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20851.

underset(("Hydrated salt"))((A)) underset("heated")overset("Storngly")tounderset("glassy bead")underset("transparent")ubrace(B+C) H_(3)BO_(3) underset("heated")overset("Strongly")toC+D Find the number of water of crystallizations in (A) :-

Answer»

4
5
10
24

Answer :C
20852.

Total number of orbitals associated with third shell will be___________

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2
4
9
3

Answer :C
20853.

Thetemperaturecoefficientof a reaction in 2 and the rateof thereaction at 25^(@)C is3 mol L^(-1) min^(-1) Calculate the rate at 75^(@)C.

Answer»

Solution :According of Arrhenius, for every `10^(@)C` rise in the temperature, the rate of reaction is DOUBLED.
Rise in the temperature `=75^(@)C-25^(@)C=50^(@)C`
`=5XX10^(@)C`
The increase in the rate of the reaction `=2^(5)=2` times
The rate of reaction `75^(@)C=3xx32`
`=96"MOL"L^(-1)"MIN"^(-1)`.
20854.

Which of thefollowing is co-polymer ?

Answer»

ORLON
teflon
PHBV
polyisoprene

ANSWER :C
20855.

Which of the following reagents is useful to distinguish between chlorobenzene and benzyl chloride

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`AgNO_3` SOLUTION
Fehling's solution
`NaHSO_3` solution
Any ONE of the above

Solution :A) Benzyl chloride `(1^@)` when treated with `AgNO_3`solution to give white ppt but to the for most stable benzyl carbocation while chloro BENZENE doesnot react with `AgNO_3`solution
20856.

Which of the following amines forms a yellow oil with NaNO_2//conc.HCl ?

Answer»

`C_6H_5-NH-CH_3`
`C_6H_5-NH-C_6H_5`
`C_2H_5-NH-C_2H_5`
All of these

Answer :D
20857.

The reagents that can be used to convert benzenediazonium chloride to benzene are_____

Answer»

`SnCl_(2)//HCl`
`CH_(3)CH_(2)OH`
`H_(3)PO_(2)`
`LiAlH_(4)`

Solution :Only `CH_(3)CH_(2)OH and H_(3)PO_(2)` convert benzenediazonium chloride to benzene while `SnCl_(2)//HCl` GIVES `C_(6)H_(5)NHNH_(2)`. Since diazonium salts are PREPARED and used in aqueous solution, therefore, `LiAlH_(4)` cannot used to reduce diazonium salts.
20858.

Which of the following can converts acidified Cr_(2)O_(7)^(2(-)) to green product?

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`SO_(2), H_(2)SO_(3) and H_(2)SO_(4)`
`SO_(3), H_(2)S and H_(2)SO_(3)`
`SO_(3), H_(2)S and Fe^(+3)`
`SO_(3)^(-2), H_(2)S and Fe^(+2)`

SOLUTION :`2CuCl_(2)+SO_(2)+2H_(2)O to 2CuCl (s) +2HCl(g)+H_(2)SO_(4)(d)`
20859.

XeF_4 disproportionates in water giving reduced and oxidised products What is the ratio of oxidation states of Xe in the reduced and the oxidised products:

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SOLUTION :`OVERSET(+4)(6XeF_4)+12H_2Otooverset(0)(4Xe)+overset(+6)(2XeO_3)+24HF+3O_2`
20860.

What happens when ethylene glycol is treated with periodic acid?

Answer»

Solution :Ethylene GLYCOL on TREATMENT with periodic acid gives formaldehyde. This REACTION is selective for vicinal 1,2- diols and it PROCEEDS through a cyclic periodate ester INTERMEDIATE.
20861.

Which one of the followingdeficientdisease of Vitamin A ?

Answer»

Cheilosis
XEROPHTHALMIA
CONVULSIONS
Perncious Anaemia

ANSWER :B
20862.

What is the distance between Na^(+) and Cl^(-) ions in NaCl crystal if its density is 2.165g cm^(-3) ? NaCl crystallizes in fcc lattice.

