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.

Noble gases are prepared by the:

Answer»

CONDENSATION of gases of the air
Fractionation of LIQUID OXYGEN
Removal of nitrogen and oxygen from air
Fractionation of liqud air

Answer :D
Discussion
2.

Noble gases are only sparingly soluble in water due to :

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DIPOLE - dipole INTERACTIONS
INDUCED dipole-induced dipole interactions
dipole-induced dipole interactions
hydrogen bonding

Answer :C
Discussion
3.

Noble gases are mostly inert. Assign reasons. Or Explain the causes of inertness of noble gases.

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Solution :Noble GASES are CHEMICALLY inert because of the following reasons :
(i) They have completely filled `ns^(2)np^(6)` ELECTRONIC configuration in their respective valence shells.
(ii) They have HIGH IONIZATION enthalpies.
(iii) Electon gai enthalpies of noble gases are positive.
Discussion
4.

Noble gases are mostly chemically inert. Give reason.

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Solution : Noble GASES are CHEMICALLY inert because they have complete octets and possess STABLE structure. Their IONISATION enthalpies are LARGE.
Discussion
5.

Noble gases are monoatomic in nature. Explain.

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Solution :All the orbitals (which are occupied by electrons) are completely filled in noble gases. THEREFORE, they have no tendency to share electrons with other atoms to FORM diatomic molecules and hence are MONOATOMIC in NATURE.
Discussion
6.

Noble gases are chemically inert. This is due to ………………..

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UNSTABLE ELECTRONIC configuration
stable electronic configuration
only FILLED p-orbital
only filled s-orbital

Answer :B
Discussion
7.

Noble gases are chemically inert. Give reasons.

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Solution :Noble gases are chemically inert because they have their octet complete EXCEPT HELIUM, i.e. they have a STABLE ELECTRONIC configurations.
Discussion
8.

Noble gases are chemically inert. Give one reason

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SOLUTION :OCTET STRUCTURE or They have `ns^2np^5` CONFIGURATION
Discussion
9.

Noble gases are adsorbed by............

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anhydrous `CaCl_2`
`FE(OH)_3`
CONC.`H_2SO_4`
activated charcoal

Solution :activated charcoal
Discussion
10.

Noble gases are absorbed on

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ANHYDROUS `CaCl_(2)`
Charcoal
Conc. `H_(2)SO_(4)`
Coconut

Solution :Noble GASES are found in very minute AMOUNT in ATMOSPHERE . These are separated form each other by using cocunut charcoal. Which adsorb different gas at different TEMPEARTURE.
Discussion
11.

Noble gases and metals both have closed packed structures, yet the melting point of noble gases are exceptionally low. Why?

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Solution :Although both have the closed packed structures, the atoms in noble GASES are held together by weak VAN DER WAALS. forces and thus show lower MELTING points
Discussion
12.

Noble gases are:

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Colourless
Ordouless
Tasteless and non-inflammable
All

Answer :D
Discussion
13.

Noble gases A, B, C, D, E are passed through Dewar's flask at-100^@C A, C & E gases are adsorbed. Unadsorbed gases B, D are passed through another Dewar's flask at-180^@C. Gas B is adsorbed. The Dewar's flask al-100^@C is put in contact with another Dewars flask al-193^@C. The gas Eis diffused in it. Finally A & C are seperated by warming the flask from-100^@C w -90^@C. Ilere 'C' comes out He and Ne gases respectively are

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Ne
Ar
Kr
Xe

Solution :
`He, Ne underset("Deewars flask "-180^@C)overset("another")to`Ne is absorbed. He is unadsorbed, He has WEAK vander wall forces.
Charcoal - 1 contains Ar, Kr, Xe is cooled to `-180^@C`. At this TEMP. Ar. Gets LIQUIFIED while Kr & Xe remain adsorbed.

(E) = Ar, (A)= Kr,(C )= Xe
Discussion
14.

Noble gas used in the miner's cap lamp is

Answer»

krypton
argon
helium
radon

Answer :A
Discussion
15.

Noble gas that is not present in air

Answer»

He
Ar
Kr
Rn

Answer :D
Discussion
16.

Noble gas atoms have very high ionization energles. Explain.

