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.

All the following elements show both positive and negative oxidation states, except

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N
H
O
F

ANSWER :D
2.

All the elements in a group in the periodic table have the same

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ELECTRONIC configuration
number of ELECTRONS in the VALENCE shell
atomic number
atomic weight

Answer :B
3.

All the C - O bonds in carbonate ion (CO_(3)^(2)-) are equal in length - Explain.

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Solution :Carbonate ion `(CO_(3)^(2)-)` = 3 bond PAIR + 1 lone pair `RARR` trigonal PLANAR
DUE to RESONANCE all C - O bond length are equal.
4.

All the C-O bonds in carbonate ion (CO_(3)^(2-)) are equal in length. Explain .

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SOLUTION :EXPLAIN on the BASIS of RESONANCE.
5.

All the alkali metals give characteristic flame test. The decreasing order of the frequency of light emitted by them is

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`LI gt Na gt K gt RB gt CS`
`Li gt Na =K = Rb gt Cs`
`Li =Na gt K gt Rb =Cs`
`Cs gt Rb gt K gt Na gt Li`

SOLUTION :As the IE decreases, the frequency of light emitted goes on increasing.
6.

All the antural processes in this universe produce

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a DECREASE in entropy of the universe
an increase in entropy of the universe
An increase in entropy of system and surroundings
sometimes increase or sometimes decrease in entropy

Solution :Spontaneous `rArrDelta S_("total") gt 0`
`rArr DELTA S_("sys") + Delta S_("surr.) gt 0`
7.

All reactions involving chemical decomposition are

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REVERSIBLE
reversible and ENDOTHERMIC
exothermic
may be reversible or IRREVERSIBLE and endothermic or exothermic

Answer :D
8.

(All products from P to related to Boron. The by products are not included) The ionic structure [BH_(2)(NH_(3))_(2)]^(+)[BH_(4)]^(-) corresponds to

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Z
Y
X
S

Answer :B
9.

(All products from P to related to Boron. The by products are not included) The compound X is

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ANSWER :A
10.

(All products from P to related to Boron. The by products are not included) The following is called inorganic benzene

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P
X
Y
Z

Answer :D
11.

All photons of light have same energy.

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ANSWER :F
12.

All ortho and para directional group are activator ? Give examples.

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Solution :All ORTHO and para directional group are not ACTIVATORS. HALOGEN (X=F, CL, BR, I) group are deactivator.
13.

All of the following substances react with wter. The pair that yeilds the same gaseous product is

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K and `KO_(2)`
Ca and `CaH_(2)`
`Na and Na_(2)
Ba and `BaO_(2)`

SOLUTION :(B) Ca and `CaH_(2)` yield the same product
`Ca+2H_(2)O to Ca(OH)_(2)+H_(2)`
`CaH_(2)+2H_(2)O to Ca(OH)_(2)+2H_(2)`
14.

All molecules of an ideal gas move with the same speed There is no attraction between the molecules in an ideal gas.

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Solution :All molecules of the ideal GAS does not MOVE with same SPEED their AVERAGE speed is same .
15.

All members of groups 14 when heated in oxygen form oxides. Which of the following is the correct trend of oixdes?

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DIOXIDES `CO_(2), SiO_(2)` and `GeO_(2)` are ACIDIC while `SnO_(2)` and `PbO_(2)` are amphoteric
CI,GeO, SnO and PbO are amphoteric
Monoxides react with haemoglobin to form TOXIC compounds.
All oxides burn with blue flame.

Answer :A
16.

Allmatershas nature? Why

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SOLUTION :No thewavelengthassociatedwithordinaryobjectedare soshrtbecauseof theirlargemassestheirwavepropertiescannotbe DETECTED.
17.

All Lewis acids are not Bronsted acids. Why?

