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.

The basicity of phosphorus acid is :

Answer»

Two
Three
One
Zero

Answer :A
Discussion
2.

The basicity of orthophosphoric acid is:

Answer»

2
3
4
5

Answer :B
Discussion
3.

The basicity of hydroxides of lanthanides ........... with increase in atomic member due to ............... .

Answer»

SOLUTION :DECREASES , LANTHANOID CONTRATION.
Discussion
4.

The basicity of HClO_(4) is 'X'. What is the value of 'X' ?

Answer»


SOLUTION :BASICITY of `HClO_(4) :` NUrepleacable .H. is .I.
Discussion
5.

The basicity of aniline is less than that of cyclohexylamine. This is due to

Answer»

`+R` - effect of `-NH_(2)` group
`-I` effect of `-NH_(2)` group
`-R` effect of `-NH_(2)` group
HYPERCONJUGATION effect

Solution :`-NH_(2)` has a `+R` effect, it DONATES electrons to the benzene ring. As a RESULT of resonance, the lone pair of electrons on the N atom gets delocalized over the benzene ring and thus it is less readily AVAILABLE for protonation. Hence aniline is a weaker base than cyclohexyl amine.
Discussion
6.

What is the basicity of H_3PO_2 and why ?

Answer»

1
2
3
4

Answer :A
Discussion
7.

The basicity of aromatic (aryl) amines follows the order :

Answer»

`3^@gt 2^@ gt 1^@ gt` NH_3`
`3^@lt2^@lt1^@ltNH_3`
`2^@ lt 3^@ lt 1^@lt NH_3`
NONE

Answer :B
Discussion
8.

The basicstruturalunit of silicates is

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`(SiO_(3))^(2-)`
`(SiO_(4))^(2-)`
`(SIO)^(-)`
`(SiO_(4))^(4-)`

ANSWER :D
Discussion
9.

The basic structural unit of silicates is …………

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`(SiO_(3))^(2-)`
`(SiO_(4))^(2-)`
`(SIO)^(-)`
`(SiO_(4))^(2-)`

ANSWER :D
Discussion
10.

The basic structural unit of silicates is :

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`SiO_(3)^(2-)`
`SiO_(4)^(2-)`
`SIO^(-)`
`SiO_(4)^(4-)`

ANSWER :D
Discussion
11.

Basic unit fo zeolite, mica, asbestos and feldpar is

Answer»

`(--overset(R)overset(|)underset(R)underset(|)Si-O--)_(n)(R=Me)`
`(SiO_(3))^(2-)`
`(SiO_(4))^(4-)`
`SiO_(2)`

Solution :`(SiO_(4))^(4-)`, orthosilicates
Discussion
12.

The basic repeating structural unit of a crystalline solid is called a………………

Answer»

SOLUTION :UNIT CELL
Discussion
13.

The basic strength "…................" as we proceed fromLa(OH)_(3) to Lu(OH)_(3) ( decreases or increases or remains constant).

Answer»


ANSWER :DECREASES
Discussion
14.

The basic sterength of amine is in the order of………….

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`NH_3 GT CH_3 NH_2 gt (CH_3)_2NH`
`(CH_3)_2NH gt CH_3NH_2 gt NH_3`
`CH_3NH_2 gt (CH_3)_2NH gt HN_3`
`NH_3 gt (CH_3)_2NH gt CH_3NH_2`

ANSWER :B
Discussion
15.

The basic principle of Cottrell 's precipitator is

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neutralisation of charge on colloidal particles
scattering of light
Le chatelier 's principle
peptisation

Solution :In Cottrell smoke precipitator , the smoke is allowed to pass through a chamber having a series of PLATES charged to very high POTENTIAL (20000 to 70000V). Charged particles of smoke GET attracted and by charged plates and the gases coming out of CHIMNEY preciptated become free of charged particles.
Discussion
16.

The basic oxide among the following is

Answer»

`N_(2)O_(3)`
`As_(2)O_(3)`
`Sb_(2)O_(3)`
`Bi_(2)O_(3)`

Answer :4
Discussion
17.

Explain the principle behind the Hydrogen bomb.

Answer»

NUCLEAR fusion
Nuclear fission
Nuclear disintegration
None of these

Answer :A
Discussion
18.

The basic character of the transition metal monoxides follows the order (AtomicNos.,Ti = 22, V=23, Cr= 24, Fe=26)

Answer»

TiOgt VOgtCrOgt Fe
YOgtCrOgtTiOgt FeO
CrOgtYOgtFeOgtTiO
TiOgtFeOgtYOgtCrO

Solution :The basic CHARACTER of the transition metal monoxide isTiO gt YOgt CrO gt FeO because basic character of oxidesdecrease with increase in atomic number. OXIDES oftransitional metals in low oxidation state i.e., + 2 and+ 3 are GENERALLY basic except `Cr_(2)O_(3)`
Discussion
19.

