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.

8751.

Vanadium monoxide shows ………………….defect.

Answer»

SOLUTION :SCHOTTKY DEFECT
Discussion
8752.

Which one of the halogens is radioactive ?

Answer»

SOLUTION :ASTATINE (At) is RADIOACTIVE HALOGEN.
Discussion
8753.

Which of the following has maximum number of unpaired electrons

Answer»

 `Mg^(2+)`
`TI^(3+)`
`V^(3+)`
`FE^(2+)`

ANSWER :D
Discussion
8754.

Which one does not belong to ligand:

Answer»

`PH_3`
`NO^+`
`BF_3`
`CI^-`

ANSWER :C
Discussion
8755.

Which of the following are not used as food preservatives ?

Answer»

Table salt
SODIUM HYDROGEN carbonate
CANE SUGAR
Benzoic acid

Solution :Sodium hydrogen carbonate and cane sugar are not PRESERVATIVES.
Discussion
8756.

Which of the following sets contain only copolymers?

Answer»

SBR, Glyptal, Nylon 6,6
Nylon 6, Butyl rubber, NEOPRENE
POLYTHENE, Polyester, PVC
MELMAC, Bakelite, Teflon

Answer :A
Discussion
8757.

What are (X) and (Y) in the following reaction? Give the mechanism of formation of (X) and (Y).

Answer»

SOLUTION :
Discussion
8758.

The total number of fundamental particles in one atom ofoverset 14(6C) is:

Answer»

6
8
14
20

Answer :D
Discussion
8759.

Zn(s)+Cl_(2)(1atm)toZn^(2+)+2Cl^(-). E_(cell)^(o) of the cell is 2.12 V. To increase E

Answer»

`[Zn^(2+)]` should be INCREASED
`[Zn^(2+)]` should be decreased
`[Cl^(-)]` should be decreased
`P_(Cl_(2))` should be decreased

Solution :According to Nernst's EQUATION
`E_(CELL)=E_(cell)^(o)(-0.059)/(2)"LOG"(["Anode"])/(["Cathode"])`
At anode`toZntoZ_(n)^(2+)+2e^(-)`
At cathode`toCl_(2)+2e^(-)to2Cl^(-)`
So, `E_(cell)=2.12(-0.059)/(2)"log"([Zn]^(++))/([Cl^(-)]^(2))`
E will be increase when CONCENTRATIONS of both `[Zn]^(++)` and `Cl^(-)` is decreased.
Discussion
8760.

Which of the following reactions depict the nucleophilic substsitution of C_(2)H_(5)Br ?

Answer»

`C_(2)H_(5)Br+C_(2)H_(5)SN a to C_(2)H_(5)SC_(2)H_(5)+NABR`
`C_(2)H_(5)Br+2HtoC_(2)H_(6)+HB r`
`C_(2)H_(5)Br+AgCN to C_(2)H_(5)NC+AgBr`
`C_(2)H_(5)Br+KOH to C_(2)H_(5)OH+KB r`

Answer :A::C::D
Discussion
8761.

Which of the following pairs represents linkage isomers?

Answer»

<P>`[Cu(NH_(3))_(4)][PtCl_(4)] and [Pt(NH_(3))_(4)][CuCl_(4)]` <BR>`Pd(P Ph_(3))_(2)(NCS)_(2)] and [Pd(P Ph_(3))_(2)(SCN)_(2)]`
`[Co(NH_(3))_(5)(NO_(3))]SO_(4) and [Co(NH_(3))_(5)(SO_(4))]NO_(3)`
`[PtCl_(2)(NH_(3))_(4)]Br and [PtBr_(2)(NH_(3))_(4)]Cl_(2)`.

Answer :B
Discussion
8762.

Which of the following will give a primary amine on hydrolysis

Answer»

ALKYL isocyanide
Alkyl cyanide
Oxime
Nitroparaffin

Answer :A
Discussion
8763.

Which one is the strongest bond?

Answer»

H - CI
CI-CI
C-CI
B-CI

Answer :A
Discussion
8764.

Which of the following act as an antiseptic and disinfectant respectiely?

Answer»

0.2% PHENOL, 1% phenol
1% phenol, 0.2% phenol
2% phenol, 20% phenol
20% phenol, 2% phenol

SOLUTION :0.2% phenol, 1% phenol
Discussion
8765.

Which of the following has the largest size

Answer»

`S^(2-)`
`Se^(2-)`
`O^(2-)`
`TE^(2-)`

SOLUTION :Size increases down the GROUP.
Discussion
8766.