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Solution :Because all the ELECTRONS in the outermost shell are paired and a large amount of energy is needed to unpair them. So, they POSSESS HIGH ionization ENERGIES
Discussion
17.

Noble gas forms compounds with

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Fluorine
Oxygen
Fluorine and Oxygen
Fluorine and Sulphur

Solution :NOBLE gases react with `F_2` and `O_2` and FORM compounds LIKE `XeF_2, XeF_4 , XeO_3` etc.
Discussion
18.

Noble gases do not react with other elements because

Answer»

The size of their AOMS are very small
They are not found in abundance
They are monoatomic
They have vert STABLE ELECTRONIC configuration

Answer :D
Discussion
19.

NO_(3)^(-)toNO_(2) (acid medium) E^(0)=0.790V NO_(3)^(-)toNH_(3)OH^(+) (acid medium) E^(0)=0.731V. At what pH, the above two will have same E value ? Assume the concentration of all other species NH_(3)OH^(+) except [H^(+)] to be unity.

Answer»


Solution :`overset(+5)(N)O_(3)^(-)+2H^(+)+e^(-)tooverset(+4)(N)O_(2)+H_(3)O,0.79"VOLT"`
`overset(+5)(N)O_(3)^(-)+8H^(+)+6e^(-)toNH_(3)OH^(+)+2H_(2)O,0.731"volt"`
`E=0.79-(0.0591)/(1)log((1)/([H^(+)]^(2)))=0.731-(0.591)/(6)log((1)/([H^(+)]^(8)))`
`0.059=0.059(2-(8)/(6))pHimpliespH=(6)/(4)=1.5`
Discussion
20.

Nobel's oil is :

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FIRE extinguisher
Insecticide
Explosive
Detergent

Answer :C
Discussion
21.

NO_3^(-) is reduced to NH_(4)^(+)What is the change in oxidation number of nitrogen .......

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SOLUTION :`OVERSET(+5)(NO_3^(-))+8barerarroverset(-3)(NH_4^(+)),`CHANGE in oxidation state =[-3-5]=8
Discussion
22.

NO_(3^(-)), is isoelectronic with

Answer»

`CO_(2)`
`CO_(3)^(2-)`
NO
`NO_(2)^(-)`

ANSWER :B
Discussion
23.

NO_(3)^(Θ) isdetected by (a) _____when (b)_____is formed on the addition of FeSO_(4) and cone H_(2)SO_(4)

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Answer :a.Ring TEST ,b.`[Fe(H_(2)O)_(3)NO]^(2)`
Discussion
24.

NO_(3)^(-) is isoelectronic with,

Answer»

`CO_(2)`
`SO_(2)`
`CO_(3)^(2-)`
`SO_(3)^(2-)`

ANSWER :C
Discussion
25.

NO_2 readily forms a dimer. Explain.

Answer»

Solution :`NO_2`is unstable due to the PRESENCE of one unpaired ELECTRON. Therefore, it dimerises to FORM a stable compound.
Discussion
26.

NO_(2) is not obtained when is heated

Answer»

`PB(NO_(3))_(2)`
`AgNO_(3)`
`LINO_(3)`
`KNO_(3)`

ANSWER :D
Discussion
27.

NO_2 is obtained by heating:

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`CsNO_3`
`KNO_3`
`LiNO_3`
`NaNO_3`

ANSWER :C
Discussion
28.

NO_2 cannot be obtained by heating:

Answer»

`KNO_3`
`PB(NO_3)_2`
`CU(NO_3)_2`
`AgNO_3`

ANSWER :A
Discussion
29.

No reaction occurs in which of the following equations

Answer»

`I^(ɵ)+Fe^(2+) rarr`
`F_(2)+2NaCl rarr`
`Cl_(2) +2NaF rarr`
`I_(2)+2NaBr rarr`

Solution :Reduction POTENTIAL of `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. So `F_(2)` can DISPLACE `CL^(Θ), Br^(Θ) and I^(Θ)` but not vice VERSA. Similarly, `Cl_(2)` can displace `Br^(Θ) and I^(Θ)` but not vice versa and `Br_(2)` can displace only `I^(Θ)` but not vice versa. In (a) `Fe(" not " Fe^(2+))` is a better reducing agent tahn `I^(Θ)`.
Discussion
30.

