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 dot at the end of this sentence has a mass of about one microgram. Assuming that the black stuff is carbon, calculate the approximate number of atoms of carbon needed to make such a dot. (1 microgram = 1 xx 10^(-6) g)

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ANSWER :`5 xx 10^(16)` ATOMS
2.

The ascending order of stability of the carbanion CH_(3)^(-) (P), C_(6)H_(5)CH_(2)^(-)(Q), (CH_(3))_(2)CH^(-) (R) and CH_(2) = CHCH_(2)^(-) (S) is

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`P lt R lt S lt Q`
`R lt P lt S lt Q`
`R lt P lt Q lt S`
`P lt R lt Q lt S`

Solution :`(CH_(3))_(2)CH^(-)(R) lt CH_(3)^(-) (P) lt CH_(2) = CHCH_(2)^(-)(S) lt C_(6)H_(5)CH_(2)^(-) (Q)`
THUS, option (b) is correct.
3.

The As-Cl bond distance in AsCl_3 is 2.200^@. Estimate the single bond covalent radius of Arsenic. (Covalent radius of Cl is 0.99A^@).

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SOLUTION :Internuclear distance - RADIUS of chlorine atom = radius of arseinc atom
`2.20-0.99=1.21Å`
COVALENT radius of As=1.21Å`
4.

The AsF_(3) molecule is trigonal bipyramidal. The hybrid orbitals used by the As atoms for bonding are-

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<P>`3_(x^(2)-y^(2)),d_(z^(2)),s,p_(x),p_(y)`
`d_(xy),s,p_(x),p_(y),p_(z)`
`d_(x^(2)-y^(2)),s,p_(x),p_(y)`
`s,p_(x),p_(y),p_(z),d_(z)^(2)`

Solution :`AsF_(5)` has `sp^(3)d` hybridisation. In `sp^(3)d` hybridisaton, the `d_(z^(2))` orbital is used along with the 's' and three 'p' orbitals to form three EQUATORIAL BONDS and two equally strong axial bonds for a trigonal BIPYRAMID.
5.

The ashes of plants contain Alkali metal, 90% of which is

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K
Li
Cs
Ca.

Answer :A
6.

The aryl group left after the removal of a hydrogen atom from benzene is called..............group while that obtained from methyl group of toluene is called............group.

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ANSWER :PHENYL, BENZYL
7.

The artificial sweetener containing chlorine that has the appearance and taste as that of sugar and is stable at cooking temperature is :

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Aspartame
Saccharin
Sucrolose
Alitame

Answer :A
8.

The arrangement of X^(-)ions around A^(-)ion is solid AX is given in the fig (not drawn to scale) If the radius of X^(-) is 250 pm, the radius of A^(+)is

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104 pm
125 pm
183 pm
57 pm

Solution :Cation`A^(+)`OCCUPIES octahedral VOID FORMED by anion `X^(-)`
RADIUS of octahedral void = `0.414times"Radius of the anion packing"`
=`0.414times250 "pm"`
=`103.5 "pm"cong104 "pm"`
9.

The arrangement of orbitals on the basis of energy is based upon their (n +l) value. Lower the value of (n +l), lower is the energy . For orbitals having same value of (n +l), the orbital with lower value of n will have lower energy. I. Based upon the above information, arrange the following orbitals in the increasing order of energy (a) 1s, 2s, 3s, 2p (b) 4s, 3s, 3p, 4d (c) 5p, 4d, 5d, 4f, 6s (d) 5f, 6d, 7s, 7p II. Based upon the above information, solve the questions given below: (a) Which of the orbital out of 4d, 4f, 5s, 5p has the lowest energy ? (b) Which of the orbital out of 5p, 5d, 5f, 6s, 6p has the highest energy ?

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Solution :I (a) `(n +L)` values are `2s =1 +0 =1, 2s = 2 + 0= 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3`
Hence, increasing order of their energy is `1s lt 2s lt 2p lt 3s`.
(B) `4s = 4 + 0 = 4, 3s = 3 + 0 = 3, 3p = 3 +1 = 4, 4d = 4 + 2 = 6` Hence, `3s lt 3p lt 4s lt 4d`
(c) `5p = 5 +1 = 6, 4d = 4 + 2 = 6, 5d= 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6` Hence, `4d lt 5p lt 6s lt 4f lt 5d`
(d) `5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0= 7, 7p = 7 + 1 = 8`. Hence, `7s lt 5f l 6d lt 7p`
II. (a) `4d = 4 + 2 = 6, 4f = 4 + 3 = 7, 5s = 5 + 0 = 5, 7p = 7 + 1 = 8`. Hence, 5s has the lowest energy.
(b) `5p = 5 + 1 =6, 5d = 5 + 2 = 7, 5f = 5 + 3 = 8, 6s = 6 + 0 = 6, 6P = 6 + 1 = 7` Hence, 5f has highest energy.
10.

