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.

On the basic of data given below, predict which of the following gases shows least adsoprtion on a definite amount of charcoal ? {:("Gas", CO_(2), SO_(2), CH_(4), H_(2)), ("Critical temp/K", 304, 630, 190, 33):}

Answer»

`CO_(2)`
`SO _(2)`
`CH_(4)`
`H_(2)`

Solution :HIGHER the CRITICAL temperature, greater is the adsorption, `H_(2)` has lowest critical temperature and HENCE is least adsorbed.
2.

On the addition of mineral acid to an aqueous solution of borax, the compound formed is :

Answer»

ORTHOBORIC acid
Borohydride
Metaboric acid
Pyroboric acid

Answer :A
3.

On the addition of a solution containing CO_(4)^(2–) ions to the solution of Ba^(2+), Sr^(2+) and Ca^(2+) ions, the ppt obtained first will be of-

Answer»

`CaCO_(4)`
`SrCrO_(4)`
`BaCrO_(4)`
a MIXTURE of all the THREE

Answer :C
4.

On t reatingPCl_(5)with conc. H_(2)SO_(4), SO_(2)Cl_(2) is formed as the final product this showsthat H_(2)SO_(4)

Answer»

is DERIVATIVEOF `SO_(2)`
is a monobasic ACID
has great affinity for `H_(2)O`
has TWO hydroxyl groups in its structure

Solution :`PCl_(3)+H_(2)SO_(4) to SO_(2)Cl_(2)+2POCl_(3)+2HCl`
5.

On sulphonation of C_(6)H_(5)Cl

Answer»

m-Chlorobenzenesulphonic ACID is formed
Benzenesulphonic acid is formed
o-Chlorobenzenesulphonic acid is formed
0- and p-Chlorobenzenesulphonic acid is formed

Solution :`-CL` is o,p-directing
6.

On sulphonation of C_6H_5CI

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m-chlorobenzenesulphonic ACID
Benzenesulphonic acid is FORMED
o-chlorobenzenesulphonic acid is formed
o-and p-chlorobenzenesulphonic ACIDS are formed

ANSWER :D
7.

On sulhonation , phenol gives two products, ortho and para phenol sulphonic acids . The variation in productsdepend upon the __________.

Answer»

CONCENTRATION of sulphuric acid
CHANGE in temperature
change in PRESSURE
change in VOLUME

Answer :B
8.

On strongly heating anhydrous calcium acetate gives :

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ACETONE
Ether
Acetic ANHYDRIDE
CO

Answer :A
9.

On strongly heating AgNO_3 we get :

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`N_2O` and NO
`NO_2` and `o_2`
NO and `O_2`
`NO_2` and NO

Answer :B
10.

On strongly heatingAgNO_3 the gases evolved are:

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`N_2O` and NO
`NO_2` and `o_2`
NO and `O_2`
`NO_2` and NO

Answer :B
11.

On strong heating lead nitrate gives :

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`PBO,NO,O_2`
`PbO,NO,NO_2`
`PbO_2,PbO,NO_2`
`PbO,NO_2,O_2`

ANSWER :D
12.

On strong heating CaO and C the products formed are:

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CA and CO
`CaC_2 and CO`
`Ca(OH)_2`
`CaC_2 and CO_2`

ANSWER :B
13.

On strong heating 4 gm of a solid compound produced 528 ml of a diatomic gas (A_(2)) at NTP condittion and 2.68 gm of solid residue. The atomic mass of element A is -

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56 gm
112 gm
28 gm
14 gm

Solution :`X overset(DELTA)(rarr) A_(2) (g) + Y`
`4 gm "" 2.68 gm`
`A_(2) (g) = 4 - 2.68 = 1.32gm`
`n_(A_(2)) (g) = (V ("gas") NTP "in litre")/(22.4) = (w ("gas") H_(2))/(M.Wt.A_(2))`
`n_(A_(2)) (g) = (528)/(22400) = (1.32)/(M.Wt.A_(2))`
`M.Wt.A_(2) = 56 gm = 2 xx` Atomic mass of A
Atomic mass of `A = 28 gm`
14.