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Solution :Lewis base is attracted by any ELECTRON DEFICIENT PARTICLE. It is called a nucleophile, Bronsted base, by DEFINITION, is attracted by proton only. It is called a protophile. A protophile (attract proton) is considered as a nucleophile, (ATTRACTS all positive species) but a nucleophile need not be a protophile.
18.

All elements in the third period have

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THREE COMPLETE shells
three complete subshells
three valence electrons
three electrons less than octet.

Solution :For FILLING of third period, the complete subshells are 1s, 2S and 2P.
19.

All elements commonly exhibit an oxidation state of

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`+1`
`-1`
ZERO
`+2`

ANSWER :B
20.

All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents . Explain .

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Solution :Because of the small size and high ELECTRONEGATIVITY and higher ionization enthalpy , LITHIUM compounds have CONSIDERABLE COVALENT character while compounds of other alkali metals are ionic in nature . As a result , compounds of lithium are more soluble in ORGANIC solvents while those of other alkali metals are more soluble in water .
21.

All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.

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SOLUTION :Smallest size of `LI^(+)`ion and high polarizing power give it the COVALENT CHARACTER therefore, lithium compounds are covalent and soluble in organic solvents whereas, other alkali metal compounds are ionic and soluble in water.
22.

All collidal dispersion have

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very HIGH OSMOTIC pressure
low osmotic pressure
high osmotic pressure
no osmotic pressure.

Answer :A
23.

All carbon-carbon bond length is same in molecule

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Butene-1
Benzene
Butene-2
Propyne-1

Answer :B
24.

All cell do not contain

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an anode
a cathode
ions
a POROUS parition

Solution :All CELLS CONTAIN anode, cathode and ions but porous parition is not NECESSARY in all the cells .
25.

All C-C bonds in benzene are of equal length.

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ANSWER :T
26.

All Bronsted bases are also lewis bases, but all bronsted acids are not lewis acids. Why?

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Solution :Because, bronsted BASES are proton acceptors and LEWIS bases CONTAIN lone pair/s of electrons available for coordination and thus they can accept PROTONS.
Example: `NH_(3),H_(2)O`.
27.

All bonds in benzene are equal due to

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Tautomerism
Inductive EFFECT
Resonance
ISOMERISM

Answer :C
28.

All atoms are spa hybridised in the following substance

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Methane
Ethane
Diamond
Graphite 

ANSWER :D
29.

What are aromatic compounds ? Give at least two examples.

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ANSWER :F
30.

All are primary pollutants except

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`SO_2`
`H_2SO_4`
`NO_2`
PARTICULAR matter

Answer :B
31.

All ammonium salts liberate ammonia when :

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heated alone
heated with caustic soda
heated with `H_(2)SO_(4)`
heated with `NaNO_(2)`

SOLUTION :It is the correct answer.
`(NH_(4))_(2) CO_(3) + 2NAOH overset("heat")rarr Na_(2)CO_(3) + 2NH_(3) + 2H_(2)O`
(Typical ammoniacal smell)
32.

All ammonium salts on heating with sodium hydroxide produce ammonia gas. From which ammonium salt, can maximum amount of ammonia be obtained ?

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`0.5 mol (NH_(4))_(3)PO_(4)`
`0.5 mol (NH_(4))_(2)SO_(4)`
`1.0 mol NH_(4)Cl`
`1.0 mol NH_(4)NO_(3)`

Solution :`{:(underset("0.5 mol")((NH_(4))_(3)PO_(4)),-=,underset("1.5 mol")(2NH_(3))),(underset("0.5 mol")((NH_(4))_(2)SO_(4)),-=,underset("1.0 mol")(2NH_(3))),(underset("1.0 mol")(NH_(4)Cl),-=,underset("1.0 mol")(NH_(3))),(underset("1.0 mol")(NH_(4)NO_(3)),-=,underset("1.0 mol")(NH_(3))):}`
`(NH_(4))_(3)PO_(4)` can give MAXIMUM mass of ammonia GAS.
33.

