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10301.

Which is (are) correct out of the following:

Answer»

`CH_(2)=CH_(2)LTH-C-=C-h` (rate of electrophilic addition)
`ltH_(3)C-C-=C-CH_(3)` (rate of catalytic hydrogenation)
(rate of electrphilic SUBSTITUTION reaction)
(order of acidic strength)

Answer :B::C::D
10302.

What is enzyme catalysis? Give the characteristics of enzyme catalysed reaction?

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Solution :Enzymes are complex protein molecules with three dimensional structures. They catalyse the chemical reaction in living organism. They are often present in colloidal state and extremely specific in catalytic action. This process is called enzyme catalysis. Special characteristics of enzyme catalysis.
(i) Effective and efficient conversion is the special characteristic of enzyme catalysed reactions. An enzyme may transform a million molecules of reactant in a minute
. For e.g., `2H_2O_2 to 2H_2O + O_2`
For this reaction, activation energy is 18k.cal/MOLE without a catalyst.
With colloidal platinum as a catalyst, the activation energy is 11.7 k.cal/mole.
But with the enzyme catalyst, the activation energy of this reaction is less than 2 k.cal/ mole
(II) Enzyme catalysis is highly specific in nature.
`H_2N-CO-NH_2 + H_2O to 2NH_3 + H_2O`
The enzyme urease which catalyses the reaction of urea does not catalyse the HYDROLYSIS of methyl urea `(NH_(2)-CO-NHCH_(3))`
(iii) Enzyme catalysed reaction has maximum rate at optimum temperature. At first rate of the reaction increases with increase of temperature, but above a particular temperature, the activity of enzyme is DESTROYED. The rate may even drop to zero. The temperature at which enzymic activity is high (or) maximum is called as optimum temperature.
e.g., enzyme involved in human body have an optimum temperature `37^(@)C//98^(@)F`.
(iv) The rate of enzyme catalysed reactions VARIES with the pH of the system. The rate is maximum at a pH called optimum pH.
(v) Enzymes can be inhibited i.e., poisoned. Activity of an enzyme is decreased and destroyed by a poison. The physiological action of drugs is related to their inhibiting action. e.g., sulpha drugs, penicillin inhibits the action of bacteria and used for curing diseases like pneumonia, dysentery, cholera.
(vi) Catalytic activity of enzymes is increased by coenzymes or activators. A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.
10303.

When phenol is treated with propan -2-ol in the presence of HF, Friedel-Craft reaction takes place. Identify the products.

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SOLUTION :
10304.

Which of the following enhances leathering property of soap ?

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SODIUM carbonate
Sodium rosinate
Sodium stearate
Trisodium phosphate.

Solution :Sodium rosinate ENHANCES the LEATHERING PROPERTY of SOAP.
10305.

The species have bond angles of 120^(@) is

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`ClF_(3)`
`NCl_(3)`
`BCl_(3)`
`PH_(3)`

Solution :B in `BCl_(3)` is `sp^(2)`-hybridized. It has trigonal planar geometry and hence has bond angles of `120^(@)`.
10306.

The solution of (F) on treatment with oxalicacid and then with an excess of potassium oxalate gives blue crystals of compound (G). Identify (A) to (G) and give balanced chemical equations for reactions for reactions at step. (i) to (v).

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ANSWER : The ORS is chromite `FeOCr_(2)O_(3)`.
`K_(2)Cr_(2)O_(7)+4H_(2)SO_(4)+3H_(2)C_(2)O_(4) rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+6CO_(2)+7H_(2)O`
`Cr_(2) (SO_(4))_(3)+underset("(G) Blue crystal")(6K_(2)C_(2)O_(4) rarr) 2K_(3)[Cr(C_(2)O_(4))_(3)]+3K_(2)SO_(4)`
10307.

The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. The KE of this electron is:

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13.6 eV
Zero
-13.6 eV
6.8 eV

Answer :A
10308.

Which of the following has S-S bond

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`H_(2) S _(2) O _(8)`
`H_(2) S _(2) O _(7)`
Mustard GAS
`H_(2) S _(2) O _(6)`

Answer :D
10309.

