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.

Mention any two biological effects of D_2O

Answer»

Solution :(a) `D_2O` retards the growth of living organisms LIKE PLANTS and animals
(B) Pure heavy water kills SMALL fishes, TADPOLES and mice when fed upon it
Discussion
2.

Mention any two anomalous properties of second period elements.

Answer»

SOLUTION :(i) In the 1" GROUP, LITHIUM forms compounds with more COVALENT character while the other elements of this group form only ionic compounds.
(ii) In the 2ND group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.
Discussion
3.

Write any three postulates of Bohr's model for hydrogen atom.

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Solution :Postulates of bohr's theory.
Electron revolve ROUND the nucleus in CERTAIN orbits of definite ENERGY.
The quantum mechanically allowed orbits are those for which the angular momentum of the morning electron is an INTEGRAL multiple of quantity `h/(2pi)`
Energy is emitted of Absorbed only when an electron jumps from one orbit to another. When an electron jumps from HIGHER energy to loiner energy the energy difference is emitted in form Cf Radiation.
`E_(2)-E_(1)=DeltaE=hv`
Discussion
4.

Bonellia feeds with the help of

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SOLUTION :INTERFERENCE, DIFFRACTION, POLARISATION.
Discussion
5.

Mention any three methods of preparation of haloalkanes from alcohols.

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SOLUTION :Haloalkanes are prepared from alcohols by treating with (i) HCl(ii) `PCl_5` (iii) `SO_2Cl_2`
(i) Reaction of alcohol with hydrogen halide:
`underset("Ethanol")(CH_3-CH_2OH + HCl) underset(ZnCl_2)OVERSET("ANHYDROUS")to underset("CHLORO ethane")(CH_3 - CH_2 Cl) + H_2O`
(ii) Reaction of alcohol with phosphorous halide:
`underset("Ethanol")(CH_3-CH_2OH)+underset("Pentachloride")underset("Phosphorus")(PCl_5) to underset("Chloro ethane")(CH_3 - CH_2Cl) + POCl_3 + HCl`
(iii) Reaction of alcohol with thionyl chloride:
`underset("Ethanol")(CH_3-CH_2OH) + SOCl_2 to underset("Chloroethane")(CH_3-CH_2Cl + SO_2uarr) + HCluarr`
Discussion
6.

Mention any three applications of equilibrium constant.

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Solution :(a) Predicting the extent of a REACTION.
(b) Predicting the direction of a reaction.
( C) Calculating the EQUILIBRIUM CONCENTRATIONS.
Discussion
7.

Mention an example in which H_2O acts as reducing agent.

Answer»


ANSWER :`2F_(2)+2H_(2)O to O_(2)+4HF`
Discussion
8.

Mention an industrial application of silicones.

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SOLUTION :Siliconesare USED for makingwater-proof papers, wool, textiles, wood, etc. by coating them with a thin filmof SILICONS.
Discussion
9.

Mention any 4 redox reaction that takes place in our daily life.

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SOLUTION :(a) Burning of COOKING gas, wood
(B) Rusting of iron articles
(c) Electroplating
(d) Galvanic and electrolytic cells
Discussion
10.

Mention about the uses of barium.

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SOLUTION :In metallurgy, its compounds are used in pyrotechnics, petroleum mining and radiology.
(ii) Deoxidizer in copper REFINING.
Its alloys with nickel which readily emits electrons hence used in electron tubes and spark plug ELECTRODES.
(iv) As a scavenger to remove last traces of oxygen and other gases in television and on ELECTRONIC tubes.
(V) An isotope of `"barium"^(133)`Ba, used as a se isotope of barium "Ba, used as a source in the calibration of gamma ray detector nuclear chemistry.
Discussion
11.

Mention about the purification of (a) aniline and (b) naphthalene

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Solution :(a) Aniline can be purified by steam distillationbecause It is immiscrible with water and steam volatile,
(b) Napthalene can be purified by sublimation it change on heating directly to VAPOUR state and on COOLING, it changes back into SOLID FORM.
Discussion
12.

