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.

Co^(2+)(aq)+SCN^(-)(aq)toComplex (X) Ni^(2+)(aq.)+"Dimethylglyoxime"overset(NH_4OH)toComplex (Y). The coordination number of cobalt and nickel in compleses X and Y are four. The IUPAC names of the complexes (X) and (Y) are respectively :

Answer»

Tetrathiocyanato-S-cobalt(II) and Bis(dimethylglyoximate) nickel(II)
Tetrathiocyanato-S-cobaltate(II) and Bis(dimethylglymato) nickel(II)
Tetrathiocyanato-S-cobaltate(II) and Bis(dimethylglymato) nickelate(II)
Tetrathiocyanato-S-cobaltate(III) and Bis(dimethylglymato) nickel(II)

Solution :`(X)=[CO(SCN)_4]^(2-),(Y)=[Co(DMG)_2]`
Discussion
2.

Clarify stoichiometry of reaction occurs on the electrodes of electrolytic cell and its relationship with quantity of electricity.

Answer»

Solution :* The amount of electricity(or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
For example, in the reaction : `Ag_((aq))^(+)+e^(-) to Ag_((S))`
* One mole of the electron is required for the reduction of one mole of silver ions.
* We know that charge on one electron is equal to `1.6021xx10^(-19)C`.
* So, the charge on one mole of electrons is equal to :
`=N_(A)xx1.6021xx10^(-19)C`
`=6.02xx10^(23)mol^(-1)xx1.6021xx10^(-19)C`
`=96487C" "mol^(-1)`
* This quantity of electricity is called Faraday and is represented by the symbol F. for approximate calculations, we use
`1F=96500C" "mol^(-1)`
* 1F=electric quantity of 1 mol electron is 96487 (0.96500) COULOMB is known as 1 Faraday.
Generally, in electrolytic reactions current is measured in ampere and time in second.
* So TOTAL electricity=quantity of electricity=Q=It
* To calculate required quantity of electricity to produced product on electrode following STEPS should be followed:
(i) WRITE oxidation/reduction reaction occurs on the electrode.
(ii) Determine the required quantity of Faraday based on reaction occurs for 1 mol of product.
(iii) Calculate current in ampere by using Faraday value.
e.g., (i) reaction occurs for the reduction of Ag metals:
Reaction : `Ag_((aq))^(+)+e^(-) to Ag_((S))`
1 mole electron=1F, 1 mol electricity will produced 1 mol of Ag.
(ii) required electricity for reduction of 1 mol Mg:
Reaction : `Mg_((aq))^(2+)+2e^(-) to Mg_((S))`
2 mole electron=2F , 1 mole Mg is produced by 2 F electricity.
(iii) required electricity for reduction of 1 mol of Al:
`Al^(3+)+3e^(-) to Al`
3 mol electron=3F, for 1 mole Al 3 F electricity is required.
Discussion
3.

Classical smog consists of dust particles, _______and ________.

Answer»
Discussion
4.

Co^(2+)(aq.)+SCN^(-) (aq.) to "Complex"(X) Ni^(2+)(aq.) +"Dimethylglyoxime" overset(NH_(4)OH) to "Complex"(Y) The coordination number of cobalt and nickel in complexes X and Y are four. The geometry of complexes (X) and (Y) are respectively :

Answer»

TETRAHEDRAL and SQUARE PLANAR.
both tetrahedral
square planar and tetrahedral
both square planar.

ANSWER :A
Discussion
5.

Co^(2+)(aq.)+SCN^(-) (aq.) to "Complex"(X) Ni^(2+)(aq.) +"Dimethylglyoxime" overset(NH_(4)OH) to "Complex"(Y) The coordination number of cobalt and nickel in complexes X and Y are four. Select the correct statement for the complexes (X) and (Y) .

Answer»

(X) is paramagnetic with two unpaired electrons.
(Y) is DIAMAGNETIC and shows INTERMOLECULAR H-bonding .
(X) is paramagnetic with THREE unpaired electrons and (Y) is diamagnetic
(X) and (Y) both are diamagnetic.

Answer :C
Discussion
6.

Claisen condensation is not given by:

Answer»




ANSWER :B
Discussion
7.