Which of the following crystal system has only primitive Bravais lattice ?

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TETRAGONAL
Orthorirombic
Monoclinic
Rhombohedral

Answer :D
10310.

Which amongst the following are ambidentate as well as flexidentate ligand?

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`NO_(2)^(-)`
`SCN^(-)`
`EDTA^(4-)`
None of these

Answer :D
10311.

Which of the following are biodegradable polymers?

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NYLON -6, 6
PHBV
Nylon-2-nylon-6
Polychloroprene

ANSWER :B::C
10312.

The temperature of 20 litres of nitrogen was increased from 100 K to 300 K at a constant pressure. Change in volume will be

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80 litres
60 litres
40 litres
20 litres

Answer :C
10313.

Which of the following reactions form benzylamine:

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`C_6 H_5 CONH_2 OVERSET(NAOBR)to`
`C_6H_5CN overset(H^(+)//H_2 O)to`

ANSWER :A::D
10314.

The vapour pressure of benzene at a certain temp. is 640 mm Hg. A non-volatile-non-electrolyte and weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance ?

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6.96
65.3
63.8
none of these

Answer :C
10315.

Which of the following statements are correct regarding glucose ? A) It is a dextrose B) It forms osazone C) It forms gluconic acid with bromine - water D) It is a ketohexose

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A,D
B,D
C,D
A,B,C

Answer :D
10316.

Write the IUPAC name of (i)Zn_(2)[Fe(CN)_(6)] (ii) Pt[Cl_(2)(NH_(3))_(2)]

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SOLUTION :(i) ZINC HEXACYANOFERRATE (II) (ii) diamminedichloridoplatinum (II).
10317.

Which is the correct representation of the solubility product constant of Ag_(2)CrO_(4)

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`[AG^(+)]^(2)[CrO_(4)^(-2)]`
`[Ag^(+)] [CrO_(4)^(-2)]`
`[2AG^(+)] [CrO_(4)^(-2)]`
`[2Ag^(+)]^(2) [CrO_(4)^(-2)]`

Solution :`Ag_(2)CrO_(4) hArr [2Ag^(+)] + CrO_(4)^(--)]`
Hence `K_(sp) = [Ag^(+)]^(2) [CrO_(4)^(--)]`.
10318.

Which one of the following statements is wrong about zinc blende type structure ?

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Each `Zn^(2+)` ion is surrounded tetrahedrally by four `S^(2-)` IONS and each `S^(2-)` ion by four `Zn^(2+)` ions
`S^(2-)` ions form fcc arrangement
AgBr has zinc blende type STRUCTURE
Cuprous halides have zinc blende type structure

ANSWER :C
10319.

Which of the following order is not correct?

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`MeBr GT Me_2CHBr gt Me_3CBr gt Et_3CBr(S_N2)`
`PhCH_2Br gt PhCHBrMe gt PhCBrMe_2 gt PhCBrMePh(S_N1)`
`Mel gt MeBr gt MeCl gt MEF(S_N2)`
All the above are correct

Solution :The more is the stability of intermediate carbonium ion, the more is the chance of `S_N1` mechanisın. The intermediates obtained will be `PH overset(+)CH_2` (i),`Phoverset(+)CH-Me` (ii), `Phoverset.(+)C-Me_2` (iii), `Phoverset(+)C-MePh`(IV). The stabilty is of the ORDER iv > iii > ii > i.
10320.

Which of the following is a chain-growth polymer?

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Nylon
Dacron
Glyptal
Polypropylene

Answer :D
10321.

Whattypeoforescan beconcentratedbymagneticseparationmethod ?

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Solution :Themagneticseparationmethodfor concentrationisemployedwhen either the ore ortheimpuritiesassociatedwith itare magneticinnature.For example,chromite`(FeO.Cr_2 O _3 =FeCr_2O_4) `- an oreofchromium, MAGNETIC`(Fe_3O _4) `- an oreof ironandpyrolusite(`MnO_2) `- an oreof maganesebeing magneticareseparatedfrom non-magneticsiliciousgangueby thismethod.Similarly ,tinstoneorcasiterite(`SNO _2`) beingnon- magneticcan beseparatedfrom magneticimpurities like tungstates ofironandmanganese`(FeWO _ 4 andMnWO _ 4 ) `whicharegenerallyassociatedwith it,BYTHIS method.
10322.