Answer»

Solution :Let the edge length of unit CELL = a cm
For fcc structure, number of MOLECULES per unit cell, Z=4
Gram formula mass of NaCl, M = 23 +35.5=58.5g `mol^(-1)`
Mass of unit cell = `(Z xx" Gram formula mass")/("Avogadro.s number")`
= `(4 X (58.5 g mol^(-1)))/((6.022 xx 10^(23) mol^(-1)) = 3.886 xx 10^(-22)` g
Density of unit cell = `("Mass")/("Volume")`
or `2.165 g cm^(-3) = (3.886 xx 10^(-22)g)/(a^3)`
`:. a^3 =((3.886 xx 10^(-22)g))/((2.165 g cm^(-3)) = 1.795 xx 10^(-22) cm^3`
`= 179.5 xx 10^(-24) cm^3`
Edge length, `a=(179.5)^(1//3) xx 10^(-8) cm^3`
= `5.64 xx 10^(-8)` cm=564 pm
Edge length of `Na^(+)Cl^(-)` unit cell = `2 (""^(r)N_a^(+) + ""^(r)Cl_a^(-))`=564 pm
or `(""^(r)Na_a^(+) + ""^(r)Cl_a^(-)) =(564)/(2) = 282` pm
Distance between `Na^+` and `Cl^(-)` ions = 282 pm
20863.

Which has the highest freezing point at one atmoshpere ?

Answer»

0.1 M NACL solution
0.1 M SUGAR solution
0.1 M BaCl_2 solution
0.1 M FeCl_3 solution

Answer :B
20864.

What role does the molecular interaction play in solution of alcohol and water ?

Answer»

Solution :There is strong hydrogen BONDING in alcohol molecules as well as water molecules. On mixing, the molecular interactions are weakened. Hence, their solution will show positive DEVIATION from ideal behaviour. As a result, the solution will have higher vapour PRESSURE and LOWER boiling point than that of water and alcohol.
20865.

Which of the following ligand can act as chelating agent but does not have chiral centre?

Answer»

`NTA^(3-)`
bn
pn
None of these

Solution :`nta^(3-)`
20866.

Why are aldehydes more reactive than ketones?

Answer»

Solution :ALDEHYDES are more reactive than ketones due to the following TWO reasons:
(i) Due to smaller +I-effect of one ALKYL group in aldehdyes as compared to larger +I-effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes than in ketones. As a result, nucleophilic addition reactions occur more READILY in aldehydes than in ketones.
(ii) Due to presence of a H-atom on the carbonyl group, aldehydes can be more easily oxidised than ketones. As a result, aldehydes act as reducing agents and thus reduce TOLLEN' reagent, fehling's solution, etc.
20867.

WhenKMnO_4 reacts with acidified FeSO_4:

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Only `FeSO_4` is oxidised
Only `KMnO_4` is oxidised
`FeSO_4` Is oxidised and `KMnO_4` is REDUCED
NONE of the above

Answer :C
20868.

The wedge_(m)^(@) values for NaCl and KCl are 126.5 and 149.9Omega^(-1)cm^(2)mol^(-1) respectively. The ionic conductances of Na^(+) at infinite dilution is 50.1 Omega^(-1)cm^(2)mol^(-1). Calculate the ionic conductance at infinite dilution for K^(+) ion.

Answer»


SOLUTION :`wedge_(m)^(@)(KCL)-lamda_(m)^(@)(NaCl)=lamda_(K^(+))^(@)-lamda_(NA^(+))^(@)=149.9-126.5=23.4Omega^(-1)cm^(2)mol^(-1)`
`lamda_(K^(+))^(@)=23.4+lamda_(Na^(+))^(@)=23.4+50.1=73.5Omega^(-1)cm^(2)mol^(-1)`.
20869.

What volume of 10M HCl and 3M HCl should be mixed to get 1L of 6M HCl solution ?

Answer»

428ml, 572ml
500ml, 500ml
572ml, 428ml
492ml,508ml

Answer :A
20870.

Which of the following statement is/are false for [TiCl_(4)]"- and "[FeCl_(4)]^(-) ?

Answer»

Both tetrahedral and diamagnetic.
Both tetrahedral but former is diamagnetic and later is paramagnetic.
SQUARE planar and tetrahedral RESPECTIVELY and both are paramagnetic.
Solutions of both COMPLEXES are coloured.

