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.

In the reactions AlCl_3 +Cl to AlCl_4 ^(-)Cl^(-)acts as

Answer»

Bronsted acid
Bronsted BASE
LEWIS base
Lewis acid

Solution :`Cl^(-) ` DONATES lone PAIR
Discussion
2.

In the reaction X_((g))+ Y_((g)) hArr 2Z_((g)),2 moles of X, I mole of Y and I mole of Z are placed in a 10 lit vessel and allowed to reach equilibrium. If final concentration of Z is 0.2 M, then K_(c). for the given reaction is

Answer»

1.6
`(80)/(3)`
`(16)/(3)`
`(3)/(16)`

SOLUTION :
but `(Z) = 1+alpha=0.2, alpha=0.8`
`K_(C)=(((1+alpha)/(10))^(2))/(((2-2)/(10))((1-alpha)/(10)))=(16)/(3)`
Discussion
3.

In the reaction with HCl, an alkene reacts in accordance with the Markovnikov's rule, to give a product 1-chloro-1- methylcyclohexane. The possible alkene is:

Answer»



(A) and (B)

SOLUTION :
Discussion
4.

In the reaction with HCl , an alkene reacts in accordance with the Markovnikov's rule to give a product 1-chloro-1-methylcyclohexane. The possible alkene is

Answer»



(A) and (B)

SOLUTION :
Discussion
5.

In the reaction with HCl, an alkene reacts in accordance with Markovnikov's rule to give a product 1-chloro-1-methylcyclohexane. The possible alkene is :

Answer»




(A) and (B)

SOLUTION :
Discussion
6.

In the reaction the stoichiometry coefficients of Cr_(2)O_(7)^(2-), NO_(2)^(-) and H^(+) respectively are Cr_(2)O_(7)^(2-)+NO_(2)^(-)+H^(+)rarrCr^(3+)+NO_(3)^(-)+H_(2)O

Answer»

1,3 and 8
1,4 and 8
1,3 and 12
1,5 and 12

Answer :A
Discussion
7.

In the reaction the stoichiometry coefficients of Cr_(2)O_(7)^(2-) , NO_2^(-) and H^(+) respectively are Cr_(2)O_(7)^(2-) +NO_(2)^(-) +H^(+) rarr Cr^(3+) +NO_(3)^(-) +H_(2)O

Answer»

1,3,8
1,4,8
1,3,12
1,5,12

Solution :`Cr_(2)O_(7)^(-2) +3NO_2^(-) +8H^(+) rarr 2Cr^(3+) +3NO_(3)^(-)+4H_(2)O`
Discussion
8.

In the reaction The product X is.

Answer»




ANSWER :D
Discussion
9.

In the reaction, the product (C ) is: C_(6)H_(5)NH_(2)underset(0-5^(@)C)overset(NaNO_(2)+HCl)rarr (A) underset(KCN)overset(CuCN)rarr (B) overset(H^(+)//H_(2)O)rarr(C )

Answer»

`C_(6)H_(5)CH_(2)CH_(2)NH_(2)`
`C_(6)H_(5)COOH`
`C_(6)H_(5)OH`
None of these

Solution :N//A
Discussion
10.

In the reaction SO_(2)+1/2 O_(2) overset(V_(2)O_(5))to SO_(3) the change in the oxidation state of vanadium is form V^(5+) to

Answer»

`V^(2+)`
`V^(4+)`
`V^(3+)`
`V^(6+)`

ANSWER :B
Discussion
11.

In the reaction SnO_2 + nC rarr Sn + nCOthen the value of n is _____

Answer»


Solution :`SnO_(2)+nC rarr SN +nCO`
Balance O by multiplying R.H.S.
CO by 2 which will balance the number of C atoms on boths ides also. HENCE n = 2
Discussion
12.

In the reaction sequence, What is the relationship between A & B.

Answer»


ANSWER :POSITIONAL ISOMERS
Discussion
13.

