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.

Among the electrolytes Na _(2) SO_(4), CaCl_(2), Al_(2) (SO_(4)) _(3) and NH_(4)Cl, the most efective coagulating agent for Sb _(2) S_(3) sol is

Answer»

`Na _(2) SO_(4)`
`CaCl_(2)`
`Al _(2) (SO_(4)) _(3)`
`NH_(4)Cl`

Solution :`Sb _(2) S_(3)` is a NEGATIVELY charged sol. Hence `Al_(2) (SO_(4)) _(3)` is the most effective coagulating agent because higher the magnitude of OPPOSITE charge `(Al ^(3+)),` higher is the coagulating power.
Discussion
2.

Among the electrolysis of Na_(2)SO_(4), CaCl_(2), Al_(2)(SO_(4))_(3) and NH_(4)Cl, the most effective coagulating agent for Sb_(2)S_(3) sol is

Answer»

`Na_(2)SO_(4)`
`CaCl_(2)`
`Al_(2)(SO_(4))_(3)`
`NH_(4)Cl`

ANSWER :C
Discussion
3.

Among the d-block elements, the highest melting point is shown by "…..............."

Answer»


ANSWER :TUNGSTEN
Discussion
4.

Among The correct order of increasing acid strength is:

Answer»

IltIIIltIIlt IV
IltIIltIIIltIV
IIltIltIIIltIV
IIltIIIltIltIV

Answer :B
Discussion
5.

Among the compounds of BF_(3), NCl_(3), H_(2)S, SF_(4) and BeCl_(2) identify the ones in which the central atoms has the same type of hybridisation.

Answer»

`BF_(3), NCl_(3)` and `H_(2)S`
`H_(2)S` and `BeCl_(2)`
`NCl_(3)` and `H_(2)S`
`SF_(4)` and `BeCl_(2)`

SOLUTION :N ATOM in `NCl_(3)` and S atom in `H_(2)S` are `sp^(3)` hybridised.
Discussion
6.

Among the complex ions [Co(NH_(2)CH_(2)CH_(2)NH_(2))_(2)Cl_(2)]^(+), [CrCl_(2)(C_(2)O_(4))_(2)]^(3-),[Fe(H_(2)O)_(4)(OH)_(2)]^(+), [Fe(NH_(3))_(2)(CN)_(4)]^(-) [Co(NH_(2)CH_(2)CH_(2)NH_(2))_(2)(NH_(3))Cl]^(2+)and[Co(NH_(3))_(4)(H_(2)O)Cl]^(+) the number of complex ion(s) that show (s) cis-trans isomerism is

Answer»


SOLUTION :`[Co(NH_(2)CH_(2)CH_(2)NH_(2))Cl_(2)]^(+)=[M(A A)_(2)B_(2)]rarr` shows cis-trans isomerism
`[CrCl_(2)(C_(2)O_(4))_(2)]^(3-)=[MA_(2)(BB)_(2)] or [M(A A)_(2)B_(2)]rarr` shows cis-trans isomerism
`[Fe(H_(2)O)_(4)(OH)_(2)]=[MA_(4)B_(2)]rarr` shows cis-trans isomerism
`[Fe(NH_(3))_(2)(CN)_(4)]^(-)=[MA_(2)B_(4)]rarr` shows cis-trans isomerism
`[Co(NH_(2)CH_(2)CH_(2)NH_(2))_(2)(NH_(3))Cl]^(2+)= [M(A A)_(2)BC]rarr` shows cis-trans isomerism
`[Co(NH_(3))_(4)(H_(2)O)Cl]^(2+)=[MA_(4)BC]rarr` shows cis-trans isomerism
Discussion
7.

Among the compounds as following which will react with CH_3-overset(O)overset(|)C-CH_3 to give product containing gt C = N - ? (i) C_(6)H_(5)NH_(2) (ii) (CH_(3))_(3)N (iii) C_(6)H_(5)NHC_(6)H_(5) (iv) C_(6)H_(5)NH_(2) and C_(6)H_(5)NHNH_(2)

Answer»

Solution :`C_(6)H_(5)NHC_(6)`
Discussion
8.

