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.

Convert 27^@C to Kelvin (K) and choose the correct option

Answer»

210K
27K
310K
300K

Answer :D
Discussion
2.

Convert 25^@C to Fahrenheit(F) and choose the correct alternative?

Answer»

77F
25F
30F
100F

Answer :A
Discussion
3.

Convert 0.5 atmospheres pressure into mm of mercury.

Answer»

Solution :1 atm = 760 mm of HG
`:. 0.5` atm PRESSURE = `(760 mm XX 0.5 atm)/(1 atm)`
`= 380.0 mm of Hg. `
Discussion
4.

Conversion of MnO_(4)^(-) into MnO_(4)^(2-) is oxidation

Answer»


ANSWER :F
Discussion
5.

Conversion of ethyl alcohol into acetaldehyde is an example of

Answer»

REDUCTION
Molecular rearrangement
Hydrolysis
Oxidation

Solution :`CH_(3)CH_(2)OH overset([O])rarrCH_(3)CHO+H_(2)O`.
Discussion
6.

Conversion of benzene to chlorobenzene in the presence of CuCl_2//HClis named as ……………….. .

Answer»

SOLUTION :GATTERMANN REACTION
Discussion
7.

Conversion of benzene to chlorobenzene in the presence of CuCl_2//HCl is named as .........

Answer»

FITTIG REACTION
Raschig process
Dow's process
GATTERMANN reaction

SOLUTION : Gattermann reaction
Discussion
8.

Conversion of benzene to acetophenone can be brought by

Answer»

WURTZ reaction
Wurtz-Fitting reaction
Friedel-Crafts alkylation
Friedel-Crafts acylation

Answer :D
Discussion
9.

Conversation of graphite into diamond is .. ........ reaction.

Answer»

Endothermic
Exothermic
Heat conductor
Electric

Answer :A
Discussion
10.

Controlled oxidation of alkanes.

Answer»

Solution :Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalyse give a variety of oxidation products.
(i) `UNDERSET("Methane")(2CH_(4))+O_(2) underset("100 atm.")overset(cu 573 K)rarr underset("Methanol")(2CH_(3)OH)` ...(i)
(ii) `CH_(4) + O_(2) overset(Mn_(2)O_(3), Delta)rarr underset("Methanol")(HCHO)+H_(2)O` ...(ii)
(iii) `underset("Ethane")(3CH_(3)CH_(3))+3O_(2) underset(Delta)overset((CH_(3)COO)_(2)Mn)rarr underset("ETHANOIC ACID")2CH_(3)COOH + 2H_(2)O` ...(iii)
(iv) Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to corresponding alcohols by potassium permanganate.
`underset("2-Methylpropane")((CH_(3))_(3)CH)underset(+(O))overset(KMnO_(4))rarr underset("2Methylpropane-2-ol")((CH_(3))_(3)C-OH)` ..(iv)
Discussion
11.

Control rods used in nuclear reactor are made of

Answer»

Iron
Graphite
Cadmium
Boron

Answer :C::D
Discussion
12.

Contrast the action of heat on the following with reason: a. Na_(2)CO_(3) and CaCO_(3) b. MgCl_(2).6H_(2)O and CaCl_(2).6H_(2)Oc. Ca(NO_(3))_(2) and NaO_(3)

Answer»

SOLUTION :(a). On heating, the DECAHYDRATE loses water of crystallisation to form monohydrate. Above `373 K`, they completely become anhydrous and become a powder.
`Na_(2)CO_(3).10H_(2)Ooverset(373 K)(rarr)Na_(2)CO_(3).H_(2)O+9H_(2)O`
`NaCO_(3).H_(2)O overset(above)underset(373 K)(rarr)underset((soda ash))(Na_(2)CO_(3))+H_(2)O`
The `Na_(2)CO_(3)` or soda ash is thermally very stable and does not decompose UPON heating.
When `CaCO_(3)` is heated above `1200 K`, it decompose to evolve carbon dioxide.
`CaCO_(3)overset(1200 K)(rarr)CaO+CO_(2)`
(b). When `MgCl_(2).6H_(2)O` is heated, it decomposes to give magnesium oxide `(MgO)`.
`MgCl_(2).6H_(2)OrarrMgO+5H_(2)O+2HCl`
The halide of calcium is hyproscopic. The `CaCl_(2)6 H_(2)O` get dehydrated upon heating.
`CaCl_(2).6 H_(2)O overset(heat)(rarr)CaCl_(2)+6 H_(2)O`
On this basis it is used as a dehydrating agent.
(c ). When calcium nitrate is heated, it decompose and forms the oxide.
`2Ca(NO_(3))_(2)overset(Heat)(rarr)2CaO+4NO_(2)+O_(2)`
When sodium nitrate is heated strongly it gives nitrates and evolve oxygen.
`2NaNO_(3)overset(Heat)(rarr)2NaNO_(2)+O_(2)uarr.`
Discussion
13.