Mendeleev's periodic table is based on atomic masses of elements. It was the first successful attempt to classify all the known elements (63) that time. One of the most important advantage of this classification was that mendeleev predicted the physical and chemical properties of three elements : Eka-boron, Eka-silicon and Eka-aluminium. These elements were discovered as:

Answer»

GERMANIUM, SCANDIUM, GALLIUM
scandium, germanium, gallium
iron,SULPHUR and germnaium
iron,sulphur and scandium

Answer :B
Discussion
13.

Melting point of covalent molecule is less because....

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Molecules are attracted by VAN DER Waals FORCES in covalent molecule.
Covalent bond are EXOTHERMIC.
Covalent bond are WEAK than ionic bond.
Covalent molecule has definite shape

Solution :Molecules are attracted by van der Waals forces in covalent molecule.
Discussion
14.

Melting point of calcium halides decreases in the order

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`CaF_(2) gt CaCl_(2) gt CaBr_(2) gt CaI_(2)`
`CaI_(2) gt CaBr_(2) gt CaCl_(2) gt CaF_(2)`
`CaBr_(2) gt CaI_(2) gt CaF_(2) gt CaCl_(2)`
`CaCl_(2) gt CaBr_(2) gt CaI_(2) gt CaF_(2)`

SOLUTION :`CaF_(2) gt CaCl_(2) gt CaBr_(2) gt CaF_(2)` . As the size of the anion increases, the covalent character increases and hence the MELTING points decrease.
Discussion
15.

Melting point is highest for

Answer»

B
AI
Ga
In

Solution :B
Discussion
16.

Melting point, enthalpy of vapourisation and viscosity data of H_(2)O and D_(2)O is given below:{:(,H_(2)O,D_(2)O),("Melting point/k",373.0,374.4),("Enthalpy of vapourisation at (373K)/KJ mol"^(-1),40.66,41.61),("Viscosity/centipoise",0.8903,1.107):} On the basis of this data explain in which of these liquids intermolecular forces are stronger?

Answer»

Solution :The MELTING point, enthalpy of vaporization and viscosity values of all these items depend upon the intermolecular foreces of ATTRACTION . Since their values are highter for `D_(2)O` as compared to those of `H_(2)O`, THEREFORE, intermolecular forces of attraction are stronger in `D_(2)O` than in `H_(2)O`.
Discussion
17.

Melting point and boiling point increase as the molar masses …………..

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ANSWER :INCREASE
Discussion
18.

Melting point is higher for

Answer»

B
Al
Ga
In

Solution : .B. has GAINT MOLECULAR STRUCTURE.
Discussion
19.

Melting and boiling points of alkynes are lower than those of the corresponding alkenes.

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ANSWER :F
Discussion
20.

Melting & boiling point of NaCl respectively are 1080 K & 1600K. Delta S for stage -I & II in NaCl_((s)) underset(Delta H_("fus") = 30kJ)overset(I)rarr NaCl_((l)) underset(Delta H_("vap") = 160kJ)overset(II)rarr are

Answer»

`{:(Delta S (I) (KJ//mol//K),Delta S (II) (KJ//mol//K)),(1//36,1//10):}`
`{:(Delta S (I) (KJ//mol//K),Delta S (II) (KJ//mol//K)),(36,100):}`
`{:(Delta S (I) (KJ//mol//K),Delta S (II) (KJ//mol//K)),(1//36,10):}`
`{:(Delta S (I) (KJ//mol//K),Delta S (II) (KJ//mol//K)),(36,1//10):}`

Solution :Stage-I is FUSION
`Delta S_("fusion") = (Delta H_("fusion"))/(M.P) = (30 xx 10^(3))/(1080) = (1)/(36)` KJ
State -II is vapouraization
`Delta S_("vap") = (Delta H_("vap"))/(B.P) = (160 xx 10^(3))/(1600) = (1)/(10)KJ`
Discussion
21.

Melting and boiling point of ionic compounds are higher then covalent compounds .

Answer»

SOLUTION :Ionic COMPOUNDS possess strong ELECTROSTATIC force of attraction between the oppositely charged ions, hence their melting and boiling points are higher than covalent compounds.
Discussion
22.

