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 correct equation for the degree of an associating solute, 'n' molecules of which undergoes association in solution, is ………….

Answer»

1) `ALPHA = (n (i-1))/(n -1)`
2) `alpha = (n(1-i))/((n -1))`
3) `alpha = (n (I -1))/(1-n)`
4) `alpha = (n (1-i))/(n (1-i))`

SOLUTION :`alpha = ((1-i n ))/((n -1))(or) (n (i -1))/( (1-n))`
2.

The correct electronic configuration of Ti (Z=22) atom is

Answer»

`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2`
`1s^2 2s^2 2p^6 3s^2 3p^6 3d^4`
`1s^2 2s^2 2p^6 3s^2 3p^63d^4`
`1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^3`

ANSWER :A
3.

The correct electronic configuration of alkaline earth metal is_________

Answer»

[noble GASES]ns1
[noble gases]NS2
[noble gases]ns2np6
NS(n-1)d1-10

Solution :(noble gases] `ns^(2)`
4.

The correct electronic configuration of ._(29)Cu^(2+) is

Answer»

`[Ar] 4s^(1) 3D^(10)`
`[Ar] 4s^(2) 3d^(8)`
`[Ar] 4s^(2) 3d^(9)`
`[Ar] 4s^(0) 3d^(9)`

Answer :D
5.

The correct differential form of van't Hoff equation is ………….

Answer»

SOLUTION :`(d(ln k))/(dT) = (Delta H^(@))/(RT^(2))`
6.

The incorrect electronic arrangment is

Answer»

2,8,13,1
2,8,12,2
2,8,8,1
2,8,8,2

Answer :B
7.

The correct descending order of the heat liberated (in kJ) during the neutralization of the acids CH_(3)C O O H(W),HF(X), HCO OH (Y) and HCN(Z) under indentical conditions (K_(a) " of" CH_(3) CO OH = 1.8 xx 10^(-5), HCO OH = 1.8 xx 10^(-4), HCN = 4.9 xx 10^(-10)and HF = 3.2 xx 10^(-4)) is

Answer»

`Y GT X gt Z gt W`
`X gt Y gt W gt Z`
`Z gt X gt Y gt Z`
`Z gt W gt X gt Y`

Solution :From the `K_(a)` values, strengthof theacids is in theorder :
`HF gt HCO OH gt CH_(3)CO OH gt HCN`
Weaker the acid, greater is the energy used for ionization and less is the HEAT liberated or stronger the acid, more the heat is liberated. Hence, order of liberation of heat is
`HF gt HCO OH gt CH_(3) CO OH gt HCN`
or `X gt Y gt W gt Z`
8.

The correct decreasing order of stability of I,II and III carbocations is: (I) CH_(3)-overset(+)(C)H-CH_(3) (II) CH_(3)-overset(+)(C)H-COCH_(3) (III) CH_(3)-overset(+)(C)H-OCH_(3)

Answer»

`I GT II gt III`
`II gt III gt I`
`III gt I gt II`
`II gt I gt III`

ANSWER :C
9.

The correct descending order of the heat librated (in kJ) during the neutralisation of the acids CH_(3)COOH(W), HF(X), HCOOH(Y) AND HCN(Z) under identical conditions (K_(a) of CH_(3)COOH = 1.8 xx 10^(-5), HCOOH = 1.8xx 10^(-4), HCN = 4.9 xx 10^(-10) and HF = 3.2 xx 10^(-4)) is

Answer»

`Y gt X gt Z gt W`
`X gt Y gt W gt Z`
`W gt X gt Y gt Z`
`Z gt W gt Y gt X`

Solution :From the `K_(a)` values, strength of acids is in the order :
`HF(X)gt HCOOH(Y) gt CH_(3)COOH(W) gt HCN(Z)`
Stronger the acid, higher is the heat liberated during the neutralisation because lesser ENERGY is required for ionisation.Therefore, correct order of heat liberated is `X gt Y gt W gt Z`.
10.

The correct decreasing order of surface tension for water , ethanol and n - hexane is

Answer»

n hexane `lt` ETHANOL `lt` water
Water `gt` ethanol `gt` n -hexane
n- hexane `lt` water `lt` ethanol
ethanol `lt` water `lt` n - hexane

Solution :Intermolecular forces of ATTRACTIONS are MAXIMUM in water and MINIMUM in n- hexane
11.

