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.

Identify the product in the following reaction CH_(3)-CH_(2)-CH_(2)Broverset(alc.KOH)underset(Delta)(to)underset(("major"))("Product")

Answer»

`CH_(3)CH_(2)OH_(2)`
`CH_(2)=CH_(2)`
`CH_(3)CH_(2)-CH_(2)-OH`
`CH_(3)CH=CH_(2)`

Answer :B::C::D
2.

Identify the product for the following reaction: CH-=CH+HOCl to

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`Cl_(2)CHCHO`
`CH(OH)=CHCl`
`ClCH_(2)CH_(2)OH`
`CH_(3)COCL`

Solution :
3.

Identify the product C in the series CH_(3)CN overset(Na// C_(2)H_(5)OH)rarr A overset(HNO_(2))rarr B overset(Cr_(2)O_(7)^(2-) //H^(+))rarr

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`CH_(3)COOH`
`CH_(3)CH_(2)NHOH`
`CH_(3)CONH_(2)`
`CH_(3)CHO`

SOLUTION :`CH_(3)CN underset(C_(2)H_(5)OH)overset(Na)rarr CH_(3) CH_(2)NH_(2) overset(HNO_(2))rarr C_(2)H_(5) OH overset(Cr_(2) O_(7)^(2-) //H^(+))rarr CH_(3)COOH`
4.

Identify the product (B)

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Both (A) and (B) formed

SOLUTION :Where as REDUCES Both keto & ester GROUP into alcohols
5.

Identify the product and mention the type of organic reaction CH_3-overset(Br)overset(|)CH-CH_3 underset(C_2H_5ONa)overset(C_2H_5OH)to ?

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SOLUTION :
This is ONE of the ELIMINATION REACTION.
6.

Identify the possible structure of X and Y

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ANSWER :AD
7.

Identify the paramagnetic species among the followingusing molecular orbital theory.

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SOLUTION :`O_(2)^(2-) , N_(2), C_(2)^(+)`

`C_(2)^(+)` is PARAMAGNETIC.
8.

Identify the possible set of quantum numbers for outer shell electron in element K.

Answer»

`n=4,l=0,m=0 , s = PM 1/2`
`n=4,l=1,m=1 , s = pm 1/2`
`n=3,l=2,m=-1 , s = pm 1/2`
`n=3,l=0,m=0 , s = pm 1/2`

ANSWER :A
9.

Identify the pairs which are not of isotopes? (1) ""_(6)^(12) X,_(3)^(16) (2) ""_(17)^(35) X,_(17)^(37) (3) ""_(6)^(14) X,_(7)^(14) (4) ""_(4)^(8) X,_(5)^(8)Y

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`1&2`
`2&4`
`3&4`
`1&4`

ANSWER :C
10.

Identify the pairs which are not of isotopes

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`._(6)^(12)X ,._(6)^(13)X`
`._(17)^(35)X ,._(17)^(37)X`
`._(6)^(14)X ,._(7)^(14)X`
`._(4)^(8)X ,._(5)^(8)X`

Solution :(A) `._(6)^(12)X` and `._(6)^(13)Y` have same atomic number but different mass number. (B) `._(17)^(35)X` and `._(17)^(37)Y` have same atomic number but different mass number. So, these both pairs are isotopes to each other.(C) `._(6)^(14)X` and `._(7)^(14)Y` have different atomic number but same mass number. (D) `._(4)^(8)X` and `._(5)^(8)Y`have different atomic number but same mass number. So, these both pairs are isobars to each other.
11.

Identify the pairs which are not of isotopes ?

Answer»

`._(16)^(12)X, ._(6)^(13)Y`
`._(17)^(35)X, ._(17)^(37)Y`
`._(16)^(14)X, ._(7)^(14)Y`
`._(4)^(8)X, ._(5)^(8)Y`

Solution :(c) and (d) are not isotopes because in each, atom have different atomic number
12.

Identify the pair of species that possess same bond order .

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`O_(2)^(2-), B_(2)`
`NO, CO`
`N_(2) , O_(2)`
`He_(2), H_(2)`

ANSWER :a
13.

Identify the pair of isomers which are geometrical isomers of each.

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`CH_2 = CH = CH_2 & CH_3 - CH = CH_2`
`CH_3 - CH = CH - CH_3 & CH_2 = CH - CH_2CH_3`

SOLUTION :Arrangment of methyl GROUPS is DIFFERENT
14.