{:("Medium","Equivalent weight of "KMnO_(4)),("A) Acidic","a) 158"),("B) Neutral","b) 79"),("C) Strongly basic","c) 52.6"),("D) Weakly basic","d) 31.6"):} The correct match is

Answer»

A - d, B - c, C - a, D - c
A - d, B - c, C - a, D - b
A - d, B - b, C - a, D - c
A - d, B - c, C - a, D - a

Solution :`{:("ACIDIC",:,MnO_(4)^(-)rarrMn^(+2)),("strongly BASIC",:,MnO_(4)^(-)rarrMnO_(4)^(2-)),("neutral/weakly basic",:,MnO_(4)^(-)rarrMnO_(2)):}`
Discussion
23.

Mechanism of reductive ozonolysis is given below for an alkene Which is correct for the above mechanism

Answer»

Ozone ACT as electrophile and as well as nucleophile in this reaction
First step of this reaction is an ELECTROPHILIC addition
ozonide is formed in the step-II
When ozonide is cleaved in the PRESENCE of reducing agent such as ZN or `Me_(2)S` the products will be aldehydes and/or ketones.

Answer :A::B::C::D
Discussion
24.

Meation the condition favouring ionic bond formation.

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Solution :(1) Lower IONISATION energy of METALS.
(2) Higher ELECTRON affinity of non-metal
(3) Higher magnitude of lattice energy of BOND
(4) Higher electromagnativity difference between metal and non-metals.
(5) Larger radius of cation of smaller radius of anion.
Discussion
25.

Meaning of Alkali word is...

Answer»

ASH of Coal
Ash of Herb
Ash of WOOD
NONE of the above

Answer :C
Discussion
26.

Me_3SiCl is used during polymerization of organosilicons because....

Answer»

chain length of ORGANO silicon polymers can be controlled by adding `Me_3SiCl`.
`Me_3SiCl` blocks the END TERMINAL of SILICONE polymer.
`Me_3SiCl` improves the QUALITY and yield of the polymer.
`Me_3SiCl` actsas a catalyst during polymerization.

Solution :The chain length of the polymer can be controlled by adding `(CH_3)_3SiCl` which blocks the ends.
Discussion
27.

Mean free path decreases if

Answer»

`H_2` is REPLACED by He 
He is replaced by `H_2`
He is replaced by `O_2` 
`O_2` is replaced by `CO_2` 

Solution :`LAMBDA alpha 1/(sigma) , sigma_(He) < sigma_(H_2) < sigma_(O_2) < sigma_(CO_2)`.
Discussion
28.

Me_(3)SiCl isused during polymerisation of organosilicones because

Answer»

the chain lengths of organosilicone polymers can be controlled by ADDING`Me_(3)SiCl`
`Me_(3)SiCl` BLOCKS the end terminalof siliconepolymer
`Me_(3)SiCl` IMPROVES the quality and YIELD of the polymer
`Me_(3)SiCl`ACTS as acatalyst duringpolymeriation

Solution :`Me_(3)SiCl`is used to control the length of silicone polymers by blockingthe end terminalof the siliconepolymer. Thus, options(a) and (b)are correct.
Discussion
29.

Me_(2)SiCl_(2) on hydrolysis will produce

Answer»

`Me_(2)SI(OH)_(2)`
`Me_(2)Si=O`
`[--O-Me_(2)Si-O--]_(N)`
`Me_(2)SiCl(OH)`

Answer :A
Discussion
30.

Me_(2)C=CH_(2)underset(Delta)overset(conc.H_(2)SO_(4))tounderset(("major"))(A)underset(Delta)overset(H^(+)//KMnO_(4))toBoverset(NaOI)toCunderset(CaO,Delta)overset(NaOH)to product. Incorrect about the product.

Answer»

It GIVES two MONOCHLORO products.
Two isolated dibromides are theoretically possible.
Product is LIQUID at ROOM temperature
It can be formed by Wurtz REACTION in good yield

Solution :
Discussion
31.

Maximum value (n+l+m) for unpaired electrons in second excited state of chlorine ._(17)Cl is:

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28
25
20
none of these

Solution :
CONFIGURATION in SECOND EXCITED STATE may be GIVEN as:
Discussion
32.