The correct descending order of solubility of sulphate salts of alkaline earth metals in water is...

Answer»

`BeSO_(4) gt CaSO_(4) gt MgSO_(4) gt BaSO_(4)`
`BaSO_(4) gt CaSO_(4) gt MgSO_(4) gt BeSO_(4)`
`BeSO_(4) gt MgSO_(4) gt CaSO_(4) gt BaSO_(4)`
`MgSO_(4) gt CaSO_(4) gt BaSO_(4) gt BeSO_(4)`

ANSWER :C
12.

The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is…….

Answer»

`-SO_(3)H, -COOH, -CONH_(2), -CHO`
`-CHO, -COOH, -SO_(3)H, -CONH_(2)`
`-CONH_(2)-CHO, -SO_(3)H, -COOH`
`-COOH, -SO_(3)H,-CONH_(2), -""^(-)C HO`

ANSWER :D
13.

The correct decreasing order of priority for the functional groups of organic compounds in the IUPACsystem of nomenclature is

Answer»

`-SO_(3)H,-COOH,-CONH_(2),-CHO`
`-CHO,-COOH,-SO_(3)H,-CONH_(2)`
`-CONH_(2),-CHO,-SO_(3)H,-COOH`
`-COOH,-SO_(3)H,-CONH_(2),-CHO`

SOLUTION :It is based on IUPAC system.
14.

The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is

Answer»

`-CONH_(2),-CHO,-SO_(3)H,-COOH`
`-COOH,-SO_(3)H,-CONH_(2),-CHO`
`-SO_(3)H,-COOH,-CONH_(2),-CHO`
`-CHO,-COOH,-SO_(3)H,-CONH_(2)`

Answer :B
15.

The correctdecreasingorder of firstionizationenthalpiesof fiveelementsof thesecondperiodis

Answer»

`Be gtB gt C gt N gt F `
`N gt F gt C gt B gt Be `
`F gt N gt C gt Be gt B`
`N gt F gt B gt C gt Be`

Solution :Aswe MOVE fromleftto rightacrossa PERIOD`Delta_(i) H_(1)` generallyincreases.ThusF has THEHIGHEST`Delta_(i)H_(1)` .But asdiscussedin Ans. 55 above in spiteof highernuclear charge of B,`Delta_(i) H_(1)` of Beis higherthan thatof B. Combiningthesetwo TRENDS, the`Delta_(i)H_(1)` of thefiveelementsdecreasesin theorder :
`F gt N gt C gt Be gt B` i.e.,option( c) is correct.
16.

The correct decreasing order of ease of acid catalysed estrification is :

Answer»

`CH_(3)[CH_(2)]_(3)OH gt CH_(3)CH_(2)C(OH)CH_(3) gt (CH)_(3)COH`
`CH_(3)C(OH)C_(2)H_(5) gt CH_(3)[CH_(2)]_(3)OH gt (CH_(3))_(3)COH`
`(CH_(3))_(3)COH gt CH_(3)CH_(2)C(OH)CH_(3) gt CH_(3)[CH_(2)]_(3)OH`
None of these is correct

Solution :ESTERIFICATION of alcohols INVOLVE the cleavage of `RO-H` BOND. The roder of reactivity in this reaction is `1^(@) gt 2^(@) gt 3^(@)`.
17.

The correct decreasing order for the function groups of organic compounds in the IUPAC system of nomenclature is

Answer»

`-CONH_(2), -CHO, -SO_(3)H, -COOH`
`-COOH, -SO_(3)H, -CONH_(2), -CHO`
`-SO_(3)H, -COOH, -CONH_(2), -CHO`
`-CHO, -COOH, -SO_(3)H, -CONH_(2)`.

SOLUTION :is the CORRECT sequence.
18.

The correct decreasing order for acid strength is :

Answer»

`CHCH_(2)COOH gt O_(2)NCH_(2)COOH lt FCH_(2)COOH lt ClCH_(2)COOH`
`FCH_(2)COOH gt N"CC"H gt NO_(2)CH_(2)COOH gt ClCH_(2)COOH`
`NO_(2)CH_(2)COOH gt N"CC"H_(2)COOH gt FCH_(2)COOH gt ClCH_(2)COOH`
`NO_(2)CH_(2)COOH gtFCH_(2)COOH gt CNCH_(2)COOH gt ClCH_(2)COOH`

ANSWER :C
19.