What is redox reactions ? Identity the substance oxidised and the substance reduced in the following reactions : MnO_(2)+4HCl to MnCl_(2)+2H_(2)+Cl_(2)

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Solution :
As it is clear, MANGANESE decrease its OXIDATION number from +4 to +2. HENCE, `MnO_2` gets reduced and it is an OXIDISING agent. Chlorine atom in HCl increases its oxidation number from -1 to 0. Thus, HCl gets oxidised and it is reducing agent.
15.

Identify the oxidising agent, reducing agent, substance oxidised and substance reduced in the following reactions. MnO_2 + 4HCl to MnCl_2 + Cl_2 + 2H_2O

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SOLUTION :
As it is CLEAR, manganese decrease its oxidation number from +4 to +2. Hence, `MnO_2` gets reduced and it is an oxidising agent. Chlorine atom in HCl increases its oxidation number from -1 to 0. Thus, HCl gets oxidised and it is REDUCING agent.
16.

Identify the oxidant and resultants in the following reactions: (a) CH_4(g)+4Cl_2(g)rarr"CCl"_4(g)+2HCl(g) (b) C_2H_4O_4(aq)+2H^(+)+MnO_2(s) rarrMn^(2+)(aq)+2CO_2(g) +2H_2O(l)(c) I_2(aq)+S_2O_3^(-2)(aq)rarr2I^(-)(aq)+S_4O_6^(2-)(aq) (d) Cl_2(g)+2Br^(-)(aq)rarr2Cl^(-)(aq)+Br_2(aq)

Answer»


ANSWER :Oxidants and RELUCTANT are RESPECTIVELY
(a) `Cl_2, CH_4` (B) `MnO_2, C_2H_4O_4` , (c) `I_2, S_2O_3^(2-)`
(d) `Cl_2,Br^(-)`.
17.

Identify the oxidant and the reductant in the following chemical reactions: 2I^(-)(aq)+Cl_2(g) to 2Cl^(-) (aq)+I_2(s)

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SOLUTION :`2I^(-)(AQ)+Cl_2(G) to 2Cl^(-) (aq)+I_2(s)`
OXIDANT is `Cl_2(g)` and the reductant is `I^(-)(aq).`
18.

Identify the oxidant and reduction in the following reactions. (a) Zn(s) +1/2 O_2(g)rarrZnO(s) (b) CH_4(g) + 4Cl_2(g) rarr "CCl"_4(g) + 4HCl(g) (c) I_2(aq) +2S_2O_3^(2-)(aq)rarr2I^(-) (aq)+S_4O_6^(2-)(aq) (d) Zn(s)+2H^(+)(aq) rarr Zn^(2+) +(aq) +H_2(g)

Answer»

Solution :(a) `Zn(s) +1/2O_2(G) rarr ZnO(s)`
In this reaction, ZINC donates electrons to O to give zinc ions and oxide ions. Thus, Zn acts as reluctant while oxygen acts as OXIDANT.
(b) `CH_4(g)+4Cl_2(g)rarr"CCl"_4 (g) + 4HCl(g) CH_4` is oxidised and acts as reluctant while `Cl_2` is reduced and acts as oxidant.
(c) `I_2(aq)+2S_2O_3^(2-)(aq) rarr 2I^(-)(aq) +S_4O_6^(2-)(aq)`
`I_2` gains electrons and is reduced . Therefore , it acts as oxidant. `S_2O_3^(2-)` acts as reluctant.
(d) `Zn(s)+2H^(+)(aq) rarr Zn^(2+) (aq) +H_2(g)`
Zinc loses electrons and gets oxidised . Therefore , it acts as reluctant . Hydrogen gets reduced and acts as an oxidant.
19.

Identify the oxidant and reductant in the folowing rection : (a) 10 H^(+)(aq)+4Zn(s)+NO_(3)^(-)(aq)rarr4Zn^(2+)(aq)+NH_(4)^(+)(aq)+3H_(2)O(l) (b)I_(2)(g)+H_(2)S(g)+H_(2)S(g)rarrHI(g)+S(s)