The basic nature of the Lanthanide hydroxides decreases from Ce(OH)_(3), to Lu(OH)_(3)This is due to

Answer»

DECREASE in IONIC radius
Due to LANTHANOID contraction
Increase in METALLIC character
All of these

Answer :A
Discussion
20.

The basic character of the transition metal oxides follows the order…

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`CRO GT VO gt FEO gt TIO`
`VO gt CrO gt TiO gt FeO`
`TiO gt FeO gt VO gt CrO`
`TiO gt VO gt CrO gt FeO

Answer :D
Discussion
21.

The basic character of transition metal monoxides follow the order :

Answer»

`VO gt CrO gt TiO gt FeO`
`CrO gt VO gt FeO gt TiO`
`TiO gt FeO gt VO gt CrO`
`TiO gt VO gt CrO gt FeO`.

SOLUTION :As he size of METAL ION decreases from TI to Fe. The basic CHARACTER of their monoxide decreases.
Discussion
22.

The basic character of the transition metal monoxides follows the order. (Atomic Nos, Ti = 22,V = 23, Cr=24,Fe = 26)

Answer»

`Ti O gt VO gt CrO gt FeO `
`VO gt CrO gt TiO gt FeO`
`CrO gt VO gt FeO gt TiO`
`TiO gt FeO gt VO gt CrO`

Solution :The ORDER of basic character of the transition metal MONOXIDE is `TiO gt VO gt CrO gt FeO` because basic character of oxides DECREASES with INCREASE in ATOMIC number.
Discussion
23.

The basic character fo the transition metal monoxides follows the order (Atomic no's. Ti = 22, V = 23, Cr = 24 , Fe = 26)

Answer»

TiO GT VO gt CrO gt FeO
VO gt CrO gt TiO gt FeO
CrO gt VO gt FeO gt TiO
TiO gt FeO gt VO gt CrO

Solution :Basic character of oxide DECREASES from LEFT toright in a PERIOD of periodic table.
Discussion
24.

The basic character of hydrides of the VB group elements decreases in the order:

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`SbH_3 GT PH_3 gt AsH_3 gt NH_3`
`NH_3 gt SbH_3 gt PH_3 gt AsH_3`
`NH_3 gt PH_3 gt AsH_3 gt SbH_3`
`SbH_3 gt AsH_3 gt PH_3 gt NH_3`

ANSWER :C
Discussion
25.

The basic character of methylamins in vapour phase is :

Answer»

`3^@ GT 2^@ gt 1^@ gt NH_3`
`2^@ gt 3^@ gt 1^@ gt NH_3`
`1^@ gt 2^@ gt 3^@ gt NH_3`
None

Answer :A
Discussion
26.

The basic character of the transition metal monoxides follows the order

Answer»

`VO gt CRO gt TiO gt FEO`
`CrO gt VO gt FeO gt TiO`
`TiO gt FeO gt VO gt CrO`
`TiO gt VO gtCrO gt FeO`

Answer :D
Discussion
27.

The basic character of hydrides of the 15th-group elements decreases in the order

Answer»

`SbH_(3) gt PH_(3) gt AsH_(3) gt NH_(3)`
`NH_(3) gt SbH_(3) gt PH_(3) gt AsH_(3)`
`NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)`
`SbH_(3) gt AsH_(3) gt PH_(3) gt NH_(3)`

SOLUTION :`NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)`
On MOVING down the group atomic size increases and AVAILABILITY of lone pair decreases. Hence, basic CHARACTER decreases,
Discussion
28.

The basic character of an amine is due to presence of _________on nitrogen atom.

Answer»

SOLUTION :A LONE PAIR of ELECTRONS.
Discussion
29.

The basic character of hydrides of nitrogen family............ on moving down the group.

Answer»

SOLUTION :DECREASE
Discussion
30.

The basic character of amines is due to

Answer»

presence of NITROGEN atom
lone PAIR of electrons on nitrogen atom
TETRAHEDRAL structure
high ELECTRONEGATIVITY of nitrogen

Solution :Basictyof aminedepends uponthesharingof lone pairof electrons.
Discussion
31.

The basic character of amines is due to :

Answer»

PRESENCE of nitrogen atom
Tetrahedral structure
Lone PAIR of ELECTRONS on nitrogen atom.
High ELECTRONEGATIVITY of nitrogen

Answer :C
Discussion
32.

The basic character of amines is because

Answer»

They produce `OH^(-)` ions when treated with water
They have replaced H atoms on H atom
They have lone pair of ELECTRON on N atom
None of the reason is correct.

Solution :`R-OVERSET(..)(N)H_(2)+H^(+) rarr R- overset(+)(N)H_(3)`
Discussion
33.

The basic character of amine can be explained :

Answer»

In terms of lewis and Arrhenius CONCEPT
Only in terms of LOWRY BRONSTED concept
In terms of Lewis and Lowry Bronsted concept
Only in Lewis concept

Answer :C
Discussion
34.

