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.

NO is purified by

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ABSORPTION in `(NH_4)_2 SO_4` solution
Passing into CONC. `H_2 SO_4`
ABSORBING in `FeSO_4` solution
Electrolysis method.

Answer :C
2.

No gas is liberated when the following HX is treated with MnO_(2)and con .H_(2)SO_(4)

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HCl
HF
HBr
HI

Answer :B
3.

No correlation exists between the (R) and (S) designation and the direction of rotation of planepolarised light. Justify the statement.

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Solution :The rotation of a compound, `[(+) or (-)]` is something that we measure in the laboratory with a polarimeter and it depends on how the molecule interacts with light. On the other hand, the (R) and (S) designation is our own artifical way of describing how the atoms or groups are ARRANGED in space around a chiral CENTRE, i.e., the configuration of a particular chirality centre. Therefore, no necessary correlation exsists between the (R) and (S) designation and the direction of rotation of plane-polarised light. DEXTROROTATORY or (+)- compounds may have R or S configuration. Similarly, LEVOROTATORY or (-)-compounds may have R or S configuration.
.
4.

No chemical compound of helium is known Or Helium forms no real chemical compound.

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Solution :Firstly, He does not contain d-orbitals in the valence shell and HENCE electrons cannot be promoted to higher energy levels like that in Xe to form bonds. Therefore, helium does not form chemical compounds. SECONDLY, IONIZATION ENTHALPY of He is sufficiently higher (2372 kJ `mol^(-1)`) than those of Xe (1170 kJ `mol^(-1)`) and `O_(2)` (1175 kJ `mol^(-1)`).
5.

No compounds of Kr and Xe are known.

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SOLUTION : No COMPOUNDS of He and NE are KNOWN.
6.

NO and NO_2 are:

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EVEN ELECTRON molecules
Odd electron molecules
Diamagnetic
Paramagnetic

Answer :B
7.

No. alums are formed by sulphates of

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Li
`NH_(4)`
`MG^(2+)`
`Fe^(3+)`

ANSWER :B::C::D
8.

N,N-dimethyl butanamine-2 contains

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six `SP^(3)` hybridised carbon atoms
seven `sp^(3)` hybridised atoms
two `sp^(3)` hybridised nitrogen atoms
1 and 2 are correct

Answer :A
9.

N,N - dimethylanilinium acetate obtainedfrom

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SOLUTION :
10.

N,N dimethyl acetamide is

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`(CH_(3))_(2)NCOCH_(3)`
`CH_(3)N(COCH_(3))_(2)`
`CH_(3)CONH_(2)`
`CH_(3)N(COCH_(3))_(2)`

ANSWER :A
11.

N,N- dimethyl ethanamide on reduction with LiAIH_4 gives

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N- methylethanamine
N,N -dimethylethanamine
ETHANAMINE
Trimethyl AMINE

Answer :B
12.

N,N-dimethyl aniline on coupling with C_(6)H_(5)N_(2)Cl yields

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4-(N,N-dimethyl) amino azo benzene
4-(N,N-dimethyl) NITROSO benzene
4-(N,N-dimethyl) amino AZOXY benzene
4-(N,N-dimethyl) amino hydrazo benzene

Answer :A
13.

N,N-Diethyl-3-methylbenzamide (also called N,N-diethyl-m-toluamide. Or DEET) is used many insect repellants. Write its structure.

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SOLUTION :
14.

Nitrous acid reacts with all classes of amines. The product obtained from these reactions depend on whether primary, secondary or tetriary and ,wheather the amine or aliphatic or aromatic. Aliphalic primary amines react with acid (NaNO_(2)+HCl) to form alcohol as major product. In , addition to alcohol, alkene and alkyl halides are also formed as minor product. Certain cyclic primaryamines can undergo either ring expansion or ring contraction reactions on treatment with acid. This reaction is called Demajanov ring expansion or contraction. Two compounds (A) and (B) are treated with nitrous acid.

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ANSWER :B
15.

Nitrous oxide may easily be distinguished from nitric oxide by:

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adding water and shaking
introducing a GLOWING SPLINT
mixing with some air
adding some hydrogen

Answer :C
16.

Nitrous acid reacts with H_2SO_4 to give:

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`NO_2 + SO_2`
`NO + SO_2`
`NO+SO_3`
NONE of the above

Answer :A
17.

Nitrous acid reacts with H_(2)SO_(4) give

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`NO + SO_(2)`
`NO + SO_(3)`
`NO_(2) + SO_(2)`
None

Answer :C
18.

