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.

Separation of petroleum into its components is mostly done by

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CHROMATOGRAPHY
Sublimation
Distillation under reduced pressure
Fractional distillation

SOLUTION :DIFFERENCE in boiling point.
Discussion
2.

Separate out the following into extensive and intensive : Volume, Temperature, Pressure,Boiling point, Free energy

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SOLUTION :Volume and FREEENERGY are EXTENSIVE , OTHERS are intensive
Discussion
3.

Semiconductors possess conductivity in the range ______ to _________

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SOLUTION :`10^(-6)` to `10^4 OHM^(-1) m^(-1)`
Discussion
4.

Select wrong statement:-

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If a very small amount of `AlCl_(3)` is added to gold sl, coagulation occurs, but if a large quantity of `AlCl_(3)` is added, there is no coagulation
Organic ions are more strongly adsorbed on charged surfaces in comparison to INORGANIC ions
Both emulsifier and peptising agents stabilise colloids but their ACTIONS are different
Colloidal solutions are thermodynamically stable

Solution :When excess of an electrolyte is added, the colloidal PARTICLES are precipitated. This is because colloidal particles take up ions carrying CHARGE opposite to that present on themselves. This causes neutralization leading to their coagulation.
Discussion
5.

Select wrong clientical reaction aniong the following

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`8NH_(3)+3Cl_(2) to 6NH_(4)CL+N_(2)`
`2Ca(OH)_(2)+2Cl_(2) to Ca(Ocl)_(2) +CaCl_(2)+2H_(2)O`
`2NaOH+Cl_(2) to 2NaCl+H_(2)+O_(2)`
`MnO_(2)+4HCL to MnCl_(2)+Cl_(2)+2H_(2)O`

Answer :C
Discussion
6.

Select true statement.

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Gauche from of 3-hydroxy propanol is more stable than it's anti form
Anti from of n-butane (across `C_(2)-C_(3)`) is more stable than it's gauche from
At room temperature mole fraction of ECLIPSED from is almost zero
Doipole MOMENT of 1,-2-dichloro ethance decreases on heating

Solution : due to INTRAMOLECULAR H-bonding in gauche form


Staability of eclipsed from is less than staggered at room temperature.
On heating rate of intercoversion from anti-gauche increases therefore `mu` increases.
Discussion
7.

Select the types of redox reaction form the following on the basis of type of tedox changes: (a) intermolecular redox, (b) intramolecular redox (c) auto redox. If none, write none. (i)C_(6)H_(5) CHO overset(NaOH)(rarr) C_(6)H_(5)CH_(2) OH + C_(6) H_(5) COONa (iI) Cr_(2)O_(7)^(2-) + 2OH^(-) rarr 2CrO_(4)^(2-)+ H_(2)O (iii) 2Mn_(2)O_(7) rarr 4MnO_(2) + 3O_(2) (iv) NO_(3)^(-) + H_(2)S + H_(2)O + H^(+) rarr NH_(4)^(+) + HSO_(4)^(+-) (v) Fe + N_(2) H_(4) rarr NH_(3) + Fe(OH)_(2) (vi) 2KOH + Br_92) rarr KBr + KBrO (vii) 2Cu^(+) rarr Cu + Cu^(2+) (viii) Ag(NH_(3))_(2)^(+) overset(2H^(+))(rarr) Ag^(+) + 2NH_(4)^(+) (ix) 5KI+ KIO_(3) + 6HCl rarr 3I_(2) + 6KCI + 3H_(2)O

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ANSWER :(a) `(iv),(V),(IX)`, (B) `(iii)`, (c) `(i),(vi),(vii)`
Discussion
8.

Select the structure with correct numbering for IUPAC name of the compound

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SOLUTION :
Discussion
9.

Select the structures that are not resonance structures of the same molecule .

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ANSWER :D
Discussion
10.

Select the species which is not aromatic.

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Solution :It is a conjugated SYSTEM but has only `4 PI`elec-trons, does not SATISFY Huckel rule.
Discussion
11.