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Solution :(a) Writing the O.N of all the atoms above their symbols we have
`10 overset(+1)H^(+)(aq)+4overset(0)Zn(s)+overset(+5)BO_(3)^(-)(aq) to 4 overset(+2)Zn^(2+)(aq)+ overset(-3)NH_(4)^(+)(aq)+3overset(+1)H_(2)O(l)`
here there is no change in the O.N of H toms But hte O.N of Zn changes from zero in Zn to +2 in `Zn^(+)` therefore it is oxidised and hence Zn acts as a reductant
The O.N of N decresses FORM +5 in `NO_(3)^(-) to -3 in NH_(4)^(+)` and therefore it is reduced and hence `NO_(3)^(-)` acts as the oxidant
(b) Writing the O.N of all hte atoms above their symbols we have
`I_(2)^(0)(g) + H_(2) overset(-2)H_(2)(g) to 2 overset(+1) overset(-1)I(g)+overset(0)S(s)`
Here O.N of H does not change The O.N of `I_(2)` DECREASE frm zero in `I_(2)` to -1 in HI therefore `I_(2)` is REDUCING and hence it acts as an oxidant The O.N of S increases form -2 in `H_(2)S` to zero in S therefore `H_(2)S` is oxidation and hence it acts as th reductant
20.

Identify the oxidant and reductant in the following reactions: I_2(g)+H_2S(g)to2IH(g) +S(s).

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Solution :Writing the oxidation number of all the atoms above their symbols, we have
`overset0(I_2)(g)+overset(+1)(H_2)overset(-2)(S)(g)tooverset(+1)(2H)overset(-1)I(g) +overset(0)S(s)`
The oxidation number of `I_2` decreases from ZERO in `I_2` to -1 in HI, THEREFORE, `I_2` is reduced and HENCE it acts as an oxidant.
THe O.N of S INCREASE from -2 in `H_2S` to zero in S, therefore , `H_2 S` is oxidised and hence it acts as the REDUCTANT.
21.

Identify the oxidant and reductant in the following reactions: 10H^(+)(aq)+4Zn(s)+NO_3^(-) (aq)to4Zn^(2+) (aq)+NH_4^(+) +3H_2O(l)

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Solution :Writing the oxidation number of all the atoms above their symbols, we have
`OVERSET(+1) 10H^(+)(AQ)+overset0 4Z(s)+overset(+5)(NO_3^(-))(aq) to overset(+2)(4Zn^(2+))(aq)+overset(-3)(NH_4^(+))+overset(+1)(3H_2) overset(-2)O(L)`
Oxidation number of Zn changes from zero in Zn to +2 in `Zn^(2+)` and therefore, it is oxidized and hence Zn acts as a reductant.
THe oxidation number of N decrease from +5 in `NO_3^(-)` to -3 to `NO_4^(+)` and therefore, it is REDUCED and hence `NO_3^(-)` acts as the oxidant.
22.

Identify the oxidant and reductant in the following reaction: Zn(s)+2H^+(aq)toZn^(2+)(aq)+H_2(g)

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SOLUTION :Zn is the REDUCING agent (reductant) where as `H^+` is the oxidising agent (OXIDANT).
23.

Identify the organic compound X formed in the following reaction . CH_(3)COOAg + I_(2) rarr X + AgI + Y (the equation is unbalanced)

Answer»

Iodomethane
Ethyl iodide
METHYL ethanoate
None of these

Solution :`CH_(3)COOAg + I_(2) rarr CH_(3)I + AgI + CO_(2)`
`CH_(3)COOAg + CH_(3)I rarr CH_(3)COOCH_(3) + AgI`
(Methyl iodide is most reactive of all the alkyl HALIDES i.e., `CH_(3)CL, CH_(3)Br, CH_(3)I`)
`2CH_(3)COOAg + I_(2) rarr 2AgI + CO_(2) + underset("Methyl ethanoate")(CH_(3)COOCH_(3))`
This reaction is known as Birbaum -Simonini reaction
24.

Identify the oxidant and reductant in the following reaction: Zn(s)+1/2O_2(g) toZnO(s)

Answer»

SOLUTION :ZN is the reducing agent (REDUCTANT) and `O_2` is the oxidising agent (oxidant)
25.

Identify the one which does not come under the organic addition reaction.

Answer»

Hydration
Dehydration
Halogenation
Hydro halogenation

Answer :B
26.

Identify the one which does not come under the organic additiori reaction.

Answer»

HYDRATION
DEHYDRATION
HALOGENATION
Hydro halogenation

SOLUTION :Dehydration
27.

Identify the number of compounds in which positive charge will be delocalised?

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Answer :5(I, iii, V, VI, VII)
28.

