attached to halogen

Answer»

1,3,5
2,4,5
2,3,5
1,2,5

Answer :D
Discussion
50.

Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution :

Answer»

Solution :Presence of strong electrong electron withdrawing groups at o - and p - positions increases the REACTIVITY of ARYL halides towards nucleophilic substitution reaction. ORDER of reactivity : (III) `gt` (II) `gt` (i).
Discussion