Continuous use of which fertilizer increase the acidity of soil ?

Answer»

Urea
Super PHOSPHATE of LIME
AMMONIUM sulphate
None of these

Answer :C
Discussion
14.

Containers A and B have same gas. Pressure, volume and temperature of A are all twice that of B. Then the ratio of the number of molecules of A and B are

Answer»

<P>`1:2`
`2:1`
`1:4`
`4:1`

Solution :For B,`P_(1)=P,V_(1)=V,T_(1)=T`.
For A, `P_(1)=2P,=2P,V_(2)=2P,T_(2)=2T`
Applying ideal gas EQUATION, Pv=nRT.
`(P_(1)V_(1))/(n_(1)RT_(1))=(P_(2)V_(2))/(n_(2)RT_(2))" or " (PV)/(n_(1)RT)=(2pxx2V)/(n_(2)R(2T))`
or `(n_(2))/(n_(1))=(2)/(1)`
Discussion
15.

Containers A, B and C of equal volume contain oxygen, neon and methane respectively at the same temperature and pressure. The correct increasing order of their masses is

Answer»

`A lt B lt C`
`B lt C lt A`
`C lt A lt B`
`C lt B lt A`

Solution :Under similar conditions of temperature and PRESSURE, equal volumes of different GASES CONTAIN equal number of moles.
`:.` Masses of `O_(2)`,NE and `CH_(4)` will be in the ratio `32:20:16`.
Discussion
16.

Construct a Born-Haber cycle and use it to calculate the first electron affinity of chlorine. DeltaH_(atom(CI))= +122 kJ mol^(-1) "" DeltaH_(atom(Mg))=+148 kJ mol^(-1) DeltaH_("1st ionisation energy" (Mg))=+738 KJ mol^(-1) "" DeltaH_(2nd " ionisation cnergy"(MgCI_(2))= +1451 kJ mol^(-1) DeltaH _("lattice energy "(MgCI_(2))= -2526 kJ mol^(-1) "" DeltaH_(2nd " formation "(MgCI_(2)) =-641 kJ mol^(-1)

Answer»

SOLUTION :-348 kJ/mol
Discussion
17.

Constitutional isomers is possible between memebers of which of the following types of compounds?

Answer»

amines and AMIDES
amides and carboxylic ACIDS
PRIMARY amines and SECONDARY amines
amines and aldehydes

Solution :primary amines and secondary amines are functional isomers . Same M.F.
Discussion
18.

Constitutional isomerism is also known as structural isomerism. The isomers which differ in the connectivity of their atoms are called constitutional isomers. Which of the compound will show tautomerism?

Answer»

acetoxime
cyclohexanol
cyclohexanone
hexanol

Solution :
Discussion
19.

Constitutional isomerism is also known as structural isomerism. The isomers which differ in the connectivity of their atoms are called constitutional isomers. Which pair is correctly matched?

Answer»

n-butane and isobutane, METAMERS
2-PENTANOL and 3-pentanol, position isomers
CYCLOHEXENE and cyclopentene, chain isomers.
methyl propyl ETHER and diethyl ether, functional isomers

Solution :n-butane and Isobutane are chain isomers 2 pentanol 3-pentanol are position isomers cyclohexene and cyclopentene are not isomers methyl propyl ether and diethyl ether are Metamers.
Discussion
20.

Considerting x-axis as the internuclear axis, which out of the following atomic orbitals will form a sigma bond? a. 1s and 1s b. 1s and 2p_(x) c. 2p_(3) and 2p_(y) d. 1s and 2s

Answer»

Solution :The sigma bond is formed by axial overlap along INTERNUCLEAR axis and is present in the following CASES.
In the case a,b and C can form sigma bond, and d canot form sigma bond.
a. 1s and 2s
b. 1s and `2p_(x)`
d. 1s and 2s
`2p_(y)` and `2p_(y)` atomic orbitals are INVOLVED in the sidwise overlap leading to the formation of `PI` bond.
Discussion
21.