Identify the nature of hydrides that are formed by the elements of atomic numbers 15, 19, 23, 44 with dihydrogen. Compare their reactivity towards water.

Answer»

Solution :(i) Element of ATOMIC number 15 belong to p-block and fonns covalent hydride `(PH_3)`.
(ii) Element of atomic number 19 belong to s-block and forms saline hydride (KH).
(iii) Element of atomic no. 23 belong to d-block and fonns interstitial hydride `VH_(1.6)` which is a non-stiochiometric hydride.
(IV) Element of atomic number 44 belong to d-block :-it is ruthenium. It does not from any hydride
(v) Only ionic hydride KH REACTS violetly with water and involved in hydrogen gas PRODUCTION.
`KH+H_2OtoKOH+H_2uparrow`
29.

Identify the most stable species in the following set of ions giving reasons : (i) overset(+)(C)H_(3), overset(+)(C)H_(2)Br, overset(+)(C)HBr_(2), overset(+)(C)Br_(2)(ii) overset(Θ)(C)H_(3), overset(Θ)(C)H_(2)Cl, overset(Θ)(C)HCl_(2),overset(Θ)(C) Cl_(3)

Answer»

Solution :(i) `overset(C)H_(3)` is the most STABLE species because the -I-effect of Br intensifies the +ve charge and hence destability the species. Further, more the number of Br atoms, less stable is the species. Thus, the stability of the species decreases in the order : `overset(+)(C)H_(3) gt overset(+)(C)H_(2)Br gt overset(+)(C)HBr_(2) gt overset(+)(C)Br_(3)`
(ii) -I effect of the Cl atom DISPERSES the -ve charge and thus stabilizes the species. Further, more the number of Cl atoms, more is the DISPERSAL of the -ve charge and hence more stable is the species. Thus, `.^(-)C Cl_(3)` is the most stable species. The stability of other species decreases in the order `.^(-)CHCl_(2) gt .^(-)CH_(2)Cl gt .^(-)CH_(3)`.
30.

Identify the most stable species in the following set of ions giving reasons. (a) overset(+)(C )H_(3), overset(+)(C )H_(2) Br, overset(+)(C )HBr_(2), overset(+)(C )Br_(3) (b) overset(Θ)(C )H_(3), overset(Θ)(C )H_(2)Cl, overset(Θ)(C )HCl_(2), overset(Θ)(C )Cl_(3)

Answer»

Solution :(a) `overset(+)(C )H_(3)` is the most STABLE SPECIES because the -I effect of Br intensifies the +ve charge and hence destabilise the species. Further, more the number of Br atoms, less stable in thespecies. Thus, the STABILITY of these species decreases in the order.
`overset(+)(C )H_(3) gt overset(+)(C )H_(3)Br gt overset(+)(C )HBr_(2) gt Br_(3)`
(b) `-I` effect of the Cl atom disperse the -ve charge and thus stabilizes the species. Further, more th number of Cl atoms, more is the dispersal of the -ve charge and hence more stable is the species and hence more stable is the species. Thus, `-C Cl_(3)` is the most stable species. Thus stability of other species decreases in the order.
`-CHCl_(2) gt -CH_(2)Cl gt -CH_(3)`
31.

The most stable hydride of the following

Answer»

NaH
LiH
KH
CsH

Answer :B
32.

Identify the most acidic hydrogen containing compound from the following

Answer»




SOLUTION :It BECOMES AROMATIC
33.

The most acidic among the following compounds is

Answer»




SOLUTION :It BECOMES AROMATIC
34.

Identify the molecule with zero dipole moment.

Answer»

`"CCL"_(4)`
`NF_(3)`
`CHCl_(3)`
`H_(2)` S

ANSWER :A
35.

Identify the molecule with bond order 1

Answer»

1) `N_(2)`
2) `O_(2)`
3) `H_(2)`
4) `C_(2)H_(4)`

SOLUTION :`H_(2)`
36.

Identify the missing quantum numbers and the sub energy level

Answer»

SOLUTION :
37.

Identify the major products in the following reactions ? CH_3-CH_2-undersetunderset(""CH_3)(""|)oversetoverset(""CH_3)(""|)(.^(o+)N)-CH_2-CH_2-CH_3 overset(overset(Theta)(O)H//Delta)to

Answer»

`CH_3-CH=CH_2`
`CH_2-CH_2`
`{:(""CH_(3)),("|"),(CH_(3)CH_(2)-N-OH),(""o+"|"),(""CH_(3)):}`
`C_2H_5-OH`

ANSWER :B
38.