Solution :`[FeCl_(4)]^(-)` is paramagnetic.
20871.

Which major product is obtained on heating phenol with KBr and KBrO_(3) ?

Answer»

2 - BROMOPHENOL
3 - bromophenol
4- bromophenol
2, 4, 6 TRIBROMOPHENOL

Solution :2, 4, 6 Tribromophenol
20872.

Which alkaline earth metals nitride is volatile:

Answer»

`Be_3N_2`
`Mg_3N_2`
`Ca_3N_2`
None

Answer :A
20873.

Which is not correct for Frenkel defects ?

Answer»

Low co-ordination number 
Electrical CONDUCTIVITY increases 
Low radius RATIO 
Equal number of cations and ANIONS are MISSING from lattice sites.

Answer :D
20874.

Which one of the following is used as Antihistamine?

Answer»

Omeprazole
Chloranphenicol
DIPHENHYDRAMINE
Norethindrone

Solution :Diphenhydramine is USED as ANTIHISTAMINE.
20875.

Which of the following order is correct in spectrochemical series of ligands?

Answer»

`Cl^(-) lt F^(-) lt C_(2)O_(4)^(2-) lt NO_(2)^-) lt CN^(-)`
`NO_(2)^(-) lt C_(2)O_(4)^(2-) lt Cl^(-) lt F^(-) lt CN^(-)`
`C_(2)O_(4)^(2-) lt F^(-) lt Cl^(-) lt NO_(2)^(-) ltCN^(-)`
`F^(-) lt Cl^(-) lt NO_(2)^(-) lt CN^(-) ltC_(2)O_(4)^(2-)`

ANSWER :A
20876.

Which of the following reactions is showing the acidic property of carboxylic acid ?

Answer»




all of these

Answer :D
20877.

Vitamin A, Vitamin D, Vitamin C, Vitamin C, Vitamin E , Vitamim K .

Answer»

SOLUTION :VITAMIN C iswatersoluble vitaminswhere otherare fatsolube VITAMINS .
20878.

Which one of the following is isomeric with sucrose?

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Lactose
Ribulose
Glucose
Fructose

Answer :A
20879.

What is slag?

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Solution :It is the WASTE product FORMED when a flux combines with GANGUE during the EXTRACTION of METALS.
20880.

Which of the non metals reacts with sulphur ?

Answer»

NOBLE gases
Chlorine
Iodine
Nitrogen.

Solution :`2S+Cl_2toS_2Cl_2`
20881.

The standard oxidation potentials of the electrodes Ag|Ag^+,Sn|Sn^(2+),Ca|Ca^(2+),Pb|Pb^(2+) are -0.8,0.136,2.866 and 0.126V respectively. The most powerful oxidising agent among these metal ions is :

Answer»

`PB^(2+)`
`Ca^(2+)`
`Sn^(2+)`
`AG^+`

Answer :D
20882.

What are the important features of valence bond theory?

Answer»

Solution :(i) The ligand ` to`metal bond in a coordination complex is covalant in nature.It is formed by sharing of electrons (provided by the ligands ) between the central metal atom and the ligand.
(ii) Each ligand should have at least one FILLED orbital containing a lone pair of electrons (iii) In order to axxommodate the electron pairs donated by the ligands ,the central metal ion present in a complex provides required number (coordination number) of vacant orbitals
(iv) These vacant orbitals of central metal atom undergo hybridisation the process of mixing of atomic orbitals of comparable energy to form equal number of new orbitals called hybridised orbitals with same energy.
(v) The vacant hybridisedorbitals are directional and their orientation in SPACE givesa definite geometry to the complex ion.
(vii)In the octahedral complexes, if the ( n-1) d orbitals are involved in hybridisation then they are called inner orbitals complexes or low spin complexes or spin paired complexes . If the nd orbitals are involved in hybridisation ,then such complexesare called outer orbitals or high spin or spin free complexes. Here n represents the principle quantum number of the outermost shell.
(viii) The complexes containing a central metal atom with unpaired electron (s) are paramagnetic. If all the electrons are paired then the complexes will be DIAMAGNETIC .
(ix ) Ligands such as ` CO,CN^(-) ` en and ` NH_3` present in the complexes cause pairing of electrons present in the central metal atom.Such ligands are called strong field ligands
(x) Greater the overlapping between the ligands orbitals and the hybridised metal orbitals greater is the bond strength.
20883.

