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.

Cresol and benzyl alcohol are isomers.

Answer»


ANSWER :T
Discussion
2.

[Cr(CN)_(6)][Co(NH_(3))_(6)] contains

Answer»

CATIONIC complex
ANIONIC complex
Neutral LIGANDS
Anionic ligands

Answer :A::B::C::D
Discussion
3.

Cr^(3+), Mn^(4+), V^(2+) are paramagnetic. Calculate their magnetic moment values.

Answer»

SOLUTION :`CR^(3+), Mn^(4+), V^(2+)`Configuration is `d^(3)`. DUE to 3 unpaired electrons, they are paramagnetic.
`mu= sqrt(3(3+2))= sqrt(15)= 3.87 mu_(B)`
Discussion
4.

Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+) E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(o) is

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`-0.10V`
`+0.18V`
`-0.54V`
`0.54V`

Solution :`I^(-)` get OXIDIZED to `I_(2)` hence will FORM anode and `Cr_(2)O_(7)^(2-)` get REDUCED to `Cr^(3+)` hence will form CATHODE.
`E_(cell)^(o)=E_("cathode")^(o)-E_("anode")^(o),E_(cell)^(o)=E_(Cr_(2)O_(7)^(-2))-E_(I_(2))^(o)`.
`0.79=1.33-E_(I_(2))^(o),E_(I_(2))^(o)=1.33-0.79,E_(I_(2))^(o)=0.54V`.
Discussion
5.

Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+) E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(@)=?

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`0.54V`
`-0.054V`
`+0.18V`
`-0.18V`

Solution :In the given RACTION, `I^(-)` has been oxidized to `I_(2)` and `Cr_(2)O_(7)^(2-)` ions have been REDUCED to `Cr^(3+)`
`thereforeE_(cell)^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)`
i.e., `0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=0.54V`
Discussion
6.

Cr_(2)O_(7)^(2-) + I^(-) rarr I_(2)+Cr^(3+) E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V. Then E_(I_(2))^(@)=?

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0.54V
`-0.54V`
`+0.18V`
`-0.18V`

ANSWER :A
Discussion
7.

Cr_(2)O_(7)^(2-) + C_(2)H_(4)O overset(H^(+))rarr Cr^(3+) + C_(2)H_(4)O_(2). How many H^(+) ions are in the balance equation ?

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ANSWER :7
Discussion
8.

Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)toCr^(++ +)+7H_(2)O,E^(@)=1.33V,3xx[2I^(-)toI_(2)+2e^(-)],E^(@)=-0.54V Find out the value of the equilibrium constant and Gibbs free energy change in the reaction given above.

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SOLUTION :`E_(CELL)^(@)=1.33+(-0.54)V=0.79V,n=6`.
Discussion
9.

Cr_(2)O_(3) is......... and CrO is in nature

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SOLUTION :AMPHOTERIC, BASIC
Discussion
10.

Cr_(2)O_(3) dissolves in aqueous NaOH in the presence of an oxidant to form :

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`Cr(OH)_(3)`
`CrO_(4)^(2-)`
`Cr^(3+)`
`Cr_(2)O_(7)^(2-)`

ANSWER :B
Discussion
11.

Cr_(2)O_(3) can be reduced by an……process

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SOLUTION :ALUMINOTHERMIC PROCESS
Discussion
12.

Cr_2O_(7)^(2-)+X overset(H^+) toCr^(3+) +H_2O +oxidised productof X ,Xin theabovereaction cannotbe

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`C_2 O_(4)^(2-)`
`FE^(2+)`
`SO_(4)^(2-)`
`S^(2-)`

SOLUTION :`SO_(4)^(2-)`is NOTA reducingagentand hencedoes notreduce`Cr_2 O_(7)^(2)`
Discussion
13.

Cr_2O_23 on raction with HCl and the product on reaction with NaOH(aq) give respectively

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`CrO_2Cl,CrO_4^(-2)`
`CR(OH)_2,CrO_(4)^(2-)`
`Cl_2,Cr_2O_(7)^(-2)`
`Cl_(2).Cr(OH)_3`

Solution :`Cr_2O_7^(-2) + 3Sn^(+2) rarr 3Sn^(+4) + 2Cr^(+3)`
Discussion
14.