Considering x- axis as the internuclear axis,which out of the following will not form a sigma bond and why?

Answer»

`1S & 1s`
`1s & 2P_x`
`2P_y & 2P_y`
`1s & 2S`

ANSWER :C
Discussion
22.

Considering x-axis as the internuclear axis, which out of the following will not form a sigma bond and why ? (a) s and1s(b) 1s and 2p_(x) (c) 2p_(y) and 2p_(y) (d)1s and 2s.

Answer»

Solution :Only (C)will not form a `sigma ` - bond because taking x-AXIS as the intermolucular axis , there will be LATERAL
(sideway) overlap between the TWO ` 2p_(y)` orbitals forming `pi` bond .
Discussion
23.

Considering X-axis as the internuclear axis which out of the following will not form a sigma bond and why ? (a) 1s and 1s(b) 1s and 2p_(x) (c) 2p_(y) and 2p_(y)(d) 1s and 2s

Answer»

Solution :`{:((a) 1s and 1s),((B) 1s and 2p_(x)),((d)1s and 2s):}} {:("All these THREE"),("overlap an intrnuclear"),("axis so it FORM" sigma"bond"):}`.
(c) `2p_(y)` & `2p_(y)` are not overlap on intranuclear axis but remain parallel so it form `PI` bonds.
Discussion
24.

Considering x - axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2p_(x) (c ) 2p_(y) and 2p_(y) (d) s and 2s.

Answer»

Solution :Only (c ) will not form a `sigma` BOND because taking x - axis as the internuclear axis, there will be SIDEWISE OVERLAP between the TWO `2p_(y)` orbitals forming a `pi-` bond . All other will form `sigma-` bond.
Discussion
25.

Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why ?

Answer»


`sigma`-bond is formed between two atoms by the END to end (head-on) overlap of atomic orbitals, hence (1), (2) & (4) form `sigma`-bond. In CASE of (3), if x-axis CONSIDERED as INTERMOLECULAR axis, the `pi`-bond will form due to lateral OVERLAPPING between two `p_(y)` orbitals.

Solution :(1) 1s, 1s
(2) 1s, `2p_(x)`
(3) `2p_(y),2p_(y)`
(4) 1s, 2s.
Discussion
26.

Considering x-axis as molecular axis, which out of the following will form a sigma bond. (i) 1s and 2p_(y) (ii) 2p_(x) and 2p_(x) (iii) 2p_(x) and 2p_(y) (iv) 1s and 2p_(z)

Answer»

Solution :Along X-AXIS as molecular axis only `2p_(x)` and `2p_(x)` can from a SIGMA bond

`2p_(x)+2p_(y)` and 1s and `2p_(2)` ALSO cannot FORM `sigma` bond.
Discussion
27.

Considering x-axis as molecular axis, which out of the following will form a sigma bond. (i) 1s and 2p_(y) (ii) 2p_(x)and 2p_(x) (iii) 2p_(x)and 2p_(z) (iv) 1s and 2p_(z)

Answer»

SOLUTION :Along X-axis as molecular axis, only `2p_(x) and 2p_(x)` can form a sigma BOND
`2p_(x)+2p_(y)and 1_(s)and 2p_(z)` also cannot form `sigma` bond.
Discussion
28.

Considering x- axis as molecular axis, which out of the following will form a sigma bond. (i) 1s " and " 2p_(y)""(ii) 2p_(x)" and " 2p_(x) (iii) 2p_(x)" and " 2p_(z)""(iv) 1s " and " 2p_(z)

Answer»

SOLUTION :Along x-AXIS as molecular axis, only `2p_(x)" and " 2p_(x)` can from a SIGMA bond.


`2p_(x) + 2p_(y)" and " 1s " and " 2p_(z)` ALSO cannot form `sigma` bond.
Discussion
29.

Considering the van der Waals equation of state (P+a//V^(2))(V-b) =RT for ammonia (NH_(3)) and nitrogne (N_(2)) the value of a for NH_(3) is larger than that of N_(2) Ammonia has a lower molecular weight than nitrogen .

Answer»

SOLUTION :Both are CORRECT but does not GIVE correct EXPLANATION .
Discussion
30.

Considering van der Waals equation of state for a real gas (P+n^(2)a//V^(2))(V-nb) =nRT the constant 'a' for O_(2) is less than that for H_(2)O_(g) The molar mass of O_(2) is almost twice that of H_(2)O .

