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.

Schiff's reagent gives pink colour with

Answer»

Aldehydes
Ethers
Ketones
CARBOXYLIC acid

Solution :Schiff.s regent is rosaniline hydrochloride DYE decolorised by passing `SO_(2)` . In the presence of a COMPOUND containing -CHO group the PINK color of the dye is restored back.
Discussion
2.

Schiff's bases or anils are formed when aniline reacts with :

Answer»

ALCOHOLS
AROMATIC aldehydes
Aliphatic KETONES
Aromatic ketones

Answer :B
Discussion
3.

Schiff's reagent gives pink colour with :

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ALCOHOLS
Acetaldehyde
KETONE
Mesitylene CHLORIDE

Answer :B
Discussion
4.

Schiff's bases are formed when aniline reacts with

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Aromatic aldehydes
ARYL KETONES
Arylhalides
Aryl ALCOHOLS

ANSWER :A
Discussion
5.

Schiff's base is substituted

Answer»

imine
amine
nitro
nitrile

Answer :A
Discussion
6.

Schiff's base and Schiff's reagent are:

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Same compounds
Different compounds
Physically same but CHEMICALLY different
None

Answer :B
Discussion
7.

Schemes 1 and 2 decribe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes. The product X is :

Answer»




SOLUTION :
Discussion
8.

Schiff bases or anils are formed, when aniline reacts with

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ALIPHATIC ALDEHYDES
Aromatic aldehydes
Aliphatics ketones
Aromatic ketones

Solution :Aromatic aldehydes condense with primary AMINES (aromaic as well as aliphatic) to form anils or schiff-bases
`{:(ArCHO+RNH_(2)RARR ArCH(OH)NHR),(""darr-H_(2)O),(""ArCH=NR),("""Schiff-base"):}`
Discussion
9.

Schemes 1 and 2 decribe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes. The correct statement with respect to product Y is :

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It gives a positive TOLLEN's test and is a functional isomer of X.
It gives a positive Tollen's test and is a GEOMETRICAL isomer of X.
It gives a positive iodoform test and is a functional isomer of X.
It gives a positive iodoform test and is a geometrical isomer of X.

Solution :
Both (X) and (Y) are functional isomers. COMPOUND (Y) gives positive iodoform test (yellow ppt.)
Discussion
10.

Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.

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SOLUTION :FERROMAGNETIC
Discussion
11.

Scheel's green, formerly used as a green pigment for colouring wall paper is.

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SODIUM ARSENITE `(Na_3AsO_3)`
CUPRIC arsenite `(CuHAsO_3)`
SILVER arsenite `(Ag_3 AsO_3)`
None

Answer :A
Discussion
12.

Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each of them.

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SOLUTION :FERRIMAGNETIC
Discussion
13.

Schematic alignment of magnetic moments of ferromagnetic, antiferromagnetic and ferrimagnetic substances are given below. Identify each ofthem.

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Solution :i)antifero magnetism ii)Ferrimagnetism III)FERROMAGNETISM
Discussion
14.

Scattering of light by colloidal particles is called………..

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SOLUTION :TYNDALL EFFECT
Discussion
15.

Scarlet flame colour of bunsen flame is characteristic of :

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Sn
K
Sb
Sr

Answer :D
Discussion
16.

Scandium resembles with aluminium in the following properties except :

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the NATURE of bonding in both `Al^(3+)` and `Sc^(3+)` compounds in MAINLY ionic
both liberate hydrogen vigorously from water
both form basic hydroxides
bothform VOLATILE HALIDES.

Answer :C
Discussion
17.

Scandium forms no coloured ionis, yet it is regarded as a transition elements. Explain why ?

Answer»

Solution :Scandium in the ground state has ONE ELECTRON int eh 3d-subshell . HENCE,it is regarded as a transitio elements. However, in its common oxidation state `+3`, it has no electron in 3dsusshell `( 3d^(0))` .Hence, it does not FORM coloured ion.
Discussion
18.

Sc^(3+), Ti^(4+), V^(5+) are diamagnetic. Give reason.

Answer»

SOLUTION :(i) `Sc^(3+), Ti^(+4), V^(5+)`have `d^(0)` ELECTRONIC configuration. n = 0
(ii) `mu = sqrt(0(0+2)) = 0 mu_(B)`.So they are diamagnetic.
Discussion
19.

Sc (Z=21) is a transition elements but Zn(Z=30) is not because….

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Both `Sc^(3+) and Zn^(2+)` ions are colourless and form with compounds
In case of Sc, 3d orbitals are partially filled but in Zn these are completely filled
Last ELECTRON is assumed to be added to 4s level in case of zinc
Both Sc and Zn DONOT exhibit VARIABLE oxidation states

Answer :B
Discussion
20.

Sc^(3+) ions are colourless where as V^(3+) ions are coloured. Give reason.

Answer»

Solution :`SC^(3+)` ions does not contain any UNPAIRED electrons HENCE d-d transition does not TAKE place.
`V^(3+)` ION contain two unpaired electrons, hence d-d transition takes place. Therefore `V^(3+)` ions are coloured.
Discussion
21.

