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29001.

The law of multiple proportions are illustrated by which pair of compounds?

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NACL and NABR
`H_2O` AND D_2O`
NAOH and KOH
`SO_2 and S_O3`

ANSWER :D
29002.

The law of multiple proportions is illustrated by the pair of compounds.

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SODIUM CHLORIDE and sodium BROMIDE
water and HEAVY water
sulphur DIOXIDE and sulphur trioxide
magnesium hydroxide and magnesium oxide.

Answer :C
29003.

The law of mass action was enunciated by

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Guldberg and WAAGE
Badenstein
Birthelot
GRAHAM

ANSWER :A
29004.

The law of multiple proportion is lillustrated by the pair of compounds

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sodium chlordie and sodium BROMIDE
water and heavy water
sulphur DIOXIDE and sulphur TRIOXIDE
magnesium hydroxide and magnesium OXIDE

Solution :The law of multipw proportion is illustrated by the pair
of coompounds : `SO_(2) and SO_(3).`
29005.

The law of multiple proportion may be illustrated by

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KBR, KI
`H_2O , D_2O`
`CO , CO_2`
`CAO , CaCO_3`

SOLUTION :`H_2O , D_2O`
29006.

The law of constant proportions was enunciated by:

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Dalton
Berthelot
Avodadro
Proust.

ANSWER :D
29007.

The law formulated by Dr. Nernst is

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First LAW of THERMODYNAMICS
SECOND law of thermodynamics
Third law of thermodynamics
Both (a) and (B)

Answer :C
29008.

The laughing gas is

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`N_(2)O_(5)`
`N_(2)O_(3)`
`NO_(2)`
`N_(2)O`

ANSWER :D
29009.

The lattice site in a pure crystal cannot be occupied by ........

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MOLECULE 
ion 
electron 
atom 

Solution :In pure CRYSTALS, there is perfect arrangment of constituent particles such as atoms, ions or molecules. ELECTRONS only occupy LATTICE SITE if there is defect in a crystal.
29010.

The lattice site in a pure crystal cannot be occupied by

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molecule
ion
ELECTRON
atom

Solution :electron
29011.

The lattice enthalpy of KI will be, if the enthalpy of (I) Delta H^- (KI) = -78.0 kcal mol^-1 (II) Ionisation energy of K to K^+ is 4.0 eV(III) Dissociation energy of I_2 is 28.0 kcal mol^-1 (IV) Sublimation energy of K is 20.0 kcal mol^-1(V) electron gain enthalpyfor I to I^- is -70.0 kcal mol^-1(VI) Sublimation energy of I_2 is 14.0 kcal mol^-1 (1 eV =23. 0 kcal mol^-1)

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`+14.1 KJ mol^-1 `
`-14.1 kJ mol^-1 `
`-141 kJ mol^-1 `
`+141 kJ mol^-1 `

ANSWER :C
29012.

The lattice points of a crystal of hydrogen iodide are occupied by :

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HI molecules
H atoms and I atoms
`H^(+)`cations andanions
`H_(2)` molecules and I2 molecules

Solution :Since, HI is a covalent MOLECULE and crystallizes in face centred cubic (FCC) structure, HI molecules are present at the lattice points of the CRYSTAL.
29013.

The lattice points of a crystal of hydrogen iodide are occupied by-

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HI molecules
H atoms and I atoms
`H^(+)` CATIONS and `I^(-)` anions
`H_(2) ` molecules and `I_(2)` molecules

Answer :A
29014.

The lattice enthalpy and hydration enthalpy of four compounds are given below {:("Compound","Lattice enthalpy","Hydration enthalpy"),(,("in kJ mol"^(-1)),("in kJ mol"^(-1))),(P,+780,-920),(Q,+1012,-812),(R,+828,-878),(S,+632,-600):}The pair of compounds which is soluble in water is

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P and Q
Q and R
R and S
P and R

Solution :A compount is SOLUBLE in water when its hydration enthalpy is GREATER than its lattice enthalpy.
29015.

The lattice energy order for lithium halide is :

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`LiFgtLiClgtLiBrgtLil`
`LIClgtLiFgtLiBrgtLiI`
`LiBrgtLiClgtLiFgtLil`
`LilgtLiBrgtLiClgtLiF`

ANSWER :A
29016.