Which of the following is a secondary pollutant

Answer»

`NO_2`
`CO`
`O_3`
`SO_2`

SOLUTION :OZONE `(O_3)` is SECONDARY pollutants which is formed by the interaction of PRIMARY air pollutants.
Discussion
8767.

Which one of the vitamin is syntbesised in. our body by using sun rays ?

Answer»

A
B complex
C
D

Answer :C
Discussion
8768.

Which of the following is an addictive drug ?

Answer»

Papaverine
Pencilline
Sulphadiazine
ASPIRIN

ANSWER :A
Discussion
8769.

What are Wade's Rule?

Answer»

Solution :Wade.s rules are USED to rationalize the SHAPE of borane dclusters by calculating the TOTAL number of SKELETAL eletron PAIRS (SEP) available for cluster bonding.
Discussion
8770.

When excess amount of NH_(3) is reagent with Cl_(2)gives :-

Answer»

`NH_(4)CL`
`N_(2)`
`NCl_(3)`
(1) & (2) both

Answer :D
Discussion
8771.

XeF_2 has a bent shape.

Answer»

SOLUTION :FALSE
Discussion
8772.

Which of the following is a liquid?

Answer»

`SCl_(4)`
`SF_(6)`
`SF_(4)`
`OF_(2)`

Answer :A
Discussion
8773.

Whichis to be used to prevent chemical reaction in food ?

Answer»

LOW temperature
High temperature
Salting
All of these

Answer :D
Discussion
8774.

Which of the following compounds is not chiral

Answer»

`DCH_(2)CH_(2)CH_(2)Cl`
`CH_(3)CH_(2)CHDCl`
`CH_(3)CHDCH_(2)CH_(2)Cl`
`CH_(3)CHClCH_(2)D`

SOLUTION :`DCH_(2)-CH_(2)-CH_(2)-Cl` OTHERS are CHIRAL
`{:(""H),("|"),(CH_(3)-CH_(2)-C^(**)-D,),("|"),(""Cl):}``{:(""H),("|"),(CH_(3)-C^(**)-CH_(2)-CH_(2)-Cl),("|"),(""D):}`
`{:(""H),("|"),(CH_(3)-C^(**)-CH_(2)D),("|"),(""Cl):}`
Discussion
8775.

What are P and Q? Name the reaction occuring in step 1. (i) C_(6)H_(5)-overset(O)overset(||)C- NH_(2)overset(Br_(2)//NaOH)toP (ii) Poverset(NaNO_(2)//HCl,0^(@)C)toQ

Answer»

SOLUTION :`P=CH_(3)NH_(2)` (Methanamine) `Q=CH_(3)OH` (METHYL ALCOHOL)
STEP- 1 reaction is Hoffmann bromomide degradation reaction.
Discussion
8776.

Ultra microscope works on the principle of

Answer»

LIGHT reflection
Light absorption
Light scattering
Light polarization

Answer :C
Discussion
8777.

Which elements in first transition series has maximum third ionization enthalpy?

Answer»

ZN
Cu
Mn
Cr

Solution :`Zn^(2+)` has `d^(10)` CONFIGURATION
Discussion
8778.

Write the empirical formulae of the following : (i) N_(2)O_(4) (ii)C_(6)H_(6)(iii)C_(6)H_(12)O_(6)(iv)H_(2)O_(2)(v)H_(2)O (vi) Na_(2)CO_(3) (vii)CH_(3)COOH

Answer»

Solution :`(i)NO_(2) (II)CH (iii)CH_(2)O (iv)HO (v) H_(2)O (vi)Na_(2)CO_(3)`
Discussion
8779.

Which of the following reactions are disproportionation reactions? (a) Cu^(+) rarr Cu^(2+) + Cu (b) MnO_(4)^(2-) + 4H^(+) rarr 2MnO_(4)^(-) + MnO_(2) + 2H_(2)O (c ) 2KMnO_(4) rarr K_(2)MnO_(4) + MnO_(2) + O_(2) (d) 2MnO_(4)^(-) + 3Mn^(2+) + 2H_(2)O rarr 5MnO_(2) + 4H^(+)

Answer»

a,b
a,b,C
b,c,d
a,d

Solution :`{:(Cu^(+), rarr, Cu^(2+) ,+, Cu),((+1),,(+2),,(0)):}`
`{:(MnO_(4)^(2-),+,4H^(+),rarr,2MnO_(4)^(-),+,MnO_(2),+,2H_(2)O),((+6),,,,(+7),,(+4),,):}`
Discussion
8780.

Whichof thefollowingstands TURE formolecularityof a reaction?