The correct bond order in the following specie is :

Answer»

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

Solution :`O_(2)^(-) lt O_(2)^(+) lt O_(2)^(2+)`
ACCORDING to molecular ORBITAL theory (MOT)
`"" O_(2)^(-) O_(2)^(+) O_(2)^(2+)`
NUMBER of `E^(-)` : 1715 14
Bond order :1.5 2.53.0
20.

The correct arrangenment of O, P and N in order of increasing radii is

Answer»

<P>O < N < P
P LT O lt N
O lt P lt N
N lt O lt P

ANSWER :A
21.

Thecorrectarrangementfor theionsin theincreasingorder oftheirradii is

Answer»

`Na^(+), C1^(-) , Ca^(2+) `
`Ca^(2+), K^(+), S^(2-)`
`Na^(+), A1^(3+),Be^(2+)`
`C1^(-) ,F^(-) , S^(2-)`

Solution :forisoelectronicions , FOLLOWINGTHE sameargumentas DISCUSSED in`Ca^(2+), K^(+) , S^(2-)` i.e.,OPTION( b) iscorrect.
22.

The correct arrangement for decreasing order of electrophilic Substitution reactions of I)Toluene II) Benzene III) Phenol IV) Chlorobenzene

Answer»

`I gt II gt III gt IV`
`IV gt I gt II gt III`
`III gt I gt II gt IV`
`II gt IV gt III gt I`

Answer :C
23.

The correct acidity order of the following is

Answer»

`(III) gt (IV) gt (II) gt (I)`
`(IV) gt (III) gt (I) gt (II)`
`(III) gt (II) gt(I) gt (IV)`
`(II) gt (III) gt (IV) gt (I)`

SOLUTION :A carboxylic acid is stronger acid then phenol, hence both III and IV are stronger ACIDS than both I and II. Also IV has a methyl group that gives electron donating inductive effect and decreases the acid strength. Therefore, III is stronger acid than IV.Between I and II, the DOMINANT electron withdrawing inductive effect of chlorine increases acid strength of phenol slightly, hence II is stronger acid than I. THUS, the overall order is : (A) `III gt IV gt II gt I`.
24.

The correct acidic order of the following is

Answer»

`I GT II gt III`
`III gt I gt II`
`II gt III gt I`
`I gt III gt II`

SOLUTION :p-Nitrophenol (III) `gt` PHENOL (I) `gt` p-Cresol (II)
25.

The coordination number of each sphere in hexagonal close packing is ______ while that of body-centred cubic packing is _______

Answer»


ANSWER :12,8
26.

The coordination number of CI^- ion in NaCL structure is ______________whereas that in CsCl structure _____________

Answer»


ANSWER :6,8
27.

The coordination number of cation and anion in Fluorite CaF_(2) and Zinc blende ZnS are respectively x : y and a : b. Find (x + y + a + b)

Answer»


ANSWER :`20 = (8 + 4 + 4 + 4 )`
28.

The coordination number of Al in the crystalline state of AlCl_3 is

Answer»


Solution :At low temperature , `AlCl_3` EXISTS as a close packed lattice of `Cl^-` ions with `AL^(3+)` ions occupying octahedral voids. Thus, `Al^(3+)` ions `Cl^-` ions AROUND it . Hence, its coordination number is 6.
29.

The coordination number of a tetrahedral void is _________ while that of an octahedral void is _________

Answer»


ANSWER :4,6
30.

The coordination number of Alin the crystalline ofAlCl_(3) is

Answer»


SOLUTION :At low temperature . `AlCl_(3)`EXISTS as a CLOSE paked lattice of ` Cl^(-)` ions with ` Al^(3+)`ions occupying octahedral VOIDS. Thus, ` Al^(3+)`ions has 6 `Cl^(-)` ions AROUND it. Hence, its corrdination number is 6.
31.

The coordinate of the three corners of a shaded face on a cubic unit cell are(1/2 , 1/2 ,1) , ( 0,1, 1/2) and ( 1,1, 1/2) as shown in the figure. Determine the Miller indices of the plane.