The lattice energy of the lithium halides is in the following order

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LiFgtLiClgtLiBrgtLiI
LiClgtLiFgtLiBrgtLiI
LiBrgtLiClgtLiFgtLiI
LiIgtLiBrgtLiClgtLiF

Solution :`LiF GT LiCl gt LiBr lt LII `
Lattice energy depends on the SIZE and charge of the ION.
29017.

The lattice energy of solid NaCl is 180 kcal/mol. The dissolution of the solid in water in the form of ions is endotermic to the extent of 1 kcal/mol. If the solution energies of Na^(+) and Cl^(-) are in the ratio 6:5, What is the enthalpy of hydration of Na^(+) ion.

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`-85.6` kcal/mol
`-97.5` kcal/mol
`82.6` kcal/mol
`+100` kcal/mol.

ANSWER :B
29018.

The lattice energies of the oxides of Mg, Ca, Sr and Ba follow the order-

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`CaOlt SrOlt BaOlt MgO`
`CaOgt BaOgt SrOgt MgO`
`MgO GT CaO gt SrOgt BAO`
`BaO gt SrO gt CaOgt MgO`

ANSWER :C
29019.

The latent heats of fusion in J g^(-1) of five substances a (mol.mass = 18) , b(mol. mass = 20) , c (mol. mass = 30), d (mol. mass = 60) and e (mol. mass = 30) are respectively 80, 45, 90, 45, 45. Which of the following pair has same value of Delta H_("fusion ")

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a ,C
B ,E
d ,e
c, d

ANSWER :D
29020.

The latest technique used for the purification of organic compounds containing minute quantities is:

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DISTILLATION
Sublimation
Chromatography
Crystallisation

Answer :C
29021.

The latent heat of vapourisation of water is 9700 Cal/mole and if the b.p. is 100^(@)C, equllioscopic constant of water is

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`0.513^(@)C` kg/mol
`1.026^(@)C` kg/mol
`10.26^(@)C` kg/mol
`1.832^(@)C` kg/mol

SOLUTION :`K_(b)=(M_(1)RT_(b)^(2))/(1000 DELTA H_(V))=(18xx1.987xx(373)^(2))/(1000xx9700)=0.513^(@)C`
29022.

The latent heat of vaporisation of water at 100^(@) is 2257 (kJ)/(kg). The Delta H for the process H_(2)O(g) rarr H_(2)O(l) is very nearly

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`+ 2257 J`
`- 2257 J`
`+ 40.7 kJ`
`- 40.7 kJ`

SOLUTION :`DELTA H` for process `= -2257 XX 18`
`= - 40626`
`= - 40.7 kJ`
29023.

The latent heat of fusion of ice is 80 "cal"//g at 0^(@)C what is the freezing point of a solution of Kcl in water containing 7.45 grm of solute in 500 grm of water , assuming the salt is dissociated to the extent of 95%

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`0.73^(@)C`
`-0.73^(@)C`
`272.27K`
`273.73K`

ANSWER :C
29024.

The last member of inert gas elements is

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HELIUM
NEON
ARGON
RADON

ANSWER :D
29025.

The last member o inert gas family is

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Krypton
Radon
Xenon
Argon

Answer :B
29026.

The last member in each period of the periodic table is

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An INERT GAS element
A TRANSITION element
A halogen
An ALKALI metal

Answer :A
29027.

The last filled orbital in lanthanides is

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6S
5s
4s
3s

Answer :A
29028.

The last filled orbital in actinides is

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6S
7s
5s
4s

Answer :B
29029.

The last energy level of radon contains

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10 electrons
8 electrons
2 electrons
15 electrons

Answer :B
29030.

The last element of the P - block in 6^(th) period is representedby theouter most electronicconfiguration :

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`7s^(2)7P^(6)`
`5f^(14)6d^(10) 7s^(2)7p^(0)`
`4f^(14)5D^(10)6s^(2)6p^(6)`
`4f^(14)5d^(10)6s^(2)6p^(4)`

Answer :C
29031.

The last electron which entersthe atom of transition element is called

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s - electron
p - electron
f- electron
d - electron

Solution :LAST `E^-` ENTERS into d-orbital , HENCE d-electron
29032.

The last electron placed in the third(n=3) quantum shell for:

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Kr
Zn
Cu
Ca

Answer :C
29033.