Among the complex ions, [Co(en)_(2)CI_(2)]^(+) , [CrCI_(2)(C_(2)O_(4))_(2)]^(3-) , [Fe(H_(2)O)(OH_(2)]^(+) , [Fe(NH_(3))_(2) (CN)_(4)]^(-) [Cp(en)_(2)(NH_(3))CI]^(2+) and [Co(NH_(3))_(4)(H_(2)O)CI]^(2+) , the number of complex ion(s) that show (s) cis-trans isomerism is

Answer»

5
1
4
6

Solution :
Discussion
9.

Among the complexes of Ni listed below, the following is the only correct statement : I-[Ni(CO)_(4)] II-[Ni(CN)_(4)]^(2-) III-[Ni(Cl_(4))]^(2-)

Answer»

All of them have tetrahedral geometry
I and III are tetrahedral and II has square PLANAR geometry
I and II are tetrahedral and III is square planar
All of them have square planar geometry.

Answer :B
Discussion
10.

Among the butane conformers, which occur at energy minima on a graph of potential energy versus dihedral angle?

Answer»

GAUCHE only
ECLIPSED and TOTALLY eclipsed
Gauche and anti
Eclipsed only

Answer :C
Discussion
11.

Among the C-X bond (where, X = Cl, Br, I) the correct bond energy is

Answer»

`C-ClgtC-BrRgtC-I`
`C-IgtC-ClgtC-Br`
`C-BrgtC-ClgtC-I`
`C-IgtC-BrgtC-Cl`

ANSWER :A
Discussion
12.

Arrange the following in increasing order of their basic strength. C_(2)H_(5)NH_(2),C_(6)H_(5)NH_(2), NH_(3),C_(6)H_(5)CH_(2)NH_(2)and(C_(2)H_(5))_(2)NH

Answer»

EgtAgtDgtCgtB 
E gtAgtCgtBgtD 
DgtEgtAgtCgtB 
DgtBgtCgtAgtB 

ANSWER :A
Discussion
13.

Among the aromatic compound with molecular formula C_(7)H_(7)Cl, how many isomers are possible ? Which of these is maximum reactive in nature?

Answer»

Solution :
four isomers are possible as shown below:
Out of these, the cleavage of `C-Cl` bond can easily TAKE place in benzyl chloride to FORM a benzyl CARBOCATION that is resonance STABILISED. Therefore, benzyl chloride is maximum reactive in NATURE.
.
Discussion
14.

Among the alkenes which one produces tertiary butyl alcohol on acid hydration?

Answer»

`(CH_3)_2C = CH_2`
`CH_3 - CH =CH -CH_3`
`CH_3-CH = CH -CH_2`
`CH_3 - CH = CH_2`

SOLUTION :`(CH_3)_2C = CH_2`
Discussion
15.

Among the alkaline earth metals, the element forming predominantly covalent compounds is

Answer»

Ba
Sr
Ca
Be

Answer :D
Discussion
16.

Among the alkaline earth metals, the element forming predominantly covalent compounds is :

Answer»

Ba
Ca
Sr
Be

Answer :D
Discussion
17.

Among the alkali metals cesium is the most reactive because

Answer»

its incomplete shell is nearest to the nucleus
it has a SINGLE ELECTRON in the valence shell
it is the heaviest alkali metal
the outermost electron is more loosely bound than the outermost electron of the other alkali metals.

Solution :Alkali metals are highly reactive SINCE they have a STRONG tendency to lose the single valence s-electron to form unipositive ions having inert gas configuration. In case of CESIUM, the distance of the valence electron is maximum (among the alkali metals) so force of attraction by the nucleus is least, hence it is most reactive.
Discussion
18.

Among the aicd which have lowest pKa value .

Answer»

`CH_(3) COOH`
`CH_(3) - CH_(2) - COOH`
`CH_(3) - underset(NO_(2)) underset(|) (CH) - COOH`
`CH_(3) - underset(COOH) underset(|) (CH) - COOH`

ANSWER :C
Discussion
19.

