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.

Freon-12 is commonly used as

Answer»

an insecticide
a refrigerant
a solvent
fire extinguisher.

Answer :B
Discussion
2.

Frenkel defect is shown by crystals having __________ coordination number and ___________difference in the size of the cations and the anions.

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ANSWER :LOW, LARGE
Discussion
3.

Frenkel defect is shown by crystals having…… coordination number and ………. Differencein the size of the cations and the anions.

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Answer :` 10^(6) "to" 10^(4) OHM^(-1) m^(-1)`
Discussion
4.

Frenkel defect is also known as………

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STOICHIOMETRIC defect
dislocation defect
impurity defect
non- stoichiometric effect

Answer :B::D
Discussion
5.

Frenkel defect is also known as _____

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STOICHIOMETRIC DEFECT
dislocation defect
impurity defect
non-stoichiometric effect

Answer :a,B
Discussion
6.

Free radicals can undergo

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rearrangement to a more STABLE FREE radical
decomposition to give another free radical
combination with other free radical
all are correct

Answer :D
Discussion
7.

Free radical reactions

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Occur in gas phases
Are OFTEN autocatalytic
Are initiated by LIGHT, OXYGEN or peroxides
All are correct

Answer :D
Discussion
8.

Free energy change ( DeltaG) is related to the total entropy change (DeltaS_("total") vizDeltaS_("system") + DeltaS_("surrounding")) as ".....................".

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SOLUTION :`DELTA G = - T Delta S_("TOTAL")`
Discussion
9.

Free energy change for a reversible process is

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`GT0`
`LT0`
EQUAL to zero
unpredictable

Answer :C
Discussion
10.

Fractional crystallization is carried out to separate

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Organic solids MIXED with inorganic solids
Organic solids highly soluble in water
Organic solids having small DIFFERENCE in their solubilities in a suitable solvent
Organic solids having GREAT difference in their solubilities in a suitable solvent

ANSWER :C
Discussion
11.

Fractiona bond order is in

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`O_(2)`
`O_(2)^(+)`
`O_(2)^(2-)`
`N_(2)`

ANSWER :B
Discussion
12.

Fraction of the total octahedral voids occupied will be :

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`(1)/(2)`
`(1)/(4)`
`(1)/(8)`
`(1)/(6)`

ANSWER :A
Discussion
13.

Fraction of molecules (eta) are related with velocity according to relation eta'=-(3)/(4)v^(2)+3v-(9)/(4)(1levle3) Then find most probable speed?

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0
1
2
3

Answer :C
Discussion
14.

Fraction of oxygen molecules ((dN)/(N)) in the range U_("mps") to U_("mps")+fU_("mps") whre fltlt1d:

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`(4f)/(2sqrt(PI))`
`(4)/SQRT(pi)((M)/(2RT))^(1//2)e^(-1)`
`(f)/(epi)`
`1`

Answer :a
Discussion
15.

Foure gases P, Q, R and S have almost same values of 'b' but their 'a'values (a, b are Vander Waals Constants) are in the order QltRltSltP. At a particular temperature, among the foure gases the most easily liquefiable one is

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P
Q
R
S

Answer :A
Discussion
16.

Four ten litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same temperature and pressure. The ratio of the total number of atoms of these gases present in different flasks would be

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`1:2:3:2`
`2:1:2:3`
`1:3:2:2`
`1:1:1:1`

ANSWER :B
Discussion
17.

Four structures are given in options (A) to (D), Examine them and selet the aromatic structure.

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Solution :Aromaticity required FOLLOWING conditions :
Planarity
(ii) complete DELOCALIZATION of the electrons in the ring
(iii) Presence of (4n+2) electrons in the ring where N is an integer (n=0,1,2,…). This is often REFERRED to as Huckel Rule.
(i) THEREFORE, (i) and (iii) are aromatic structures.
Discussion
18.

Four structures are given in options (a) to (d) . Examine them and select the aromatic structures .

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Solution :Cyclopropenyl cation (a) has COMPLETELY delocalized `2pi`-electrons. Therefore , in accordance with Huckel RULE, it is aromatic.
Cyclooctatetraene is not planar and also it has `8pi`-electrons. Therefore, in accordance with Huckel rule , it is not aromatic.
(c )DIPHENYL has two benzene rings each one of which is planar and has 6 `pi`-electrons . Therefore, like benzene, it is also aromatic.
(d)Cyclopropenyl anion although planar has only 4 `pi`-electron. Therefore, in accordance with Huckel rule, it is not aromatic.
Discussion
19.

Four structures are given in option (1) to (4). Examine them and select the aromatic structure.

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SOLUTION :`(4n+2)PI-` HUCKEL RULE
Discussion
20.