Which of the following cations has the strongest tendency towards complex formation?

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`Sm^(3+)`
`Lu^(3+)`
`Gd^(3+)`
`YB^(3+)`

ANSWER :B
10323.

What is a unit cell ? What are its parameters ? Show them by drawing a unit cell .

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Solution :Unit cell is the smallest PORTION of a crystal lattice which when REPEATED in different directions generates the entire lattice PARAMETERS give the dimensions along B the three EDGES , a , b , c and the ANGLES between the edges as shown in the diagram .
10324.

Which of the following is a secondary alcohol ?

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`CH_3CH_2CH(CH_3)CH_2OH`
`CH_3CH_2CH(CH_3)OH`
`(CH_3)3COH`
`CH_3CH_2CH_2COOH`

ANSWER :B
10325.

What is the free energy change for the half reaction Li^(+) + e^(-) to Li ? Given E_(LI^(+) | Li)^(@) = -3.0 V , F = 96500 C mol^(-1) and T = 298 K

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`298.5 kJ mol^(-1)`
`-298.5 kJ mol^(-1)`
`32.166 CV^(-1) mol^(-1)`
`-289500 CV mol^(-1)`

Solution :`DELTA G ^(@) = - N FE^(@) = - 1 xx 96500 xx (-3.0) J = 289.5kJ`.
10326.

Use the following reagents in the correct order and bring about the conversion of benzene to aniline. Reagents -HCl.NH_3," heat, alk " KMnO_4,CHCl_3"/"AlCl_3,Br_2"/"KOH that can be used.

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SOLUTION :
10327.

Time required to decompose SO_(2)Cl_(2) to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

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Solution :For a first order reaction,
`K=(0.693)/(t_(1//2))=(0.693)/("60 minutes")=1.155xx10^(-2)"minutes"^(-1)`
or `k=(0.693)/(60xx60s)=1.925xx10^(-4)s^(-1).`
10328.

The weight percentage of deuterium in heavy water is:

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22
11.11
4
20

Answer :D
10329.

Treatment of butanal with dilute NaOH solution gives

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`CH_(3)CH_(2)CH_(2)COOCH_(2)CH_(2)CH_(2)CH_(3)`
`CH_(3)CH_(2)CH_(2)CHOHCH_(2)CH_(2)CHO`
`CH_(3)CH_(2)CH_(2)CHOHCH(C_(2)H_(5))CHO`
`CH_(3)CH_(2)COCH_(2)CH_(2)CH_(2)CHO`

Solution :This is ALDOL condensation reaction
`CH_(3)CH_(2)CH_(2)CHO + CH_(3)CH_(2)CH_(2)CHO overset("dil.NaOH") to underset("2-ethyl 3- hydroxy hexanal ")(CH_(3)CH_(2)CH_(2)CHOHCH(C_(2)H_(5))CHO)`
10330.

Which of the following statement is/are correct? I. The ligand thiosuphao, S_(2)O_(3)^(2-) can give rise to linkage isomers. II. In metallic carbonyls the ligand CO molecule acts both as donor and acceptor. III. The complex [Pt(Py)(NH_(3))(NO_(2))ClBr] exists in eight different geometrical isomeric forms IV. The complex ferricyanide ion does not follows effective atomic numer (EAN) rule.

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I and II only
II and IV only
I, II and III
I, II and IV

Answer :D
10331.

Which of the following is metamer of diethyl amine ?

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N-Methylpropane-2-amine
N-Methylpropan-1-amine
N-Methylbutan-2-amine
Both (A) and (B).

