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 S_(N)1, the rate of the reaction depends on the____

Answer»

nucleophile
medium
CONCENTRATION of the SUBSTRATE
none of these

Solution :concentration of the substrate
2.

In SiO_(2), each silicon atom is surrounded by

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4 oxygen ATOMS in a square PLANAR MANNER 
4 oxygen atoms in a tetrahedral manner 
6 oxygen atoms in a OCTAHEDRAL manner 
3 oxygen atoms in a planar fashion 

Answer :B
3.

In SI system, the units of coefficient of viscosity, eta are

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`KG s^(-1)m^(-2)`
`Kg m^(-1)s^(-1)`
`Kg CM^(-1)s^(-1)`
`g m^(-1)s^(-1)`

Answer :B
4.

In SF_(6) , the bond angle is …………………..

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SOLUTION :`90^(@)`
5.

In SF_6 molecules

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BONDS are POLAR but MOLECULE is NON polar
bonds are polar and molecule is ALSO polar
bonds are non polar but molecule is polar
bonds are non polar and molecule is non polar

Answer :A
6.

In SF_(4) molecule , the lone pair ofelectrons occupies an equatorial position rather than axial position in the overall trigonalbipyramidal arrangment . Why?

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Solution :The LP - bp REPULSIONS are less if it occupies EQUATORIAL position than if it occupies AXIAL position . As a
result , energy is less and stability is more
7.

In SF_4 molecule, lone pair of electrons occupies equatorial position but not axial position why?

Answer»

Solution :
In a the LP is PRESENT at AXIAL position so THER are threelp - bp repulsions at `90^(@)`. In b the lp is present at equatorial position, and t here are only two lp-bp repultions at `90^(@)`.Hence arrangement b. is more stable i.e. lp occupies EQUITORIAL position.
8.

In separation by fractional crystallisation method, crystallisation of which of the following compounds occurs first ?

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compound HOSE MELTING point is highest
compound hose boiling point is highest
compound hose solubility is minium
compound hose molecular WEIGHT is miniumum

Solution :In fractional CRYSTALLISATION, the compound with minimum solubility crystallises FIRST.
9.

In self ionization of water the equilibrium is in which direction ?

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SOLUTION :Toward reactant, [Reactants] `GT`[Products]
`[H_2O] gt gt [H^+]` and `[OH^-]`
10.

In sample of hydrogen atomsin ground state, electron make transition from ground state to particular excited state, where path length isfour times of debroglie wavelenght . Electrons make back transitiontothe ground state prouducing all possible photons.

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WAVELENGHTOF photon used for this transition is 121.5 NM
Final excitd state of `Li^(+2)`is 6
kinetic energy of electronin intaial excited state of `Li^(+2)`is 13.6 eV
All are correct

Solution :`3^(rd)` highest energetic photons`rArr(13.6-3.4)eV=10.2 eV`, transition is `2 to 1`
`lambda=(hc)/(E)=(1240)/(10.2)nm=121.5nm`
For H atoms =gor `Li^(+)`ION
`13.6xx1[1-(1)/(2^(2))]=13.6xx9[(1)/(3^(2))-(1)/(6^(2))]`
`therefore`for `Li^(+)`ion transition is `3 to 6`
Initial excited state=3
Final excited state =6
K.E. ofelectronin intial excited state of `Li^(+2)`=`13.6xx(3^(3))/(3^(3))`=`13.6 eV`.
11.

In saturated solution at definite temperature how the ions are arrange ?

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Solution :In saturated solution IONS are in EQUILIBRIUM of sparingly SOLUBLE salt.
`BaSO_(4(s)) hArr Ba_((AQ))^(2+) +SO_(4(aq))^(2-)`
12.

In Schrodinger’s wave equation V indicates

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WAVE function
total ENTHALPY of ELECTRON and PROTON
VELOCITY of electrone
potential energy

Answer :D
13.