Four statements for Cr and Mn are given below: (i) Cr^(2+) and Mn^(3+) have the same electronic configuration. (ii) Cr^(2+) is a reducing agent while Mn^(3+) is an oxidising agent. (iii) Cr^(3+) is an oxidizing agent while Mn^(3+) is a reducing agent. (iv). both Cr and Mn are oxidizing agents. The correct statemetns are

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`i`,III,IV
`i`,II
`i`,ii,iv
`i`,iv

Answer :B
Discussion
21.

Four species are listed below: (i) HCO_3^- (ii) H_3O^+ (iii)HSO_4^- (iv)HSO_3F Which one of the following is the correct sequence of their acid strength ?

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`iii LT I lt iv lt ii`
`iv lt ii lt iii lt i`
`ii lt iii lt I lt iv`
`I lt iii lt ii lt iv `

Solution :`HSO_3F` is the super acid. Its ACIDIC strength is greater than any given SPECIES. The `pK_a` value of other species are given below:
`HSO_3^(-) to` 10.25
`H_3O_(+) to` -1.74
`HSO_4^(-) to` 1.92
Lesser the `pK_a` value, higher will be its acidic strength. Hence SEQUENCE of acidic strength will be,`HSO_3F gt H_3O^(+) gt HSO_4^(-) gt HSO_3^(-)`
Discussion
22.

Four samples of water A. B, C and D have the D.O. levels of 4 ppm. 3.8 ppm, 2.1 ppm and 4.9 ppm respectively 1he most polluted sample of water is

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D
B
C
A

Answer :C
Discussion
23.

Four processes are indicated below : The processes that do not produce 1-methylcyclohexanol are

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II , IV
I, II
III, IV
I, III

Answer :D
Discussion
24.

Four particles have speed 2,3,4 and 5 cm/s respectively. Their rms speed is:

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3.5 cm/s
`((27)/(2))cm//s`
`SQRT(54)cm//s`
`((sqrt(54))/(2))cm//s`

ANSWER :d
Discussion
25.

Four particles have speed 2,3,4 and 5 cm/s respectively. Their rms speed is :

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3.5 cm/s 
(27/2) cm/s 
54 cm's 
(54/2) cm/s 

Solution :`U_(RMS) = SQRT((2^2 + 3^2 + 4^2 + 5^2)/(4))`
Discussion
26.

Four moles of NaNO_(3) when heated to 800^(@) C , totally how many moles of paramagnetic gas molecules liberated ?

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ANSWER :2
Discussion
27.

Four one litre flasks are seperately filled with gases O_2,F_2, CH_4 and CO_2 under same conditions. The ratio of number of molecules in these flasks is

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`2:2:4:3` 
`1:1:1:1`
`1:2:3:4 `
`2:2:3:4 `

ANSWER :B
Discussion
28.

Four moles of acidified KMnO_(4) reacts with excess of H_(2)O_(2)to produce 2X moles of O_(2) Then X is.....

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SOLUTION :`2KMnO_(4)+3H_(2)SO_(4)+4H_(2)O_(2) to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2)`
`2-5`
`4-?`
Discussion
29.

Four metals A,B,CD have standard electrode potentials of -3.05,-1.66,-0.40 and +0.8V respectively.The metal which is most easily reduced is :

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A
B
C
D

Solution :D is most EASILY REDUCED
Discussion
30.

Four hydroxy compounds have functional groups as shown X : -CH_(2)OH , Y : -overset(|)(C)HOH Z : Ph-OH, W : Ph-underset(|)(C)HOH The purple colouration with FeCl_(3) will be given by

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X only
X and Y
Z only
X, Y, Z and W

Solution :`3Ph-OH+FeCl_(3) rarr UNDERSET(("VIOLET"))((PHO)_(3))Fe+3HCl`
Discussion
31.

Four isomeric carboxylic acids are formed from four different area ethyl benzene rarr benzoic acid , xylene rarr pthalic acid m- Xylene rarr isophthalic acid, and p- xylene rarr terphthalic acid

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Discussion
32.

four grams of hydrocarbon (C_(x)H_(y)) on complete combustion gave 12grams of CO_(2). What is the empirical formula of the hydrocarbon ? (C=12 , H=1)

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`CH_(3)`
`C_(4)H_(9)`
CH
`C_(3)H_(8)`

ANSWER :D
Discussion
33.

Four grams of hydrocarbon (C_(x)H_(y)) on complete combustion gave 12grams of CO_(2). What is the empirical formula of the hydrocarbon ? (C = 12, H = 1)

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`CH_(3)`
`C_(4) H_(9)`
`CH`
`C_(3) H_(8)`

Answer :D
Discussion
34.

Four grams of copper chloride on analysis was found to contain 1.890 g of copper and 2.110 g of chlorine (Cl). What is the empirical formula of copper chloride ?

