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 compound in which C uses only sp^(3) hybrid orbitals for bond formation is

Answer»

HCOOH
`(H_(2)N)_(2)CO`
`(CH_(3))_(3)COH`
`CH_(3)CHO`

Answer :C
Discussion
2.

The compound having the highest melting point and lowest boiling point is a) n-Hexane

Answer»

n-Hexane
n-Pentane
2-Methylbutane
2,2-Dimethylpropane

Solution :2,2-Dimethylpropane being highly branched has the LOWEST boiling point. Being most SYMMETRICAL, it has the highest MELTING point.
Discussion
3.

The compound having longest C-Cl bond is

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SOLUTION :In 4-chloroyelohexane bond length is LARGEST because there is no RESONANCE. In other compounds, formation of double bond TAKES place between C-Cl bond because of resonance.
Discussion
4.

Which of the following has highest dipole moment?

Answer»

`UNDERSET(CH_3) underset(|) OVERSET(Cl) overset(|)C= underset(Cl) underset(|) overset(CH_3) overset(|)C`
`underset(CH_3) underset(|) overset(H) overset(|)C= underset(Cl) underset(|) overset(CH_3) overset(|)C`

`underset(CH_3) underset(|) overset(CH_3 ) overset(|)C= underset(H) underset(|) overset(H) overset(|)C`

Answer :C
Discussion
5.

The compound having both sp and sp^(2) hybridised carbon atoms is……

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propene
propyne
3-ene 1-butyne
butadiene-1, 3

Answer :C
Discussion
6.

The compound (G) is :

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Both (B) and (C)

ANSWER :D
Discussion
7.

The compound formed when heavy water reacts with CaC_2 is

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acetylene
deuteroacetylene
hydrogen
CALCIUM HYDROXIDE

ANSWER :B
Discussion
8.

The compound formed when heavy water reacts with Ca_3P_2is 'X'. Its molecular mass is

Answer»

34
37
74
56

Answer :B
Discussion
9.

The compound formed when enamel on the surface of the teeth reacts with Fions

Answer»

`CaF_(2)`
`(--OVERSET(F)overset(|)C-underset(F)underset(|)C-overset(F)overset(|)underset(F)underset(|)C)_(n)`
`3Ca_(3)(PO_(4))_(2).CA(OH)_(2)`
`3Ca(PO_(4))_(2).CaF_(2)`

Answer :D
Discussion
10.

The compound formed in the brown ring test has the formula [Fe(H_(2)O)_(5)NO]SO_(4). The oxidation state of iron in it is

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`+1`
`+2`
`+3`
zero

Solution :`[overset(+1)(FE)(H_2O)_(5)overset(+1)(NO)]SO_(4)^(-2)`
`H_2O` is NEUTRAL ligand .
Discussion
11.

The compound formed in the BoraxBead Test of Cu^(2+) ions in the oxidising flame is:

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CU
`Cu(BO_(2))_(2)`
`CuBO_(2)`
NONE of these

Solution :COPPER META borate, `Cu(BO_(2))_(2)` is formed and has blue colour in the OXIDISING flame
Discussion
12.

The compound (F) is :

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`ET- -= -Et`
Both (A) and (B)

ANSWER :B
Discussion
13.

The compound formed as a result of oxidation of ethyl benzene by KMnO_(4) is

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BENZYL alcohol
benzophenone
acetophenone
BENZOIC acid

Answer :D
Discussion
14.

The compound 'D' is used as nitrogeneous fertilizer.

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Urea
Slakedlime
`NH_(4)OH`
NONEOF these

Solution :`NH_(2)CN+H_(2)StoNH_(2)-overset(S)overset(||)underset("Thiourea")(S)-NH_(2)`
Discussion
15.

The compound(E) is :

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TWO MOLES of
Two moles of
ONE MOLE of (A) and one mole of (B)
No reaction.

Answer :A
Discussion
16.