Among the aicd (1) HC -= C - COOH (2) H_(2)C = CH - COOH (3) CH_(3) - CH_(2)- COOH The acidic strength follows the order

Answer»

`3 LT 2 lt 1`
` 3 = 2 lt 1`
` 1 lt 2 lt 3`
`1LT 2 = 3`

Answer :A
Discussion
20.

Among statements (a) – (d), the correct ones are :(a) Cryolite Na_(3)[AIF_(6)]lowers the melting point of Al_(2)O_(3)in Hall-Heroult process (b) The method of zone refining of metals is based on the principle of solubility of theimpurity in the molten state than in the solid state (c) Extraction of Pb is possible by carbon reduction method (d) In Hall-Heroult process, Aluminium is obtained at cathode andCO_(2)at anode

Answer»

(a), (B), (C ) and (d)
(a), (c ) and (d) only
(c ) and (d) only
(b), (c ) and (d) only

ANSWER :A
Discussion
21.

Among the 15^(th) group elements, as we move from nitrogento bismuth, the pentavallency becomes less pronounced and trivalency becomes more pronounced due to

Answer»

Non metalliccharacter
INERT pair EFFECT
HIGH electronegativity
Large ionization ENERGY

ANSWER :2
Discussion
22.

Among sea water and river water which has high vapour pressure.

Answer»

SOLUTION :RIVER WATER.
Discussion
23.

Among PbS,CuS,HgS ,MnS,AgS, NiS,CoS,Bi_(2)S_(3) and SnS_(2) the total number of BLACK coloured sulphides is

Answer»


ANSWER :7
Discussion
24.

Among physisorption or chemisorption which one has higher enthalpy of adsorption?

Answer»

SOLUTION :Chemisorption/Chemical ADSORPTION.
Discussion
25.

Among phenol (I), o-nitrophenol (II), p-nitrophenol (III),m-nitrophenol (IV), the correct order of decreasing acid strength is

Answer»

IIIgtIIgtIVgtI
IgtIIgtIIIgtIV
IIIgtIVgtIgtII
IVgtIIIgtIIgtI

Answer :A
Discussion
26.

Among paraffins, it is ganerally observed that with an increase in molecular weight-

Answer»

FREEZING POINT decreases
Boilingpoint increases
Boiling point decreases
Specific GRAVITY decreases

Answer :B
Discussion
27.

Among (P) water (Q) ethanol and (R) mercury the correct order of vapour pressure at room temp is

Answer»

PgtQgtR
QgtPgtR
RgtQgtP
QgtRgtP

Solution :If the LIQUID is more VOLATILE,more is its V.P.
Discussion
28.

Among p-diphenylamine (I) p-anisidine (II) and p-toluidine (III), the correct order of decreasing basic strength is

Answer»

`I gt II gt III`
`II gt III gt I`
`I gt III gt II`
`III gt II gt I`.

Solution :
In these SUBSTITUTED anlines the substituents are all electron RELEASING .
`[CH_(3)-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(N)-CH_(2)CH_(3)]^(+) I^(-) underset("Elimination")overset(AgOH)(rarr) (CH_(3))_(3)N+CH_(2)=CH_(2)`
Discussion
29.

Among of electrolytes N_(24)a_2 S_2 and NH_4O the most effective agent for Sb_(23)_S sol is:

Answer»

`N_2a_4S`
`CaC_2`
`A_2l_2C_3S`
`NH_4O`

Solution :For negatively CHARGED sol, like `Sb_2S_3` the coagulating power of electrolyte is directly proportional to the fourth power of the valency of the EFFECTIVE ION (cation). In this case the ORDER of effectiveness is `A^3F^3C`
Discussion
30.

Among of the following the most soluble in water is

Answer»

SUPERPHOSPHATE
triple phosphate
calciumphosphate
CALCIUM SULPHATE

ANSWER :B
Discussion
31.