Why is Cr^(2+) reducing and Mn^(3+) oxidising when both have d^(4) configuration?

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SOLUTION :`Cr^(3+) // Cr^(2+)` has a negative reduction potential or positive oxidation potential. Therefore, `Cr^(3+)` is most stable. Hence, `Cr^(2+)` can easily CHANGE to `Cr^(3+)` by the LOSS of an electron. Thus, `Cr^(2+)` acts as a strong REDUCING agent. On the other hand, `Mn^(3+)//Mn^(2+)` has large positive reduction potential. Hence, `Mn^(3+)` can be easily reduced and thus acts as a strong oxidising agent.`
Discussion
15.

Cr^(2+) is a strong reducing agent whereas Mn^(3+) with the same d^4 configuration is an oxidising agent.

Answer»

Solution :`CR^(3+) // Cr^(2+)` has a negative REDUCTION potential or positive oxidation potential. Therefore, `Cr^(3+)` is most stable. Hence, `Cr^(2+)` can EASILY change to `Cr^(3+)` by the loss of an electron. THUS, `Cr^(2+)` acts as a strong reducing agent. On the other hand, `Mn^(3+)//Mn^(2+)` has large positive reduction potential. Hence, `Mn^(3+)` can be easily reduced and thus acts as a strong oxidising agent.`
Discussion
16.

Cr and Cu are more stable. Give reason.

Answer»

Solution :(i) The electronic configuration of Cr and Cu are `[Ar] 3d^(5) 4s^(1)` and `[Ar]3d^(10) 4s^(1)` respectively, The extra stability of half FILLED and fully filled d orbitals, as already EXPLAINED in XI STD, is due to symmetrical DISTRIBUTION of electrons and EXCHANGE energy.
(ii) The extra stability of half filled and fully filled d orbitals is due to symmetrical distribution of electrons and exchange energy.
(iii) When the d orbitals are considered together, they will constitute a sphere. So the half filled and fully filled configuration leads to complete symmetrical distribution of electron density
(iv) On the other hand, an unsymmetrical distribution of electron density will result in building up of a potential difference. To decrease this and to achieve a tension free state with lower energy, a symmetrical distribution is preferred.
Discussion
17.

Covert the following into basis units : "(i) 28.7 pm(ii) 15.15 "mus"(iii) 25365 mg"

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Solution :(i) `"28.7 pm = 28.7 pm"XX(10^(-12)m)/("1 pm")=2.87xx10^(-11)m`
(ii) `15.15 mu s=15.15musxx(10^(-6)m)/(1mus)=1.515xx10^(-5)s`
(iii) `" 25365 mg = 25365 mg "xx (1g)/(1000mg)xx(1kg)/(1000g)=2.5365xx10^(-2)KG`
Discussion
18.

Covert C_(6)H_(6)to C_(6)H_(5)D

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SOLUTION :
Discussion
19.

Covalent solid among the following is

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SOLID ar
MGO
FE
BN

Answer :D
Discussion
20.

Covalent molecules are usually held in a crystal structure by

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dipole-dipole interactions
electrostatic attraction
HYDROGEN BONDS
VAN der Waals' attraction.

Solution :van der Waals. attraction.
Discussion
21.

Pure covalent molecules are ususlly held in a crystal structure by

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Dipole-dipole attraction
Electrostatic attraction
HYDROGEN BONDS
VANDER waal's attraction

ANSWER :D
Discussion
22.

Covalent compounds have low melting point because

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COVALENT molecules are held by weak Van der Waal's force of attraction
covalent bond is less exothermic
covalent bond is weaker than IONIC bond
covalent molecules have definite shape.

Solution :Covalent molecules are held together by weak Van der Waal.s forces. So most of them have LOW melting POINTS.
Discussion
23.

Covalent compounds generally have :

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low MELTING POINTS and are SOLUBLE in polar solvents
low melting points and are insoluble in polar solvents
HIGH melting points and are soluble in polar solvents
high melting points and are insoluble in polar solvents.

Answer :B
Discussion
24.

Covalency of phosphorushas a maximum value

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6
5
4
3

Answer :2
Discussion
25.

Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH_(3)-Br is

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ANSWER :B
Discussion
26.

Covalency of B in BF_4 is :

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5
4
3
2

Answer :B
Discussion
27.