On strong heating ammonium acetate gives -

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Acetamide
Methyl Cyanide
Urea
Formamide

Answer :A
15.

On solidification of molten silver gives ………….. crystals.

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SHOWS ccp type of arrangement 
Shows 6 coordination NUMBER 
Shows 10 coordination number 
Shows HCP type of arrangement 

ANSWER :A
16.

On reduction with Na-Hg and water a carbohydrate gives a mixture of sorbitol and mannitol. The carbohydrate can be

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Glucose
FRUCTOSE
CANE sugar
LACTOSE

Solution :The carbohydrate can be fructose
17.

On (reductive) ozonolysis of natural rubber, 4-oxo-pentanal is obtained the monomer of natural rubber would be :

Answer»

`H_(2)C=UNDERSET(CH_(3))underset(|)(CH)-CH=CH_(2)`
`H_(3)C-CH=CH-CH=CH_(2)`
`H_(3)C=underset(CH_(3))underset(|)(C)-CH=CH_(3)`
`H_(2)C-CH-CH_(2)-CH=CH_(2)`

SOLUTION :N//A
18.

On reduction with LiAlH_4 a ketone yields :

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PRIMARY alcohol
Secondary alcohol
Tertiary alcohol
All

Answer :B
19.

On reduciton with Sn+conc. Hclof C_(2)H_(5)NO_(2)yields

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ESTERS
secondaryalchohal
PRIMARY AMINE
secodary amine

ANSWER :C
20.

On reduction , primary amine is formed by

Answer»

NITROETHANE
Ethylnitrite
AZOBENZENE
ETHYLCARBYLAMINE

ANSWER :A
21.

On reaction with water, calcium phosphide produces

Answer»

`Ca_(3)(PO_(4))_(2)`
`H_(3)PO_(4)`
`COCl_(2)`
`PH_(3)`

ANSWER :D
22.

On reaction with Cl_(2), phosphorus forms two types of halides 'A' and 'B'.Halide A is yellowish-white powder but halide 'B' is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.

Answer»

Solution :Since `Cl_(2)` reacts with phosphorus to form two halides (A and B), therefore, these halides must be `PCl_(5) and PCl_(3)`. Since halide 'A' is a yellowish-white solid but halide 'B' is a colourless oily liquid, therefore, due to HIGHER molecular mass and hence stronger FORCES of attraction, halide 'A' must be a solid, i.e., `PCl_(5)` and
`{:(P_(4)+10Cl_(2) rarr 4 underset((A))(PCl_(5))("yellowish-white solid")),(P_(4)+6Cl_(2) rarr underset((B))(4 PCl_(3))("Colourless oily liquid")):}`
halide 'B' must be a liquid, i.e., `PCl_(3)`.
Their hydrolysis products are shown below :
`{:(underset("Phosphorus pentachloride")(PCl_(5))+4H_(2)O rarr underset("Orthophosphoric acid")(H_(3)PO_(4))+5HCl),(underset("Phosphorus trichloride")(PCl_(3))" "+3H_(2)O rarr underset("Phosphorus acid")(H_(3)PO_(3))""+3HCl):}`
23.

On reaction with ketone with hydroxyl amine give ketoxime which on reduction produces

Answer»

carboxylic ACID
`1^(@)` AMINE
`2^(@)` - amine
amide

Answer :B
24.

On reaction with dil. H_(2)SO_(4) which of the following salts will give out a gas that turns an acidified dichromate paper green?

Answer»

`Na_(2)CO_(3)`
`Na_(2)S`
`ZnSO_(3)`
FeS

Solution :`Na_(2)S+H_(2)SO_(4) to Na_(2)SO_(4)+H_(2)S uarr`, `ZnSO_(3)+H_(2)SO_(4) to ZnSO_(4)+H_(2)O+SO_(2) uarr`
`FeS+H_(2)SO_(4) to FeSO_(4)+H_(2)S uarr`
`H_(2)S` and `SO_(2)` can REDUCE dichromate PAPER to green `Cr^(3+)`
25.