Answer»

It ISTHE SUMOF exponentsof themolarconcentrationof thereactantsin therateequation
It mayhavea fractionalvalue
It isthe numberof moleculesof thereactantstakingpartin a singlestepchemicalreaction
It isdeterminedexperimentally .

Answer :C
Discussion
8781.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV ul("Calculate") the speed u (m s^(-1)) of the hydrogen atoms generated in the above reaction. H_(2) is assumed to be at rest. If you were unable to determine the value for E_(C), then use 5.0 eV for E_(C)

Answer»


ANSWER :The excess ENERGY is `16.7 eV (=21.2 eV-4.5 eV)`. Because two hydrogen atoms are GENERATED upon dissociation, half of this excess is released as translation energy of the hydrogen atoms.
`1/2 m u^(2)=8.35 eV=1.34xx10^(-18) J`
Discussion
8782.

What is true for acetophenone ?

Answer»

It REACTS with alkaline `KMnO_(4)` FOLLOWED by acidic hydrolysis and forms benzoic acid
It reacts with iodine and NaOH to form triidomethane
It is prepared by the reaction of benzene with benzoyl chloride in presence of anhydrous ALUMINIUM chloride
It does not react with freshly prepared ammoniacal silver nitrate solution.

Answer :B
Discussion
8783.

Which of the following volume (V) - temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure?

Answer»




ANSWER :A
Discussion
8784.

Which statement is incorrect about D?

Answer»

D can produce yellow ppt. on reaction with `OVERSET(Theta)" OH"//I_(2)`
D can be prepared by `Ph-C-=CH` on reaction with `Hg^(2+)//H^(+)//H_(2)O`
D can form geometrically isomeric oximes on reaction with `NH_(2)-OH`
D can be converted into ALCOHOL on reaction with `Zn(Hg)//HCl`.

Answer :D
Discussion
8785.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV Considering an energy cycle, ul("determine") the bond energy E_(D)( eV) of H_(2)^(+) to the first decimal place. If you were unable to determine the values for E_(B) and E_(C), then use 15.0 eV and 5.0 eV for E_(B) and E_(C), respectively.

Answer»


Answer :From Fig. `3` below
`E_(D)=E_(B)+E_(C)-DeltaE_(A1)=13.6+4.5-15.4=2.7 EV`
Discussion
8786.

Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism ? Explain your answer. (i) underset((A))(CH_3CH_2Br) " or " underset((B))underset(Br)underset(|)(CH_3CH_2CHCH_3) (ii) underset((C))underset(Br)underset(|)(CH_3CH_2CHCH_3) "or" underset((D))underset(CH_3)underset(|)(H_3C-C-Br) (iii) underset((E))underset(CH_(3))underset(|)(CH_3CHCH_2CH_2Br) "or" underset((F))underset(CH_3)underset(|)(CH_3CH_2CHCH_2Br)

Answer»

SOLUTION :(i) A, (II) C and (III) E
Discussion
8787.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV ul("Calculate") the threshold energy E_(E) (eV) of the following dissociative reaction to the first decimal place: H_(2) rarr H^(**) (n=2) +H^(+)+e^(-) If you were unable to determine the values for E_(B) and E_(C), then use 15.0 eV and 5.0 eV for E_(B) and E_(C) respectively. When H_(2) absorbs monochromatic light of 21.2 eV, the following dissociation process occurs at the same time. H_(2) overset(21.2 eV)(rarr)H(n=1)+H(n=1) Two hydrogen atoms move in opposite direction with the same speed.

Answer»


ANSWER :`(##RES_PHY_CHM_V01_XI_C02_E01_411_A01##)`
From figure `3` above, the threshold energy `E_(E)` for the dissociative IONIZATION reaction
`H_(2) RARR H^(**)(n=2)+H^(+)+e^(-)` is `E_(B)+E_(C)+10.2 EV=13.6+4.5+10.2=28.3 eV. E_(E)=28.3 eV`
Discussion
8788.

Which is the possible number of stereo isomers for the following formula ? CH_(2)OH(CHOH)_(3)CHO

Answer»

4
8
12
16

Solution :`2^(3)=8`
Discussion
8789.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV ul("Determine") the bond energy E_(C)(eV) of H_(2) to the first decimal place.

Answer»


ANSWER :`24.9 eV`= BINDING energy of a HYDROGEN molecule `+10.2 eV+10.2 eV`
Thus, the binding energy of a hydrogen molecule `E_(C)=4.5 eV`
Discussion
8790.

Which is the correct formula of freon 12?

Answer»

`"CCl"_(2)F_(2)`
`CF_(3)`Cl
`CHCl_(2)`F
`"CCl"_(3)`F

Answer :A
Discussion
8791.