Claisen condensation is not given by

Answer»




Solution :In Claisen condensation aromatic aldehydes having no `alpha`-HYDROGEN atom react with aldehydes, KETONES or esters having `alpha`-hydrogen atoms in presence of dilute alkali to FORM `alpha-beta-` unsaturated carbonyl compound. It is not given by
as it does not have `alpha`-hydrogen atom.
Discussion
8.

Cl_2O_7 is the anhydride of

Answer»

`HOCl`
`HClO_2`
`HClO_3`
`HClO_4`

Solution :In order to find out the CORRESPONDING ACID of an anhydride , add water MOLECULE (ONE mole) to the anhydride.
`Cl_2O_7 + H_2O to 2HClO_4`.
Discussion
9.

Cl_(2)O_(6) + NaOH rarr products, how many possible products are among the following "___________" NaClO_(4), NaOCl, NaClO_(2),NaClO_(3)

Answer»


Solution :`Cl_(2)O_(6)+2NaOH RARR NaClO_(3) + NaClO_(4) + H_(2)O`
Discussion
10.

Cl_(2)O_(7) dissolves in water to give :

Answer»

`HClO_(3)`
`HClO_(4)`
`HCLO+ClO_(2)`
`ClO_(2)+CLO`.

SOLUTION :`Cl_(2)O_(7)+H_(2)Orarr2HClO_(4)`
Discussion
11.

Cl_2O_6 is an anhydride of:

Answer»

`HClO_3`
`HClO_2`
`HClO_4`
MIXED ANHYDRIDE of `HClO_3 and HClO_4`

ANSWER :D
Discussion
12.

Cl_(2)O_(6) is an anhydride of

Answer»

`HClO_(3)`
`HClO_(2)`
`HClO_(4)`
`HClO`

SOLUTION :Chlorinehexoxide is ALSO a MIXED anhydride of HClO & `HClO_(4)`
`Cl_(2)O_(6) + H_(2) O rarr HClO_(3) + HClO_(4)`
Discussion
13.

Cl_2O is an anhydride of:

Answer»

`HClO_4`
HOCL
`Cl_2O_3`
`HClO_2`

ANSWER :B
Discussion
14.

Cl_(2(g))+Ba(OH)_(2)toX_((aq))+BaCl_(2)+H_(2)OX+H_(2)SO_(4)toY+BaSO_(4),Yunderset(Delta365K)overset(Delta)toZ+H_(2)O+O_(2) Substances Y and Z are respectively

Answer»

`HClO_(4),ClO_(2)`
`HClO_(3),ClO_(2)`
`HClO_(3),ClO_(6)`
`HClO_(4),Cl_(2)O_(7)`

ANSWER :B
Discussion
15.

Cl_(2)(g)+Ba(OH)_(2) to X(aq.)+BaCl_(2)+H_(2)O X+H_(2)SO_(4) to Y+BaSO_(4) Y underset(Tgt365K)overset(Delta)to Z+H_(2)O+O_(2) Y and Z are respectively:

Answer»

`HClO_(4), ClO_(2)`
`HClO_(3), ClO_(2)`
`HClO_(3), ClO_(6)`
`HClO_(4), Cl_(2)O_(7)`

ANSWER :B
Discussion
16.

Cl_(2) underset("(cold and dilute)")(NaOH) overset(-NaCl)overset(-H_(2)O)to X Cl_(2) + underset("(Hot and concentrated)")(NaOH) overset(-NaCl)overset(-H_(2)O)to Y In these reactions, X and Y are _________ respectively.

Answer»

NAOCL and `NaClO_3`
`NaCl_3 and NaOCl`
`HOCL and HOClO_(3)`
`HOCl_(3) and HOCl`

ANSWER :A
Discussion
17.

Cl_2 reacts with water and forms

Answer»

HCL` +` HOCL
HCl` + O_(2) + O_(3)`
HCl + HOCl + `O_(3)`
`HOCl + O_(2)`

ANSWER :A
Discussion
18.

Cl_(2) reacts with hot aqueous NaOH to give

Answer»

NaCl
`NaClO_(3)`
`NaClO_(2)`
`NaClO_(4)`

Solution :`3Cl_(2) + 6NaOH RARR 5NaCl + NaClO_(3) + 3H_(2)O`
Discussion
19.

Cl_(2) reacts with CS_(2) in presence of AlCl3 to form:

Answer»

`CHCl_(3)`
`"CCl"_(4)`
`C_(3)HCl_(3)`
`C_(2)H_(6)`

ANSWER :B
Discussion
20.