Sc (Z = 21) is a transition element but Zinc (z = 30) is not because

Answer»

both `Sc^(3+)` and `Zn^(2+)` ions are colourless and form white compounds,
in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
ast electron as assumed to be added to 4s level in case of zinc
both Sc and Zn do not EXHIBIT variable OXIDATION states

Solution :in case of Sc, 3d orbital are partially filled but in Zn these are completely filled
Discussion
22.

Sc (Z = 21) is a transition element but Zn(Z = 30) is not because

Answer»

Both `SC^(3+)` and `ZN^(2+)` ions are colourless and form white compounds
In case of Sc, 3d orbitals are partially filled but in Zn these are filled
Last electron is ASSUMED to be added to 4s level in case of Zn
Both Sc and Zn do not exhibit variable oxidation states

Answer :B
Discussion
23.

SbF_(5) reacts with XeF_(4) and XeF_(6) to form ionic compounds [XeF_(3)^(+)][SbF_(6)^(-)] and [XeF_(5)^(+)][SbF_(6)^(-)\ then molecular shape of [XeF_(3)^(+)] ion and [XeF_(5)^(+)] ion respectively :

Answer»

SQUARE PYRAMIDAL, T-shpaed
Bent-T-shape, square pyramidal
SEE-saw, square pyramidal
Square pyramidal, see -saw

Solution :
Discussion
24.

SBR(GRS,Buna-S,Cold Rubber)is obtained by free radical intiator.The most commenly used free radical initiator is:

Answer»

Buta`-1,3-` dinene `(70%)` and `30%` phenyl ethene (STYRENE)
Chloroprene and styrene
VINYL acetylene and styrene
ISOPRENE and `1,3-` butadiene

Answer :A
Discussion
25.

SbF_(5) reacts with XeF_(4) to form an adduct. The shapes of cation and anion in the adduct are respectively :

Answer»

square PLANAR, TRIGONAL bipyramidal
T - SHAPED, octahedral
Square PYRAMIDAL, octahedral
Square planar, octahedral

Answer :B
Discussion
26.

[Sc (H_(2)O)_(6)]^(3+)ion is :-

Answer»

COLOURED and paramagnetic
Colourless and paramagnetic
Colourless and diamagnetic
Coloured & octahedral

Answer :C
Discussion
27.

Sb_2S_3 is the example of

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POSITIVE SOL
NEGATIVE sol
neutral sol
None of these.

Answer :B
Discussion
28.

Saturated solution of KNO_(3) is used to make 'salt-bridge' because

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Velocity of `K^(+)` is greater than that of `NO_(3)^(-)`
Velocity of `NO_(3)^(-)` is greater than that of `K^(+)`
Velocities of both `K^(+) and NO_(3)^(-)` are nearly the same
`KNO_(3)` is highly soluble in water

Solution :The salt used to MAKE 'salt-bridge' must be such that the ionic mobility of CATION and anion are of comparable ORDER so that they can keep the anode and cathode half cells neutral at all TIMES.
Discussion
29.

Saturated solution of NaCl on heating becomes:

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SUPER SATURATED
UNSATURATED
REMAINS saturated
None

Answer :B
Discussion
30.

To form a super saturated solution of salt one must:

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SUPER SATURATED
UNSATURATED
REMAINS saturated
None

Answer :B
Discussion
31.

Saturated solution of AgCl at 25^(@)C has specific conductance of 1.12xx10^(-6)ohm^(-1)cm^(-1). The lamda_(oo)Ag^(+) and lambda_(oo)Cl^- and 54.3 and 65.5 ohm^(-1)cm^2// equi. Respectively. Calculate the solubility product of AgCl at 25^(@)C

Answer»


ANSWER :A::B::D
Discussion
32.

Saturated solution of KNO_(3) is used to make salt bridge because:

Answer»

velocity of `K^(+)` is greater than that of `NO_(3)^(-)`
velocity of `NO_(3)^(-)` is greater than that of `K^(+)`
velocity of both `K^(+)` and `NO_(3)^(-)` are nearly the same
`KNO_(3)` is higly SOLUBLE in water

Solution :The SALT BRIDGE possesses the electrolyte having nearly same ionic mobilities of its cation and anion.
Discussion
33.

Saturated solution of KCl (or) Na_2SO_4 is used to make salt bridge because

Answer»

VELOCITY of `K^+` is greater than that of `CL^-`
velocity of `Cl^-` is greater than that of `K^+`
velocity of both `K^+` and `Cl^-` are NEARLY the same
KCL is highly soluble in water

Answer :C
Discussion
34.

Saturated solution of KNO_3 is used to inake salt bridge because

Answer»

VELOCITY of `K^(+)` is GREATER than that of `NO_(3)^(-)`
velocity of `NO_3^(-)` is greater than that of `K^(+)`
velocities of both `K^(+)` and `NO_3` are NEARLY the same
`KNO_3` is highly SOLUBLE in water

Answer :C
Discussion
35.