Among octahedral and tetrahedralcrystal fields, in which case the magnitude of crystal field splitting is larger ?

Answer»

SOLUTION :It is larger in case of OCTAHEDRAL crystal FIELD `(Delta_(t)=4//9 Delta_(0) or Delta_(0) = 9//4 Delta_(t))`.
Discussion
32.

Among octahedral and tetrahedral crystal fields, in which case the magnitude of crystal field splitting is higher ?

Answer»

SOLUTION :In octahedral field, the MAGNITUDE of crystal field SPLITTING is HIGHER.
Discussion
33.

Among Noble gases which one is most abundant in air?

Answer»

SOLUTION :ARGON (AR).
Discussion
34.

Among [Ni(CO)_(4)],[NiCl_(4)]^(2-),[Co(NH_(3))_(4)Cl_(2)]ClNa_(3)[CoF_(6)],Na_(2)O_(2) and CsO_(2) the total number of paramagnetic compounds is

Answer»

2
3
4
5

Solution :`[NI(CO)_(4)]`:

`Na_(2)O_(2)`
contains `O_(2)^(2-)` (peroxide) which is DIAMAGNETIC.
`CsO_(2)` contain `O_(2)^(-)` (superoxide ion) which is paramagnetic.
Discussion
35.

Among [Ni(CO)_(4)], [Ni(CN)_(4)]^(2-), [NiCl_(4)]^(2-) species, the hybridisation states at the Ni atom are, respectively (Atomic numberof Ni = 28)

Answer»

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

Solution :SEE Hint 35 above.
`[NiCl_(4)]^(2-)rarrsp^(3)`hybridisation `RARR` tetrahedral.
Discussion
36.

Among [Ni(CO)_4], [Ni(CN)_4]^(-2)and [NiCl_4]^(-2)species the hybridisation state of the Ni atom are respectively?

Answer»

`sp^3,dsp^2` and `dsp^2`
`sp^3,dsp^2` and `sp^3`
`sp^3,sp^3` and `dsp^3`
`dsp^2, sp^3` and `sp^3`

ANSWER :B
Discussion
37.

Among [Ni(CN)_(4)]^(4-), [Ni(PPh_(3))_(3)Br] and [Ni(dmg)_(2)] species, the hybridisation states of the Ni-atoms are respectively :

Answer»

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

Solution :`[overset(0)(NI)(CN)_(4)]^(4-) implies d^(10) implies sp^(3)`
Discussion
38.

Among [Ni(CO)_(4)], [NiCl_(4)]^(2-), [Co(NH_(3))_(4)Cl_(2)]Cl, Na_(3)[CoF_(6)], Na_(2)O_(2) and CsO_(2), the total number of paramagnetic compounds is

Answer»

2
3
4
5

Solution :`{:(,,"Unpaired electrons"),([Ni(CO)_(4)],3d^(10)(sp^(3)),0),([NiCl_(4)]^(2-),3d^(8)(sp^(3)),2),([Co(NH_(3))_(4)Cl_(2)]CL,3d^(6)(d^(2)sp^(3)),0),(Na_(3)[CoF_(6)],3d^(6)(sp^(3))d^(2),4),(Na_(2)O_(2),O_(2)^(2-),0),(CsO_(2),O_(2)^(-),1):}`
Discussion
39.

Among [Ni(CN)_4]^(2-), [NiCl_4]^(2-) and [Ni(CO)_4]:

Answer»

`[NI(CN)_4]^(2-)` is SQUARE PLANAR and `[NiCl_4]^(2-)` and `Ni(CO)_4` are tetrahedral
`[NiCl_4]^(2-)` is square planar and `[Ni(CN)_4]^(2-)` and `Ni(CO)_4` are tetrahedral
`Ni(CO)_4` is square planar and`[Ni(CN)_4]^(2-)` and `[NiCl_4]^(2-)` are tetrahedral
None

Answer :A
Discussion
40.

Among NH_3, PH_3, AsH_3, SbH_3, BiH_3, which one is the best reducing agent ?

