Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

8051.

Which of the following is an example of absorption?

Answer»

Water on SILICA gel
Water on `CaCl_2`
Hydrogen on FINELY DIVIDED Nickel
Oxygen on metal surface

Answer :A
8052.

What are constituent amines formed when the mixture of (i) and (ii) undergo Hofmann bromamide degradation?

Answer»




Solution :Hofmann - bromamide is an EXAMPLE of intramolecular REARRANGEMENT, THEREFORE, deuterated amine is obtained from deuterated AMIDE and `""^(15)N` amine is obtained from `""^(15)N`-amide.
8053.

Which solution has the highest pH ?

Answer»

0.10 M `CH_3COOH`
`0.10 M HCN`
`0.10 M CH_3COOK`
`0.10 M NABR`

ANSWER :C
8054.

Write the Nernst equation for the cell reaction in the Daniel cell. How will the E_(cell) be affected when concentration of Zn^(+) ions is increased?

Answer»

Solution :`Zu+CU^(2+)toZn^(2+)+Cu,""E_(cell)=E_(cell)^(@)-(0.059)/(2)"LOG"([Zn^(2+)])/([Cu^(2+)])`
When `[Zn^(2+)]` is INCREASED, `E_(cell)` will decrease.
8055.

Which of the following option is correct regarding [Fe(H_(2)O)_(5)NO]SO_(4) ? (P) Fe in +2 oxidation state (Q) It obey's Sidgwick EAN rule It is an outer orbital complex (S) NO ligand has +1 oxidation state.

Answer»

FFTT
FTTT
FFFF
FTFT

Solution :`[OVERSET(+1)FE(overset(""0)H_(2)O)_(5) overset(" "+1)NO]overset(" "-2)SO_(4)
EAN=26-1+12=37`
`SP^(3)d^(2)` hybridisation.
8056.

Which of the following will not react with ammoniacal solution of silver nitrate ?

Answer»

`HC -= HC`
`CH_(3) - C -= CH`
`CH_(3) - UNDERSET(CH_(3))underset(|)(CH) - C -= CH`
`CH_(3) - C -= C - CH_(3)`

ANSWER :D
8057.

When phenol reacts with ammonia is presence of ZnCl_(2) at 300^(@)C, it gives

Answer»

PRIMARY amine
Secondary amine
Tertiary amine
Both (B) and (C)

SOLUTION :
8058.

The volume (in mL) of 0.1 M AgNO_(3)required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H_(2)O)_(5)Cl]Cl_(2), as silver chloride is close to

Answer»


Solution :m MOLES of `[Cr(H_(2)O)_(5)CL]Cl_(2)=0.01xx30=0.3`
`RARR` m mole of `Cl^(-)=0.3xx2=0.6`
`rArr` m mole of `Ag^(+)` = m moles of `Cl^(-)`
`rArr 0.1xx V=0.6`
`rArr V=6 mL`.
8059.

Which of the following statements ragarding a chemical equilibrium wrong

Answer»

An equilibrium can be shifted by altering the temperaturee of pressure
An equilibrium can be shifted by altering the temperaturee or pressure
The same STATE of equilibrium is reacted WHETHER one starts with the REACTANTS or the products
Thet FORWARD reaction SI fovoured by the addition of a catalyst

Solution :At equlibrium rate of forward reaction is equal to the rate of backward reaction.
8060.

Which one of the following is a cationic detergent ?Sodium lauryl sulphate,cetyltrimethyl ammonium bromide,sodium dodecylbenzene sulphonate,Glyceryl palmitate.

Answer»

SODIUM lauryl sulphate
cetyltrimethyl AMMONIUM bromide
Glycerol oleate
sodium stearate

Answer :A
8061.

Which of the following statements are correct ? (I)Both melting and boiling points of H_2O are higher than those of H_2Te. (II)In both N_2O_5 and N_2O_4 all N-O bond lengths are equivalent (III)In both crystalline NaHCO_3, HCO_3^(-) forms only dimeric anion through hydrogen bond. (IV)Amongst B_2, C_2, N_2^(-) and O_2 on further ionization (losing single electron) form thermodynamically more stable species.

Answer»

(I) and (II)
(III) and (IV)
(II) and (III)
(I) and (IV)

Solution :(I)On account of H-bonding `H_2O` has higher melting and boiling points.
`{:(,H_2O,H_2Te),("m.p/K",273,222),("b.p/K",373,269):}`
(II)
(III)(a) In `NaHCO_3` , the `HCO_3^-` ions are linked into an infinite CHAIN and (b) in `KHCO_3, HCO_3^-` FORMS a dimeric anion.