In the reaction sequence, CH_(2)=CH_(2)underset("acid")overset("hypochlorous")rarrA overset(R)rarrunderset(CH_(2)OH)overset(CH_(2)OH)("|") A and R are respectively

Answer»


`CH_(3)CH_(2)Cl` and `NAOH`
`CH_(3)CH_(2)OH` and `H_(2)SO_(4)`
`CH_(2)Cl.CH_(2)OH` and `NaHCO_(3)`

Solution :`underset(CH_(2))OVERSET(CH_(2))("|| ") overset(HOCl)rarrunderset(CH_(2)Cl)overset(CH_(2)Cl)("|") overset(aq.NaHCO_(3))rarr underset(CH_(2)OH)overset(CH_(2)OH)("|")`
Discussion
14.

In the reaction : S + (3)/(2)O_(2)rarr SO_(3)+ 2 x kJand SO_(2)+(1)/(2)O_(2) rarr SO_(3)+y kJheat of formation of SO_(2) is

Answer»

`x-y`
`2x+y`
`x+y`
`y-2x`

Solution :Given `: S+(3)/(2) O_(2)rarr SO_(3), DELTAH = -2 x KJ `...(i)
`SO_(2)+(1)/(2) O_(2) rarr SO_(3) , DeltaH = - y kJ `.....(ii)
Aim `: S + O_(2) rarr SO_(2), DeltaH =?`
EQN. (i) - Eqn. (ii) gives
`S+ O_(2) -SO_(2) rarr 0,DeltaH = - 2 x - ( - y) = y - 2x`
Discussion
15.

In the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) , the equilibrium is established by to take first Cl_2 then decomposition of PCl_5 will be increase or decrease ?

Answer»

SOLUTION :DECREASE
Discussion
16.

In the reaction PCl_(5)(g)hArrPCi_(3)(g)+Ci_(2)(g) a graph in plotted to show the variation of rate of forward and backward reactions against time. Which of the following is correc? QgtK Q=K QltK

Answer»

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

SOLUTION :N//A
Discussion
17.

In the reaction P_(4)+3OH^(-)+3H_(2)Orarr 3H_(2)PO_(2)^(-)+PH_(3) phosphorus is undergoing

Answer»

oxidation
reduction
Disproportionation
hydrolysis

Answer :C
Discussion
18.

In the reactionoverset(NaOH//200^(@) C)tooverset(H^(oplus)//H_(2)O)tothe major product is :

Answer»




SOLUTION :INTRAMOLECULAR Cannizzarro 's REACTION
Discussion
19.

In the reaction of sodium thiosulphate with I_(2)in aqueous medium the equivalent weight of sodium thiosulphate is equal to

Answer»

Molar MASS of sodium thiosulphate
The average of molar masses of `Na_(2)S_(2)O_(3)` and `I_(2)`
Half the molar mass of sodium thiosulphate
Molar mass of SODIU thiosulphate `xx 2`

Solution :`2Na_(2)S_(2)O_(3)*5H_(2)O + I_(2) rarr 2NaI + Na_(2) S_(4) O_(6)` or
`:.2S_(2)O_(3)^(3-) rarr S_(4)O_(6)^(2-) +2e^(-)`
Equivalent weight of `Na_(2)S_(2)O_(3) . 5H_(2)O`
`= ("Mol. Wt.")/(1) = (248)/(1) = 248`
Discussion
20.

In the reaction of sodium thiosulphate with 12 in aqueous medium the equivalent weight of sodium thiosulphatc is equal to

Answer»

MOLAR mass of sodium this sulphate
the average molar masses of `Na_(2)S_(2)O_(3) and I_(2)`
HALF the molar mass of sodium this sulphate
twice of molar mass of sodium this sulphate

Answer :C
Discussion
21.

In the reaction of phenol with CHCl_(3) and aqueous NaOH at 70^(@)C (343K), the electrophile attacking the ring is :

Answer»

`CHCl_(3)`
`CHCl_(2)`
`:C Cl_(2)`
`COCl_(2)`

SOLUTION :See mechanism of Reimer Tiemann reaction.
Discussion
22.