In Rutherford's experiment, generally thin foils of heavy atoms like gold, platinum etc. have been used to be bombarded by the alpha-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

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Solution :Heavy atoms have a heavy nucleus carrying a large AMOUNT of positive charge. Hence, some `alpha-`PARTICLES are easily DEFLECTED BACK on hitting the nucleus. ALSO a number of `alpha`-particles are deflected through small angles because of large positive charge on the nucleus. If light atoms are used, their nuclei will be light and moreover, they will have small positive charge on the nucleus. Hence, the number of particles deflected back and those deflected through some angle will be negligible
14.

In Rutherford's gold foil experiment, a thin gold foil was bombarded with a stream of fast moving

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B PARTICLES
`ALPHA`- particles
`GAMMA` - particles
`DELTA` particles

Answer :B
15.

In Rutherford's experiment, generally thin foil of heavy atoms, like gold, platinum etc. have been used to bombard the alpha-particles. If the thin foil of light atoms like aluminium is used, what difference would be observed from the above results?

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SOLUTION :Heavy atoms like gold, platinum have heavy nucleus. Heavy nucleus contains large AMOUNT of POSITIVE charge.
When a beam of a `alpha` - particles is shot at a thin gold foil, most of them pass through without MUCH effect.
Some are deflected back due to enomous repulsive force of heavy nucleus.
If thin foil is made of light ALUMINIUM, then the number of `alpha`-particles deflected back will be negligible.
16.

In Rutherfordexperiment if light atomsusedinsteadof heavyelementsthanwhatchangeis observedin results ?

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Solution :Deviated `ALPHA`-PARTICLESARE less andbounce back`alpha`-PARTICLES ARELEAST. Because inheavyatom thenuclear canscatter more `alpha`- particle .
Iflighter atoms are usedlikealuminium than theirnuclearpossess lessmassless chargeso derivationis alsooloss.
17.

In reversible reaction, initilly the reaction proceed towards the _________ .

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FORMATION of the product
Formation of REACTIONS
Decompose of product
Equilibrium state

ANSWER :A
18.

In resonance structure of phenol…… (i) lone pair of electron of oxygen atom of -OH group is transferred to benzene ring. (ii) ti gives resonance structures having lone pair of electrons in second, fourth and sixth position of -OH group. (iii) structure with negative charge & 1 electron pair on third position of -OH group. (iv) phenol becomes polar.

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ANSWER :(i-T), (ii-T), (iii-F), (iv-T)
19.

In a reaction vessel, 100 g H_(2) and 100 g Cl_(2) are mixed and suitable conditions are provided for the reaction: H_(2(g))+Cl_(2(g))to2HCl_((g))The amount of HCl formed (at 90% yield) will be

Answer»

`36.8g`
`62.5g`
`80g`
`92.53g`

SOLUTION :`underset(2g)(H_(2(g)))+underset(71g)(Cl_(2(g)))rarrunderset(73g)(2HCl_((g)))`
`100g underset("REAGENT")underset("limiting")(100g)`
100% yield of HCl `=(100xx73)/(71)=102.8g`
90% yield of HCl = `(90)/(100)xx102.8=92.53g`
20.

In reaction X_((s)) hArr 4Y_((g)) + 3Z_((g)) , If in this equilibrium do the partial pressure of Z double then partial pressure of Y is....

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`2SQRT2` times of ACTUAL PRESSURE.
`1/2` times of initial pressure.
2 times of initial pressure.
`1/(2sqrt2)` times of initial pressure.

Answer :D
21.

In a reaction vessel, 100 g H_(2) and 100 g Cl_(2) are mixed and suitable conditions are provided for the reaction: H_(2(g))+Cl_(2(g))to2HCl_((g)) The amount of HCl formed in this reaction (at 100% yield) will be

Answer»

`102.8g`
`73g`
`36.5g`
`142g`

Solution :`UNDERSET(2g)(H_(2(g)))+underset(71g)(Cl_(2(g)))rarrunderset(73g)(2HCl_((g)))`
`100G underset("reagent")underset("limiting")(100g)`
100% YIELD of HCl `=(100xx73)/(71)=102.8g`
90% yield of HCl = `(90)/(100)xx102.8=92.53g`
22.

In reaction of alcohols with alkali metla, which of the following alcohol will react fastest?