Answer»

SOLUTION :The intercepts of the shaded plane with the AXES are ` oo, ` 2b and 2 C respecitvely , WEISS indices are ` oo`2,2,.
Reciprocals of We3iss indices ar ` 1/oo, 1/2 , 1/2 or 0, 1/2 ,1/2` or expressed.
32.

The cooling caused by the expansion of a compressed gas below its inversion temperature without doing external work is called

Answer»

JOULE Thomson effect 
Aciabatic DEMAGNETISATION 
Tyndall effect 
Compton effect 

SOLUTION :Cooling by expansion `implies J- T ` effect .
33.

The conversion KMnO_(4)rarrK_(2)MnO_(4) is an example of

Answer»

OXIDATION HALF reaction
REDUCTION half reaction
oxidation and reduction
nither oxidation nor reduction

ANSWER :B
34.

The conversion of fresh precipitate to colloidal state is called peptization. (r) It is caused by addition of common ions.

Answer»

IF both (A) and (R) are correct and (r) is the correct EXPLANATION for (a).
If both (a) and (r) are correct but (r) is not the correct explanation for (a).
IF (a) is correct but (r) is incorrect.
If (a) is incorrect but (r) is correct.

ANSWER :B
35.

The conversion of ethylene to ethyne involves

Answer»

`Br_(2)` treatment FOLLOWED by AQUEOUS KOH reaction <BR>ALCOHOLIC KOH treatment followed by `Br_(2)` reaction
`Br_(2)` treatment followed by alcoholic KOH reaction
Oxidation followed by reduction

Answer :C
36.

The conversion of ethylene to ethyl alcohol is known as

Answer»

a)ELIMINATION
B)Substitution
c)Hydrogenation
d)Hydration

Answer :D
37.

The conversion of -COOH group to -NH_(2) group can be made by :

Answer»

SANDMEYER's REACTION
Claisen condensation
Stephens reduction
Schmidt reaction

Solution :SEE reaction of ACIDS
38.

The conversion of chlorobenzene to phenol by the action of NaOH is called ……………….. .

Answer»

SOLUTION :Dow.s PROCESS
39.

The conversion of CH_(3)OH into CH_(3)COOH can be brought about by the following reagents

Answer»

`K_(2)Cr_(2)O_(7)//H^(+)`
`CO + RH`
`KMnO_(4)`
`H_(3)PO_(4)`

Solution :`CH_(3)OH + CO overset("Rh catalyst")rarr CH_(3)COOH`
40.

The conversion of bromoalkane to fluroalkane by heating with AgF is called ………………….. .

Answer»

SOLUTION :SWARTZ REACTION
41.

The conversion of benzene diazonium chloride to chlorobenzene in the presence of Cu_2Cl_2 + HCl is named as ………………….. .

Answer»

SOLUTION :SANDMEYER REACTION
42.

The conversion of atomic hydrogen to dihydrogen is a ______________ change.

Answer»

endothermic
exothermic
photochemical
nuclear

Answer :B
43.

The conversion of aldehyde having no alpha hydrogen to a mixture of carbonylc acid and primary alcohol is known as cannizzaro reaction. The most important features of this reaction is the conjugate base of hydrate of aldehye.Which of the following cannot undergo intramolecular cannizzaro reaction ?

Answer»

`H-underset(O)underset(||)(C)-underset(O)underset(||)(C)-H`
`H-underset(O)underset(||)(C)-underset(O)underset(||)(C)-Ph`
`Ph-underset(O)underset(||)(C)-underset(O)underset(||)(C)-Ph`
All

Solution :STEP II is rate determing step.
Iosn with more electron DENSITY arebetter hybride donors.
In`PH-underset(O)underset(||)(C)-underset(O)underset(||)(C)-Ph` ALDEHYDIC - Hisabsent
44.

The conversion of aldehyde having no alpha hydrogen to a mixture of carbonylc acid and primary alcohol is known as cannizzaro reaction. The most important features of this reaction is the conjugate base of hydrate of aldehye. Write order of best hydride ion donar in cannizzaro reaction

Answer»

III gt IIgt IV gt I
II gt IV gtIII gt I
IIgt III gt I gt IV
III gt I gt II gt IV

Solution :Step II is rate determing step.
Iosn with more ELECTRON density arebetter HYBRIDE donors.
In`PH-underset(O)underset(||)(C)-underset(O)underset(||)(C)-Ph` ALDEHYDIC - Hisabsent
45.