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO_(2) is ___

Answer»

10
1
5
2

Solution :`2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to2Mn^(2+)+10CO_(2)+8H_(2)O`
`therefore5C_(2)O_(4)^(-2)to10CO_(2)+10bare`
Total 10 ELECTRONS for 10 `CO_(2)` molecules so 1 ELECTRON per `CO_(2)` MOLECULE.
Discussion
23.

In the reaction of KMnO_(4) with ferrous ion in an acidic medium KMnO_(4) oxidised ferrous ion to ferric ion and itself gets reduced to manganous salt.The amountof ferrous ions oxidised 100 mL of 0.2 (N) KMnO_(4) solution

Answer»

1.117 G
1.562 g
2.173 g
1.934 g

ANSWER :A
Discussion
24.

In the reaction of chlorine with propene at 450^@C, the major product is

Answer»




ANSWER :C
Discussion
25.

In the reaction of chlorine with dry slaked lime, the oxidation number of chlorine changes i) from -1 to +1 ii) from + 1 to -1 iii) from zero to -1 iv) from zero to +1 The correct combination is

Answer»

Only II and III are correct
Only ii and IV are correct
Only i and ii are correct
All are correct

Answer :B
Discussion
26.

In the reaction NO_(2)^(-) +OCl^(-) rarr NO_(3)^(-) +Cl^(-) the oxidate state of chlorine

Answer»

Does not CHANGE
CHANGES from `+1` to `-1`
Changes from `-2` to `-1`
Changes from 0 to `-1`

Solution :
Discussion
27.

In the reaction NH_2+H_2O to NH_(4)^(+)+Ooverline(H) NH_(3) is acidic in nature. The reason for its acidity is ________.

Answer»

Acceptance of ONE `H^(+)` from WATER
Release of one `OH^(-)` ion
DUE to the NITROGEN ion
All the above.

ANSWER :A::C
Discussion
28.

In the reaction, NaOH + H_(2)O + NaOH + H_2

Answer»

`H^(-)`is oxidised
`Na^(+)`is REDUCED
both NaH and `H_2O`are reduced
NONE of the above

Solution :`Naoverset(-1)H+H_2OrarrNaOH+overset(0)(H_2),H^(-)` is oxidised .
Discussion
29.

In the reaction NaOH+Al(OH)_(3)rarrNaAlO_(2)+H_(2)O,"the "("M.wt")/("Eq.wt")" of "Al(OH)_(3) is________

Answer»


Solution :`Al(OH)_(3)` acts as ACID and ACCEPTS 1 NaOH IMPLIES `n-f=1`
Discussion
30.

In the reaction N_(2(g)) + O_(2(g)) hArr 2NO_((g)) , Delta H=+ 180 kJ On increasing the temperature the production ofNO

Answer»

Increases
Decreases
Remains constant
Cannot be predicted

Solution :Le-chatelier.s PRINCIPLE
Discussion
31.

In the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the value of the eqilibrium constant depends on .........

Answer»

SOLUTION :the TEMPERATURE
Discussion
32.

In the reaction N_2 + 3H_2 iff 2NH_3 + x kCal, one mole of N_2 reacts with 3 moles of H_2 at equilibrium.. Then the valueofalpha( degree of dissociation ) is approximately ………. . P is the pressure at equilibrium

Answer»