On reaction with hot conc.H_2SO_4, which of the following compounds loses a molecule of water :

Answer»

`CH_3COCH_3`
`CH_3COOH`
`CH_3OCH_3`
`CH_3CH_2OH`

ANSWER :D
26.

on reaction with cold conc. HBr gives :

Answer»




ANSWER :A
27.

On reaction with Cl_2, phosphorus forms two types of halides 'A' and 'B' Halide A is yellowish-white powder but halide 'B' is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.

Answer»

Solution :`A: PCl_(5)`
`B: PCl_(3)`
`PCl_(3)` Hydrolyse to give `H_(3)PO_(3)` and `PCl_(5)` hydrolysis to give `H_(3)PO_(4)`
`P_(4) + 10Cl_(2) to 4PCl_(5)`
`PCl_(5) + 4H_(2)O to H_(3)PO_(4) + 5HCl`
`P_(4) + 6Cl_(2) to 4PCl_(3)`
`PCl_(3) + 3H_(2)O to H_(3)PO_(3) + 3HCL`
28.

On reaction with Cl_2, phosphorus forms two types of halides A and B. Halide A is yellowish white powder but halide B is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.

Answer»

SOLUTION :A is `PCl_5` ( It is yellowish-white POWDER)
`P_4 + 10 Cl_2 to 4PCl_5`
B is `PCl_3` ( It is a colourless OILY liquid )
`P_4 + 6Cl_2 to 4PCl_3`
Hydrolysis products are formed as follows:
`PCl_3 + 3H_2O to H_3PO_3 + 3HCL`
`PCl_5 + 4H_2O to H_3PO_4 + 5HCl`
29.

On reacting with aqueous bromine at room temperature phenol forms which of the following ?

Answer»

meta-Bromophenol
2, 6-Dibromophenol
2, 4, 6-Tribromophenol
3,5-Dibromophenol

Answer :C
30.

On reacting with neutral ferric chloride, phenol gives

Answer»

REDCOLOUR
VIOLET COLOUR
darkgreen colour
no colourations

Solution :violet colour
31.

On reacting with electropositivemetal like Na, alcohols give sodium alkoxides on liberates H_2. In this reaction alcohols behave as _________.

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weak ACIDS
weak BASES
neutral
STRONG acids

ANSWER :A
32.

On reacting copper with dilute nitric acid gives a colourless gas (A). The gas (A) on reacting with Fe^(2+) ionin concentrated sulphuric acid and water forms a brown coloured complex. The EAN of Iron in the brown coloured complex is : [Given: Atomic No.: Fe = 26]

Answer»

26
37
25
36

Solution :`Fe^(2+)(aq)+NO+underset(("CONC."))(H_(2)SO_(4))to underset(("Brown"))([Fe(H_(2)O)_(5)(NO)]^(2+))+H_(2)O`
The oxidation number of Iron in `[Fe(H_(2)O)_(5)(NO)]^(2+)rArr x + (5xx0)+1=+2 rArr x = +1`
EAN of Fe = Atomic no. of metal - no. of electrons lost + no. of electrons gained from ligands `= 26-1+12rArr 37`
33.

On reacting with halogens, H_2S is

Answer»

oxidized
reduced
neutralized
converted into SULPHUR halide

Answer :A
34.

On Pluto, where everything is frozwn, astronauts discovered two froms of butane gauche and anti. Assuming that there are no rotations around single bonds, which statement about the two forms is correct ?

Answer»

They are enantiomers
They are diastereoisomers
They are meso compounds
The gauche from has TWO stereogenic CENTERS, and the anti has only ONE

Solution :
If no ROTATION then these two are not INTER convertible and are diastereoisomer.
35.

On prolonged exposure to air, sodium finally changes to:

Answer»

`Na_2CO_3`
`Na_2O`
NAOH
`NaHCO_3`

ANSWER :A
36.

On Pauling scale which of the following does not have electronegativity ge3.0

Answer»

Oxygen
Nitrogen
Chlorine
Bromine

Answer :D
37.

On performing a borax-bead test with a given inorganic mixture for qualitative analysis, the colour of the bead was found to emerald green both in oxidising and reducing flame. It indicates the possibility of the presence of

Answer»

`Co^(+2)`
`NI^(+2)`
`Cr^(+3)`
`CU^(+2)`

Solution :Chromium ion gives in HOT and COLD OXIDISING and reducing gren-colour flame.
38.