All alums

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CONTAIN same ions
Have similar crystaly STRUCTURE
Contain same atoms
Have the same molecular WEIGHT

SOLUTION : ALUMS have similar crystal structure to
34.

All alkanes possess sp^3 hybridised carbon atoms.

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ANSWER :T
35.

All alkali halides are soluble in water expect Lif. The low solubility of LiF in water is due to its underline ((i)) the low solubility CsI is due to underline ((ii)). LiF is soluble in underline ((iii)) solvent.

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`{:(""(i), (ii), (iii)),("low lattic", " larger hydrration" , "polar"),("ENTHALPY", "enthalpy","SOLVENTS"):}`
`{:(""(i), (ii), (iii)),("high lattice", " smaller" , "non-polar"),("enthalpy", "hydration enthalpy","solvents"):}`
`{:(""(i), (ii), (iii)),("high hydration", " high lattice" , "non-polar"),("enthalpy", "enthalpy","SOLUTION"):}`
`{:(""(i), (ii), (iii)),("smaller", " high lattice" , "polar"),("hydration enthalpy", "enthalpy","solvents"):}`

Solution :The low solubility of liF is due to its high lattice ENETHALPY whereas the low solubility of CsI is due to smaller hydration enthalpy of TWO large size ions. LiF is soluble in non-polar solvents like ethanol, acetone, etc.
36.

All alkali halides are soluble in water except LiF. The low solublity of LiF in water is due to its (i)______ the low solubility of CsI is due to (ii)_____. LiF is soluble in (iii)_________ solvents.

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(i) low lattice enthalpy, (ii) LARGE HYDRATION enthalpy, (iii) polar solvents
(i) high lattice enthalpy, (ii) smaller hydration enthalpy, (iii)-non-polar solvents
(i) high hydration enthalpy, (ii)-high lattice enthalpy, (iii)- non-polar solvents
(i)-smaller hydration enthalpy, (ii)-high lattice enthalpy, (iii)-polar solvents

Solution :The low solubility of LiF in WATER is due to its high lattice enthalpy whereas the low solubility of CsI is due to smaller hydration enthalpy of TWO large size ions. LiF is soluble in non-polar solvents LIKE ethanol, acetone, etc.
37.

All alcohols are

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COMPLETELY soluble in water
ionised in water
not soluble in water
soluble in ORGANIC solvents

Solution :All alcohols are not soluble in water. Alcohols with LARGE size of R-are ALMOST immiscible with water.
38.

Alkynes undergoes reduction using lindlar catalyst to give ……………….

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SOLUTION :CIS - ALKENES
39.

Alkynes undergoes reductionusing sodium in liquid ammonia to give ……………..

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SOLUTION :TRANS- ALKENES
40.

Alkynes on reduction with sodium in liquid ammonia from trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism ?

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Solution :SINCE 2-butene is CAPABLE of showing geometrical isomerism, THEREFORE, REDUCTION of 2-butyne with SODIUM in liquid ammonia will give trans-2-butene.
41.

Alkynes on reducton with sodium in liquid ammonia forms trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism ?

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Solution :On REDUCTION of Butyn-2, Butene-2-is formed in which EITHER both them ethyl GROUPS are in the same side or OPPOSITE side to show geometrical isomers.
42.

Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of but-2-yne show the geometrical isomerism?

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SOLUTION :
But-2-ene is CAPABLE of SHOWING GEOMETRICAL ISOMERISM.
43.

Alkynes exhibit A) Chain isomerism B) Position isomerism C) Functional group isomerism

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A only
A and B
A and C
A, B and C

Answer :D
44.

Alkynes are less reactive than alkenes towards addition of Br_2 because :-

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Both are EQUALLY stable
Original STATEMENT is incorrect.

Answer :A
45.