CI_2 (or) Br_2 (or) I_2 reacts with hot conc. Alkali solution to form

Answer»

HALIDE + Hypohalite
Halide + Hypohalite `+H_(2)O`
Halide + Halite
Halide + HALATE `+H_(2)O`

ANSWER :D
Discussion
21.

Cl_2 (or) Br_2 (or) I_2 reacts with cold and dilute alkali solution to form

Answer»

Halide `+ ` Hypohalite
Halide `+ `Hypohalite `+ H_(2)O`
Halide `+ ` Halite
Halide `+ `Halate `+ H_(2)O`

ANSWER :B
Discussion
22.

Cl_2 on passing through Na_2S_2O_3 solution gives

Answer»

`Na_2S`
`Na_2SO_4`
`NaHSO_4`
NaHS

Answer :B
Discussion
23.

Cl_2 is used in the manufacture of:

Answer»

CHLOROFORM
`C Cl_4`
Westron
All

Answer :D
Discussion
24.

Cl_2 is used in the extraction of:

Answer»

Pt
Au
BOTH (A) AND(B)
None

Answer :C
Discussion
25.

'Cl_(2)' gas is obtained by various reaction but not by

Answer»

`KMnO_(4) +` conc. `HCl overset( Delta ) ( RARR)`
`KCl + K_(2) Cr_(2) O_()7) + ` conc. `H_(2) SO_(4) overset( Delta ) ( rarr)`
`MnO_(2) = `conc.`HCl overset( Delta ) ( rarr)`
`KCl + F_(2) rarr`

Solution :`4KCl+K_(2)Cr_(2) O_(7) "conc."3H_(2)SO_(4) rarr 2CrO_(2)Cl_(2) + 3K_(2)SO_(4) +3H_(2)O`
Discussion
26.

Cl_2 is more reactive than I_2 but when KCIO_3 reacts with I_2, Cl,_2 is liberated. Why?

Answer»

SOLUTION :Reaction of chlorate with iodine is a REDOX reaction and `I_(2)` is oxidised.
`2KClO_(3)+I_(2)rarr 2KIO_(3)+Cl_(2)`
Discussion
27.

Cl_2 gas is evolved as byproduct in the manufacture of all the following elements except:

Answer»

Mg
Na
Al
K

Answer :C
Discussion
28.

Cl_(2) gas is dried over

Answer»

 CaO
NaOH
`H_(2)SO_(4)`
HBr

Answer :C
Discussion
29.

Cl_(2) can't replace

Answer»

F from NaF
Br from NaBr
I form NaI
Br from KBr

Solution :`Cl_(2)` can not displace F from NaF
Discussion
30.

Cl_(2) form addition compounds with

Answer»

CO
`SO_(2)`
`H_(2)S`
NO

Answer :C
Discussion
31.

Cl_(2) can disproportionate in.

Answer»

`NaOH,HF`
`K_(2)MnO_(4),NaOH`
`K_(2)MnO_(4),HF`
`Ca(OH)_(2)H_(2)O`

SOLUTION :`Cl_(2)+K_(2)MnO_(4)rarrKMnO_(4)+KCL`
`Cl_(2)+HFrarr"No reaction"`
`Cl_(2)+H_(2)OrarrHCl+HOCl`
`Cl_(2)+NaOHrarrNaCl+NaOCl+H_(2)O`
Discussion
32.

Cl - oversetunderset(||)(O)(C ) - OC_2H_5 underset(NH_4Cl)overset((1)xRMg X)(to)3^@ alcohol find our value of x

Answer»


SOLUTION :
Discussion
33.

Cl^(-) is the conjugate base of

Answer»

`HClO_(4)`
HCl
HOCl
`HClO_(3)`

Solution :Acid `overset(-H^(+))RARR` conjugate BASE.
Base `overset(+H^(+))rarr` conjugate acid.
Discussion
34.

Cl in ClO^(-) undergoes ....... hybridisation

Answer»

`SP^(2)`
`sp^(3)`
`sp^(3)d`
`sp^(3)d^(2)`

ANSWER :B
Discussion
35.

Cl-CH_(2)-overset(14)CH_(2)-S-CH_(2)-CH_(3)overset(NH_(3))to number of amines produced.

Answer»


ANSWER :2
Discussion
36.