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Secondary
Tertiary
PRIMARY
All equal

Solution :Primary ALCOHOL because the REACTION involves cleavage of strongest acidic group `(O-H)` of various of alcohols.
23.

In reaction H_2 + I_2 hArr 2HI, at equilibrium the concentration of products is double than reactants then what is the value of K_c ?

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2
4
`1/4`
`1/2`

ANSWER :B
24.

In qualtiative analysis when H_(2)S is passed through aqueous solution of salt acidified with dil. HCl, a black precipitate with dill. HNO_(3), it forms a solution of blue colour. Addition of excess of aqueus soltion of ammonia to this solution gives........ .

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DEEP blue PRECIPITATE of `Cu(OH)_(2)`
deepblue solution of `[Cu(NH_(3))_(4)]^(2+)`
deep blue solution of `Cu(NO_(3))_(2)`
deep blue solution of `Cu(OH)_(2).Cu(NO_(3))_(2)`

Solution :`underset(CuSO_(4)+H_(2)Soverset(Dil.HCl)tounderset("Blck PPT.")(CuS+H_(2)SO_(4)),Cus+2HNO_(3)tounderset("Blue solution")(Cu(NO_(3))_(2))+H_(2)S`
`Cu(N_(3))_(2)+4NH_(3)tounderset(("Deep blue solution"))[[Cu(NH_(3))_(4)]^(2+)+2NO_(3)^(-)`
25.

In Quartz watches, quartz is used

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in place of glass
as a hard SUBSTANCE to MAKE JEWELS for the watch
as a ferroelectric substance
as a piezoelectric substance.

Solution :In QUARTZ WATCHES, a small piece of quartz crystal acts as peiezoelectric substance.
26.

In qualitative estimation of any element using an oxidising agent is essential to predict which substance (s) gets oxidised by the oxidising agent. In case more than one substance is getting oxidised then the oxidising agent gets distributed in all the reactions taking place. From this information and the data given below answer the following questions: Fe^(2+) rarr Fe^(3+) +e^(-) E^(@) =- 0.77V MnO_(4)^(-) +5e^(-) rarr Mn^(2+) + 4H_(2)O E^(@) = + 1.51V 2Cl^(-) rarr Cl_(2) +2e^(-) E^(@) =- 1.36V 2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-) E^(@) =- 2.0V C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-) Cr_(2)O_(7)^(2-) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_(2)O E^(@) = + 1.33 volt Millimol of FeC_(2)O_(4) in the solution if 50ml of 0.1M KMnO_(4) is used for its oxidation in presence dilute HCl if 3.5 millimoles of Cl_(2) is obtained along with other products.

Answer»

`(25)/(3)` m. MOL
6m. Mol
`12.5` m mol
Quanitative ESTIMATION can be made from the give DATA

Solution :No of meq of `KMnO_(4) =` no of meq `Cl^(-)+` no of meq of `FeC_(2)O_(4)`
No of meq of `Cl^(-) =` no of meq of `Cl_(2) = 3.5 xx 2 = 7`
(n-factor =2)
`:.` no of meq of `FeC_(2)O_(4) = 25 - 7 = 18`
`:.` no of m mol of `FeC_(2)O_(4) = 18//3 = 6`
27.

In qualitiveanalysis, anorganiccompound'A' is heatedDry CuO ,CO_(2)gasis liberatedandH_(2)Ois formed. Thesamecompound'A'whenheatedwithNa_(2) O_(2), followedbyboilingwithHNO_(3)and ammoniummolybdategivesyellowprecipitate. Basedon tehtests , theelementspresentin compound'A' are __________.

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C,H ANDN
`C,H ANDP`
C and P
C and H

ANSWER :B
28.