Identify the magnetic nature of the anion of Na_(2)O_(2).

Answer»

Solution :The ANION of `Na_(2)O_(2) " is "O_(2)^(2-)` (peroxide ion ).
No. of electrons in `O_(2)^(2-) : 8 + 8 + 2 = 18 e^(-)`
Electronic configuration: `( sigma_(1s))^(2) (sigma_(1s)^(**))^(2) ( sigma_(2)^(s))^(2)(sigma_(2)^(s^(**)))(sigma_(2)p_(Z))^(2)(pi_(2)p_(x))^(2)(pi_(2)p_(y))^(2)(pi_(2)^(**)p_(x))^(2) (pi_(2)^(**)p_(y))^(2)`
Bond ORDER : `1/2 (N_(b) - N_(a)) = 1/2 (10 - 8) = 1`
Magnetic Nature : Has no unpaired electrons. HENCE diamagnetic.
39.

Identify the major formed in the following reaction

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ANSWER :B
40.

Identifythe leststable ion amongst the following?

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`Li^(-)`
`Be^(-)`
`B^(-)`
`C^(-)`

SOLUTION :Be alreadyhas fullyfilled 2s- subshell`(2 s^(2))`and thereforeshowstittletendencyto acceptan extraelectronto form `Be^(-)`.
41.

Identify the least stable ion amongst the following.

Answer»

`LI^(-)`
`Be^(-)`
`B^(-)`
`C^(-)`

ANSWER :b
42.

Identify the Lanthanide series.

Answer»

`Z = 90 " to " Z = 103`
`Z = 58 " to " Z = 71`
`Z = 40 " to " Z = 48`
`Z = 72" to " Z = 80`

Answer :b
43.

Identify the isotopes and isobars from the following list of atoms with given number of protons and neutrons. A:8p+9nB:8p+8nC:18p+22nD:20p+20n.

Answer»

Solution :A and B have the same NUMBER of PROTONS but different number of neutrons. Hence A and B are isotopes. C and D have different number of protons, but the sum of the protons annd neutrons, in both the CASES, are the same. Hence, C and D are isobars.
44.

Identify the intermediates of the following reaction.

Answer»




SOLUTION :N//A
45.

Identify the incorrectly matched set ffrom the following

Answer»


ANSWER :D
46.

Identify the incorrect statement(s):

Answer»

ALKYNES are more REACTIVE than ALKENES towards electrophilic ADDITION reaction
Alkynes are less reactive than alkenes towards electrophilic addition reaction
Alkynes decolourise `Br_2` water
Addition of HBr to alkynes in PRESENCE of peroxide proceeds via Markownikoff's addition.

Answer :A::D
47.

Identify the incorrect statements

Answer»

`H_(2)O_(2)` gives yellow colour on REACTION with `Cr(OH)_(3)`
`H_(2)O_(2)` gives BLUE COLOURATION with titaniun salt solution
`H_(2)O_(2)` decolourises pink colour of `KMnO_(4)`
`H_(2)O_(2)` turns starch iodide paper to blue

Answer :B
48.

Identify the incorrect statements.

Answer»

At BOYLE's temperature, there exists a pressure where compressibility of a real gas is more than that of ideal gas.
The parent and daughter nuclei in a `alpha`-decay are isodiaphers.
The magnetic MOMENT of `Cr^(+1)` is more than that of Cr.
The EQUIVALENT weight of `Na_(2)CO_(3)`in its titration with HCl is 106 when PHENOLPHTHALIEN is used as an indicator.

Answer :a,C
49.

Identify the incorrect statement with respect to hybridisation .

Answer»

Hybridisation is intermixing of orbitals of nearly EQUAL energies .
Shape of molecule depends upon type of hybridisation only .
HYBRID orbitals are identical in all aspects .
Hybrid orbitals can form `sigma` and `PI` bond .

ANSWER :d
50.

Identify the incorrect statement regarding the volumetric estimation of FeSO_(4) :

Answer»

`KMnO_(4)` can be used in aqueous HCL
`K_(2)Cr_(2)O_(7)` can be used in aqueous HCl
`KMnO_(4)` can be used in aqueous `H_(2)SO_(4)`
`K_(2)Cr_(2)O_(7)` can be used in aqueous `H_(2)SO_(4)`

Answer :A