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Solution :Step I. Percentage of the element
Percentage of `Cu=(1.89)/(4.0)xx100=47.25,` Percentage of `CL =(2.11)/(4.0)xx100=52.75`
Step II. Empirical formula of copper CHLORIDE
`{:("Element","Percentage","ATOMIC MASS","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("Cu",47.25,63.5,(47.25)/(63.5)=0.74,(0.74)/(0.74)=1,1),("Cl",52.75,35.5,(52.75)/(35.5)=1.48,(1.48)/(0.74)=2,2):}`
Empirical formula of the COMPOUND `= CuCl_(2)`.
Discussion
35.

Four gases P, Q , R and S have almost same values of 'b' but their 'a' values (a,b are Van der Waals Constants) are in the order Q lt R lt S lt P. At a particular temperature, among the four gases the most liquifiable one is ……………. .

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<P>P
Q
R
S

Solution :P
Discussion
36.

Four molecules have the velocities 3 xx 10 cms^(-1), 4 xx 10^4cms^(-1), 2 xx 104cms^(-1)and 5 xx 10^4cms^(-1).. Find the RMS velocity of the molecules.

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SOLUTION :`3.685 XX 10^(4) CMS^(-1)`
Discussion
37.

Four gas balloons P, Qlt R and S of equal volumes containing H_(2),N_(2)O,CO,CO_(2) respectively were pricked with needle and immersed in a tank containing CO_(2). Which of them will shring after sometime-

Answer»

<P>`P`
Q
R
`S`

ANSWER :A::C
Discussion
38.

Four elements A, B, C and D form a series of compounds havingthe formulae AB,B_(2), CB_(3), DB_(2)andDB_(3). If the jumbled up atomic numbers of A, B, C and D are 13, 19 , 26 and 35 , What are the ordered atomic numbers of A, BC and C ?

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Solution :` 13 = Al = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(1)` (can lose `1 e ^(-))`
` 19 = K = 1ss^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(1)(can lose `1e ^(-))`
26 =` Fe= 1s^(2)2s^(2) 2p^(6) 3s^(2) 3p^(6)3d^(6) 4s^(2) `(can lose ` 2e^(-) or 3 r^(-))`
35 ` = Br = 2s^(2) 2s^(2) 3p^(6) 3s^(2)3p^(6) 3d^(10)4s^(2) 4p_(x)^(2) 4p_(y)^(2) 4p_(z)^(1) ` (can gain or SHARE ` 1 e ^(-)` )
Compunds formed will be ` KBR , Br_(2), AlBr_(3), Fe Br_(2) and FeBr_(3).`
Hence ,` A = K = 19 , B= Br = 35 , C = Al = 13 , D = Fe = 26.`
Discussion
39.

Four different identical vessels at same temperature contains one mole each of C_(2)H_(6),CO_(2),Cl_(2) and H_(2)S at pressure P_(1),P_(2),P_(3) and P_(4) respectively. The value of van der Waal's constnat 'a' for C_(2)H_(6),CO_(2),Cl_(2) and H_(2)S is 5.562,3.640,6,579 and 4.490 atm L^(2) "mol"^(-2) respectively. Then:

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`P_(3)ltP_(1)ltP_(4)ltP_(2)`
`P_(1)ltP_(3)ltP_(2)lt_(4)`
`P_(2)ltP_(4)ltP_(1)ltP_(3)`
`p_(1)=p_(2)=p_(3)=p_(4)`

ANSWER :a
Discussion
40.

Four elements A ,B ,C and Dalongwith theirelectonic configurationare givenbelow : Elements:ABCD Electronic configuration :2,12,82,8,12,8,8 Nowanswerthe followingquestions : (i)Which two elements belongto thesameperiod ? (ii)Which two elements belongto thesamegroup ? (iii) Whichtwo elements belongto the 18thgroup ? (iv) Which element out of Aand C ismorereactiveand why ? (v) Whichelementout of Aand B formsthemaximumnumber tocompounds ?

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Solution :(i) Elementsin aperiodhave samenumberof shells . Thuselements A and Beach withtwo shellsbelongto secondperiod whileelements C and Dwhichhavethreeshells each belongto thirdperiod.
(ii)elements in a grouphavesamenumber of valenceelectrons. thuselements A and Cbelongto THESAME group .
(iii) Elementshaving all shellscompletely filledbelongto group 18 (noblegases. ) ThuselementsB and Dbelongto 18thgroup .
(iv) BECAUSEOF biggersizeand weakerforcesof attraction of thenucleus on thevalenceelectronelementC (2,8,1)is morereactivethanelement A.
(V) Bbeinga noblegas doesnot formcompounds . Butelement Ahas onlyelectronin thevalenceshellwhich ITCAN loseveryeasily. therefore, ELEMENT A formsmaximumnumberof compounds.
Discussion
41.

