Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

The bond angle H-O-H in ice is closest to

Answer»

`120^(@)28^(')`
`60^(@)`
`90^(@)`
`109^(@)`

Answer :D
Discussion
2.

The bond angle formed by different hybrid orbitals are in the order :

Answer»

` sp^(2) GT sp^(3) gt sp^(3)`
`sp^(3)gt sp^(2)gt sp `
` sp gt gp^(3) gt sp^(2)`
`spgtsp^(2) gt sp^(3)`

Solution :` sp = 180^(@), sp^(2) = 120^(@), sp^(3) = 109^(@) 28 `
Discussion
3.

The bond angle between two hybrid orgials is105^(@) . Calculate the percentege of s-character of the hybrid orbital .Given that cos75^(@) = 0 .2588.

Answer»

Solution : ` cos alpha = - (1)/(m) ` where m = hybridisation index
` therefore - (1)/(m) cos 105^(@) = cos (180 - 75 ) = - cos ^(@) = - 0.2588 orm = (1)/(0.2588) = 3.86`
% s-character ` = (1)/(1+m) xx10- = (1)/(1+3.86) XX100 = 20.58%`
Alternatively , s-character decreases as the bond angle decreases. For example ,
`{:(" Hybrid orbital" ""sp^(3)"" sp^(2) ""sp),("s-character " ""25% ""33.3% ""50%),(" Bond angle """ 109.5^(@) "" 120^(@) ""180^(@)):}`
THUS, when bond angle decreases below ` 109.5^(2)`, the s-character will dcrease accordingly. Although the
decrease is not linear , even then approsimate value can be calculated as follows :
Decrease in angle= `120 - 109.5^(@) = 10.5^(@)`
Decrease in s-character =` 33.3 - 25 = 83 `
ACTUAL decrease in bond angle=` 109.5^(@) - 105^(@) = 4.5^(@)`
` therefore ` EXPECTED decrease in s-character `= (8.3)/(10.5) xx4.5 = 3.56%`
Thus, s-character should decrease by about 3.56%
HENCE, s-character= ` 25 - 3.56 = 21 .44%`
Discussion
4.

The bond angle and dipole moment of water respectively are

Answer»

`109.5^(@), 1*84 ` D
`107*5^(@) , 1*56` D
`104*5^(@), 1*84` D
`102.5^(@), 1*56` D

Answer :C
Discussion
5.

The boiling points of two miscible liquids, which do not form azeotropic mixture, are close to each other. Their separation is best carried out by :

Answer»

Vacuum distillation
FRACTIONAL distillation
STEAM distillation
redistillation

Solution :If boiling POINTS are closer then best separation is DONE by fractional distillation.
Discussion
6.

The boiling point of water = ............""^(@)F.

Answer»

210
212
373
-40

Answer :A::B
Discussion
7.

The boiling point of water in pressure cooker is

Answer»

`100^@C`
`gt 100^@C`
`lt100^@C`
`25^@C`

ANSWER :B
Discussion
8.

The boiling point of water at (i) sea shore (ii) 5000 ft. height is 373 K and 370 K respectively ? Why ?

Answer»

Solution :The boiling is the TEMPERATURE at which vapour pressure is EQUAL to ATMOSPHERE pressure. At HEIGHT atmosphere pressure is less so boiling point is also lower.
Discussion
9.

The boiling point of water at a pressure of 50 atm is 538 K.Compare the theoretical efficiencies of a stem engine operating between the boiling point of water at 50 atm pressure , assuming the temperature of the sink to be 35^(@)C in each case.

Answer»

Solution :Boiling point of water at 50 atmpressure `(T_b)=265^@C`
=265+273=538 K
(i)Boiling point of water at 1 atm PRESSURE `(T_b)=100^@C`
=100+273=373 K
% Efficiency of STEAM engine `=[(T_1-T_2)/T_1]xx100`
`=[(538-373)/538]xx100`
=0.3066 x 100 = 30.66%
(ii) Boiling point of water at 50 atm pressure `(T_b)=35^@C`
=35+273 =328 K
% Efficiency of steam engine =`[(T_1-T_2)/T_1]xx100`
`=[(538-328)/538]xx100=(210xx100)/538`
=39.03 %
Discussion
10.

The boiling point of water (100^(@)C) is much higher than that of HF(19.5^(@)C), even though they have similar molecular masses. Explain.

Answer»

SOLUTION :Each water MOLECULE is INVOLVED in intermolecular H-bonding with four other water molecules but each HF molecule is involved in intermolecular H-bonding with two other HF molecules. Therefore, the degree of molecular association in water is much HIGHER than that in HF.
Discussion
11.