Answer»

SOLUTION :`BiH_3` is the best reducing AGENT (when reducing nature increases down the group `NH_3 to BiH_3`, as in case of other HYDRIDES of Group IVA, VIA and VIIA).
Discussion
41.

Among NCl_(3), PCl_(3), AsCl_(3), the decreasing order of bond angle is :

Answer»

`NCl_(3), PCl_(3), AsCl_(3)`
`NCl_(3), AsCl_(3), PCl_(3)`
`AsCl_(3), PCl_(3), NCl_(3)`
`AsCl_(3), NCl_(3), PCl_(3)`

ANSWER :A
Discussion
42.

Among Na,Hg ,S,Pt and graphite which can be used as electrodes in electrolytic cells having aqueous solutions:

Answer»

`NA` and `S`
`HG` and `PT`
`Na, Hg and S`
`Hg,Pt and graphite

Answer :D
Discussion
43.

Among NaF, NaCl, BaBr and Nal,the naF has highest melting point because

Answer»

It has MAXIMUM ionic character
It has minimum ionic character
It has ASSOCIATED MOLECULES
It has molecular weight

ANSWER :A
Discussion
44.

Among N_(2)O, SO_(2), I_(3)^(+) and I_(3)^(-) the linear species are ............. .

Answer»

SOLUTION :`N_(2)O and I_(3)^(-)` while `SO_(2) and I_(3)^(+)` are BENT molecules
Discussion
45.

Among Na^(+) Na, Mg and Mg^(2+), the largest particle

Answer»

`Mg^(2+)`
Na
Mg
`Na^(+)`.

Solution : As we MOVE from left to right across a PERIOD, there is regular decrease in atomic radii of the REPRESENTATIVE elements. Thus Na is LARGER in size than Mg.
The ionic radii of a cation is always LESS than the atomic radii of that very atom.
`underset(1.54 Å)(Na) underset(0.95A)( Na^(+)) underset(1.36A)(Mg) underset(0.65A)(Mg^(2+))`
Discussion
46.

Among MgCl_(2) , RbCl, BeCl_(2) and LiCl. The compounds with highest and lowest % of ionic characters are :-

Answer»

`RBCL and BeCl_(2)`
`MgCl_(2) and BeCl_(2)`
`BeCl_(2) and MgCl_(2)`
`RbCl and LICL`

ANSWER :A
Discussion
47.

Among Me_(3)N, C_(5)H_(5)N and MeCN (Me = methyl group). the electroncgativity ofN is in the order

Answer»

`MeCN gt C_(5)H_(5)N gt Me_(3)N`
`Me_(5)N gt Me_(3)N gt MeCN`
`Me_(3)N MeCN gt C_(5)H_(5)N`
electronegativity is SAMEIN all .

SOLUTION :`Me_(3)N,C_(5)H_(5)N, MeCN`

As the % s-character increases, electronegativity increases.
`:.` Electronegativity order is :
`sp gt sp^(2) gt sp^(3) implies MeCN gt C_5 H_(5) N gt Me_3N`
Discussion
48.

Among methonic acid and ethanoic acid which is more acidic and why?

Answer»

SOLUTION :Methanoic acid. Ethanoic acid contain `E^(-N)` donating group.
Discussion
49.

Among LiI, NaI, KI the one which is more ionic and more soluble in water :

Answer»

`KI`
`NAI`
`LiI`
NONE of these

Solution :Larger cation `(K^(+))` DEVELOPS less POLARISATION in anion and thus KI has more ionic NATURE and more soluble in water.
Discussion
50.

Among KO_(2),NO_(2)^(-),BaO_2 and NO_(2)^(+) unpaired electron is present in

Answer»

`NO_(2)^(+)` and `BaO_(2)`
`KO_(2)` and `BaO_(2)`
`KO_(2)` only
`BaO_(2)` only

Solution :`KO_(2)` because in `O_(2)^(-)` (superoxide ION) ONE UNPAIRED electron is present in the antibonding ORBITAL.
Discussion