The arrangement of decreasing sizes of H^(+), H^(-) and H is.....

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`H^(+) GT H gt H^(-)`
`H gt H^(-) gt H^(+)`
`H^(-) gt H^(+) gt H`
`H^(-) gt H gt H^(+)`

ANSWER :d
11.

The arrangement of (CH_(3))_(3)C-, (CH_(3))_(2)CH-, CH_(3)CH_(2^(-)) when attached to benzene or unsaturated group in increasing order of inductive effect is

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`(CH_(3))_(3)C- lt (CH_(3))_(2)CH- lt CH_(3)CH_(2^(-))`
`CH_(3)CH_(2) - lt (CH_(3))_(2)CH- lt (CH_(3))_(3)C-`
`(CH_(3))_(2)CH - lt (CH_(3))_(3)C-lt CH_(3)CH_(2)-`
`(CH_(3))_(3)C- lt CH_(3)CH_(2)- lt (CH_(3))_(2)CH-`

Answer :A
12.

The arrangemen t of orbitals on the basis of energy is based upon th eir (n + Z) value. Lower the value o f (n + Z), low er is th e energy. For o rb itals h aving sam e values of (n + Z), th e o rb ital w ith low er value of n will have lower energy. (I) Based upon the above inform ation, arrange th e fo llo w in g o r b ita ls in th e in c re a s in g order of energy. (a) Is, 2s, 3s, 2p (b) 4s, 3s, 3p, 4d (c) 5p, 4d, 5d, 4f, 6s (d) 5f, 6d, 7s, 7p (II) Based u p o n th e above inform ation, solve the questions given below : (a) W hich of th e following orbitals has the lowest energy ? 4d, 4f, 5s, 5p (b) W hich of the following orbitals has the highest energy ? 5p, 5d, 5f, 6s, 6p

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Solution :(I) (a) `Is lt2s lt 2p lt 3s (B) 3s lt 3p lt 4s lt 4d (c) 4dlt 5p lt 6s lt 4F lt 5d (d) 7S lt 5f lt 6d lt 7p`
(II) (a) 5S (b) 5f
13.

The aromatic species is/are

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tropylium ion
cyclopentadienyl ANION
azulene
PYRROLE

SOLUTION :Cyclopentadienyl anion is non-aromatic
14.

The aromatic compounds present as particulates are

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BENZENE
TOLUENE
NITROBENZENE
POLYCYCLIC AROMATIC hydrocarbons

Answer :D
15.

The area of 10 km to 50 km from the sea level is called .........

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SOLUTION :STRATOSPHERE
16.

The area between 10 km to 50 km from sea level is called stratosphere.

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SOLUTION :TRUE STATEMENT
17.

The aqueous solution of which of the salts has pH close to 7 ?

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`FeCl_(3)`
`CH_(3)CO ON a`
`CH_(3)COONH_(4)`
KCN

Answer :C
18.

The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride ?

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SOLUTION :Explanation for the given statement on the basis of lonisation and effect upon the concentration of SODIUM chloride is given below
(i) Sugar is non-electrolyte so it does not ionise in water while NaCl ionises completely in water and produces `Na^+` and `Cl^-` ion which help in the conduction of electricity.
(II) As concentration of NaCl is increased, more `Na^+` and `Cl^-` ions will be produced. HENCE, conductance or conductivity of the solution increases.
19.

The aqueous solution of salt ofa transition metal ion changes colour from pink to blue, when concentrated hydrochloric acid is added to it. The change in colour is due to:

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EVOLUTION of HYDROGEN the changes the oxidation state of the METAL ion
Change ini the coordination number of the metal ion from 6 to 4 and formation of new species in solution
formation of a coordination complex of the metal ion with HYDROCHLORIC acid
protonation of the metal ion.

Answer :C
20.

The aqueous solution of potash alum [K_(2)SO_(4)*Al_(2)(SO_(4))_(3)*24H_(2)O] is acidic due to

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hydrolysis of `K^+`
hydrolysis of `Al^(3+)`
hydrolysis of `SO_(4)^(2-)`
Presence of ACID in its CRYSTAL as impurity.

Answer :B
21.

The aqueous solution of potash alum is

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Basic
Acidic
Neutral
None of these

SOLUTION :The aqueous solution of potash alum is acidic DUE to cationic hydrolysis of `Al^(3+)` ION PROVIDED by potash alum.
`Al^(3+)+3H_(2)Orarr Al (OH)_(3)+3H^(+)`.
22.