The compound (D) is :

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Both (B) and (C)

ANSWER :A
Discussion
17.

The compound CuCl has ZnS stucture and the edge length of its unit cell is 500 pm. Calculate its density ( Atomic masses : Cu = 63 , Cl= 35.5 , Avogadro 's constant =6.02 xx 10^(23)"mol"^(-1) ?

Answer»


ANSWER :`2.48 " G CM"^(-3)`
Discussion
18.

The compound CuCl has ZnS structure and the edge length of its unit cell is 500 pm. Calculate its density Atomic masses Cu=63, CI=35.5. Avogadro's constant =6.02 xx 10^23 mol^(-1))

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ANSWER :`5.24 "g/cm"^3`
Discussion
19.

The compound CuCl has Zns (cubic) structure. Its density is 3.4 cm^(-3) What is the length of the edge of the unit cell ? (At masses Cu =63.5, Cl=35.5).

Answer»


ANSWER :578 PM
Discussion
20.

The compound CuCl has ZnS (cubic) Structure. Its density is3.4 g cm^(-3) , what is the length of the edge of the unit cell ?(At masses Cu= 63. 5, Cl = 35.5)

Answer»


ANSWER :400 PM
Discussion
21.

The compound CsAuCl_3 is found to be diamagnetic. Gold (Au) in this compound is in (there could be a covalent bond between two gold atoms)

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AU oxidation state only
Au(II) oxidation state only
Au(I) and Au(III) oxidation states
None of these

Solution :Au(Z=79), `[XE]^(54) 4F^(14) 5d^(10) 6s^(1)`. As `CsAuCl_(3)` is diamagnetic . It could not CONSIST of simple (Au(II)) ions because Au(II) has an oddnumber of electrons. Similarly, Au(0) is alos ruled out. Calculation of oxidation state of Au indicates that it can either be Au(II) or an equimolar mixture of Au(I) and (III). therefore, correct answer is C. the actual structure consists of an equal number of `[AuCl_(2)]^(-)` ions and `[AlCl_(4)]^(-)` ions each balanced by `Cs^(+)` ion.
Discussion
22.

The compound, CH_(3)CH_(2)-overset(CH_(3))overset("|")"C "=CH-CH_(3) on reaction with NaIO_(4) in the presence of KMnO_(4) gives

Answer»

`CH_(3)COCH_(3)+CH_(3)CHO`
`CH_(3)CHO+CO_(3)`
`CH_(3)COCH_(3)`
`CH_(3)COCH_(3)+CH_(3)COOH`

SOLUTION :
Discussion
23.

The compound CH_(3)-underset(CH_(2)-CH_(3))underset("|")(""C=CH-CH_(3)) is

Answer»

2-ethyl-2-butene
3-methyl-3-pentene
3-methyl-2-pentene
3-ethyl-2-butene

SOLUTION :Long chain MUST be SELECTED including with double BONDS.
Discussion
24.

The compound , , may be named as :

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2-ethyl-2-methyl oxirane
1,2-epoxy-2-methylbutane
1,2-oxapentane
2-methyl-2-butoxide

Answer :A::B
Discussion
25.

The compound C_7H_8undergoes the following reactions :C_7H_8 overset(3Cl_2//Delta)(to) A overset(Br_2//Fe)(to) B overset(Zn//HCl)(to)C The product C is: 1) m- bromotoluene2) o-bromotoluene3)3-bromo-2, 4, 6-trichlorotoluene4) p-bromotoluene

Answer»

m-bromotoluene
o-bromotoluene
3-bromo-2, 4, 6-trichlorotoluene
p-bromotoluene

Answer :A
Discussion
26.