Which of the following is chemisorption

Answer»

ADSORPTION of `H_2` on NI at high temperature
adsorption of `H_2` on charcoal 
adsorption of MOISTURE on SILICA gel 
DEHYDRATION by using anhydrous `CaCl_2`

Answer :A
Discussion
8792.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV {:a) ul("Determine") the energy difference E_(A1) (eV) between H_(2)(v=0) and H^(+)(V_(ion)=0) to the first decimal place. v and v_(ion) denote the vibrational quantum numbers of H_(2) and H^(+), respectively. {:b) ul("Determine") the energy difference E_(A2) (eV) between H^(+) (v_(ion)=0) and H^(+)(v_(ion)=3) to the first decimal place. The electronic energy levels E_(n)^(H) of a hydrogen atom are given by the equation E_(n)^(H)=-(Ry)/n^(2) "" (n=1, 2, 3 ....) Here n is a principal quantum number, and Ry is a constant with dimensions of energy. The energy from n=1 to n=2 of the hydrogen atom is 10.2 eV.

Answer»


Answer :(a) The spectral peak at `5.8 eV` in Fig `2` corresponds to the ELECTRON with the highest kinetic ENERGY, which is generated by the REACTION
`H_(2)(V=0) rarr H_(2)^(+)(V_(ion)=0)+e`
Accordingly,
`DeltaE_(A1)=21.2 eV-5.8 eV=15.4 eV`
(b) One can estimate from Fig. `2` that the energy DIFFERENCE `DeltaE_(A2)` between `H_(2)^(+)(V_(ion)=0)` and `H_(2)^(+)(V_(ion)=3)` is approximately `0.8 eV`.
The answer are as follows: `DeltaE_(A1)=15.4 eV`
`DeltaE_(A2)=0.8 eV`
Discussion
8793.

Write the Haworth's structure of -D(+) Glucose.

Answer»

SOLUTION :
Discussion
8794.

When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV ul("Calculate") the ionization energy E_(B)( eV) of the hydrogen atom to the first decimal place. The energy threshold for the generation of two electronically excited hydrogen atoms H^(**) (n=2) from H_(2)(v=0) has been derived to be 24.9 eV by an experiment.

Answer»


Answer :The ionization energy corresponds to `n=oo`. Accordingly,
`DeltaE_(n=2larr n=1)=`^(3)//_(4) RY``
`DeltaE_(n=2 larr n=1)=Ry`
Thus, the energy required for the ionization is `4/3` TIMES larger than the TRANSITION energy of the Lyman `alpha`-LINE.
`E_(B)=10.2 eVxx4/3=13.6 eV`
Discussion
8795.

Which of the followingreagentswill be able to disguish between allyl bromide and n- propylbromide?

Answer»

Aqueous `AgNO_(3)`<BR>`NaOH, AgNO_(3)`
`Alk. KMnO_(4)`
Tollens reagent

Solution :
a. `Aq. AgNO_(3)` wil give test for `Br^(o-)` by both.
b. `G.R` wil react with both.
d. Tollens reagent will not react with both.
Discussion
8796.

Which of these is true regarding the 3d orbital ?

Answer»

It possesses TWO radial nodes
It possesses two conical nodes
It possesses ZERO angular nodes
All of the aboveare correct STATEMENTS

ANSWER :B
Discussion
8797.

When phenol is heated with zinc dust, the product is

Answer»


ANSWER :BENZENE
Discussion
8798.

When excess of FeSO_(4) is added to sodium extract the compound formed "is" // "are"

Answer»

`Na_(2)[Fe(CN)_(6)]`
`Fe(OH)_(2)`
`Na_(2)SO_(4)`
`Na_(2)[Fe(SCN)]`

SOLUTION :`6NaCN + FeSO_(4) to Na_(4)[Fe(CN)_(6)] + Na_(2)SO_(4)`
`Na_(2)S + FeSO_(4) to Fes + Na_(2)SO_(4)`.
Discussion
8799.

Which alkene on ozonolysis givenCH_3CH_2CHO and CH_3 COCH_3?

Answer»

`CH_3CH_2CH = C(CH_3)_2`
`CH_3CH_2CH = CHCH_2CH_3`
`CH_3CH_2CH = CHCH_3 `
`(CH_3)_2C=CHCH_3 `

ANSWER :A
Discussion
8800.

Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?

Answer»

RELATIVE lowering in vapour pressure
Elevation of boiling point
Depression in FREEZING point
OSMOTIC pressue

Solution :is the correct answer.
Discussion