TheaqueoussolutionofNaHCO _(3)is______

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ALKALINE
NEUTRAL
ACIDIC
amphoteric

Answer :A
23.

The aqueous solution of following salt will have the lowest pH

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NaClO
`NaClO_(4)`
`NaClO_(3)`
`NaClO_(2)`

SOLUTION :`NaClO+H_(2)O rarr NaOH+HClO`
`NaClO_(4) +H_(2)O rarr NaOH+HClO_(4)`
`NaClO_(3) +H_(2)O rarr NaOH+HClO_(3)`
`NaClO_(2) +H_(2)O rarr NaOH+HClO_(2)`
As `HClO_(4)` is the strongest, its solution will have the LOWEST pH .
24.

The aqueous solution of borax turns red litmus to

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Blue
No Change
Red
White

Answer :A
25.

The aqueous solution of AlCl_3 shows ……. property.

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Amphoteric
Basic
Neutral
Acidic

Answer :D
26.

The aqueous solution of AlCl_3contains Al^(+3)ions because

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Its higher HYDRATION energy compensates its HIGH ionisation energy 
Its higher ionisation energy compensates its higher hyhdration energy 
Its hydration energy is same as its ionisation energy 
Al is AMPHOTERIC metal

Answer :A
27.

An electron has wavelength 1A^@ the potential by which the electron is accelerated will be

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123 pm
12.3 pm
1.23 pm
0.123 pm

Answer :A
28.

The approximate production of sodium carbonate per month is 424xx10^(6)g. While that of methyl alcohol is 320xx10^(6) gm. Which is produced more in terms of moles ?

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ANSWER :METHYL ALCOHOL
29.

The approximate production of Na_(2)CO_(3) per month is 424 xx 10^(6) g while that of methyl alcohol is 320xx10^(6) g. Which is produced more in terms of moles?

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SOLUTION :`Na_(2)CO_(3)` mass = `424xx10^(6)` G
Molecular mass of `Na_(2)CO_(3)=(23xx2) + 12 + (16xx3)`
= 46+ 12 + 48 = 106 g
No. of moles of `Na_(2)CO_(3) = (" Mass of" Na_(2)CO_(3))/("Molecular mass of" Na_(2)CO_(3))`
=`(424xx10^(6)g)/(106 g)`
= `4xx10^(6)` moles
Methyl alcohol mass = `320xx10^(6)` g
Molecular mass of `CH_(3)OH` = 12 + (1 `xx`4)+ 16 = 32 g
= 12 + 4+ 16 = 32 g
No. of moles of Methyl alcohol = ` ("Mass of Methyl alcohol")/(" Molecular mass of Methyl alcohol")`
`(320xx10^(6)g)/(32 g)`
= `10xx10^(6) mol es.`
`:.` Methyl alcohol is more produced in TERMS of moles
30.

The approximate mass of H_(2)O_(2) present in 200mL of 100 volume H_(2)O_(2) solution is

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33 GRAMS
30 grams
60 grams
66 grams

ANSWER :C
31.

The approximate mass of an electron is 10^(-27)g. Calculate the uncertainty in its velocity if the uncertainty in its position were of the order of 10^(-11)m (h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))

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Solution :`5.25 XX 10^(6) ms^(-1)`
32.

What is metallic hydrides ? Explain with examples ?

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3
30
60
6

Answer :C
33.

The approximate energy required to break A+B~ type ionic crystal into its ions is in the range of

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10 to 100 kJ/mole 
50 to 150 kJ/mole 
500 to 1000 kJ/mole 
2 to 50 kJ/mole 

Solution :Ionic crystal `implies` lattice energy = HIGH.
34.

The appropriate reagent for the transformation is

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`NH_(2)NH_(2), OH^(-)`
`ZN(Hg),HCL`
`H_(2)//Ni`
`NaBH_(4)`

Solution :The CLEMMENSEN reduction takes place in strongly acidic media and can be used for those compounds that are sensitive to BASE. The Wolff-Kishner reduction takes place in strongly basis solution and can be used for those compounds that are sensitive to acid.
35.

The approximate size of an atom is

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`10^(-6)`m
`10^(-8)`m
`10^(-10)`m
`10^(-12)`m

Answer :C
36.

The apparatus shown consists of three bulbs connected by stopcocks of negligible volume . The temperature is constant . P_(A) = 2.13atm "" P_(B) = 0.861 atm P_(C) = 1.15 atm "" V_(A) = 1.50 L V_(B) = 1.0 L "" V_(C) = 2.0 L When all the stopcocks are opened , the pressure in the bulb 'B' will be

Answer»

<P>1.41 ATM
0.861 atm
1.38 at m
1.18 atm

Solution :`P = (2.13 xx 1.5 xx 0.86 xx 1 + 1 .15 xx 2)/(4.5)`
= 1.41 atm
37.