Answer»

SOLUTION :EXPLANTION is CORRECT REASON for STATEMENT
Discussion
31.

Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear ?

Answer»

`CH_(3) - CH= CH - CH_(3)`
`CH_(3) - C EQUIV - C -CH_(3)`
`CH_(2) = CH - CH_(2) - C equiv CH`
`CH_(3) - CH_(2) - CH_(2) - CH_(3)`

SOLUTION :`CH_(3) - C equiv C - CH_(3)`
` UNDERSET(DARR)(sp^(3)) ""underset(darr)(sp)"" underset(darr)(sp) sp^(3)`
`H_(3) C - C equiv C -underset("Linear")(CH_(3))`
Discussion
32.

Considering the following titration curve and answer the questions that follows: 20 mL of the solution being titrated is present initially the approximate value of ionization constant of the species being titrated is

Answer»

`0.1 M`
`0.2 M`
`0.3 M`
`0.001 M`

Solution :At half equivalence point `PH = pK_(a) = 4`
`:. K_(a) = 10^(-4)`
Discussion
33.

Considering the following titration curve and answer the questions that follows: 20 mL of the solution being titrated is present initially The point of optimal buffering on the curve is

Answer»

P
Q
R
S

Solution :Point P is HALF EQUIVALENCE point
`["salt"] = ["acid"]` and `pH= pK_(a)`
Discussion
34.

Considering the following titration curve and answer the questions that follows: 20 mL of the solution being titrated is present initially The species being titrated is

Answer»

weakbase
WEAK acid
dilute solution of strong acid
`CH_(3)COONA`

Solution :From the graph it is EVIDENT that the titrant is a weak acid.
Discussion
35.

Considering the following titration curve and answer the questions that follows: 20 mL of the solution being titrated is present initially The initial concentration of the species being titrated is

Answer»

`0.1 M`
`0.2 M`
`0.3 M`
`0.001 M`

Solution :At EQUIVALENCE point,
`pH = 7 + 1/2 pK_(a) + 1/2"log" C` and `N_(A)V_(A) = N_(B)V_(B) = C`
Discussion
36.

Considering the fact that N_(2) makes up about 79% of the atmosphere, why don't animals use the more abundant N_(2) instead of O_(2) for biological reactions.

Answer»

Solution :Animals need large amount of energy to move around and maintain the body TEMPERATURE. Therefore, to obtain the required energy, it is much easier for them to break WEAKER double bond `(493.4 KJ mol^(-1))` of `O_(2)` than breaking the much STRONGER triple bond `(941.4 kJ mol^(-1))` of `N_(2)`.
Discussion
37.

considering the elements F,Cl, O and N, the correct order their chemical reactivity in terms of oxidising property is

Answer»

`F gt Cl gt O gt N`
`F gt O gt Cl gt N`
`Cl gt F gt O gt N`
`O gt F gt N gt Cl`

ANSWER :B
Discussion
38.

Considering the elements F, CI, O and N, the correct order of their chemical reactivity in terms of oxidizing property is

Answer»

SOLUTION :OXIDATION ABILITY is in the order : F > Cl > O > N
Discussion
39.

Considering the elements F, CI, O and N, the correct order of their chemical reactivity in terms of oxidizing property is ........

Answer»

F>Cl> O>N
F>O >N >Cl
F>O >Cl >N
O>F>N >CI

Solution :Within a period, the oxidizing character increases from left to right. Therefore, among F. O and N.oxidizing POWER decreases in the order: FgtO gt N. However, within a group, oxidizing power decreases from top to BOTTOM. Thus, F is a STRONGER oxidizing agent than CL. Further because is more electronegative than Cl, therefore, O is a stronger oxidizing agent than Cl. Thus, overall decreasing order of oxidizing power is: Fgt gt CIgtN, i.e., OPTION (b) is CORRECT.
Discussion
40.

Considering the elements E CI, O and N, the correct order of their chemical reactivity in terms of oxidizing property is :

Answer»

`FGT Cl gt O gt N `
`F gt O gt Cl gt N`
`Cl gt F gt O gt N `
`O gt F gt N gt Cl`

Answer :a
Discussion
41.