Covalence of carbon in the functional group of C and D are

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3, 3
4,4
4,3
3,4 

ANSWER :C
Discussion
28.

Coupling reaction takes place when benzene diazonium chloride is treated with :

Answer»

BENZALDEHYDE
`ALPHA`-naphthol
N, N-dimethyl aniline
Phenol

Answer :B::C::D
Discussion
29.

Coupling reaction generally occurs at ………….. Position of Benzene ring.

Answer»


ANSWER :PARA
Discussion
30.

Coupling of diazotised _____with____gives methyl orange.

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SOLUTION :Sulphanilic ACID, N,N-dimethylaniline.
Discussion
31.

Coupling of diazonium salts of following takes place in order

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`IV LT II lt III lt I`.
`IV gt III lt II lt I`.
`II lt IV lt I lt III`.
`I lt II lt III lt IV`.

SOLUTION :ELECTRON relasing power :
`C_(6)H_(5)-BAR(O) gt C_(6)H_(5)NH_(2) gt C_(6)H_(5)OH gt C_(6)H_(5)overset(+)(N)H_(3)`
Discussion
32.

Write the reactions of benzenediazonium chloride with : Aniline

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ANSWER :F
Discussion
33.

Coupling of benzene diazonium chloride and aniline is carried out at pH

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7
`10-14`
`4-5`
`6-8`

Solution :COUPLING of benzediazonium CHLORIDE and aniline is DONE under ACIDIC CONDITIONS (pH=4-5)
Discussion
34.

Coupling of benzene diazonium chloride and phenol to form p- hydroxy azobenzene (orange dye) is an example of

Answer»

ELIMINATION reaction
electrophilic SUBSTITUTION reaction
nucleophilic substitution reaction
electrophilic ADDITION reaction

Answer :B
Discussion
35.

Count number of test (s) given by this compound. (a) Liebermann's test(b) Brady's test (c ) Litmus test(d) Bromine water test (e ) Iodoform test(f) NaHCO_(3) test (g) Esterification test(h) Na metal test

Answer»


SOLUTION :N//A
Discussion
36.

Coupling is an example

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ELECTROPHILIC SUBSTITUTION reaction
nucleophilic substitution reaction
free radical substitution reaction
electrophilic ADDITION reaction.

Solution :Coupling is an example of electrophilic substitution reaction
Discussion
37.

Coulomb is the quantity of charge defined as :

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One ampere of current passing for 1 sec
The charge which deposits 0.001118 g of Ag on cathode
The charge which deposits electrochemical equivalenceof metal
1/2 ampere current for two SECOND.

SOLUTION :One empere of current passing for 1 sec
The charge which deposits 0.0118 gm mole
The charge which deposits electrochemical EQUIVALANCE of Metal
Discussion
38.

Coulomb is equal to

Answer»

AMPERE`XX`SECOND
Ampere`xx`minute
Watt`xx`second
Volt`xx`second

Solution :Coulomb=ampere`(A)xx`second(S).
Discussion
39.

Cottrell's smoke precipitator is fitted at the mouth of the chimneys used in factories.

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Solution :The CHARGED colloidal particles of carbon, after coming into CONTACT with oppositely charged ELECTRODE in Cottrell PRECIPITATOR lose their charge and settle down at the bottom.
Discussion
40.

Cotrell precipitator works on the principle of :

Answer»

DISTRIBUTION law
addition of electrolate
Le-chattelier principle
NEUTRALISATION of charge on collids

ANSWER :D
Discussion
41.

Cotton is almost pure …………..

Answer»

SOLUTION :CELLULOSE
Discussion
42.

Cottonis

Answer»

CELLULOSE FIBRE
polyamide fibre
polyester fibre
NONE of these

Answer :A
Discussion
43.

Cotton fibers consist of cellulose polymers with neighboring polymers chains held together by hydrogen bonds between -OH groups in the glucose units. Due to these hydrogen bonds :

Answer»

cotton is insoluble in water
cotton can easily absorb ghee and OILS and therefore are used to make WICKS in traditional lamps
it is easier to iron cotton clothes when they are slightly WET or by applying steam to the clothes
cotton clothes have a high wear and tear than other fibers

Solution :When water is sprinkled on cotton clothes, cellulose forms H-bond with `H_(2)O` and when ironed, `H_(2)O` evaporates and cellulose molecules straighten and FORM bonds with each other.
Discussion
44.