The basic character of amines can be explained

Answer»

only in TERMS of Lowry-Bronsted concepT
only in TENNS of LEWIS concept
both in tenns of ARRHENIUS and Lewis CONCEPTS
both in terms of Lewis and Lowry-Bronsted concepts.

Solution :: Lewis and Lowry-Bronstcd concepts.
Discussion
35.

The base that is not present in DNA is

Answer»

GUANINE
URACIL
ADENINE
Thymine

Answer :B
Discussion
36.

The bases that are common in both RNA and DNA are

Answer»

ADENINE, GUANINE, Thymine
Adenine, URACIL, Cytosine
Adenine, Guanine, Cytosine
Guanine, Uracil, Thymine

Answer :C
Discussion
37.

The base present only in RNA and not in DNA is :

Answer»

URACIL
CYTOSINE
Thymine
Guanine

ANSWER :A
Discussion
38.

The base present in nucleic acids _______ .

Answer»

purine base
pyrimidine base
an INORGANIC base
both (a) & (B)

Solution :both (a) & (b)
Discussion
39.

The base present in DNA, but not in RNA is

Answer»

uracil
thymine
guanine
adenine

Answer :B
Discussion
40.

The base present in DNA but not in RNA is………. .

Answer»

GUANINE
ADENINE
URACIL
THYMINE

ANSWER :D
Discussion
41.

The base present in Cytidine :

Answer»




ANSWER :D
Discussion
42.

The base adenine occurs in

Answer»

DNA only
RNA only
DNA and RNA both
protein

Solution :DNA and RNA both.
Discussion
43.

Thebaseadenine presentin

Answer»

DNA only
RNA only
Both DNA & RNA
Protein

Answer :C
Discussion
44.

The base pairing occurs in double helix of DNA is

Answer»

A to T and G to C
A to G and T to C
A to C and G to T
G to T and A to C

ANSWER :A
Discussion
45.

The bapour density of completely dosscisted NH_(4)Cl would be

Answer»

Singht LESS then half the of `NH_(4)Cl`
Half the of `NH_(4)Cl`
Double that of `NH_(4)Cl`
Determined by the amount of solid `NH_(4)Cl` in the EXPERIMENT

Solution :`("Normal molecular WEIGHT")/("Expreimental molecular weight")=1+alpha`
`NH_(4)ClhArrNH_(3)+HCl`
`becausealpha=therefore"Experimental Molecular WT"=("nor. mol. wt.")/(2)`
Discussion
46.

The balanced chemical equation for the reaction of potassium permanganate and sulphuric acid with ferrous sulphate to produce ferric sulphate , potassium sulphate , maganese sulphate and water is :

Answer»

`2KMnO_(4)+16H_(2)SO_(4)+10FeSO_(4)toK_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+16H_(2)O`
`2KMnO_(4)+8H_(2)SO_(4)+10FeSO_(4)toK_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(O_(4))_(3)+10H_(2)O`
`2KMnO_(4)+8H_(2)SO_(4)+20FeSO_(4) toK_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O`
`2KMnO_(4)+8H_(2)SO_(4)+10FeSO_(4)to K_(2)SO_(4)+2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+8H_(2)O`

ANSWER :D
Discussion
47.

The balance chemical equation for the reaction Cu+HNO_(3)toCu(NO_(3))_(2)+NO+H_(2)Oinvolves

Answer»

`6HNO_(3)`
`2HNO_(3)`
`4HNO_(3)`
`8HNO_(3)`.

ANSWER :D
Discussion
48.

The balanced chemical equation for the reaction : MnO_(4)^(-)+AsO_(3)^(3-)+H^(+)toMn^(2+)+AsO_(4)^(3-)+H_(2)Oinvolves

Answer»

`4 AsO_(3)^(3-)`
`5AsO_(3)^(3-)and3H_(2)^(O)`
`5AsO_(3)^(3-)and4H^(+)`
`5AsO_(3)^(3-)and6H_(2)O`

Answer :B
Discussion
49.

Balance the following equation by oxidation number method.K2​ Cr2​ O7​ +HCl→KCl+CrCl3​ +H2​ O+Cl2​ .

Answer»

`K_(2)Cr_(2)O_(7)+8HCL to2CrCl_(3)+2Cl_(2)+2KCl+4H_(2)O`
`2K_(2)Cr_(2)O_(7)+28HCl to4CrCl_(3)+4Cl_(2)+4KCl+14H_(2)O`
`K_(2)Cr_(2)O_(7)+10HCl to2CrCl_(3)+2Cl_(2)+2KCl+5H_(2)O+2(O)`
`K_(2)Cr_(2)O_(7)_14HCl to 2CrCl_(3)+3Cl_(2)+2KCl+7H_(2)O`

Answer :D
Discussion
50.

The bakelite is prepared by the reaction between

Answer»

UREA and formaldehyde
ethylene glycol
phenol and formaldehyde
tetramethylene glycol

Answer :C
Discussion