SOLUTION :`{:(""H),("|"),(CH_(3)CH_(2)-N-CH_(2)CH_(3)):}` lies metamers.
`{:(""H""CH_(3)" "H),("|""|""|"),(CH_(3)CH_(2)CH_(2)-N-CH_(3)" and "CH_(3)-CH-N-CH_(3)),("N-Methylpropan-1-amine""N-Methylopan-2-amine"):}`
10332.

Which of the following facts are true ?

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If `E^(0)(M^(n+)//M)` is negative , `H^(+)` will be reduced to `H_2` by the metal M.
If `E^(0) (M^(n+)//M)` is positive , `M^(n+)` will be reduced to M by `H_2`
In a CELL , `M^(n+)//M` assembly is attached to hydrogen -half cell. To produce spontaneou cell reaction, metal M will act as negative electrode if the potential `M^(n+)//M` is negative. It will serve as positive electrode , if `M^(n+)//M` has a positivecell potential.
COMPOUND of active metal (ZN,Na, MG) are reducible by `H_2` where as those of noble metals (Cu , Ag , Au) are not reducible

Solution :a) `2H^(+) + 2e^(-) overset("red")to H_2 ,"B)" M^(n+) + n e^(-) overset("red")to M`
c) `M^(n+)//M` assembly attached to hydrogen half cell
To produce spotaneous cell reaction
M ` to -ve` electrode , `M^(n+)//M` is negative
It acts as +ve electrode , if `M^(n+)//M` has positive cell potential
10333.

When the temperature is increased by 10^(@) the rate of a reaction

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INCREASES 4 times
increases 2-3 times
decreases to half its original value
does not change.

Solution : The rates of REACTIONS are approximately DOUBLED or TRIPLED for every `10^(@)` rise in temperature.
10334.

Which of the following compound have covalent and ionic bond ?

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R-OH
H-O-H
R-X
R-ON a

Answer :D
10335.

Which of the following is an example of a solid solution in which the solute is a gas ?

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AMALGAM of mercury with sodium
Camphor in NITROGEN gas
Solution of HYDROGEN in palladium
Oxygen DISSOLVED in water

Answer :C
10336.

Which of the following will give white precipitate with an aqueous solution of ethylamine ?

Answer»

`[Ag(NH_(3))_(2)]^(+)`
IRON (II) sulphate
Zinc sulphate
Cupric sulphate.

Solution :Aqueous solutions of `C_(2)H_(5)NH_(2)` contains `C_(2)H_(5-) overset(+)(NH)_(3)` and `OH^(-)` IONS, which will REACT with zinc ions `(Zn^(2+))` to form white ppt. of `Zn(OH)_(2)`.
10337.

The value of electronic charge is equal to :

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`(FARADAY)/(AV. NUMBER)`
`Faraday XX Av.number`
`(Av.number)/(Faraday)`
None

Answer :A
10338.

Which of the following statement(s) is(are) incorrect ?

Answer»

In thioure test for nitric ,a green coloured SOLUTIONIS obtained
it is notnecessary tocarryoutthe chromyl CHLORIDE testin a drytest tube
In `PbNO_(3)` the brownring test can be performed with its WATER extract
Suspension of `CdCO_(4)` gives BLACK ppt , with sodium sulphide solution

Solution :It is deep red colouration due to the formationof `Fe(SCN)_(3)`
b. In presence of moisture following reaction will occure `CrCO_(2)CI_(2) + H_(2)O rarr H_(2)CrO_(4) + CHI`
So test is carried outonly in dry test tube
c. White precipitate of `PBSO4` is formed and hence brown ring in hot visible
d, Gives yellowprwecipitate of `CdS`
10339.

The type of molecular forces of attraction present in the following compounds is :

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INTERMOLECULAR H-bonding
Intramolecular H-bonding
VAN DER WAAL's force
All of these

Solution :
10340.

The unit of equivalent conductivity is

Answer»

ohmcm
`ohm^(-1)cm^(2) `(g equivalent)`""^(-1)`
ohm `cm^(2)` (g equivalent)
`SCM^(-2)`

SOLUTION :`ohm^(-1)cm^(2) (g eq)^(-1)`
10341.