`(p(sqrt(27K_p)))/(8) `
`(8P)/(K_p sqrt(27))`
`(p sqrt(27))/(8K_p)`
`(n)/(V)`

Solution :`(p(sqrt(27K_p)))/(8) `
`{:( , N_(2) , + , 3 H_(2) hArr 2 NH_(3)) , ("Initial moles" , 1 ,, 3 ""0), ("Moles of EQUILIBRIUM" , (1- alpha) ,, (3 - 3 alpha) "" 2alpha):}`
`K_(P) = ((P_(NH_(3)))^(2))/((P_(N_(2))) (P_(H_(2)))^(3)) = (((2X)/(4 - 2alpha))^(2)p^(2))/([(( 1- alpha)/(4 - alpha))^(P)][(3(1 - a)P)/(4 - alpha)]^(3)) = (4 alpha^(2) p^(2))/((4 - 2 alpha)^(2)) xx ((4 - alpha)^(4))/((1 - alpha)^(4) P^(4) 27)`
`K_(P) = (4 alpha^(2) (4 - 2 alpha)^(2))/(27 (1- alpha)^(4) P^(2)) = (16 alpha^(2) (2 - alpha)^(2))/(27 P^(2) (1 - alpha)^(4)) = (16 alpha^(2) (2)^(2))/(27 P^(2) (1)^(4)) { because alpha lt lt 1}`
`implies sqrt((27 K_(P)^(2))/(64))= alpha implies alpha = (P sqrt(27 K_(P)))/(8)`
Discussion
33.

In the reaction MnO_(4)^(-)+SO_(3)^(2-)+H^(+)toMn^(2+)+SO_(4)^(2-) the number of H^(+) ions involved is

Answer»

2
6
8
16

Solution :`2MnO_(4)^(-) + 5 SO_(3)^(2-) +6H^(+) rarr 2MN^(2+) + 5SO_(4)^(2-) +3H_(2)O`
Discussion
34.

In the reaction MnO_(2)+4 HCIrarr MnCI_(2) +CI_(2)+2H_(2)O which species is oxidised

Answer»

SOLUTION :HCI is oxidised to `CI_(2)` since the O.N since the O.N of CI INCREASES from -1 in HCI to 0 in `CI_(2)`
Discussion
35.

In the reaction M + O_(2)rarrMO_(2) (superoxide) the metal M is

Answer»

Li
Na
K
Ba.

Solution :Formation of oxides, peroxides and superoxides depend on the size of ALKALI METAL. An atom with LARGER size produces SUPER oxide. Li forms oxide, Na forms peroxide whereasK forms superoxide.
Discussion
36.

In the reaction, KMnO_(4)+Na_(2)S_(2)O_(3)+H_(2)OtoMnO_(2)+SO_(4)^(2-)+OH^(-)- (assume formula masses of KMnO_(4) and Na_(2)S_(2)O_(2)are M_(1)and M_(2) respectively) -

Answer»

the EQUIVALENT mass of `KMnO_(4)=M_(1)//(3)`
the equivalent mass of `Na_(2)S_(2)O_(3)=M_(2)`
the equivalent mass of `KMnO_(4)=M_(1)//S)`
the equivalent mass of `Na_(2)S_(2)O_(3)=M_(2)//B`

Answer :A::D
Discussion
37.

In the reaction I_(2)+I^(-) rarr I_(3)^(-), which is the Lewis base ?

Answer»

`I_(2)`
`I^(-)`
`I_(3)^(-)`
None of these

Solution :In the REACTION `I_(2)+I^(-1)toI_(3)^(-),I^(-)` is the eletron pair donor and HENCE is a Lewis base.
Discussion
38.

In the reaction, I_(2)+2S_(2)O_(3)^(2-) to 2I^(-)+S_(4)O_(6)^(2-) equivalent weight will be equal to :

Answer»

`4//6` of MOLECULAR WEIGHT
molecular weight
`2//9` of molecular weight
twice the molecular weight

SOLUTION :N//A
Discussion
39.

In the reaction, I_(2)+2S_(2)O_(3)^(2-) to 2I^(-)+S_(4)O_(6)^(2-) equivalent mass of iodine is :

Answer»

EQUAL to its MOLECULAR MASS
`1//2` the molecular mass
`1//4` the molecular mass
twice the molecular mass

Solution :N//A
Discussion
40.