The boiling point of p-nitrophenol is higher than that of o-nitrophenol because

Answer»

`NO_2` group at para position behaves in a different way from that at ORTHO position
INTRAMOLECULAR hydrogen BONDING exists in p-nitrophenol
There is intermolecular hydrogen bonding in p-nitrophenol
p-nitrophenol has a higher MOLECULAR weight than o-nitrophenol

Answer :C
Discussion
12.

The boiling point of H_(2)O is higher than that of H_(2)S . Explain.

Answer»

SOLUTION :DUE to EXTENSIVE intermolecular H-bonding in `H_(2)O` , the b.p. of `H_(2)O` is much HIGHER than that of `H_(2)S`.
Discussion
13.

The boiling point of inert gases increases with the increase in atomic number because "_______________".

Answer»

there is ABSENCE of intermolecular FORCE of attraction between the molecules
dispersion FORCES increase with the mass of the molecule
viscosity DECREASES with increase in mass of the molecular
ionisation ENERGY decreases with increase in atomic number.

Answer :b
Discussion
14.

The boiling point of glycerol is more than propanal because of:

Answer»

Hybridisation
H-bonding
Resonance
All these factors

Solution :It is because the H-Bonding in GLYCEROL is more EXTENSIVE.
Discussion
15.

The boiling point of glycerol is 563K. Before boiling point it is decomposed. So, which is the suitable technique for purification of it?

Answer»

Partitial DISTILLATION
Distillation
Distillation under REDUCED pressure
Steam distillation

Answer :C
Discussion
16.

The boiling point of chorofrom ,CHCl_(3), is 61.7^(@)C and its enthalpy of vaporization is 31.4 KJ-mol^(-1) Calculate the molar entropy of vaporization forchlorofrom :

Answer»

`10.7Jxxmol^(-1).K^(-1)`
`93.8Jxxmol^(-1).K^(-1)`
`301J xxmol^(-1).K^(-1)`
`509J xxmol ^(-1).K^(-1)`

ANSWER :B
Discussion
17.

The boiling point D_(2)O is greater than H_(2)O it if because

Answer»

`D_2O` has a lower ionic PRODUCT
`D_2O` has a lower DIELECTRIC CONSTANT
`D_2O` is an associated liquid
The molecular weight of `D_2O` is greater than `H_2O`

Solution :B.P `prop` MOL. WT
Discussion
18.

The boiling point of benzene is

Answer»

`170^@C`
`80^@C`
`250^@C`
`270^@C`

ANSWER :B
Discussion
19.

The boiling point of aniline is 457k. At which temperature aniline boils in simple distillation and in vepour distillation?

Answer»

SOLUTION :In simple distillation at 457K and in vapour distillation boils at LOWER temperature 373K
Discussion
20.

The boiling point difference the two liquids is not much then..........method used to separate them

Answer»

SIMPLE DISTILLATION
Distillation under REDUCED pressure
steam distillation
fractional distillation

Answer :D
Discussion
21.

The Bohr's energy of a stationary state of hydrogen atom is given as E_(n)=(-2pi^(2)me^(4))/(n^(2)h^(2)). Putting the values of m and e for n^(th) energy level which is not the correct value?

Answer»

`E_(N)=(-21.8xx10^(-19))/(n^(2))J atom^(-1)`
`E_(n)=(-13.6)/(n^(2))eV atom^(-1)`
`E_(n)=(-1312)/(n^(2))kJ MOL^(-1)`
`E_(n)=(-12.8xx10^(-19))/(n^(2))erg atom^(-1)`

Answer :D
Discussion
22.

The Bohr's radius of Li^(2+) of 2^(nd) orbit is………….

Answer»

Solution :0.7053A
`r_n=((0.529xx2^(3)n^(2)))/(Z)A,n=2,Z=3(For Li^(2+))`
`R=(0.259xx2^(2)=(0.529xx4)/(3))=0.7053A`
Discussion
23.

The BOD values of four samples of water A,B,C and D are 165 ppm. 120ppm, 20ppm and 5ppm respectively . The most polluted and least polluted water sample are

Answer»

A & B
B & C
C & D
A & D

ANSWER :D
Discussion
24.

The BOD of pure water is ______ ppm.

Answer»


ANSWER :5
Discussion
25.

The blue compound formed in the positive test for nitrogen with Lassaigne solution of an organic compound is

Answer»

`Na_(4)[FE(CN)_(5)(NOS)]`
`Na_(3)[Fe(CN)_(6)]`
`Fe(CN)_(3)`
`Fe_(4)[Fe(CN)_(6)]_(3)`

ANSWER :D
Discussion
26.