Consideringthe elements F , C1 , O and N thecorrectorderof theirchemicalreactivityin termsofoxidisingproperty is : ( a) F gt C1 gt O gt N ( b) F gt O gt C1 gt N ( c)C1 gt F gt O gt N ( d)O gt F gt NC1

Answer»

Solution :WITHINA periodthe oxidisingcharacter increases from leftto right . ThereforeamongF, O and Noxidisingpowerdecreasesin the order : `F gt Ogt N`. Howeverwithin a groupoxidising POWER decreases from top to bottom. Thus F is STRONGER oxidisingagentthan C1. Furtherbecause Ois moreelectronegativethan C1 thereforeO is a strongeroxidisingagent than C1 Thusoveralldecreasingorderof oxidisingpower is `: F gt O gt C1 gt N` i.e.,OPTION( b) is correct.
Discussion
42.

Considering the elements B,Al,Mgand K, the correct order of their metallic charcter is

Answer»

`B gt AL gt Mg gt K`
`Al gt Mg gt B gt K`
`Mg gt Al gt K gt B`
`K GTMG gt Al gt B`

Answer :D
Discussion
43.

Consideringthe elements B, C ,N, F andSi thecorrectorderof their non- metallic character is : (a) B gt C gt Si gt N gt F ( b) Si gt C gt B gt N gt F( c) F gt N gt C gt B gt Si ( d)F gt N gt C gt Si gt B

Answer»

Solution :In aperiodthe NON metalliccharacterincreases fromleftto right . ThusamongB ,C ,NAND f,non - metalliccharacterdecreases inthe ORDER `: F gtN gt C gt B` However withina group non- metallic character DECREASES from top to bottom. ThusC is morenon- metallic than Si.Thereforethe CORRECT sequence ofdecreasingnon- metalliccharacter is `: F gt N gt C gt B gt Si` i.e.,option ( c) iscorrect.
Discussion
44.

Considering the elements B,C,N,F and Si, the correct order of their non-metalllic character is

Answer»

Solution :F > N > C > B >SI
Discussion
45.

Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is :

Answer»

`B gt C gt SI gt N gt F `
`Si gt C gt B gt N gt F `
`F gt N gt C gt B gt Si`
`F gt N gt C gt Si gt B `

ANSWER :d
Discussion
46.

Considering the elements B, Al, Mg, and K, the correct order of their metallic character is :

Answer»

`B GT Al gt Mg gt K`
`Al gt Mg gt B gt K`
`Mg gt Al gt K gt B`
`K gt Mg gt Al gt B`

ANSWER :d
Discussion
47.

Considering the elements B,Al, Mg and K, the correct order of their metallic character is

Answer»

Solution :K > MG > AL >B
Discussion
48.

Considering the elements B, AI, Mg and K, the correct order of their metallic character is:

Answer»

B>Al>Mg>K
Al>Mg >B>K
Mg >Al>K>B
K>Mg>Al>B

Solution :In a period, metallic character decreases as we MOVE from left to right. Therefore, metallic character of lt Mg and Al decreases in the order. K GT Mg gt Al. However, within a group, the metallic character, INCREASES from top to bottom. Thus, Al is more metallic than B. Therefore, the correct sequence of decreasing metallic character is K gt Mg AlgtB, i.e.. option (d) is correct.
Discussion
49.

Considering theelements B ,A1, Mg and K, the correctorderof theirmetalliccharacteris : (a) B gt A1 gt Mg gt K ( b)A1 gt Mg gt B gt K( c)Mg gt A1 gt K gt B( d)K gt Mg gt A1 g B

Answer»

Solution :In a periodelectronegativityincreasesand HENCETHE metallic character decreases as wemove from left to right. Thereforemetalliccharacterof K, MG and A1decreasesin the order : `K GT Mg gt A1` . However, within agroupthe electronegativitydecreasesand hencethe metalliccharacterincreases from top to bottom. ThusA1 ismoremetallicthan B. Thereforethe CORRECTSEQUENCE ofdecreasing metalliccharacter is `: K gt Mg gt A1 gt B` i.e., option( d) iscorrect.
Discussion
50.

Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is

Answer»

B >C>SI>F>Si
Si >C >B >N>F
F>N>C>B>Si
F>N>C>Si>B

Solution :In a period, the non-metallic character increases from left to right. THUS, among B, C, N and F, non-metallic character decreases in the order: FgtNgtCgtB. HOWEVER, within a group, non-metallic character decreases from top to bottom. Thus, C is more non-metallic than Si. Therefore, the correct sequence of decreasing non-metallic character is: FgtNgtCgtBgtSi, i.e., option (c) is correct.
Discussion