The conversion of aldehyde having no alpha hydrogen to a mixture of carbonylc acid and primary alcohol is known as cannizzaro reaction. The most important features of this reaction is the conjugate base of hydrate of aldehye.Order of the above reaction is

Answer»

1
2
3
4

Solution :Step II is rate determing step.
Iosn with more electron DENSITY arebetter hybride donors.
In`PH-UNDERSET(O)underset(||)(C)-underset(O)underset(||)(C)-Ph` Aldehydic - Hisabsent
46.

The conversion of a keto form into enol is enolisation. It depends on structural factor, temperature and nature of solvent. Resonance and hydrogen bonding increases enol content. Enolic form of phenol is more stable than keto form by-13 kcal/mol of energy hence phenol exist exclusively as an enol. Enolic tautomer is less polar due to intramolecular hydrogen bonds than the corresponding keto form. Any polar solvent would decrease the enolisation and fovour keto form intramolecular hydrogen bonding stabilized enol form by 7kcal/mol and resonance stabilizes enol form by 15 kcal/mole The enol content of ethyl acetoacetate is highest in

Answer»

WATER
acetic acid
ethanol
hexane

Solution :Water FACILITATE HYDROGEN BONDING.
47.

The conversion of a keto form into enol is enolisation. It depends on structural factor, temperature and nature of solvent. Resonance and hydrogen bonding increases enol content. Enolic form of phenol is more stable than keto form by-13 kcal/mol of energy hence phenol exist exclusively as an enol. Enolic tautomer is less polar due to intramolecular hydrogen bonds than the corresponding keto form. Any polar solvent would decrease the enolisation and fovour keto form intramolecular hydrogen bonding stabilized enol form by 7kcal/mol and resonance stabilizes enol form by 15 kcal/mole Ketonic form predominates in

Answer»

`CH_(3)-CO-CH_(3)`
`C_(6)H_(5)COCH_(2)COCH_(3)`
`CH_(3)-CO-CH_(2)-COCH_(3)`
`CH_(3)COCH_(2)COOC_(2)H_(5)`

SOLUTION :Compound A has less percentage of enol.
48.

The conversion can be performed suitable by

Answer»

I) `BH_3 - THF` II) `H_2O_2 , OH^(-)`
I) HBR PEROXIDE II) `OH^(-)`
I) `Cl_2,H_2O` II) `LiAlH_4`
I) HBr II) `OH^(-)`

SOLUTION :a) I. `BH_3 - THF`II. `H_2O_2 , OH^(-)`
B) I. Hbr peroxideII. `OH^(-)`
49.

Tin contains only 5 - 10 % of tin as SnO_(2), the rest being Siliceous matter, wolfram and pyrites of iron copper and arsenic. It is not attached by water. Only molten tin reacts with steam liberating hydrogen. In stannous oxide, tin is +2 oxidising state. So stannous oxide is also described as tin (II) oxide Which of the following ionic species is more stable

Answer»

`Sn^(+2)`
`Sn^(+4)`
`Ge^(+2)`
`Pb^(+4)`

SOLUTION :Down the group lowest oxidation state is more STABLE. Due to inert pair EFFECT, `Sn^(+2) gt Sn^(+4)`
50.

Tin contains only 5 - 10 % of tin as SnO_(2), the rest being Siliceous matter, wolfram and pyrites of iron copper and arsenic. It is not attached by water. Only molten tin reacts with steam liberating hydrogen. In stannous oxide, tin is +2 oxidising state. So stannous oxide is also described as tin (II) oxide Tin is attacked by

Answer»

Hot KOH
Hot HCl
Conc. `HNO_(3)`
All of these

Solution :`SN+ 2KOH +H_(2)O RARR K_(2) SO_(3) + 2H_(2)`
`Sn+ 2HCl rarr SnCl_(2) +H_(2)`
`4Sn+ 4HNO_(3) rarr H_(2)SnO_(3) + 4NO_(2) + H_(2)O`