In the reaction, I_(2)+2KClO_(3)rarr2KIO_(3)+Cl_(2) i) Iodine is oxidised ii) Chlorine is reduced iii) Iodine displaces chlorine iv) KCIOP_(3) is decomposed The correct combination is

Answer»

Only i & iv are CORRECT
Only iii & iv are correct
i,II,iii are correct
All are correct

Solution :1) In the above reaction `I_(2)^(0) rarr I^(+5) implies `oxidation
2) `implies CI^(+5) rarrCl_(2)^(0) implies`reduction
3) LOWER halogen displaces higher halogen .
Discussion
41.

In the reaction, H_(2)S+H_(2)O_(2) to S+2H_(2)O:

Answer»

`H_2S` is an ACID and `H_2O_2` is a BASE
`H_2S`is base and `H_2O_2` is an acid
`H_2S` is an oxidising AGENT and `H_2O_2`is a reducing agent
`H_2S` is a reducing agent and `H_2O_2`is an oxidising agent

Solution :`H_2S` is a reducing agent, `H_2O_2` is a an oxidising agent
Discussion
42.

In the reaction, H_(2)O_(2)+PbS to H_(2)O+PbSO_(4) the increase in the oxidation number of sulphur is

Answer»

2 units
4 units
6 units
8 units

Answer :D
Discussion
43.

In the reaction :- H_2C = CH - CH = CH_2 underset(180^(@)C)overset(Br_2)toX 'X'will be

Answer»

`BR-CH_2 - CH = CH - CH_2-Br`
`UNDERSET(Br)underset(|)CH_2 - underset(Br)underset(|)CH-CH = CH_2`
`underset(Br)underset(|)CH_2 - HC= Cunderset(Br)underset(|)(-)CH_3`
`CH_3-C overset(Br)overset(|)C=C overset(Br)overset(|)(-)CH_3`

SOLUTION :at HIGH temperature `(180^@)` BROMINE gives 1,4 product
Discussion
44.

In the reaction H- C -= CH underset((ii) CH_(3)CH_(2)Br)overset((i)NaNH_(2)//liq. NH_(3))to X underset((ii) CH_(3)CH_(2)Br)overset((i)NaNH_(2)//liq. NH_(3))to Y

Answer»

X = but 2-ync, Y = HEX-2-yne
X = but-1-yne, Y = hex-2-yne
X = but-1-yne, Y = hex-3-yne
X = but-2-yne, Y = hex 3-yne.

SOLUTION :
Discussion
45.

In the reaction H-C-=CH underset((ii) CH_(3)CH_(2)Br)overset((I) NaNH_(2)//liq. NH_(3))rarr X underset((ii) CH_(3)CH_(2)Br)overset((I) NaNH_(2)//liq.NH_(3))rarr Y X and Y are .

Answer»

X=2-Butyne , Y = 3-Hexyne
X=2-Butyne , Y = 2-Hexyne
X = 1-Butyne, Y = 2-Hexyne
X=1-Butyne, Y = 3-Hexyne

Solution :
Discussion
46.

In the reaction given Mg(s)+NO_(3)^(ɵ)+H_(2)OtoMg(OH)_(2)(s)+overset(ɵ)(O)HO(aq.)+NH_(3)(g) 20 " mL of " sample of NO_(3)^(ɵ) solution is treated with Mg. The NH_(3)^(g) was passed into 50 " mL of " 0.1 M HCl. The excess HCl requried 30 " mL of " 0.1 M KOH for its neutralisation calculate the molarity of NO_(3)^(ɵ) ions in the original sample?