The blue colour of potassium solution in liquid ammonia is due to the presence of -

Answer»

solvated ELECTRON
potassium AMIDE
IMPURITIES present in potassium
potassium oxide

ANSWER :A
Discussion
27.

The blood red colour in the combination test of nitrogen and sulphur in organic compound is the to the formation of

Answer»

FERRIC THIOCYANATE
Ferric acetate
Ferrous sulpho cyanide
Ferric cyanide

Solution :The black red colour in S.F.E test for N & S is due to the formation of `Fe(CNS)_(3)`
Discussion
28.

The blood red colour in the combination test of nitrogen and sulphur in organic compound is due to the formation of:

Answer»

FERRIC sulpho cyanide
ferric acetate
ferrous sulpho cyanide
ferric cyanide

Answer :A
Discussion
29.

The blocking of capillaries due to sudden release of bubbles ofN_2 gas in blood is known as

Answer»

BENDS
BLENDS
MENDS
all of these

ANSWER :A
Discussion
30.

The bleaching action of H_(2)O_(2) is due to the following reaction

Answer»

`PbS+4H_(2)O_(2) RARR PbSO_(4)+4H_(2)O`
`H_(2)O_(2) rarr H_(2)O+(O)`
`H_(2)O_(2)+(O) rarr H_(2)O+O_(2)`
`H_(2)O_(2)+O_(3) rarr H_(2)O+2O_(2)`

ANSWER :B
Discussion
31.

The bimolecular reaction is represented by :

Answer»

(IV),(III),(Q)
(I),(ii),(S)
(III),(ii),(P)
(II),(i),(R)

Answer :D
Discussion
32.

The bicarbonate of alkali metal which does not exist in the solid state is ………… .

Answer»

SOLUTION :`LiHCO_(3)`
Discussion
33.

the biggest particulate matter is

Answer»

`HNO_(3)` DROPLETS
Fly ASH
`H_(2)SO_(4)` droplets
Soot

Solution :Fly ash
Discussion
34.

The best sequence to use to form propylbenzene from benzene is

Answer»

Friedel-Crafts ALKYLATION USING 1-chloropropane and aluminum trichloride.
Friedel-Crafts acylation FOLLOWED by REDUCTION
ELECTROPHILIC addition of 1-chloropropane.
nucleophilic aromatic substitution using propyl Grignard.

Solution :Friedel-Crafts acylation followed by reduction.
Discussion
35.

The best sequence of reactions for the following conversion is

Answer»

(i)1 mol `Br_2//FeBr_3 " " (II)KMnO_4,"HEAT" " " (iii)HNO_3+H_2SO_4`
(i)`HNO_3+H_2SO_4 " " (ii)"1 mol" Br_2//FeBr_3 " " (iii)KMnO_4, "heat"`
(i)`KMnO_4,"heat" " " (ii)HNO_3+H_2SO_4 " " (iii)"1 mol" Br_2//FeBr_3`
(i)`" 1 mol" Br_2//FeBr_3 " " (ii) HNO_3+ H_2SO_4 " " (iii)KMnO_4,"heat"`

ANSWER :A
Discussion
36.

The best sequence of reactions for preparation of the following compound from benzene is

Answer»

(i)`CH_3`COCl/`AlCl_3` (II)OLEUM (iii)`(CH_3)_2`CH-Cl(1 mole ) /`AlCl_3`
(i)`(CH_3)_2CH-Cl` (1 mole)/`AlCl_3` (ii)`CH_3COCl//AlCl_3`(iii)Oleum
(i)Oleum (ii)`CH_3COCl//AlCl_3` (iii)`(CH_3)_2CH-Cl`(1 mole) /`AlCl_3`
(i)`(CH_3)_2CH-Cl`(1 mole) /`AlCl_3`(ii) Oleum (iii)`CH_3CO Cl//AlCl_3`

ANSWER :B
Discussion
37.

The best route for the preparation of BeF_(2) is :

Answer»

thermal decomposition of `(NH_(4))_(2)" BeF_(4)`
`BeO+C+F_(2)OVERSET(600-800K)hArrBeF_(2)+CO`
`Be+F_(2)overset(High T)rarrBeF_(2)`
all of the above

Solution :N//A
Discussion
38.

The best representation of Lewis structure can be selected by using following guidelines . Which one is correct ?

Answer»

A structure in which all FORMAL charges are zero preferred over the ONE with charges .
A structure with small formal charges is preferred over the one with higher formalcharges.
A structure in which NEGATIVE formal charges are placed on the most ELECTRO negative atom is preferred .
All of the above are correct.