Cost of electricity for the production of x L H_2 at STP at cathode is Rs. x, then cost of electricity for the production of x L O_2 gas at STP at anode will be (assume 1 mol of electrons as one unit electricity) :

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`2x `
4x
16x
32 X

Solution :At CATHODE ` 2H^(+) + 2e^(-) to H_2 `
`{:("RS :xxL"),("At ANODE " 2OH^(-1) (1)/(2) O_2 + H_2O + 2e^(-)):}`
`(x)/(2) L " of " O_2 to Rsx , xL" of " O_2 to ? ((x)/(x) xx x)/(2) = 2x (Rs )`
Discussion
45.

Cost iron is prepared from haematite . In most of the Cases ore needs no concentration,The ore is roasted to remove the volatile impurities ,During roasting the ,mass becomes porous Iron in the form of ferrous oxide os mixed with coke and heated in a blast furnace ,pig iron ios formed and is casted in the desired shape. One of the important advantages of rosting the iron ore is

Answer»

WASTAGE of ORE in the form of `FeSiO_(3)` is avoided
Was TAGE of one in the form of `Fe_(2)(SiO_(3))_(3)` is avided
Ferrous oxide is REMOVED as mot=lten slag `FESiO_(3)`
Ferric oxide is removed as molten slag `Fe_(2)(SiO_(3))_(3)`

Answer :a
Discussion
46.

Cost iron is prepared from haematite . In most of the Cases ore needs no concentration,The ore is roasted to remove the volatile impurities ,During roasting the ,mass becomes porous Iron in the form of ferrous oxide os mixed with coke and heated in a blast furnace ,pig iron ios formed and is casted in the desired shape. the approximate wight percent of coke preesent in the charge of blast furnace is

Answer»

10
20
30
40

Answer :C
Discussion
47.

CoSO_(4)Cl.5NH_(3) exists in two isomeric forms 'A' and 'B'. Isomer 'A' reacts with AgNO_(3) to give white precipitate, but does not react with BaCl_(2). Isomer 'B' gives white precipitate with BaCl_(2) but does not react with AgNO_(3). Answer the following questions. Identify 'A' and 'B' and write their structural formulas. (ii) Name the type of isomerism involved. (iii) Give the IUPAC name of 'A' and 'B'.

Answer»

Solution :(i) As ISOMER A reacts with `AgNO_(3)` to give a white PRECIPITATE, Cl must be present in the ionization SPHERE.
As it does not react with `BaCl_(2)`, `SO_(4)^(2-)` must be present in coordination sphere.
`therefore` Formula of `A=[Co(NH_(3))_(5)SO_(4)]Cl` (coordination no. of Co=6)
As reactions are reverse for isomer B, formula `B=[Co(NH_(3))_(5)Cl]SO_(4)`
(ii) Ionisation isomerism
(iii) A = Pentaamminesulphatocobalt (III) chloride , B = Pentaamminechloridocobalt (III) SULPHATE.
Discussion
48.

Cost iron is prepared from haematite . In most of the Cases ore needs no concentration,The ore is roasted to remove the volatile impurities ,During roasting the ,mass becomes porous Iron in the form of ferrous oxide os mixed with coke and heated in a blast furnace ,pig iron ios formed and is casted in the desired shape. the composition of pig iron is shown as

Answer»

FeC
`FE_(3)C`
`FeC_(3)`
`Fe_(2)C`

ANSWER :B
Discussion
49.

Cosider the two alkenes2-methhyl-1-pentene and2-methyl-2-pentene and decide which would be most stable.

Answer»

Solution :First write the structure of the alkenes, then decide how many SUBSTITUENTS the double bond of each has. 2-Methyl-2-pentene has THREE substituents on its double bond, whereas 2-methyl-1-pentene has two, and THEREFORE 2-methyl-2-pentene is the more STABLE.
Discussion
50.

Cosider the given reaction for preparation of alkyne. (Fritsch reaction). Anti group will migrate because of less steric hindrance. Product (A) is : (C^(*)=C^(14))

Answer»




SOLUTION :ANTI MIGRATION
Discussion