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag^(+) and Pb^(2+) at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^(-) concentrations of Ag^(+) and Pb^(2+) at equilibrium ? (K_(sp) "for" AgCl=1.8xx10^(-10), K_(sp) "for" PbCl_(2)=1.7xx10^(-5))

Answer»

`[AG^(+)]=1.8xx10^(-7) M , [Pb^(2+)]=1.7xx10^(-6)M`
`[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-5)M`
`[Ag^(+)]=1.8xx10^(-9) M , [Pb^(2+)]=1.7xx10^(-3)M`
`[Ag^(+)]=1.8xx10^(-11) M , [Pb^(2+)]=8.5xx10^(-4)M`

Solution :`K_(SP) ` for `AgCl = [Ag^(+)][Cl^(-)]`
`:. 1.8xx10^(-10)=[Ag^(+)][0.1]`
or`[Ag^(+)]=1.8xx10^(-9)M`
`K_(sp)` for `PbCl_(2) = [Pb^(2+)][Cl^(-)]^(2)`
`:.1.7xx10^(-5)=[Pb^(2+)][0.1]^(2)`
or `[Pb^(2+)]=1.7xx10^(-3)M`.
29.

In qualitative analysis, the metals of Group-I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag^+ and Pb^(2+) at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^- concentration is 0.10 M. What will the concentrations of Ag^+ and Pb^(2+) be at equilibrium ? (K_(sp) for AgCl = 1.8xx10^(-10), K_(sp) for PbCl_2=1.7xx10^(-5) )

Answer»

`[Ag^+]=1.8xx10^(-7) M , [Pb^(+2)]=1.7xx10^(-6)` M
`[Ag^+]=1.8xx10^(-11) M , [Pb^(+2)]=8.5xx10^(-5)` M
`[Ag^+] = 1.8xx10^(-9)M , [Pb^(+2)]=1.7xx10^(-3)` M
`[Ag^+]=1.8xx10^(-11) M, [Pb^(+2)]=8.5xx10^(-4)` M

SOLUTION :`K_(sp)=[Ag^+][Cl^-]`
`1.8xx10^(-10)= [Ag^+][0.1]`
`[Ag^+]= 1.8xx10^(-9)` M
`K_(sp)=[Pb^(+2)][Cl^-]^2`
`1.7xx10^(-5) = [Pb^(+2)][0.1]^2`
`[Pb^(+2)]=1.7xx10^(-3)` M
30.

In qualitative analysis, to identily the HA group sulphides, HCl is added to salt solution before the addition of H_(2)S. Because

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Low `S^(2-) ` ion concentration is required to get PPT
HIGH `S^(2-) ` ion concentration is required to get ppt
IIA group metal sulphides have higher values of `K_(sp) ` than that of IVA group metal sulphides
`P^(H) ` value increases

Solution :HCl ensures low CONC. Of `S^(2-) `
31.

In qualitative analysis, to identify the IIA group sulphides, HCl is added to salt solution before the additions of H_2S . Because

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Low `S^(2-) ` ION concentration is REQUIRED to get ppt
High `S^(2-) ` ion concentration is required to get ppt
IIA group metal SULPHIDES have higher VALUES of `K_(sp) ` than that of IVA group metal sulphides
`P^(H) ` value increases

Solution :HCl ENSURES low conc. Of `S^(2-) `
32.

In qualitative analysis Cu^(2+), Cd^(2+) and Pb^(2+) ions are precipate out as sulphide of 2^(nd) group an Ni^(2+) , Zn^(2+) , Mn^(2+) are precipate out as sulphide group and of 3^(rd) B group. In this analysis the reactant HCI + H_2S and NH_4Cl + NH_4OH + H_2O successively added. What is the reason for that ?

Answer»

Solution :In SECOND group `K_(SP)` is very LESS so `CU^(2+), Cd^2, Pb^(2+)` form `PP^+` with `S^(-2)`. Because of common ion `[H^+]` to `[S^(2-)]`concentration decreases so ppt of 3rd group will not form and only 2nd group precipitant.
33.