The selection of reducing agent depends on the thermodynamic factor.Explain with an example.

Answer»

Solution :From the Ellingham diagram ,it is clear that METALS for which the STANDARD free energy of formation`(Delta_(f)G^(@))` of their oxides is more negative can reduce the metal for which the standard free energy of formation `(Delta_(f)G^(@))` of oxides is less negative.Thermodynamic factor has a major role in selecting the reducing agent for a PARTICULAR reaction.Only that reagent will be preferred which will lead to decrease in the energy `(DeltaG^(@))` at a certain SPECIFIC temprature.
E.g. Carbon reduce ZnO to Zn but not CO.
`ZnO+CtoZn+CO`
`ZnO+COtoZn+CO_(2)`
In the first case,there is increase in the magnitude of `DeltaS^(@)`while in the second case,it almost remains the same.In other words,`DeltaG^(@)` will have more negative value in the first case,when C is the reducing agent then in the second case when CO acts as the reducing agent.Therfore ,C is a better reducing agent.
10342.

When equal volume of the following solutions are mixed, precipitation of AgCl (K_(sp)=1.8xx10^(-10)) will occur only with :

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`10^(-3)M Ag^(+)` and `10^(-5)M Cl^(-)`
`10^(-5)M Ag^(+)` and `10^(-5)M Cl^(-)`
`10^(-6)M Ag^(+)`and`10^(-5)M Cl^(-)`
`10^(-4)M Ag^(+)` and `10^(-4)M Cl^(-)`

Solution :Ionic product`=[Ag^(+)][Cl^(-)]`
`=(10^(-4))/(2)xx(10^(-4))/(2)=2.5xx10^(-9)`
Since ionic product is greater than solubility product PRECIPITION will take PLACE.
10343.

Unitof firstorder rate constant is

Answer»

`mol dm^(-3)"TIME"^(-1)`
`dm^(-3)mol^(-1)"time"^(-1)`
`"time"^(-1)`
`mol ,dm^(3),"time"^(-1)`

Answer :C
10344.

Which of the following conditions is not satisfied by an ideal solution?

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`triangleH_(" mixing ")=0`
`triangleV_(" mixing ")=0`
RAOULT's law is obeyed
Formation of an AZEOTROPIC mixture

Answer :D
10345.

Which of the following statements is incorrect regarding physisorption ?

Answer»

It occurs because of VAN der Waals' forces
More easily liquefiabie gases are adsorbed readily
Under high pressure it results into multimolecular LAYER on adsorbent SURFACE
Enthalpy of adsorption is low and POSITIVE.

Solution :For physisorption. `DELTAH` is negative.
10346.

Which one of the following reagent is used in the conversion of Ethanol to ethene?

Answer»

`ZN +HG //H_2O`
`LiAlH_4`
ACIDIFIED `K_2Cr_2O_7`
CONC.`H_2SO_4`

SOLUTION :Conc.`H_2SO_4`
10347.

Zinc can be coated on iron to produce galvanized iron but thereverse is notpossible. It is because

Answer»

Zinc is lighter than iron
Zinc has LOWER melting point than iron
Zinc has lower negative electrode potential than iron
Zinc has higher negative electrode potential than iron

Solution :Zinc has higher negative electrode potential than iron
HINT : `E_(ZN^(+)|Zn)^(@)=-0.76V and E_(FE^(2+)|Fe)^(@)=-0.44V`. Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.
10348.

What is paramagnetic character due to ?

Answer»

SOLUTION :UNPAIRED ELECTRONS.
10349.

When SO_(2) is passed through acidified K_(2)Cr_(2)O_(7) solution

Answer»

the solution TURNS blue
the solution is decolourised
`SO_2` is reduced
green `Cr_2(SO_4)_3` is formed.

Solution :`K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+3SO_(2)rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O`
10350.

What is the difference between the electronic configuration of lanthanoids and actinoids?

Answer»

Solution :In CASE of lanthanoids, the 4f orbitals are FILLED PROGRESSIVELY while the 5F orbitals are filled in case of ACTINOIDS.