The compound C_(4)H_(10)O_(4) reacts with acetic anhydride to give a derivative with the formula C_(12)H_(18)O_(8). How many (OH) groups are present in the original substance

Answer»

2
3
7
4

Solution :M. mass of compound
`= 4 XX 12 + 10 xx 4 xx 6 = 122`
M. mass of polyacetate
`=12 xx 12 + 18 xx 16 = 290`
`:.` No. of OH GROUP `= (290-122)/(42) = 4`
Discussion
27.

The compound C_(4)H_(10)O can show……

Answer»

metamerism
position isomerism
functional isomerism
all the three

Solution :(i) `underset("methoxy propane")(CH_(3)-O-CH_(2)-CH_(2)-CH_(3))`
(ii) `underset("ethoxy ETHANE")(CH_(3)-CH_(2)-O-CH_(2)-CH_(3))`
(III) `underset("methyl isopropyl ether")((CH_(3))_(2)-CH-O-CH_(3))`
(iv) `underset("1-butanol")(CH_(3)-CH_(2)-CH_(2)-CH_(2)-OH)`
(v) `underset("2-butanol")(CH_(3)-CH_(2)-underset(underset(OH)(|))(C H)-CH_(3))`
(vi) `underset("2-methyl-propanol")(CH_(3)-underset(underset(CH_(3))(|))(CH)-CH_(2)-OH)`
(vii) `underset("2-methyl 2-propanol")(CH_(3)-underset(underset(OH)(|))overset(overset(CH_(3))(|))(C )-CH_(3))`
(i), (ii), (iii) metamerism (iv), (v), (vi), (vii) are position isomers
(i), (ii), (iii) and (iv), (v), (vi), (vii) are functional group isomers
Discussion
28.

The compound 'C' is used to make urea and Thio urea.

Answer»

`NH_(2)CN`
`NH_(4)CN`
`(NH_(4))_(3)PO_(4)`
`NH_(4)OH`

Solution :`CaNCN+H_(2)SO_(4)tounderset(C )(NH_(2)CN+CaSO_(4)`
Discussion
29.

Thecompound (C) is :

Answer»



`ET- -= -Et`
Both (A) and (B)

ANSWER :A
Discussion
30.

The compound Ba_(2)Cu_(3)O_(7), which shows super conductivity, has copper in oxidation state ......... Assume that the rare earth element yttrium is in its usual +3 oxidation state.

Answer»

SOLUTION :`+(7//3)`
Discussion
31.

The compound 'B' is having oxidation state of:

Answer»

zero
II
III
IV

Answer :C
Discussion
32.

The compound B formed in the following sequence of reaction CH_(3)CH_(2)CH_(2)OH overset(PCl_(5))rarrA overset(alc.KOH)rarrB is

Answer»

Propyne
Propene
Propanal
Propane

Solution :`C_(2)H_(5)CH_(2)OH OVERSET(PCl_(5))rarrC_(2)H_(5)CH_(2)Cl overset(alc.KOH)rarr UNDERSET("Propene")(H_(3)C-CH=CH_(2))`
Discussion
33.

The compound AgF_(2) is ustable however if formed the comound acts as a very strong oxidising agents why ?

Answer»

Solution :In `AgF_(2)` oxidtion state of Ag is +2 which is highly UNSTABLE THEREFORE it quickly ACCEPTS an electrons form the more STABLE +1 OXIDATION state`Ag^(2+)+E^(-)rarr Ag^(+)`
therefore `AgF_(2)` if formed will act as a strongoxidising agent
Discussion
34.

The compound AgF_(2) is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

Answer»

Solution :In `AgF_(2)`, OXIDATION number of Ag is +2 which is very unstable. Therefore it rapidly gains electron and COMES into +1 oxidation state.
`Ag^(+2)+e^(-)TOAG^(+)`
Therefore, `AgF_(2)` is strong OXIDIZING agnet.
Discussion
35.