In qualitative analysis, cations of group II as well as group IV precipitated in the form of sulphides. Due to low value of Ksp of group II sulphides, group reagent is H_(2)S in presence of dil. HCl and due to high value of Ksp of group IV sulphides, group reagent is H_(2)S in presence of NH_(4)OH and NH_(4)Cl. In 0.1M H_(2)S solution, Sn^(2+), Cd^(2+) and Ni^(2+) ions are present in equimolar concentration (0.1M). Given : Ka_(1)(H_(2)S) = 10^(-7), Ka_(2)(H_(2)S) = 10^(-14), K_(sp)(SnS) = 8 xx 10^(-29) , K_(sp) (CdS) = 10^(-28), K_(sp)(NiS) = 3 xx 10^(-21) At what pH precipitate of NiS will form

Answer»

`12.76`
`7`
`1.24`
`4`

Solution :`k_(sp) (NiS) = 3 XX 10^(-21) :. [S^(-2)] = (3 xx 10^(-21))/(0.1)`
`{:(H_(2)ShArrH^(+)+HS^(-1),k_(a_(1))=10^(-7)),(HS^(-1)hArrH^(+)+S^(-2),k_(a_(2))=10^(-14)):}`
`k_(a_(2)) = ([H^(+)][S^(2)])/([HS^(-1)])`
`k_(a_(1)).k_(a_(2)) = 10^(-21)([H^(+)]^(2)[3XX10^(-20)])/([0.1])`
`[H^(+)] =sqrt((1)/(300)):.pH = 1.2388`
34.

In propyne …….. (i) sp hybridization in two carbon atoms. all carbon has sp hybridization. (iii) one carbon has sp^(3) hybridization. (iv) all C-H bonds are of sigma type.

Answer»


ANSWER :(i-T), (ii-F), (iii-T), (iv-T)
35.

In present of dil. H_(2)SO_4the equivalent mass of KMnO_4is

Answer»

M/5
M/6
M/10
M/2

Solution :In dil `H_2SO_4, overset(+7)(MnO_4^(-))overset((5))rarrMn^(2+),E=M/5`
36.

In presence of UV light, benzene reacts with chlorine forming _________

Answer»


ANSWER :BENZENE hexachloride
37.

In presence of sodium ethoxide two molecules of ethyl acetate interact to form acetoacetic ester, this process is known as :

Answer»

ALDOL condensation
Claisen condensation
Polymerization
Schmidt reaction

Solution :SEE PROPERTIES of ESTERS
38.

In presence of peroxides , the addition of HBr to propene occurs according to ______ rule and gives _____ as the major product but in absence of peroxides, addition occurs according ______ rule and gives _______as the major product.

Answer»


ANSWER :anti-Markovnikov's , 1-bromoproane, Markovnikov's RULE ,2-bromopropane
39.

In presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because

Answer»

both are HIGHLY ionic
one is oxidising and the other is reducing
one of the steps is endothermic in both the cases
all the steps are EXOTHERMIC in both reactions

Solution :In CASE of HCL, 2nd step is endothermic while in case HI, the first step is endothermic.
40.

In presence of nickel cyanide , acetylene gives

Answer»

benzene
cyclooctatetraene
cyclohexatriene
cyclobutadiene

Solution :`4HC -= CH overset(NI(CN)_2)to underset("Cycloocatatetraene ")((CH = CH)_4)`
41.

In presence of HCI(aq), K_(2)CR_(2)O_(7) oxidises tin (Sn)inot SN^(4+) ion. The amount of tin that will be oxidised by 1 mol K_(2)Cr_(2)O_(7)

Answer»

1.0mol
1.5mol
2.0mol
2.5mol

Answer :B
42.

In presence of HCl, H_(2)S results theprecipitationof group-2 radicals but not group-4 radicals during qualitative analysis. It is due to

Answer»

higher concentration of `H^(+)`
lower concentration of `H^(+)`
higher concentration of `S^(2-)`
lower concentration of `S^(2-)`

Solution :Dissociation of `H_(2)S` is suppressed in presence of HCl due to common ION EFFECT. This decreases the `S^(2-)` ion concentration and hence only GROUP 2 radicals are precipitated as they have LOW solubility products.
43.