Cl-C-Cl angle 1, 1, 2, 2 - tetrachloroethene and tetrachloromethane respectively are about

Answer»

`120^(@)` and `109.5^(@)`
`90^(@)` and `109.5^(@)`
`109.5^(@)` and `90^(@)`
`109^(@)` and `120^(@)`

SOLUTION :`120^(@)` and `109.5^(@)`
Discussion
37.

Citrus fruits are an important source of vitamin:

Answer»

B
C
D
K

Answer :B
Discussion
38.

Citric acid is

Answer»

monocarboxylic ACID
dicarboxylic acid
tricarboxylic acid
FATTY acid

ANSWER :C
Discussion
39.

Cisplatin compound is used in the treatment of

Answer»

MALARIA
CANCER
AIDS
yellow fever

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

Cis-trans isomerism is not possible in alkynes because of :-

Answer»

SOLUTION :`180^(@)` BOND-angle at the carbon-carbon TRIPLE bond.
Discussion
41.

Cis-trans isomerism is found in squre planar complexes of molecular formula (a and b are monodentate ligands) :

Answer»

`Ma_(4)`
`Ma_(3)B`
`Ma_(2)b_(2)`
`Mab_(3)`

Answer :C
Discussion
42.

Cis-trans-isomerism is found in square planar complexes of the molecular formula: (a and b are- monodentate ligands)

Answer»

`Ma_4`
`Ma_3b`
`Ma_2b_2`
`Mab_3`

ANSWER :C
Discussion
43.

Cis-trans isomerism is exhibited by:

Answer»

`[PTCL(NH_(3))_(3)]^(+)`
`[PT(NH_(3))_(4)]^(2+)`
`[PtCl_(4)]^(2-)`
`[PtCl_(2)(NH_(3))_(2)]`

Answer :D
Discussion
44.

Cis-(or Z-)3-hexene

Answer»

SOLUTION :
Discussion
45.

Cis-(or z-)2-butanol

Answer»

SOLUTION :
Discussion
46.

Cis isoprene is used for making

Answer»

POLY PROPENE
PVC
Butyl RUBBER
dacron

Answer :C
Discussion
47.

Cis di-mu -chloro bis [chloro(triphenyl phosphine) platinum(II)] is

Answer»




SOLUTION :`MU` refers to BRIDGED Cl atoms
Discussion
48.

Cis butene dioic acid overset(K_(a_(1))(-H^(+)))overset(to)(larr)X_(1)^(-) overset(K_(a_(2))(-H^(+)))overset(to)harrX_(2)^(-) Trans-butene dioic acid overset(K_(a_(1))(-H^(+)))overset(to)(larr)Y_(1)^(-) overset(K_(a_(2))(-H^(+)))overset(to)harrY_(2)^(-) The incorrect statement regarding above information is:

Answer»

`X_(2)^(2-)` SPECIES is more basic than `Y_(2)^(2-)` species
`X_(1)^(-)` species is more basic than `Y_(1)^(-)` species
`K_(a_(1))` is GREATER than `K'_(a_(1))`
`K_(a_(2))` is greater than `K'_(a_(2))`

SOLUTION :
Discussion
49.

cis-But-2-ene reacts with D -N=N-D to yield the major product

Answer»

meso`-2,3-`dideuteriobutane
`(+)-2,3-`dideuteriobutane
`(-)-2,3-`dideuterionutane
`(+)-2,3-`dideuteriobutane

Solution :ALKENES are REDUCED toc cis-products (syn-addition) by di-imide:

This stereospecificity can be EXPLAINED by the formation a cyclie TRANSITION state.
Discussion
50.

cis-But-2-en on reaction with Br_(2) in C CI_(4) produces mainly

Answer»

`(+-)-2,3-`dibromobutane
meso`2,3-`dibromobutane
`(+)-2,3-`dibromobutane
`(-)-2,3-`dibromobutane

Solution :Addition of halogens to alkenes is steroselective (trans or ANTI) as WELL as sterospecific. The addition is also said to the anterafacial (i.e. groups of the addened BOND form opposite faces of the double bond). Cis-But`-2`-en yields race`-2,3-`dibromobutane while trans-but`-2`-en yields meso`-2,3-`dibromobutane. Therefore this type of reaction is diastereospecific.
The first step is the formation of a bridged cation, called a bromonium ion. In the second step, the NUCLEOPHILE `(Br^(-))` adds to the face away form teh bridging group to yield the anti-addition PRODUCT.

Another case:
Discussion