What happens when (i) 2-Nitropropane boiled with HCl (ii). Nitrobenezen electrolytic reduction in strongly acidic medium. (iii). Oxidation of tert- butylamine with KMnO_(4) (iv). Oxidation of acetoneoxime with trifluoreperoxy acetic acid.

Answer»

Solution :(i) 2- Nitropropane boiled with HCl, 2-nitropropane upon hydrolysis with BOILING HCl give a keton (2-propanone) and nitrous oxide.

(ii). Nitrobenzen electrolytic reduction is strongly acidic medium: Electrolytic reduction of nitrobenzene is weakly acidic medium gives aniine but in strongly acidic medium, it gives p-aminophenol obviously through the acid- catalysed rearrangement of the initially FORMED phenylhydroxylamine.

(iii) Oxidation of tert- butylamine with `KMnO_(4)`: In GENERAL, primary amines, in which the -`NH_(2)` group is attached to a tertiary carbon can be oxidised with `KMnO_(4)` to the corresponding nitro compound in excellent yield. Therefore `3^(@)` -butylamine oxidised to give 2-methyl -2-nitropropane.
`CH_(3)-underset((3^(@)-"butylamine"))underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-NH_(2)overset(KMnO_(4))(to)CH_(3)-underset((2-"methyl-2-nitropropane"))underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-NO_(2)`
(iv) Oxidation of acetoneoxime with trifluoroperoxy acetic acid: Oxidation acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.
`CH_(3)-underset(("Acetoneoxime"))underset(CH_(3))underset(|)(C)=N-OHunderset((O))underset(CF_(3)COOH)(to)CH_(3)-underset((2-"nitropropane"))underset(CH_(3))underset(|)(CH)-NO_(2)`
20884.

Which gives lactic acid on hydrolysis after reacting with HCN?

Answer»

HCHO
`CH_3CHO`
`C_6H_5CHO`
`CH_3COCH_3`

Solution :`UNDERSET"acetaldehyde"(CH_3-OVERSET(O)overset"||"C-H) overset"HCN"to CH_3-underset"CN"underset|overset(OH)overset|C-H overset"Hydrolysis"to underset"Lactic ACID"(CH_3-underset(COOH)underset|overset(OH)overset|C-H)`
20885.

The reagent required to convert 1-butyne to 2-butanone is :

Answer»

alc. KOH
`HG^(2+) ` dil `H_(2) SO_(4)`
`ZnCI_(2) + HCI`
alk . `KMnO_(4)`

ANSWER :B
20886.

Under what pressure must an equimolar mixture of C_2 and PCl_3be placed at 250^@Cin order to obtain 80% conversion of PCl_3into PCl_5? K_p forPCl_3(g) + Cl_2(g) iff PCl_5 (g) is 1.78

Answer»

SOLUTION :13.48 ATM
20887.

To halve the molarity of a solution the following should be adopted

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WEIGHT of the solute to be doubled
weight of the SOLVED to be doubled
volume of the solvent to be doubled
volume of the SOLUTION to be doubled

Answer :D
20888.

Which of the following noble gases has the highest positive electron gain enthalpy value ?

Answer»

HELIUM
KRYPTON
Argon
Neon

Solution :Neon has the highest positive (+ 116) electron GAIN enthalpy.
20889.

Which of the following is free radical ?

Answer»

`CL^+`
`Cl^-`
`Cl^.`
`NO_2^+`

ANSWER :C
20890.

What is the mass number of an element A, if A^(4-) contains 10 electrons and 6 neutrons? [Divide anser by 2]

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SOLUTION :`A^(-4)rArre^(-)=10` after GAIN of 4 electrons so in neutral ATOMS number of electrons=6
so number of protons=6
Mass number=number of p+ number of N
`=6+6=12`
20891.

Which of the following conditions will apply to the conversion of ice into water?