Answer»

Solution :Balance the EQUATION in basic medium
`cancel(8e^(-))+6H_(2)O+NO_(3)^(ɵ)toNH_(3)+9overset(ɵ)(O)H`
`underline(2(ɵ)(O)H+"Mg"to overset(+2)(M)g(OH)_(2)+cancel(2e^(-))]xx4)`
`underline(6H_(2)O+NO_(3)^(ɵ)+4"Mg"toNH_(3)+4Mg+4Mg(OH)_(2)+overset(ɵ)(O)H)`
`thereforem" Eq of "NH_(3) "formed"=m" Eq of "HCl" used for "NH_(3)`
`=(50xx0.1xx1)-(30xx0.1xx1)`
`=2 mEq`
Thus, m" Eq of "`NH_(3)` for valence factor of `8=8xx2=16`
Also, mEw of `NO_(3)^(ɵ)=m" Eq of "NH_(3)`
`=8xx2=16mEq`
`N_(NO_(3)^(ɵ))xxV_(mL)=16`
`therefore N_(NO_(3)^(ɵ))xx20=16`
`N_(NO_(3)^(ɵ))=(16)/(20)=0.8M`
`M_(NO_(3)^(ɵ))=(0.8)M`
`M_(NO_(3)^(ɵ))=(0.8)/(8)=0.1M`
Discussion
47.

In the reaction, H-C-=C-H underset((ii)CH_3CH_2Br)overset((i)NaNH_2//liq. NH_3)to X underset((ii)CH_3CH_2Br) overset((i)NaNH_2//liq. NH_3)to Y X and Y are

Answer»

X=2-butyne, Y=2-hexyne
X=1-butyne, Y=2-hexyne
X=1-butyne, Y=3-hexyne
X=2-butyne, Y=3-hexyne

Solution :`H-C-=C-H underset(-NH_3)OVERSET(NaNH_3//LIQ. NH_3) to H-C-=C^(-)NA^+ underset(-NaBr)overset(CH_3CH_2Br)to underset"1-Butyne (X)"(H-C-=C-CH_2CH_3) underset(-NH_3)overset(NaNH_2 // liq. NH_3)to Na^+ .^-C-=C-CH_2CH_3 underset(-NaBr)overset(CH_3CH_2Br)to underset"3-Hexyne"(CH_3CH_2-C-=C-CH_2CH_3)`
Discussion
48.

In the reaction given below identify the species undergoing oxidation and reactioin : (i) CH_(4)(g)+2O_(2)(g)rarr CO_(2)(g)+2H_(2)O(i) (ii)H_(2)S(g)+O_(2)(g)rarr2s(s)+2H_(2)O(l) (iii)CH_(2)=CH_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g) (iv)HgO(s)rarr2Hgl)+O_(2)(g) (v)Mg(s)+S(s)rarr MgS(s)

Answer»

Solution :(i) `CH_(4)` is oxidised to `cO_(2)` while `O_(2)` is reduced to `H_(2)O` (ii) `H_(2)S` us ixudused TI S wile `O_(2)` is reduced to `H_(2)O` (III) `CH_(2)=cH_(2)` is rduced to`CH_(3) rarr CH_(3)` while `H_(2)` is oxidised to `CH_(3)-CH_(3)`
(iv) `HG^(2+)` is reduced to Hg whle `O^(2-)` is oxidised to `O_(2)` (v)Mgis oxiised to `Mg^(2+)` while S is reduced to `S^(2-)`
Discussion
49.

In the reaction for dinitration The major product X is

Answer»




ANSWER :C
Discussion
50.

In the reaction Fe(OH)_(3)(s)hArrFe^(3+)(aq)+3OH^(-)(aq), if the concentration of OH^(-) ions is decreased by 1/4 times, then the equilibrium concentration of Fe^(3+) will

Answer»

not changed
also decreased by `1//4` time
increase by 4 TIMES
increase by 64 times

Solution :`K_(C)=([Fe^(3+)][OH^(-)]^(3))/([Fe(OH)_(3)(s)])=[Fe^(3+)][OH^(-)]^(3)]` [`because` Concentration of solids is CONSTANT) When concentration of `OH^(-)` IONS decreased by `1.4` times, then
`K_(C)=[Fe^(3+)]xx(([OH^(-)])/(4))^(3)=(1)/(64)[Fe^(3+)][OH^(-)]^(3)`
To MAINTAIN `K_(C)` as constant, comcentration of `Fe^(3+)` will increase by 64 times.
Discussion