(IV)`B_2-B.O.=1 and B_2^(+) - B.O=1/2 , C_2-B.O. =2 and C_2^(+) B.O.=1.5`
`N_2^(-)-B.O=2.5 and N_2-B.O =3 , O_2-B.O=2 and O_2^(+)B.O.=2.5`
Bond strength `prop` bond ORDER.
8062.

The value of charge of one electron is equal to ................................ .

Answer»

SOLUTION :`1.6 XX 10^(-19)C`
8063.

Two solutions have different osmotic pressures. The solution of lower osmotic pressure is called:

Answer»

ISOTONIC solution
Hypertonic solution
Hypotonic solution
None

Answer :C
8064.

Which of the following product is obtained on performing Riemer Tiemann reaction on phenol ?

Answer»

SALICYLIC acid
Phenyl acetate
SALICYLALDEHYDE
1,4 Benzoquinone

Solution :Salicylaldehyde
8065.

Which is liquid at room temperature :

Answer»

Acetamide
Formation
METHANE THIOL
`CH_3Cl`

ANSWER :B
8066.

Which of the following particle is emitted in the reaction ._(13)Al^(27) + ._(2)He^(4) rarr ._(14)P^(30) +....

Answer»

`._(0)N^(1)`
`._(-1)E^(0)`
`._(1)H^(1)`
`._(1)H^(2)`

SOLUTION :Equate atomic no. and MASS no.
8067.

Which among the following process is used to remove grease from soap

Answer»

EMULSIFICATION
Coagulation
ADSORPTION
PEPTIZATION

SOLUTION :Emulsification
8068.

The values of Delta H and Delta S for the reaction. C("graphite") + CO_(2)(g) rarr 2CO(g) are 170 kJ and 170 JK^(-1), respectively. This reaction will be spontaneous at :

Answer»

910 K
1110 K
510 K
710 K

Solution :For a SPONTANEOUS process, `DELTA G` must be NEGATIVE.
Here, `Delta G` will be negative when `T delta S gt Delta H`
`Delta H = 170 kJ = 170000 J`
`Delta S = 170 JK^(-1) mol^(-1)`
`T Delta S gt Delta H` only when T = 1110 K
8069.

Which compound yields an N-nitroso amine after treatment with nitrous acid (NaNO_(2)+HCl)?

Answer»




`CH_(3)CH_(2)-NH-CH_(3)`

Solution :`2^(@)` AMINES GIVE `"Rea"^(N)`
8070.

Write the energy currency of the cell.

Answer»

ATP
ADP
AMP
Glycolysis.

Solution :ATP is regared as ENERGY CURRENCY of the CELL.
8071.

Write the name of the linkage joining two amino acids.

Answer»

SOLUTION :PEPTIDE BOND , `-CO-NH-`.
8072.

Which of the following wil giveH_(2_((g))) at cathode andO_(2_((g))) at anode on electrolysis using platinum electrodes

Answer»


Solution :For dilute aq. Solution of NACL.
`NaCl=Na^(+)+CL^(-)`
`H_(2)O=H^(+)+OH`
At cathode
`H^(+)` will DISCHARGE SINCE the discharge potential of
`H^(+) lt Na^(+)`
`2H^(+)+2e^(-)toH_(2)(g)`
At anode Due to very lesser amount of `Cl^(-),OH^(-)` ions will discharge at anode
`4OH^(-)toO_(2)+2H_(2)O+4e^(-)`
8073.

Which of the following does not contain material particles

Answer»

ALPHA RAYS
Beta rays
GAMMA rays
Canal rays

Solution :`gamma-`rays does not contain MATERIAL particles
8074.

Which of the following methods is employed to control the corrosion of underground pipelines and tanks?

Answer»