Four diatomic species are listed below in different sequences. Which of these represents the correct order of their increasing bond order-

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`C_(2)^(2-) LT He_(2)^(+)lt NO lt O_(2)^(-)`
`He_(2)^(+) lt O_(2)^(2-) lt NO lt C_(2)^(2-)`
`O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)`
`NO lt C_(2)^(2-) lt O_(2)^- lt He_(2)^(+)`

ANSWER :B
Discussion
42.

Four diatomic species are listed below in different sepuences . Which of these presents the correct order of their increasing bond order ?

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`O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)`
`No lt C_(2)^(2-) lt O_(2)^(-) lt He_(2)^(+)`
`C_(2)^(2-) lt He_(2)^(+) lt NO lt O_(2)^(-)`
`He_(2)^(+) lt O_(2)^(-) lt NO lt C_(2)^(2-)`

SOLUTION :`He_(2)^(+) ,lt O_(2)^(-) ,lt NO ,lt C_(2)^(2-)`
`((1)/(2)) ""(1(1)/(2))""(2(1)/(2))""(3)`
Discussion
43.

Four diatomic species are listed below, identify the correct order in which the bond order is increasing-

Answer»

`NO LT O_(2)^(-) lt C_(2)^(2-)He_(2)^(+)`
`O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)`
`C_(2)^(2-) lt He_(2)^(+) lt O_(2)^(-) lt NO`
`He_(2)^(+)ltO_(2)^(-) lt NO lt C_(2)^(2-)`

ANSWER :D
Discussion
44.

Four aromatic carboxylic acids on heating with soda-lime at 630 K, give toluene as the only product. Write the structures of the four aromatic carboxylic acid.

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Solution :SINCE all the four aromatic carboxylic acids on DECARBOXYLATION with soda-lime give toluene, THEREFORE, three of the acids MUST be o-, m- and p-toluic acids while the fourth must be phenylacetic acid as shown below :
Discussion
45.

Formulation of SO_(3) takes place according to the reaction 2 SO_(3) + O_(2) hArr 2 SO_(3), DeltaH=-45*2 " kcal " Which of the following factors favours the formulation of SO_(3) ?

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INCREASE in temperature
Increase in pressure
Removal of oxygen
Increase in volume

Answer :B
Discussion
46.

Formulate possible compounds of ‘Cl’ in its oxidation state is:0, -1, +1, +3, +5, +7 H_(2)O_(2) act as an oxidising agent as well as reducing agent where as O_(3) act as only oxidizing agent. Prove it.

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Solution :(i) Cl oxidation number 0 in `Cl_(2)`.
(ii) Cl oxidation number-1 in HCl.
(iii) Cl oxidation number +3 in `HClO_(2)` .
(iv) Cl oxidation number +5 in `KClO_(3)` .
(v) Cl oxidation number +7 in `Cl_(2)O_(7)` .
In `H_(2)O_(2)` oxidation number of OXYGEN is -1 and it can vary from 0 to -2 (+2 is POSSIBLE in `OF_(2)`). The oxidation number can decrease or increase, because of this `H_(2)O_(2)` can act both oxidising and REDUCING agent. Ozone `(O_(3))` only acts as oxidising agent since it decomposes to give NASCENT oxygen.
Discussion
47.

Formulae of metaborate and borate ions respectively are

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`BO_(3)^(3-) and BO_(2)^(-)`
`BO_(2)^(-) and BO_(3)^(3-)`
`BO_(2)^(-) and BO_(3)^(-)`
`BO_(2)^(2-) and BO_(3)^(3-)`

ANSWER :B
Discussion
48.

Formula of iron (III) oxide according to stock notation nomenclature method is….

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`FeO_(2)`
`Fe_(3)O_(4)`
`Fe_(2)O_(3)`
FeO

Solution :`Fe_(2)O_(3)rArr2(X)+3(-2)=0`
`therefore2x=6`
`thereforex =+3`
Discussion
49.

formula of Gypsum is

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`CaSO_(4)2H_(2)O`
`CaSO_(4)1//2H_(2)O`
`3CaSO_(4)H_(2)O`
`2CaSO_(4).2H_(2)O`

SOLUTION :`CaSO_(4)2H_(2)O`
Discussion
50.

Formula of cane sugar is C_(12)H_(22)O_(11). No. of molecules present in 34.2 g of cane sugar is

Answer»

`6.022 xx 10^(21)`
`6.022 xx 10^(20)`
`6.022 xx 10^(22)`
`6.022 xx 10^(18)`

Solution :Gram molecular mass of cane SUGAR `(C_(12)H_(22)O_(11))`
`= 12xx12 +22 xx 1 +16 xx11`
`= 342.0 G`
342.0 g of can sugar contain
`=6.022 xx 10^(23)` MOLECULES
34.2 g of cane sugar contain
`= (34.2)/(342)xx6.022xx10^(23)`
`=6.022 x 10^(22)` molecules.
Discussion