No of water molecules of cystalization in the epsom salt.

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ANSWER :7
Discussion
31.

No. of unpaired e^(-) in K[(Ag(CN)_(2)] is

Answer»


Solution :No unpaired ELECTRON `(AG^(+):t_(2G)^(6)e_(g)^(4))`
Discussion
32.

No. of unpaired electrons in Fe^(2+) ions is :

Answer»

three
two
four
five

Answer :C
Discussion
33.

No of products and number of fractions are respectively :

Answer»

6,5
6,4
6,6
6,3

Solution :
Discussion
34.

No. of radicalnodes in 4dx^(2) -y^(2)orbitals are

Answer»


ANSWER :1
Discussion
35.

No . Of products and No . Of fractions are respectively

Answer»

6,5
6,4
5,4
6,3

Solution :
Discussion
36.

No. of moles of H_(2)O gasevolved when one mole of the following compound reacts with sodium.

Answer»


SOLUTION :N//A
Discussion
37.

No. of nuclei among the following which undergo radioactivedecay (based on number of protons only) Silicon - 30 , Germanium - 72 , Barium - 130 , Radon - 216

Answer»


SOLUTION :`RN^(216)`
Discussion
38.

No. of 'N' atoms present in Novalgin

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SOLUTION :No. of 'N ' ATOMS PRESENT in NOVALGIN
Discussion
39.

No. of lone pairs in I bar(Cl_(2))

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SOLUTION : , TOTAL ELECTRON pairs `= (7+2+1)/(2) = 5`, Lone pairs = Total -b.p.s = 5-2 =3
Discussion
40.

No. of isomers of [Co(NH_(3))_(2)Cl_(2)(en)]^(+) are

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Solution :`phicis` from 4 TRANS FORM, cis - form has OPTICAL.
Discussion
41.

No. of ions produced by dissolving PtCl_(4).2NH_(3) in aqueous media is

Answer»


Solution :COORDINATION NUMBER `6[PT(NH_(3))_(2)Cl_(4)]`
Discussion
42.

No. of hybrid orbitals present in IF_(7)

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SOLUTION :HYBRIDISATION`-sp^(3)d^(3) :.` HYBRID orbitals - 7
Discussion
43.

No. of geometrical isomersof [RhCl(CO)(PPh_(3))(NH_(3)) are

Answer»


Solution :[M ABED] can GIVE 3 geometrical isomers.
Discussion
44.

No. of Geometrical isomers for following compound is :

Answer»

8
16
32
10

Answer :B
Discussion
45.

No. of gaseous product obtain on heating of FeSO_(4).7H_(2)O

Answer»


SOLUTION :NA
Discussion
46.

No. of gaseous product obtain on heating of FeSO_4. 7H_2O

Answer»


ANSWER :B
Discussion
47.

No of compounds producing gas on hydrolysis (with H_(2)O ) is Al_(4)C_(3),BaC_(2),Mg_(2)C_(3),SiC,B_(2)H_(6),Fe_(3)C

Answer»


SOLUTION :N//A
Discussion
48.

No of allylic hydrogens in thermodynamic product is ''X'' No of allylic hydrogens atom in kinetic product is ''Y'' The value of |X+Y| is …….

Answer»


ANSWER :8
Discussion
49.

No. of acyclic isomers of the compound having the molecular formula C_(4)H_(10)O is :

Answer»

4
6
5
7

Solution :Acyclis isomers of `C_(4)H_(10)O` are :
(i) `CH_(3)CH_(2)CH_(2)CH_(2)OH (ii) CH_(3)CH_(2)UNDERSET(OH)underset(|)(CH)CH_(3)`
(iii) `CH_(3)underset(CH)underset(|)(CH_(2))OH "" (IV) C_(2)H_(5)-O-C_(2)H_(5)`
(v) `CH_(3)OCH_(2)CH_(2)CH_(3)`
Discussion
50.

No. of acidic hydrogen in Barbuteric acid

Answer»


SOLUTION :FOUR ACIDIC HYDROGENS
Discussion