Maximum racemization take place when

Answer»

`100% H_(2)O`
100% Acetone
80% H_(2)O + 20%` Acetone
80% Acetone+ 20% `H_(2)O`

Answer :C
Discussion
33.

Maximum probability of finding an electron along xy orbital is

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ALONG X-AXIS
In between x & y axis
Along y - axis
Along z axis

Answer :B
Discussion
34.

Maximum oxidation state is present in the central metal atom of which compound

Answer»

`CrO_2CI_(2)`
`MnO_2`
`[Fe(CN)_6]^(3-)`
`MnO`

ANSWER :B::C
Discussion
35.

Maximum oxidation state (+8) is exhibited by

Answer»

CO & Ni
Ru & Os
Cl & I
Te & I

Answer :B
Discussion
36.

Maximum number of sigma bonds that may be present in an isomer of C_(4)H_(8) are:

Answer»

10
11
12
13

Answer :C
Discussion
37.

Maximum number of possible stereoisomers (configurational only) with the molecular formula C_(4)H_(10)O are

Answer»

1
2
4
7

Solution :2 configurational ISOMERS
Discussion
38.

Maximum number of possible stereoisomers (configurational only) with the formula CH_(3)CH=CHCH(CH_(3))COOH

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2
3
4
6

Solution :4
Discussion
39.

Maximum number of planar atoms in SF_(5) molecule

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5
4
6
7

Answer :A
Discussion
40.

Maximum number of molecules are present in

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15 L of `H_(2)` gas at S.T.P
5 L of `N_(2)` gas at S.T.P
0.5 of `H_(2)` gas
10 g of `O_(2)` gas

Solution :(a) 15 L of `H_(2)` gas at S.T.P
`= (15L)/((22.4L))xx6.022xx10^(23)=4.033xx10^(23)`
(b) 5L of `N_(2)` gas at S.T.P.
`= ((5L))/((22.4L))xx6.022xx10^(23)`
`=1.344xx10^(23)`
(C) 0.5 g of `H_(2)=((0.5g))/((2.0g))xx6.022 xx 10^(23)`
`=1.505 xx 10^(23)`
10 g of `O_(2)=((10.0g))/((32.0g))xx6.022xx10^(23)`
`=1.882xx10^(23)`.
Discussion
41.

Maximum number of hydroge bonds that one water molecle is capable of forming is

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1
2
3
4

Answer :D
Discussion
42.

Maximum number of electrons that can be present in M and N - shells respectively are

Answer»

18, 32
8, 18
32, 50
32, 48

Answer :A
Discussion
43.

Maximum number of electrons that can be present in any molecular orbital is

Answer»

3
6
8
2

Answer :D
Discussion
44.

Maximum number of electrons present in 'N' shell is

Answer»

18
32
2
8

Solution :MAX. no. of ELECTRONS in N-shell (n=4) `=2n^2 =2 XX 4^2=32`
Discussion
45.

Maximum number of electrons may be present in one 4f-orbital

Answer»

2
4
7
14

Answer :A
Discussion
46.

Maximum number of electrons in a subshell with l = 3 and n = 4 is

Answer»

10
12
14
16

Answer :C
Discussion
47.

Maximum number of electrons in a subshell wih l=3 and n=4 is

Answer»

10
6
14
22

Answer :C
Discussion
48.

Maximum number of electrons in a shell with principal quantum number n is given by

Answer»

n
2n
`n^2`
`2n^2`

ANSWER :D
Discussion
49.

Maximum number of electrons in a subshell of an atom is determined by the following

Answer»

`2L + 1`
`4l - 2`
`2 n^(2)`
`4l + 2`

Solution :No. of orbitals in a subshell `= (2l + 1)`
`:.` Maximum no. of ELECTRONS in a subshell
`= 2 (2l + 1) = 4 L + 2`
Discussion
50.

Maximum number of electrons are present in

Answer»

2.24 lit. of `SO_(2)` at `S.T.P`
0.2 moles of `NH_(3)`
1.5 gm moles of oxygen
2 MOLE atoms of sulphur

Solution :No. of electrons =No. of moles `xx` AVOGADRO no X No. of electrons in a molecule
Discussion