The compound AgF_(2) is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Answer»

Solution :In `AgF_(2)` the O.S. of AG is +2 which is a HIGHLY unstable STATE of silver. Therefore, it quickly undergoes reduction by taking an electron and attains a stable O.S. of +1.
`underset(("unstable"))underset(O.S.=+2)(Ag^(2+))+e^(-)tounderset(("Stable"))underset(O.S.=+1)(Ag^(+))`
This is why `AgF_(2)`, if formed, ACTS as a very STRONG oxidizing agent.
Discussion
36.

The compound AgF_2 is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?

Answer»

Solution :The O.N of silver is +2 in `AgF_2` which is quit UNSTABLE. It readily ACCEPTS an electron to form stable +1 state. Hence good o0XIdising AGENT.
Discussion
37.

The compound 'A' when treated with methyl alcohols and few drops of H_(2)SO_(4) give smell of wintegreen. The compound 'A' is

Answer»

Succinic ACID
SALICYLIC acid
TARTARIC acid
Oxalic acid

SOLUTION :
Discussion
38.

The compound 'A' is a colourless solid. It gives following tests: Test-I: The addition of compound 'A' solution to a solution of KI, followed by acidification with dilute H_(2)SO_(4), results in the formation of brown coloured compound 'B'. Test-II: The acidified KMnO_(4) solution is decolourized, by a solution of compound 'A', but no gas is evolved. Compound 'A' and 'B' may be:

Answer»

`Na_(2)S_(2)O_(3)` and S
`NaNO_(2)` and `KI_(3)`
`NaNO_(2)` and `NO_(2)`
`Na_(2)S_(2)O_(3)` and `I_(2)`

ANSWER :B
Discussion
39.

The compound 'A' is a colourless solid. It gives following tests: Test-I: The addition of compound 'A' solution to a solution of KI, followed by acidification with dilute H_(2)SO_(4), results in the formation of brown coloured compound 'B'. Test-II: The acidified KMnO_(4) solution is decolourized, by a solution of compound 'A', but no gas is evolved. Which of the following reaction(s) is/are correctly matched about Test-I and Test-II?

Answer»

`2NaNO_(2)+3KI +2H_(2)SO_(4) rarr KI_(3) +2NO +2H_(2)O +2SO_(4)^(2-)+2K^(+) +2Na^(+)`
`5NaNO_(2) +2KMnO_(4)+3H_(2)SO_(4) rarr 5NaNO_(3)+2MnSO_(4)+K_(2)SO_(4)+3H_(2)O`
Test-(I) and Test-(II) both are redox reactions
Test-(I) and Test-(II) both are PRECIPITATION reactions

Answer :A::B::C
Discussion
40.

The compound 1, 2-butadiene has …. Hybridised carbon atoms

Answer»

SP
`sp^(2)` HYBRIDIZED
`sp, sp^(2)`
`sp, sp^(2), sp^(3)`

ANSWER :D
Discussion
41.

The compouhnd formed when alcoholic solution of ethylene dibromide is heated with granualted zinc is

Answer»

ethene
ethyne
ethane
bromoethane

Solution :`UNDERSET("Ethylene dibromide")({:(,CH_(2)Br),(,"|"),(,CH_(2)Br):})+Zn underset("heat")OVERSET("ALCOHOL")to {:(CH_(2),),("||",),(CH_(2),):}+ZnBr_(2)`
Discussion
42.

The composition ofthe equilibrium mixture (Cl_(2)hArr2Cl), which is attained at 1200^(@)C is determined by measuring the rate of effusion through a pin-hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as Krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (Atomic weight of Kr=84)

Answer»

Solution :`(r_("mix"))/(r_(Kr))=sqrt((M_(Kr))/(M_("mix"))),` i.e., `1.16=sqrt((84)/(M_("mix")))"or" M_("mix")=62.43`
If x moles of `Cl_(2)` dissociate at equilibrium, then
`{:(,Cl_(2),hArr,2Cl),("At.eqm".,1-x,,2x):}`
Average , molecular MASS of the MIXTURE`=((1-x)xx71+2x xx35.5)/((1-x)+2x)=(71)/(1+x)`
`:.(71)/(1+x)=62.43` WHICHGIVES x=0.137. Hence, fraction dissociated=0.137
Discussion
43.