Cathodic protection
Barrier protection
Passivation
Sacrificial protection

Solution :Such pipelines and tanks are usually made of steel (an alloy of iron), and their corrosion or rusting is an electrochemical process.
In cathodic protection, a buried steel pipeline is connected to an active metal (that is, a highly electropositive substance) such as `Mg`. Consequently, a voltaic CELL is formed, the active metal is the anode and iron becomes the cathode. Wet soil forms the electrolyte, and the electrode reactions are `"Anode :" Mg(s) rarr Mg^(2+)(aq.)+2e^(-)`
`"Cathode :" O_(2)(g)+2H_(2)O(l)+4e^(-) rarr 4OH^(-)(aq.)`
As the cathode, the iron-CONTAINING steel pile is protected from oxidation. Of course, the `Mg` rod (or strip) is eventually consumed and must be replaced, but this is cheaper than digging up the pipeline. This uise of an active metal to protect iron from corrosion is called cathodic protection. The more activemetal is preferentially oxidized and is thus called a 'sacrifical anode'.
Similar methods are used to protect bridges and the hulls of ships from corrosion. The small yellow horizontal strips attached to the ship's hull are blocks of `Ti` (coated with `Pt`). The hull is steel (mostly iron). When the ship is insalt water, the `Ti` blocks become the anode and the hull the cathode, in a voltaic cell.
Because oxidation always occurs at the anode, the ship's hull (the cathode) is protected from oxidation (corrosion). Other active matals, such as `Zn`, are also used as sacrificial anodes. Galvanizing (coating the iron with zinc) is another method of corrosion protection. It combines the two approches: sactrified protection and cathodic the two Even if the `Zn` coating is broken so the iron is exposed, the iron is not oxidized. `Zn` is still oxidized in preference to the less reactive `Fe` as long as the two metals remain in contact. Surprisingly, corrosion can also be inhibited if electrons are pumped out of metal. This methodis based on a rather complex phenomenon called passivation. Passivation occurs naturally. Some metals such as `Al` or `Cr` that are high in the `emf` series are relaticely corrosion reistant because of passivation. Iron corrodes completely in dilute nitric acid, but it does not corrode in concentrated nitric acid because of passivatikon. The rapid eraction that takes place in the concentrated acid causes formation of an oxide film that protects the iron from further attack.
The most widely used method is to use paint or a similar coating to prevent `O_(2)` and `H_(2)O` from COMING in contact with the metal surface.
Another group of methods uses what are called corrosion inhibitors, substance that form surface films that INTERFERE with the FLOW of charge needed for corrosion to occur. There are anodic inhibitors, which act at the anode and interfere withe matel dissolution reaction. These include inorganic salt such as chromates, phosphates. and carbontes, as well as compounds containing organic nitrogen or sulfur. There are also cathodic inhibitors, which interfere with the reduc-tion process that occurs at the cathodic region of the metal surface. These inculde salt of magnesium, zinc, or nickel.
8075.

What happens when H_3PO_3 is heated?

Answer»

Solution :On HEATING, `H_3PO_3` UNDERGOES DISPROPORTIONATION to FORM PHOSPHINE and orthophosphoric acid `4H_3PO_3toPH_3+3H_3PO_4`
8076.

What is the side product of soap industry ? Give reactions showing soap formation.

Answer»

SOLUTION :GLYCEROL.
8077.

The substrate will be binded to the active site of the enzyme through

Answer»

Ionic binding
Hydogen bonding
VAN DER Waals INTERACTIONS
Dipole-dipole interactions

Solution :Through ionic , vanderwall forces. H bond and dipole - dipole interactions.
8078.

The shape of [Fe(CO)_(5)]is…………..

Answer»


ANSWER :TRIGONAL bipyramidal
8079.

ThestandardreducationpotentialE^@for halfreactionsare Zn toZn^(2+) + 2e^(-, E^@=- 0.76VFetoFE^(2 +)+ 2e^(-),E^(@)=- 0.41V theEMFof the cellreaction Fe^(2+) +ZntoZn^(2+) +Fe is

Answer»

`-0.35 `V
`+0.35V`
`+1.17V`
`-1.17 V`

SOLUTION :`EMF= E^@` CATHODE-`E^@""_("ANODE")= 0.41- ( - 0.76) = 0.35V `
8080.

Which of the following is not an example of SN^2 reaction ?

Answer»



`CH_3I + (CH_3)_3 C - OVERSET(o.)(O)underset(t-BuOH)(rarr)`
`(CH_3)_3C -Br + CH_3COO^(-) overset(CH_3COOH)`

Answer :D
8081.

Which of the following statements is incorrect for 1s orbital of hydrogen atom?

Answer»

It is possible to FIND an electron at a distance `2a_(0)` (`a_(0)=` Bohr radius)
The magnitude of potential energy is twice of KINETIC energy for a given orbit
The total energy of an electron is maximum in its first orbit.
The probality DENSITY of FINDING an electron is maximum at the nucleus

Answer :C
8082.

The specific reaction rates of a chemical reaction are 2.45xx10^(-5)sec^(-1) at 273 K and 1.62xx10^(-4)sec^(-1) at 303 K. Calculate the activation energy.

Answer»

SOLUTION :`"96.101 kJ/mole"`
8083.

Which are the monomers of Nylon 66 ?

Answer»

Adipic ACID - HMDA
Adipic acid - Butadiene
Phenol - FORMALDEHYDE
MELAMINE - Formaldehyde

Solution :Adipic acid - HMDA
8084.

The shape of 2-butene is:

Answer»

Linear
Planar
Tetrahedral
Pyramidal

Answer :B
8085.

What starting material is required for the following reaction?

Answer»




ANSWER :D
8086.