Select the species having zero oxidation state at the underlined elements :

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`(CH_(3))_(2)ul(S)O`
`ul(C)_(12)H_(22)O_(11)`
`H_(2)ul(S)_(2)O_(3)`
`ul(N)_(2)H_(4)`

Answer :A::B
Discussion
12.

Select the reactio(s) that would result in the formation 2-bromopropane. (1) CH_(3)CH=CH_(2)+Hbroverset("peroxide")(to)(II) CH_(3)CH=CH_(2)+Hbroverset(C Cl_(4))(to) (III) CH_(3)CH_(2)CH_(3)+Br_(2)overset(hv)(to)(IV) CH_(3)CH=CH_(2)+Broverset((C Cl)_(4))(to)

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I and III
II and III
I, II and III
I, II and III

Solution :`IIto` Markovnikoff's addition
`IIIto` Halogenation of ALKANE TAKE PLACE.
Discussion
13.

Select the reactions in which the correct orientations have been mentioned in the major products .

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ANSWER :A::C
Discussion
14.

Select the pair of metamers from the following :

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SOLUTION :PAIR (a) are FUNCTIONAL ISOMERS.
Discussion
15.

Select the pK_(a) value of the strongest acid from the following :

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`1.0`
`3.0`
`2.0`
`4.5`

SOLUTION :SEE CHEMICAL PROPERTIES of ACIDS
Discussion
16.

Select the pair of atoms having the same number of electrons in their outermost shell

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NA, Ca
Mg, Fe
As, Bi
PB, Sb

Solution :(a) `Na = [NE] 3s^(1), Ca = [Ar] 4s^(2)`
(b) `Mg = [Ne] 3s^(2), Fe [Ar] 3d^(6) 4s^(2)`
(c) As and Bi belong to the same group with outer shell E.C. `= ns^(2) NP^(3)`
(d) Pb and Sb lie in different groups
Discussion
17.

Select the oxidant/reducant atoms in the followingchange. Also report the number of electrons involved in redox change. As_(2)S_(3) + HNO_(3) rarr H_(3) AsO_(4) + H_(2)SO_(4) + NO

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ANSWER :`84`
Discussion
18.

Select the organic compounds, which will give red colour in Lassaigne's test?

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`NaCNS`
`NH_(2)-OVERSET(O)overset(||)(C)-NH_(2)`
`NH_(2)-overset(S)overset(||)(C)-NH_(2)`

ANSWER :C::D
Discussion
19.

Select the option in which all the parametres are intensive :

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PH, volume, Electrodes potential.
MOLAR enthalpy, HEAT capacity, Resistivity.
Temperature,Molality, ELECTROMOTIVE force.
Mass, Pressure, Molar entropy.

Answer :C
Discussion
20.

Select the optically active compound among the following :

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SOLUTION :N//A
Discussion
21.

Select the optically active compound among the following:

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Solution :`A to ` PLANAR compound so OPTICALLY inactive.
`B to ` NON planar and optically active DUE to absence of POS & COS
`C to ` non planar but having POS so , optically inactive.
`D to `Planar compound.
Discussion
22.

Select the one :

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`4f`
`5d`
`3S
`6P`

Solution :`(n+l)` for REST all is `=7`
For3s it is3+ 0 =3 .
Discussion
23.

Select the natural series among the following:

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`(4N+1)`
`(4n+2)`
4n
`(4n+3)`

ANSWER :B::C::D
Discussion
24.

Select the non-polar molecules among the following :

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`NO_(2)`
`B_(2) H_(6)`
`PF_(3)CI_(2)`
`B_(2)B_(3)H_(6)`

Answer :A::B
Discussion
25.

Select the name of mineral which is used for the production of boric acid

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Chromite
Colemanite
Chalcopyrite
Calamine

Answer :B
Discussion
26.

Select the most stable structure of COCl_(2)

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ANSWER :C
Discussion
27.

Select the most stable carbocation from amongst the following.