Nitrous acid reacts with all classes of amines. The nature of products obtained from these reactions depends on whether the amine is primary, secondary or tertiary. It also depends on whether the amine is aliphatic or aromatic. Aliphatic primary amines react with nitrous acid (NaNO_(2)+HCl) to form alcohol as the major product. In addition to alcohol, alkene and alkyl halide are also formed as minor products. Certain cyclic primary amines undergo either ring expansion or ring contraction on treatment with nitrous acid. this reaction is called Demjanov ring expansion or contraction. nitrous acid also reacts with nitroalkanes having acidic hydrogen. Q. Which of the following product(s) will be obtained when n-propyl amine is treated with NaNO_(2) and HCl ?

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`CH_(3)-underset(OH)underset(|)(C)H-CH_(3)`
`CH_(3)-CH=CH_(2)`
`CH_(3)-underset(CL)underset(|)(C)H-CH_(3)`
All of these

Answer :D
19.

Nitrosoamines (R_2N-N=O) are insoluble in water. On heating with conc. H_2SO_4 , they give secondary amines. The reaction is called

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Liebermann NITROSO reaction
Etard reaction
Fries reaction
Perkin reaction

Solution :`R_2N NO+H_2O to R_2NH+ HNO_2`
20.

Nitrosonitroalkaneis obtainedby 2^(0) - nitroalkane with

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`HNO_(3)`
`HNO_(2)`
`AgNO_(2)`
`KNO_(2)`

Answer :B
21.

Nitrous acid has no action on:

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`CH_(3)CH_(2)NO_(2)`
`CH_(3)CH_(2)COOH`
`C_(6)H_(5)N(CH_(3))_(2)`
`C_(6)H_(5)CONH_(2)`.

ANSWER :B
22.

Nitrosyl chloride is:

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`NOCL`
`NOCl_2`
`NO_2Cl_2`
`N_2OCl_2`

ANSWER :A
23.

Nitrosoamines (R_2 N-N=O) are insoluble in water. On heating them with conc. H_2 SO_4, they give sec. amines. The reaction is called

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Liberman's NITROSO REACTION
Etard's reaction
Fries reaction
Perkin's reaction.

Solution :Liberman.s nitroso reaction
24.

Nitroso amines (R_2H-N=O) are water soluble .On heating with conc. HCl, they give secondary amines. The reaction is called :

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Perkin REACTION
Fries reaction
Liebermann NITROSO reaction
Etard reaction

ANSWER :C
25.

Nitrosoamines are formed byunderline(" primary amines").

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SOLUTION :NITROSOAMINES are FORMED by SECONDARY AMINES.
26.

Nitrophenol is a colorless weak monoprotic acid (pK_(a) = 7.2) whose conjugate base is bright yellow. To 2.00 mL of a solution of 0.0100M nitrophenol is added 1.00 M NaOH in 0.01mL portions, and the adsorbance of the solution at 485 nm is monitored. what does the graph of A_(485) as a function of added volume of NaOH look like ?

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ANSWER :B
27.

Nitroparaffins on reduction give :

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ANSWER :B
28.

Nitroparaffins on reduction give:

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amides
alkylamtncs
AMMONIUM salts
acetanilides

Answer :B
29.

Nitronium ion is isoelectronic with

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`CO_(2)`
CO
`NO_(2)`
NO

Answer :B
30.

Nitroparaffins are obtained by oxidation of

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`3^(@)` AMINES
a-halocarboxylic ACID
acid AMIDE
`3^(@)` alkyl amines

Answer :D
31.

nitromethane will exhibit which type of isomerism?

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METAMERISM
GEOMETRICAL ISOMERISM
tautomerism
optical isomerism

ANSWER :C
32.

Nitromethane to N-methylmethanamine

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Solution :`underset("(Nitromethane)")(CH_(3)NO_(2))overset(Sn//HCl)underset("REDUCTION")RARR CH_(3)NH_(2) overset(CH_(3)CL)rarr underset("Dimethyl a MINE")(CH_(3)-NH-CH_(3))`
33.

Nitromethane to methylamine

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Solution :Methylamine :
`UNDERSET("NITROMETHANE")(CH_(3)-NO_(2))+6[H]underset(Delta)overset("Sn/conc. HCl")rarrunderset("methylamine")(CH_(3)-NH_(2))+2H_(2)O`
34.

Nitromethane reacts with chlorine in presence of alkali to yield

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Chloromethane
Chloropicrin
Nitrosyl chloride
Chloroform

Answer :B
35.