Which of the following statements is not correct for a pseudo first order rate constant?

Answer»

Its VALUE is independent of the reactant PRESENT in SMALL amount.
Its value is DEPENDENT on the temperature.
Its value will not change if VOLUME is changed.
It is dependent on the concentration of the reactant.

Answer :C
8087.

Which of the following staements is not correct?

Answer»

Argon is USED in electric bulbs
Krypton is OBTAINED during RADIOACTIVE disintegration
Half life of radon is only 3.8 days
Helium is used in PRODUCING very low temperatures.

Answer :B
8088.

Which is not a electrophile

Answer»

`AlCl_(3)`
`BF_(3)`
`(CH_(3))_(3)C^(+)`
`NH_(3)`

SOLUTION :Electron donating species called nucleophile. `NH_(3)` have a LONE pair of electron.
8089.

What is the role of function of sulphur in vulcanization of rubber ?

Answer»

Solution :During vulcanization, natural ID HEATED with `3-5%` sulphur . The FUNCTION of sulphur is to introduce sulphur (or disulphide) BRIDGES or cross-links between polymer chainsthereby IMPARTING more tensile strength, elasticity and resistance to abrasion.
8090.

Which of the following statements is correct regarding formic acid ?

Answer»

It is a reducing agent.
It is a WEAKER acid than acetic acid.
It is an oxidising agent.
When its calcium salt is heated it FORMS ACETONE.

ANSWER :A
8091.

The wavelength of a spectral line for an electronic transition is inversely related to:

Answer»

velocity of electron undergoing transition
number of electrons undergoing transition
the difference in energy levels INVOLVED in the transition.
None of these

Answer :C
8092.

which of thefollowingreactionof aminescan beused todistinguish between1^(0), 2^(0) , 3^(0) amine ?

Answer»

Oxidation
Hydroxylation
Nitration
Acetylation

Answer :D
8093.

Which of the following statement(s) is /are correct with respect to the crystal field theory ?

Answer»

It considers only the metal ion d-orbitals and gives no consideration at all to other metal orbitals.
It cannot account for the `pi` bonding in complexes .
The LIGANDS are POINT charge which are either ions or neutral molecules
The magnetic properties can be explained in term of splitting of d-orbital in different CRYSTAL FIELD.

Answer :A::B::C::D
8094.

Which of the following coordinate complexes is an exception to EAN rule? (Given At. No. Pt =78, Fe=26, Zn=30, Cu=29]

Answer»

`[Pt(NH_(3))(6)]^(4+)`
`[FE(CN)(6)]^(4-)`
`[Zn(NH_(3))_(4)]^(2+)`
`[Cu(NH_(3))_(4)]^(2+)`

Solution :EAN = Z-X +Y
(a) EAN = 78-4+12=86
b) EAN = 26-2+12=36
C) EAN 30-2 +8 = 36
(d) EAN = 29-2+8=35
8095.

What happens to the colour of coordination compound [Ti(H_(2)O)_(6)]Cl_(3) when heated gradually ?

Answer»

Solution :On heating, its colour becomes LIGHTER as `H_(2)O` is GRADUALLY LOST. When all `H_(2)O` molecules are lost, it becomes colourless because in the ABSENCE of ligands, CRYSTAL field splitting does not occur.
8096.

The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is

Answer»

vapour phase refining
LIQUATION
ZONE refining
DISTILLATION

ANSWER :B
8097.

White phosphorus contains :

Answer»

`P_(2)` molecules
`P_(4)` molecules
`P_(6)` molecules
one `P-P` bond.

Answer :B
8098.

The time taken for the completion of 90% of a first order reaction is 't' min . What is the time (in sec) taken for the completion of 99% of the reaction ?

Answer»

`2T`
`t//30`
`120T`
`60T`

ANSWER :C
8099.

Thespecificconductance(orconductivity ) of0.014MNaBrsolutionis 3.16 xx 10^(-4) S cm^(-1) .Findthe molarconductivity of thesolution .

Answer»


Solution :Concetrationof the solution= 0.014 M NaBr
Specificconductance= conductivity = K
`=3.16 XX 10^(-4)S cm^(-1)( or Omega^(-1)cm^(-1))`
Molarconductivity = `^^_(m) = ?`
`^^_(m) = (k xx 1000)/( C)`
` =(3.16 xx 10^(-1)xx 1000)/( 0.014)`
`mol L ^(-1)`
`= 22.57 S cm^(2) mol^(-1)(or Omega^(-1)cm^(2)mol^(-1))`
8100.

Which of the following compound can not show haloform reation ?

Answer»




`PhCH_(2)-OVERSET(O)overset(||)(C)-CH_(3)`

SOLUTION :N//A