In presence of fluoride ion Mn^(2+) can be titrated with MnO_(4)^(-) both reactants being converted to a complex of Mn(II) in presence of F^(-) ions. A 0.545 g of sample containing Mn_(3)O_(4) was dissolved and all manganese was converted to Mn^(2+). The titration in presence of fluoride ion consumed 31.1 mL of KMnO_(4) that was 0.117 N against oxalate. (a) Write balanced chemical equation of titration was assuming that the complex is MnF_(4)^(-). (b) What was the % of Mn_(3)O_(4) in sample?

Answer»


ANSWER :(B) `40.734%`
44.

In presence of excess base andexces shalogen a methyl ketone is converted first into a trihalo substituted ketone and then into a carbonylc acid . After thetrihalo substituted ketone is formed hydroxide ion attacks the carboxyl carbon because the trihalo methyl ion is thegroup more easily expelled from the tetrahedral intermediate .The conversion of methyl ketone to a carboxylic acid is called a haloform reaction because one of the product is halofor (CHCl_(3)) or CHI_(3) or CHBr_(3). Which of the following compoud show haloform reaction and racemisation inOD^(-)//D_(2)O.

Answer»

`CH_(3)CH_(2)OH`


`CH_(3)-UNDERSET(O)underset(||)(C)-underset(Et)underset(|)(CH)-Ph`

SOLUTION :`CH_(3)-underset(O)underset(||)(C)-overset(**)underset(Et)underset(|)(CH)-Ph`has `alpha` -Hat CHIRAL carbon.
45.

In presence of excess base andexces shalogen a methyl ketone is converted first into a trihalo substituted ketone and then into a carbonylc acid . After thetrihalo substituted ketone is formed hydroxide ion attacks the carboxyl carbon because the trihalo methyl ion is thegroup more easily expelled from the tetrahedral intermediate .The conversion of methyl ketone to a carboxylic acid is called a haloform reaction because one of the product is halofor (CHCl_(3)) or CHI_(3) or CHBr_(3). Product" " is :

Answer»




Solution :`CH_(3)-UNDERSET(O)underset(||)(C)-overset(**)underset(Et)underset(|)(CH)-Ph`has `alpha` -HAT chiral CARBON.
46.

In presence of excess base andexces shalogen a methyl ketone is converted first into a trihalo substituted ketone and then into a carbonylc acid . After thetrihalo substituted ketone is formed hydroxide ion attacks the carboxyl carbon because the trihalo methyl ion is thegroup more easily expelled from the tetrahedral intermediate .The conversion of methyl ketone to a carboxylic acid is called a haloform reaction because one of the product is halofor (CHCl_(3)) or CHI_(3) or CHBr_(3).

Answer»

`Ph-underset(O)underset(||)(C)-underset(Et)underset(|)overset(Me)overset(|)(C)-COOH`
`Ph-C-underset(O)underset(||)(C)-underset(Et)underset(|)(CH)-Me`
`Ph-underset(O) underset(||)(C)-underset(Me)underset(|)(O)-CH-Et`
`Ph-C-underset(O)underset(||)(C)-underset(Et)underset(|)(CH)-OET`

Solution :`CH_(3)-underset(O)underset(||)(C)-overset(**)underset(Et)underset(|)(CH)-Ph`has `ALPHA` -HAT chiral carbon.
47.

In presence of excess of oxygen, sodium metal on immediate reaction produces ......

Answer»

`Na_(2)O`
`Na_(2)O`
`NaO_(2)`
NAO

ANSWER :B
48.

In preparation of Fe(OH)_3 from FeCl_3 first NH_4Cl is added and than NH_4OH is added ? Why ?

Answer»

SOLUTION :
Due to common ION effect of `NH_4^+`its CONCENTRATION increases so the REACTION mens in reverse direction and `OH^-` remain limited to `Fe_((aq))^(+)`and from `Fe(OH)_3`.
49.

In preparation of borazene, diborane is heated with which compound ?

Answer»

`NH_3`
`H_2`
`O_2`
`H_2O`

ANSWER :A
50.

In potentiometric titrations, the end point of a titration is determined

Answer»

with the HELP of an indicator
by MEASURING POTENTIAL of an ELECTRODE
by measuring conductance
by measuring density

Solution :By measuring the potential of an electrode.