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ANSWER :B
Discussion
28.

Select the most stable intermediate.

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ANSWER :A
Discussion
29.

The most acidic compound is

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SOLUTION :Electron denating groups INCREASES BASIC strength
Discussion
30.

Select the most basic compound.

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ANSWER :C
Discussion
31.

Select the most appropriate statemen. In BF_(3)

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all thebonds are COMPLETELY ionic
the B-F bond is partially ionic
B-F bond has partial double bond character
all the bondsare covalent

Solution :Due to BACK donationof a LONE pairof ELECTRONS from F to B , B-F bonds has partial double character.
Discussion
32.

Select the members of group 14 that (i) forms the most acidicdioxide (ii) is commonly found in +2 oxidation state, (iii)used as semiconductor.

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Solution :(i) As a ELECTRONEGATIVITY of element decreases from C to Pb,the acidiccharacter of theirdioxide also decreases from C to Pb.THUS, C forms the most acidic dioxidei.e., `CO_(2)`.
(ii) Since theinert paireffect becomesmore and more predominant does the group, therefore, STABILITY of +2 oxidation state increasesdown the group. thus, lead is commonly found in +2 oxidation state.
(iii) Siliconand germanium are neithertoo electronegativenor too electrpositive. Therefore they act as metalloidsand hence BEHAVEAS semiconductors.
Discussion
33.

Select the molecule which has only one pi bond.

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`CH_(3)-CH=CH-CH_(3)`
`CH_(3)-CH=CH-CHO`
`CH_(3)-CH=CH-COOH`
All of these

Answer :A
Discussion
34.

Select the member(s) of group 14 thatused as semiconductor.

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SOLUTION :SILICON and GERMANIUM
Discussion
35.

Select the major product obtained from the addition of HBr to 1-methylcyclohexene.

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1-bromo-2-methylcyclohexane
6-bromo-1-methylcyclohex-I-ene
3-bromo-1-methylcyclohex-1-ene
1-bromo-1-methylcyclohexane

Answer :D
Discussion
36.

Select the least basic compound.

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SOLUTION :Amides are less basic than AROMATIC AMINES due to Resonance aswell as REACTION with strong base
Discussion
37.

Select the least acidic compound.

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`NO_(2)-CH_(2)-OVERSET(overset(O)(||))(C)-O-H`
`F-CH_(2)-overset(overset(O)(||))(C)-O-H`
`Ph-CH_(2)-overset(overset(O)(||))(C)-O-H`
`CH_(3)-CH_(2)-overset(overset(O)(||))(C)-O-H`

SOLUTION :ELECTRON withdrawing group increases the ACIDIC strength.
Discussion
38.

Select the indicator form the given table for titratioin of 20 mL of 0.02 M CH_(3)COOH with 0.02M NaOH. Given pK_(a)(CH_(3)COOH) = 4.74

Answer»

I
II
III
IV

Solution :`PH = 1/2[pK_(W) + pK_(a) + "LOG"C]`
Discussion
39.

Select the incorrect statements

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KOH is a STRONGER base then NaOH
MILk of magnesia is an aqueous suspension of mg`(OH)_(2)`
`CaO_(2)` is LESS stable then`MgO_(2)`
`BaC_(2)O_(4)` is soluble in acetic acid

Answer :4
Discussion
40.

Select the incorrect statement(s).

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At critical CONDITIONS, VOLUME occupied by the gas 12 times the volume of 1 mole GASEOUS molecuels
A gas can be liquefied above its `T_(b)` (Boyle's temperature) by APPLICATION of pressure.
At very high pressure and moderate temperature, the volume of a moles of real gas is equal to `(nRT)/(P)+nb`
For a real gas, follwowing equation `(P+(a)/(TV^(2)))(V-BETA)=RT`,wehre `alpha` and `beta` are constant: Boyle's temperature is `sqrt((alpha)/(Rbeta))`

Answer :b
Discussion
41.