Answer»

`{:(DELTAH,DELTAS,DELTAG),(-,-,-):}` at low T
`{:(DeltaH,DeltaS,DeltaG),(-,-,+):}` at low T
`{:(DeltaH,DeltaS,DeltaG),(+,+,+):}` at low T
`{:(DeltaH,DeltaS,DeltaG),(+,+,-):}` at high T

Answer :C::D
20892.

When is mixed and reacted with (Br_(2))/(KOH) then how many products are obtained.

Answer»


Solution :It is HOFFMAN's Bromamide reaction and in this reaction cross over PRODUCTS are not OBTAINED.
20893.

What type of isomerism does it exhibit ?

Answer»

SOLUTION :It EXHIBITS GEOMETRICAL ISOMERISM.
20894.

Which element does not exhibit allotropy ?

Answer»

C
As
Bi
P

Answer :D
20895.

What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis.

Answer»

SOLUTION :
20896.

What is galvanic cell ? Discuss briefly with one example.

Answer»

Solution :The device in which electrical energy is produced from chemical reactions is called galvanic cell. The common galvanic cell is a Daniel cell. It consists of two beakers containing 10 M solution of `ZnSO_(4)`, and 1.0M solution of `CuSO_(4)`. A ZINC rod is dipped into `ZnSO_(4)` solution and a copper rod is dipped into `CuSO_(4)` solution. These rods act as electrodes. The solution in the two beakers are connected by inverted U tube containing concentrated solution of some electrolyte such as `KCI, KNO_(3)` or `NH_(4)NO_(3)`. The two openings of the U tube are plugged with some porous material such as glass wool or cotton.

The U tube which connects the two glass beakers is called salt bridge.
Salt bridge completes the cell circuit and increases the life of the cell. The arrangement is shown in Fig. When the two electrodes are connected through a voltmeter, the flow of electric current through external circuit is indicated by the voltmeter. The cell formed is called galvanic or voltaic cell.
The potential difference is due to the contribution of two half cells. At zinc electrode, zinc is oxidised to `Zn^(2+)` ions which go into the solution. The electrons released at zinc electrode move towards the other electrode (copper) through the outer circuit. These are accepted by `Cu^(2+)` ions of `CuSO_(4)` solution which are reduced to copper. Thus, OXIDATION occurs at zinc electrode and it is called anode. REDUCTION occurs at copper electrode and it is called cathode. The reactions may be represented as :
At anode :
`Zn(s)toZn^(2+)(aq)+2e""` (oxidation)
At cathode :
`Cu^(2+)(aq)+2etoCu(s)""` (reduction)
The overall cell reaction is
`Zn(s)+Cu^(2+)(aq)toZn^(2+)(aq)+Cu(s)`
The reactions occurring at anode and cathode are ALSO referred as oxidation half cell reaction and reduction half cell reaction. The cell may be represented as :
`Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`
Anode (-) Cathode (+)
20897.

Fill up the gap: When ice melts to liquid water, its entropy _____.

Answer»

BECOMES zero
decreases
increases
remains same

Answer :C
20898.

Whichof the followingcompoundsproduces alkene on heating

Answer»

`[CH_(3) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(N) - CH_(3)]^(+) OH^(-)`
`(CH_(3))_(3)C-X`
`[CH_(3) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(N) - CH_(2)-CH_(3)]^(+) OH^(-)`

Solution :`[CH_(3)- underset(CH_(3))underset(|)overset(CH_(3))overset(|)(N) - CH_(2) - CH_(3)]^(+) OH^(-) overset(DELTA)to(CH_(3))_(3) N + H_(2)O + CH_(2) = CH_(2)`
20899.

Two process are shown such that an ideal gas is taken from state 1 to state 3.Compare the following and choose the correct option(s).

Answer»

`W_(A) GT W_(B)`
`DeltaU_(A) = DeltaU_(B)`
`DeltaS_(A) = DeltaS_(B)`
`q_(A) = q_(B)`

ANSWER :B::C::D
20900.

Write the major product(s) of the following reactions : (i) (iii) (CH_(3))_(3)C-OH overset("Cu")underset(573K)to

Answer»

SOLUTION :
(III) `UNDERSET("2-Methylpropene")(CH_(3)-overset(CH_(3))overset(|)C=CH_(2))`