Answer :d
Discussion
39.

The best reagents for the ring alkylation of phenol are

Answer»

`C_(2)H_(5)CL` and anhy. `AlCl_(3)`
`CH_(3)CH=CH_(2)` and `H_(2)SO_(4)`
`(CH_(3))_(2)CHOH` and `H_(2)SO_(4)`
`CH_(3)CH=CH_(3)` and HF

Solution :`RX` and anhy. `AlCl_(3)` give poor yield beause `AlCl_(3)` coordinates with oxygen of PHENOL.
Discussion
40.

The best reagent to convert pent-3-en-2-ol into pent-3-in-2-on is

Answer»

Acidic permanganate
Acidic dichromate
Chromic ANHYDRIDE in GLACIAL acetic acid
Pyridinium chloro-chromate

Solution :Pyridinium chlorochromate `(PC C, CrO_(3), C_(2)H_(5)N.HCl)` has an advantage over other oxidising agents that the oxidation of `1^(@)` alcohol stops at aldehyde STAGE (oxidation of aldehyde to carboxylic acid is prevented) and `2^(@)` alcohol containing doube bond is oxidised to ketone retaining the double bond (double bond is not AFFECTED).
Discussion
41.

The best reagent to convert pent-2-en-2-ol into pen-2-ene-2-one is

Answer»

PYRIDINIUM chloro-chromate
Chromic ANHYDRIDE in galcial ACETIC ACID
ACIDIC dichromate
Acidic permanganate

Answer :A
Discussion
42.

The best reagent for converting an alcohol into the corresponding chloride is

Answer»

`PCl_(3)`
`PCl_(5)`
`SOCl_(2)`
`HCl//ZnCl_(2)`

SOLUTION :`SOCl_(2)`
Discussion
43.

The best reagent for converting 2-phenylpropanmide into 2-phenylpropanamine is ……………….

Answer»

excess `H_(2)`
`Br_(2)` in AQUEOUS NaOH
iodine in the presence of RED phophorous
`LiAIH_(4)` in ether

Solution :N//A
Discussion
44.

The best reagent for converting, 2-phenylpropanamide into 1-phenylethanamine is_____________

Answer»

Excess `H_(2)//PT`
`NaOH//Br_(2)`
`NaBH_(4)//` METHANOL
`LiAIH_(4)` / ether

Solution :N//A
Discussion
45.

The best method to separate the mixture of ortho and para nitrophenol (1:1) is

Answer»

steam distillation
crystallization
vaporization
color spectrum

Solution :In `o-nitrophenol`, because chelation (intramolecular hydrogen BONDING) is possible, hydrogen bonding with the solvent water molecules is hindered and hence, solubility is lowered. MOREOVER, SINCE chelation CAUSES the `o-compound` to behave as a monomer, this isomer will be more volatile than the corresponding `m-` and `p-isomers`.

Thus, the effects of chelation are lower solubility and greater volatility in the `o-nitrophenol`, thereby enabling it to be separated from its `m-` and `p-isomers` by steam distillation. Howerver, `o-isomer` may also be separated from the `p-isomer` by crystallization or by CHROMATOGRAPHY.
Discussion
46.

The best method for the separation of naphthalene and benzoic acid from their mixture is

Answer»

sublimation
chromatography
crystallisation
distillation

Solution :Both naphthalene and benzoic acid SUBLIME on HEATING and hence cannot be SEPARATED by sublimation. Benzoic acid is, however, soluble in HOT water but naphthalene and hence can be separated by crystallization.
Discussion
47.

The best method for the preparation of 2,2-dimethylbutane is via the reaction of

Answer»

`Me_3CBr` and `MeCH_2Br` in Na/ether
`(Me_3C)_2 CULI` and `MeCH_2Br`
`(MeCH_2)_2` CuLi and `Me_3CBr`
`Me_3CMgI` and `MeCH_2I`

SOLUTION :
Discussion
48.

The best method for locating the position of double bond in an unknown alkene is ...............

Answer»

SOLUTION :OZONOLYSIS
Discussion
49.

The best leaving group produces a _________ base that is a (n) ________ anion.

Answer»

WEAK, STABLE
STRONG, stable
weak, UNSTABLE
strong, unstable

ANSWER :A
Discussion
50.

The best indicator for detection of end point in litration of a weak acid and a strong base is

Answer»

methyl orange (3 to 4)
methyl RED ( 5 to 6)
bromothymol blue ( 6 to 7.5 )
phenolphthalein (8 to 9.)

Solution :PH range of end point of TITRATION between W.A and `S . B gt 7`
Discussion