Saturated monocarboxylic acid is second oxidative product of

Answer»

`1^(@)` alcohol
`2^(@)` alcohol
both a & B
ketones

Answer :C
Discussion
36.

Saturated mono carboxylic acid are also called

Answer»

Paraffin
Olefin
FATTY acid
Mineral acid

Answer :C
Discussion
37.

Saturated fatty acids are represented by which of the forumula

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`C_(n)H_(n)O_(2)`
`C_(n)H_(3N)O_(2)`
`C_(n)H_(2n+1)`
`C_(n)H_(2n)O_(2)`

SOLUTION :The monocarboxylic ACIDS are CALLED fatty acids, because some of the HIGHER members were obtained from fats. The general formula is `C_(n)H_(2n+1)COOH` or RCOOH or `C_(n)H_(2n)O_(2)`.
Discussion
38.

Saturatd solution of KNO_3 is used to mke salt bridge because .

Answer»

Velocity of `K^(+)` is GREATER than that of `NO_3^(2)`
Velocity of `NO_(3)^(-)` is greater htan that of `K^(+)`
`KNO_3` is highly soluble in qater

Solution :Velocity ALMOST same, so conductivity in cathodic and anodic COMPARTMENTS is not AFFECTED.
Discussion
39.

Sapphire is a valuable precious stone containing___________.

Answer»

Cu
Zn
Al
Mg

Answer :C
Discussion
40.

Saponiflication of ethyl acetate (CH_(3)COOC_(2)H_(5)) by NaOH is studied by titration of the reaction mixture, have 1 : 1 molar ratio of the reactants. If 10ml of 1 N HCl is required by 5ml of the solution at the start and 8ml of 1N HCl is required by another 5ml after 10 minutes, then rate constant is

Answer»

`k=(2.303)/(10)log.(10)/(8)`
`k=(2.303)/(10)log.(10)/(2)`
`k=(1)/(10)[(1)/(8)-(1)/(10)]`
`k=(1)/(10)[(1)/(2)-(1)/(10)]`

Answer :C
Discussion
41.

Sapphire is a mineral of :

Answer»

`AL`
`ZN`
`CU`
`HG`

ANSWER :A
Discussion
42.

Saponification of ethyl benzoate with caustic soda as alkali gives

Answer»

Benzyl alcohol and ethanoic ACID
Sodium benzoate and ethanol
Benzoic acid and sodium ethoxide
Phenol and ethanoic acid

Solution :`underset("Ethylbenzoate")(C_(6)H_(5)COOC_(2)H_(5))+NaOH overset(Delta)to underset(Sod" benzoate")(C_(6)H_(5)COONA)+underset("Ethanol")(C_(2)H_(5)OH)`
Discussion
43.

Saponification of ethyl acetate (CH_(3)COOC_(2)H_(5)) by NaOH (Saponification of ethyl acetate by NaOH is second order reaction) is studied by titration of the reaction mixture initially having 1:! molar ratio of the reactants. If 10 mL of 1 N HCl is required by 2 mL of the solution at the start and 8 mL of 1 N HCl is required by another 5 mL after 10 minutes, then rate constant is :

Answer»

`K=(2.303)/(10)"LOG"(10)/(8)`
`k=(2.303)/(10)"log"(10)/(2)`
`k=(1)/(10)[(1)/(8)-(1)/(10)]`
`k=(1)/(10)[(1)/(2)-(1)/(10)]`

Answer :C
Discussion
44.

Saponificationof an ester(A) followed by neutralizationgivesa compound (B), Which givesvioletcolourationwithFeCl_(3). The ester (A) is .

Answer»

ETHYL ACETATE
ethyl BENZOATE
DIETHYL phthlate
aspirin

Answer :D
Discussion
45.

Sanger's reagent reacts with functional group in a peptide ?

Answer»

FREE amino groups
The phenolic hydroxyl GROUP in tyrosine
The aromatic HETEROCYCLIC rings of histidine and tryptophan
Free carboxylic groups

Solution :
Discussion
46.

Saponification means hydrolysis of an ester with :

Answer»

Enzymc
`CH_3COOH`
`H_2SO_4`
NaOH

Answer :D
Discussion
47.

Sanger's reagent is used for the identification of

Answer»

N -terminal of a peptide chain
C-terminal of a peptide chain
SIDE chain of amino ACIDS
molecular weight of the peptide chain

Solution :1-Fluoro-2,4-dinitrobenzene is CALLED Sanger's reagent and is used for DETERMINATION of N-terminal amino acid in a polypeptide chain.
Discussion
48.

Sandmeyer'sreactionof benzene diazoniumcholdrideis usedin thepreparationof

Answer»

chlorodenzene
BENZENE
PHENOL
iodobenzen

Answer :A
Discussion
49.

Sandmeyer's reaction occurs in the presence of………and………….

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CUPROUS HALIDE and haloacid
cupric halide and halogen
copper-zinc COUPLE and `H_2`
NONE of these

Answer :A
Discussion
50.

Sandmeyer's reaction is used to prepane

Answer»

METHYL benzene
halobenzene
p-xylene
nitrobenzene

Answer :B
Discussion