Alkyne , C_7 H_(12), when reacted with alkaline KMnO_4 followed by acidification with HCl gives a mixture of (CH_3)_2 CHCOOH + CH_3HC_2 COOH. The alkyne C_7H_(12) is

Answer»

3- hexyne
2-methyl-2-hexene
2-methyl-3-hexyne
3-methyl-2-hexyne

SOLUTION :`UNDERSET("2-Methyl-3-hexyne")((CH_3)_2CH - C -= C - CH_2CH_3) overset([O])to (CH_3)_2 CHCOOH + CH_3CH_2COOH `
46.

Alkyne C_(3)H_(4) and C_(4)H_(6) forms which type of alkene isomers ? Give their structure.

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Solution :ALKYNES are isomers of CYCLOALKENES.
(a) Cyclic alkene isomers of `C_(3)H_(4)` alkynes :
`UNDERSET("propyne")(CH_(3)-C-=CH)` and
(B) Cyclic alkene isomers of `C_(4)H_(6)` alkynes :
47.

Alkylhalides have polar C-X bond and undergo nucleophilic substitution reactions. These give a variety of products with nucleophiles such as -OH, -OR, -NH_(2), -CN, -NC, -NO_(2), -ONO, RCOO^(-), etc. They undergo mainly two types of nucleophilic substitution reactions, S_(N)1 and S_(N)2. S_(N)1 reactions are two steps reactions which proceed through the formation of carbocations while S_(N)2 reactions are one step reaction which proceeds through the formation of transition state. The stability of carbocation and transition state determine the reactivity of alkyl halides. Which out of S_(N)1 or S_(N)2 results into inversion of configuration?

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SOLUTION :`S_(N)2`
48.

Alkylhalides have polar C-X bond and undergo nucleophilic substitution reactions. These give a variety of products with nucleophiles such as -OH, -OR, -NH_(2), -CN, -NC, -NO_(2), -ONO, RCOO^(-), etc. They undergo mainly two types of nucleophilic substitution reactions, S_(N)1 and S_(N)2. S_(N)1 reactions are two steps reactions which proceed through the formation of carbocations while S_(N)2 reactions are one step reaction which proceeds through the formation of transition state. The stability of carbocation and transition state determine the reactivity of alkyl halides. Write the structure of isomer of compound C_(4)H_(9)Br which is most reactive towards S_(N)1 reaction.

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SOLUTION :`CH_(3)-UNDERSET(CH_(3))underset(|)OVERSET(CH_(3))overset(|)C-Br`
49.

Alkylhalides have polar C-X bond and undergo nucleophilic substitution reactions. These give a variety of products with nucleophiles such as -OH, -OR, -NH_(2), -CN, -NC, -NO_(2), -ONO, RCOO^(-), etc. They undergo mainly two types of nucleophilic substitution reactions, S_(N)1 and S_(N)2. S_(N)1 reactions are two steps reactions which proceed through the formation of carbocations while S_(N)2 reactions are one step reaction which proceeds through the formation of transition state. The stability of carbocation and transition state determine the reactivity of alkyl halides. Arrange the following in the increasing order of reactivity towards S_(N)2 reaction:

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SOLUTION :`(i) LT (III) lt (II)`
50.

Alkylhalides have polar C-X bond and undergo nucleophilic substitution reactions. These give a variety of products with nucleophiles such as -OH, -OR, -NH_(2), -CN, -NC, -NO_(2), -ONO, RCOO^(-), etc. They undergo mainly two types of nucleophilic substitution reactions, S_(N)1 and S_(N)2. S_(N)1 reactions are two steps reactions which proceed through the formation of carbocations while S_(N)2 reactions are one step reaction which proceeds through the formation of transition state. The stability of carbocation and transition state determine the reactivity of alkyl halides. Complete the reaction: CH_(3)CH_(2)Br underset("Heat")overset(Ag_(2)O)to

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SOLUTION :`2CH_(3)CH_(2)Br+Ag_(2)O OVERSET("Heat")to CH_(3)CH_(2)OCH_(2)CH_(3)+2AgBr`