The lassaigne's extract is boiled with dil.HNO_(3) before testing for halogens because

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Silver HALIDES are soluble in `HNO_(3)`
`Na_(2)S and NaCN` are DECOMPOSED by `HNO_(3)`
`Ag_(2)S` is soluble in `HNO_(3)`
AgCN is soluble in `HNO_(3)`

SOLUTION :`Na_(2)S and NaCN`, formed during fusion with metallic sodium must be rmoved before adding `AgNO_(3)`, otherwise black ppt. due to `Na_(2)S` or white precipitate due to AgCN will be formed and thus white precipitate of AGCL will not be identified easily.
`Na_(2)S+2AgNO_(3) to 2NaNO_(3)+underset("Black")(Ag_(2)S)darr`
`NaCN+AgNO_(3) to NaNO_(3)+underset("white")(AgCN)darr`
`NaCl+AgNO_(3) to NaNO_(3)+underset("white")(AgCl)darr`
`Na_(2)S+2HNO_(3) to 2NaNO_(3)+H_(2)Suarr`
`NaCN+HNO_(3) to NaNO_(3)+HCNuarr`
29034.

The Lassaigne's extract is boiled with conc. HNO_(3)whiletesting for halogen. By doing so it

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Increase the CONCENTRATION of `NO_(3)^(-)`
Decomposes `Na_(2)S`and `NaCN` , if FORMED
Helpsin the precipitation of `AgCl`
Increase the SOLUBILITY product of `AgCl`

Solution :N//A
29035.

The largest group in Modern periodic table is

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ANSWER :3
29036.

The largest number of molecules is in:

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36 g of WATER
28 g of `CO_2`
46 g of `CH_3OH`
54 g of `N_2O_5`

SOLUTION :34 g of water
29037.

The largest number of molecules are in

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`36 G H_2O`
28 g CO
46 g `C_2H_5OH`
54G `N_2O_5`

ANSWER :A
29038.

The largest diagonal in a cubic crystal of edge length a is

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`SQRT(3A)`
`sqrt2a`
`sqrt3a`
`sqrt(2a)`

ANSWER :C
29039.

The largest bond angle exists in:

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`H_2Se`
`H_2Te`
`H_2O`
`H_2S`

ANSWER :B
29040.

The largest bond angle is in:

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`NH_3`
`PH_3`
`AsH_3`
`SbH_3`

ANSWER :A
29041.

The large difference in boiling points of alcohols and ethers is due to ________ in alcohols.

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SOLUTION :HYDROGEN BONDING.
29042.

The lanthanum exhibits which of the following oxidation state ?

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only `+2`
`+2 and +3`
only `+3`
only `+3 and +4`

ANSWER :D
29043.

The lanthanoids show a regular decrease in atomic and ionic radii but this decrease in very small due ...................

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SOLUTION :LANTHANOID CONTRACTION
29044.

The Lanthanoids contraction is responsible for the fact that

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ZR and Y have about the same radius
Zr and NB have SIMILAR OXIDATION state
Zr and Hf have about the same radius
Zr and ZN have the same oxidation state

Answer :C
29045.

The lanthanoide which shows all i.e., +2,+3 and +4 oxidation state is

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EUROPIUM
SAMARIUM
neodymium
GADOLINIUM

ANSWER :C
29046.

The lanthanoide ion with e.c. 4 f^(14) 5d ^(0) 6s^(0) has tendecny to

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under GOES REDUCTION
GAIN ELECTRON
lose electron
all of these

Answer :C
29047.

The lanthanoide ion used as oxidizing agent is

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`EU^(3+)`
`Yb^(3+)and Ce ^(3+)`
`Eu^(3+)and TB^(3+)`
none of these

Answer :D
29048.

The lanthanoid elements than has the electronic configuration. [Xe] 4f^(1) 5d^(1) 6s^(2) is

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LUTETIUM
cerium
yterbium
GADOLINIUM

ANSWER :D
29049.

The Lanthanide ion that would show colour is

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`SM^(3+)`
`LA^(3+)`
`Lu^(3+)`
`GD^(3+)`

ANSWER :A
29050.

The lanthanoid contraction refers to :

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valence electrons of the lanthanoid SERIES
ionic RADIUS of the series
the density of the series
nuclear MASS of the series.

Answer :B