On passing vapours of an organic liquid over finely divided Cu at 573K the product was an alkene. This reaction is

Answer»

CATALYTIC OXIDATION of primary alcohol
Catalytic DEHYDROGENATION of secondary alcohol
Catalytic dehydrogenation of tertiary alcohol
Catalytic dehydration of tertiary alcohol

Solution :`CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-OHunderset(300^@C)overset"Cu"to underset"ISOBUTENE"(CH_3-undersetunderset(CH_2)(||)oversetoverset(CH_3)(|)C+H_2O)`
39.

On passingle C ampere of electricity through a electrolyte solution for t second, m gram metal deposits on cathode. The equivalent weight E of the metal is

Answer»

`E=(Cxxt)/(mxx96500)`
`E=(Cxxm)/(txx96500)`
`E=(96500xxm)/(Cxxt)`
`E=(Cxxtxx96500)/(m)`

ANSWER :C
40.

On passingH_2S in II gp, sometimes a white turbidity is formed. This is due to:

Answer»

COLLOIDAL SULPHUR
`SnS_2`
`Bi_2S_3`
ZnS

Answer :A
41.

On passing through a magnetic field, the greatest deflection is experienced by :

Answer»

`ALPHA`-particle
`beta`- particle
`GAMMA`-RAYS
All EQUAL.

Answer :B
42.

On passing SO_(2) gas through an acidified solution of K_(2)Cr_(2)O_(7)

Answer»

The Solution GETS decolourised
The solution BECOMES blue
`SO_(2)` is REDUCED
Green `Cr_(2)(SO_(4))_(3)` is obtained

Solution :`K_(2)Cr_(2)O_(7)+3SO_(2)+H_(2)SO_(4) to Cr_(2)(SO_(4))_(3)+H_(2)O`
43.

What happens when H_2S passed through FeCl_3 solution ?

Answer»

`FeCl_2`
`Fe_2(SO_4)_3`
FES
`FeSO_4`

ANSWER :A
44.

On passing H_(2)S gas through a highly acidic solution containing Cd^(2+) ions, CdS is not precipitated because

Answer»

Of common ION effect
The SOLUBILITY of CDS is low
`CD^(2+)` ions do not form complex with `H_(2)S`
The solubility PRODUCT of CdS is low

Answer :A
45.

On passing H_2S through HNO_3 we get

Answer»

Colloidal SULPHUR
`O_(2)`
`O_(3)`
`NO_(3)`

SOLUTION :`H_(2)S +2HNO_(3) to 2H_(2)O +2NO_(2)+S`
46.

Draw structure of N_2O_5 and also find numbers of lone pairs in N_2O_5 :-

Answer»
47.

On passing H_2 S into saturated solution of BaCl_2, white ppt. obtained is of

Answer»

hydrogen chloride
formation of a COMPLEX
BARIUM chloride
barium sulphide.

SOLUTION :`H_(2)S` separates `BaCl_(2)` from its SATURATED solution as white ppt.
48.

On passing H_(2)S black ppt. of II group is obtained. The mixture may not contain

Answer»

`Pb^(++)`
`CD^(++)`
`Hg^(++)`
`Cu^(++)`

Solution :`Cd^(++)+H_(2)S to UNDERSET("Yellow PPT.")(CDS)+H_(2)O`
49.

On passing excess of CO_(2) in lime water, its milky appearance disappears because-

Answer»

Soluble `Ca(OH)_(2)` is formed
Soluble `Ca(HCO_(3))_(2)` is formed
Reaction becomes reversible
Calcium compound evaporated

Solution :`Ca(OH)_(2)+CO_(2)rarrunderset("MILKINESS disappear")underset(Ca(HCO_(3))_(2)+"soluble")underset(darr"Excessive CO_(2))(CaCO_(3)darr+H_(2)O)`
50.

On passing electricity through diluteH_2SO4 solution the amount of substance libeated at the cathode and anode are in the ratio:

Answer»

`1:8`
`8:1`
`16:1`
`1:16`

ANSWER :A