The apir of amphoteric hydroxide is

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`AL(OH)_(3), LiOH`
`Be(OH)_(2), Mg(OH)_(2)`
`Al(OH)_(3), Be(OH)_(2)`
`NI(OH)_(2), Zn(OH)_(2)`

Solution :Both `Al(OH)_(3)` and `Be(OH)_(2)` are amphoteric.
38.

The antiseptic solution of iodopovidone for the use of external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL.

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Solution :`10%w/v` MEANS that 10 g of SOLUTE in 100 ml solution
`therefore ` Amount of iodopovidone in `1.5 MM = (10G)/( 100 ml) xx 1.5 ml = 0.15g`
39.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side : On thePaulingscale theelectronegativityof fluorine is.

Answer»


SOLUTION :FOUR (4.0)
40.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side : How manyseriesof elements constitutef- blockelements

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SOLUTION :TWO i.e., LANTHANOIDS and actiniods.
41.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side : Thenumber ofgroupswhichconstitute p- blockelements is //are

Answer»

SOLUTION :SIX GROUPS (13-18)
42.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side :How manyof the followingelementsare s- blockelements Rb, A1 ,B , K ,S, Cd, Zn , Th, Sr .

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SOLUTION :THREE i.e., RB, K and Srare s-blockelements
43.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side : Howmanyperiodare presentin the Longform of theperiodictable

Answer»

SOLUTION :SEVENPERIODS
44.

The answer to each of thefollowingquestionsis a singledigitintegerrangingfrom 0 to 9. It thecorrectanswers tothe questionsnumbers A, B, Cand S (say ) are 4, 0,9and 2 respectivelythent thecorrectdarkeningof bubbles shouldbe us shownon the side : Totalnumber ofelementspresent in the 2ndshortperiod is

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SOLUTION :EIGHT (Z=3 - 10)
45.

The answer to each of the following questions is a single digit integer, ranging from 0 to 9 . If the correct answers to the questino number A,B,C and D (say) are 4,0,9 and 2 respectively,then the correct darkening of bubbles should be as shown on the side. The equilibrium mixture N_(2)(g) +3H_(2)(g) hArr 2NH_(2)(g)is compressed to half the volume by doubling the pressure at the same temperature. The ratio between constant K_(c) to the reaction quotient (Q_(c)) after compression at the same temperature.

Answer»
46.

The answer to each of the following questions is a single digit integer, ranging from 0 to 9 . If the correct answers to the questino number A,B,C and D (say) are 4,0,9 and 2 respectively,then the correct darkening of bubbles should be as shown on the side. The equilibrium constant for the reaction A_(3)(g)+3B_(2)(g) hArr 2AB_(2)(g)is 464 .0. Then the equilibrium constant of the reaction (1)/(3)A_(3)(g)+B_(2)(g) hArr AB_(2)(g)will be …………..

Answer»
47.

The answer to each of the following questions is a single digit integer, ranging from 0 to 9 . If the correct answers to the questino number A,B,C and D (say) are 4,0,9 and 2 respectively,then the correct darkening of bubbles should be as shown on the side. If concentration of SO_(2) and O_(2) in the equilibrium reaction 2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g) are quadrupled , the concentration of SO_(3) will now be ………………

Answer»
48.

The answer to each of the following questions is a single digit integer, ranging from 0 to 9 . If the correct answers to the questino number A,B,C and D (say) are 4,0,9 and 2 respectively,then the correct darkening of bubbles should be as shown on the side. For the reaction involving oxidation of ammonia by oxygen to form nitric oxide and water vapour, the equilibrium constant has the units("bar")^(n). Then n is ……..

Answer»
49.

The answer to each of the following questions is a single digit integer from 0 to 9. If the correct answer to the question number A,B, C and D say 3,0,8,5 respectively, then the correct darkening of bubbles should be as follow. Among the following, total number of metals which occur inthe native in the earth's crust are. Fe,Zn,Na,Au,Ni,Sb,Sn,Pt,Hg

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SOLUTION :N//A
50.

The answer to each of the following questions is a single digit integer from 0 to 9. If the correct answer to the question number A,B, C and D say 3,0,8,5 respectively, then the correct darkening of bubbles should be as follow. Amonst the following, the total number of ores, which can be concentrated by from floatation process haematite, bauxite, galena, copper pyrites, sphalerite caneterite, calamine, argentite, chalcorite.

Answer»


SOLUTION :N//A