Select the incorrect statement(s) :

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KOH is weaker base than NaOH
`Mg^(2+)`ionsareprecipitatedwiththeh additionof `NH_4OH`in thepresenceof `NH_4 Cl`
`CaO_2` is lessstablethan ` MgO_2`
Milkof magnesiais anaquesoussolutionof `Mg (OH)_2`

Solution :A) Down the group Ksp ofhydrolysis of IIA group ELEMENTS INCREASES. SoKOH is more basic NaOH
B) `Mg^(+2)` ions are not precipitated as `KIP lt Ksp`
C) Peroxides are more soluble than super OXIDES
Discussion
42.

Select the incorrect statement with respect to sodium peroxide ?

Answer»

It decolurises the acidified `KMnO_(4)` solution.
On HEATING with OXYGEN at `450^(@)`C and 300 atm PRESSURE, it becomes paramagnetic
It Is obtained ALONG with sodium metal, when sofium oxide is heated to a temperature more than `400^(@)`C.
It gives both hydrogen peroxide and oxygen gas with water as well with sulphuric acid.

Answer :D
Discussion
43.

Select the incorrect statement of the following.

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SOLUTION :
More ACTIVE H-more H- BONDING with MOLECULES.

Discussion
44.

Select the incorrect statement from the following :

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`CH_(3)-COOH` can give effervescence of `CO_(2)`


SOLUTION :N//A
Discussion
45.

Select the incorrect statement for H_2O_2 structure

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It is NON planar
O–O bond length is more in gaseous state than in solid phase
Both OH bond are in different plane
O–O–H bond ANGLE in gas phase is more than in solid phase

ANSWER :D
Discussion
46.

Select the incorrect statement for developing of an exposed camera film involving the reaction

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a. Hydroquinol ACTS as REDUCTANT
b. `AG^(+)`acts as oxidant
c.Hydroquinol and AgBr undergoes redox CHANGE
d.it is intramolecular change

Solution :Hydroquinol acts as reductant , `Ag^(+)` acts as oxidant `(Ag^(+) rarroverset(0)(Ag))`Hydroquinol and AgBr undergoes redox change .
Discussion
47.

Select the incorrect statement about the boron

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PURE form of the elements are obtained by the reduction of `BCl_(3)` with ZINE at `900^(@)C`
Crystalline boron is ATTACKED only by hot concentrated oxidising agents
Amorphous boron and ammonia at white heat GIVES `(BN)_(x)`, a slippery white SOLID witha layer structure resembling that of graphite
Boron does form `B^(3+)` cation easily

Answer :B
Discussion
48.

Select the incorrect statement about Barium

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It shows photo electric efect
It is silvery white metal
Its ionisation ENERGY is greater than radium
It forms `Ba(NO_(3)_(2)` which is used in the PREPERATION of green fire

Solution :Barium will not EXHIBIT photo electric EFFECT. Radium has greater `IP_(1)` value than barium.
Discussion
49.

Select the incorrect statement :

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K.E. of photo-electron does not DEPEND upon the WAVELENGTH of INCIDENT radiation
Photoelectric current depends on intensity of incident ratiation and on frequency
Stopping Potential depends on frequency of radiation and not on intensity
photoelectric current increases when we use photons of HIGH frequency

Solution :KE = `h(UPSILON - upsilon_0) , KE alpha (upsilon - upsilon_0)`
We know `lambda alpha 1/(upsilon)therefore KE alpha 1/(Delta lambda)`
`therefofe ` KE dependent on `lambda`
Discussion
50.

Select the incorrect order for given properties :

Answer»

Thermal stability : `BaSO_(4)gtSrSO_(4)gtCaSO_(4)`
Solubility : `BaSO_(4)gtSrSO_(4)gtCaSO_(4)`
Thermal stability : `Li_(2)CO_(3)ltNa_(2)CO_(2)ltK_(2)CO_(3)`
Solubility : `Li_(2)CO_(3)gtNa_(2)CO_(3)gtK_(2)CO_(3)`

Solution :N//A
Discussion