Nitromethane is subjected to the treatment with chlorine in the presence of sodium hydroxide,the main product is :

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Monochloronitromethane
Trichloromethane
CHLOROPICRIN
NONE of the above

ANSWER :C
36.

Nitroglycerine is

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An ester
An elcohol
A NITRO compound
An acid

Answer :A
37.

Nitrolim, a nitrgenous fertilizer is:

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`Ca_3H_2`
`CA(CN)_2`
`CaCN_2`
`CaCN_2+ C`

ANSWER :D
38.

Nitroglycerine is :

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An acid
An ALCOHOL
An ESTER
A PARAFFIN

ANSWER :C
39.

The explosive Nitroglycerine is :

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An acid
An ALCOHOL
An ESTER
A PARAFFIN

ANSWER :C
40.

Nitrogethane is acidic only towards

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`Na_(2)CO_(3)`
`NAOH`
`C_(2)H_(5)OH`
Liquid `NH_(3)`

SOLUTION :Properties of `1^(@)` and `2^(@)` nitroalkanes.
41.

Nitrogen(1) oxide is produced by:

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THERMAL DECOMPOSITION of ammonium NITRATE
 disproportionation of `N_(2)O_(4)` 
 thermal decomposition of ammonium nitrite 
 interaction of hydroxylamine and NITROUS acid.

Answer :A::D
42.

Nitrogen was discovered by:

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Cavendish
Lavoisier
Scheele
Daniel Rutherdord

Answer :D
43.

Nitrogenforms_______oxides.

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3
4
5
6

Answer :C
44.

Nitrogen tetraoxide (N_(2)O_(4)) decomposes as : N_(2) O_(4)(g) rarr 2NO_(2)(g) If the pressure of N_(2)O_(4) falls from 0.50 atm to 0.32 atm is 30 minutes the rate of appearance of NO_(2)(g) is :

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0.006 ATM `"min"^(-1)`
0.003 atm `"min"^(-1)`
0.012 atm `"min"^(-1)`
0.024 atm `"min"^(-1)`

Solution :Rate of disappearance of `N_(2)O_(4)`
`= (Delta[N_(2)O_(4)])/(Deltat)= (0.50-0.32)/(30)`
`= 0.006 atm min^(-1)`
Rate of APPEARANCE of `NO_(2)` = Twice rate of disappearance of `N_(2)O_(4)`
`2xx0.006 = 0.012 atm "min"^(-1)`
45.

Nitrogen shows oxidation state +5, but it does not form pentahalide as

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it is the first NUMBER ofgroup 15
it is a diatomic gas
it has no d-orbitals so VALENCE shell cannot be expanded
it has SMALL ATOMIC and ionic radii

Answer :C
46.

Nitrogen sequioxide is.

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`N_2 O_3`
`N_2 O_4`
`N_2 O_5`
`N_2 O`

ANSWER :A
47.

In which of the following phosphorus is in +1 oxidation state?

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`H_3PO_2`
`H_3PO_3`
`H_4P_2O_7`
`H_3PO_4`

ANSWER :B::C
48.

Nitrogen shows different oxidation states in different oxides. In which of the following oxides, nitrogen is in +4 oxidation state? NO ,N_2 O ,N_2 O_3 ,NO_2

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NO
`N_2O`
`N_2O_3`
`NO_2`

ANSWER :B
49.

Nitrogen sesquoxide colour is ……………….. .

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colourless
Brown
BLUE
Red

Solution :Blue
50.

Nitrogen pentoxide decomposes according to equation 2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g) The first order reaction was allowed to proceed at 140^(@)C and the data below were collected. a) Calculate rate constant in all the cases. b) What will be the concentration of N_(2)O_(5) after 100 minutes? c) Calculate initial rate of reaction.

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SOLUTION :a) For FIRST order reaction , k`=(2.303)/tloga/(a-x)`
In first CASE
`k=2.303/(20 min)log(0.400)/0.289 = 2.303/(20 min)log400/289 = 2.303/(20 min)[log 400 - log 289]`
`=2.303/(20 min) [2.6201 - 2.4609]=(2.303 xx 0.1412)/(20 mm) = 1.63 xx 10^(-2) min^(-1)`
In second case
`k=2.303/(40 min) log 0.400/0.209 = 2.303/(40 min) log 400/209 = 2.303/(40 min)[log 400- log 209]`
`=2.303/(40 min) [2.6021 - 2.3201] = (2.303 xx 0.2820)/(40 min)= 1.62 xx 10^(-1) min^(-1)`
In third case