The composition of mica is

Answer»

`NaAlSiO_4.3H_(2)O`
`K_(2)O.3Al_(2)O_(3).6SiO_(2).2H_(2)O`
`K_(2)Hal(SiO_4)_3`
`NaK.SiO_4. 10H_(2)O`

Solution :‘MICA’ is mixture of potassium, ALUMINIUM, silicon oxides
Discussion
44.

The composition of LPG is butane and isobutane. The amount of oxygen that would be required for combustion of 1 kg of LPG will be approximately.

Answer»

4.5kg
1.8 KG
2.7 kg
3.6 kg

ANSWER :D
Discussion
45.

The composition of magnalium Alloy is

Answer»

`Mg(1-15%) , Al (85-99%)`
`MN(25-30%)`
`Mo{20%} , W {80%}`
NONE of these

Solution :Magnalium is an alloy of `Mg to (1-5%)`, `Al to (85-99%)`
Discussion
46.

The % composition of four hydro carbons is as follows: {:((i),(ii) ,(iii),(iv),),(75,80,85.7,91.3,%C),(2520,14.3,8.7,%H,):} The date illustrates the law of

Answer»

CONSTANT proportion
Conservation of MASS
MULTIPLE Proprotion
RECIPROCAL Proportions

Solution :LAW of multiple porportions
Discussion
47.

The composition of compound A is 40% X and 60Y. The composition of compound B is 25% X and 75%Y. According to the law of multiple Proportions the ratio of the weight of element Y in compounds A and B is:

Answer»

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

SOLUTION :`{:(,X,Y),(Ararr,1,(60)/(40)),(Brarr,1,(75)/(25)):}`
Discussion
48.

The composition of compound A is 40% X and 60%Y. The composition of compound B is 25% X and 75%Y. According to the law of multiple Proportions the ratio of the weight of element Y in compounds A and B is

Answer»

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

SOLUTION :`{:(,"X,","Y"),(Ararr,1,(60)/(40)),(Brarr,1,(75)/(25)):}`
Discussion
49.

The composition of C is/are:

Answer»

`HgO.HgNH_(2)I`

`(NH_(4))_(2)[HgI_(4)]`
`[HG(NH_(3))_(4)]I_(2)`

SOLUTION :(A) `rarr NH_(4)CI`, (B) `rarr NH_(3),`
(C) `rarr HgO.Hg(NH_(2))I` (D) `rarr CrO_(2)CI_(2)`,
(E) `rarr Na_(2)CrO_(4)`
Discussion
50.

The composition of a sample of wustite is Fe_(0.93)O_(1.00). What percentage of the iron is present in the form of Fe(III) ?

Answer»

SOLUTION :The composition is `Fe_0.93 O_1.00` instead and FeO because some `Fe^(2+)` ions have been REPLACED by `Fe^(+)` ions. Let us FIRST calculate the number of `Fe^(2+)` and `Fe^(3+)` ions present. The formula `Fe_0.93 O_1.00` implies that 93 Fe ATOMS are combined with 100 O-atoms. Out of 93 Fe atoms , suppoe Fe atoms present as `Fe^(3+)=x`. Then `Fe^(2+)=93-x`.As the compound is neutral, total charge on `Fe^(2+)` and `Fe^(3+)` ions =total charge on `O^(2-)` ions. Thus,
`3 xx x +2 (93-x)=2 xx 100` or 3x+ 186-2x =200 or x=14 , i.e., `Fe^(3+)=14`
Hence, `Fe^(2+)` =93-14=79
Thus, out of 93 Fe atoms , Fe present as `Fe^(3+)`=14 `therefore` % age of Fe present of Fe (III) =`14/93xx100=15%`
Discussion