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.

Electron affinity of oxygen is less negative than sulphur. Justify this statement.

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Solution :When an electron is added to oxygen, the added electron goes to the .L. SHELL (n=2). As the .L. shell POSSESS smaller region of space, the added electron feels SIGNIFICANT repulsion from the other ELECTRONS present in this level.
E.A of O=-141 KJ `mol e^(-1)`
E.A of S=-200 KJ `mol e^(-1)`.
2.

Electron affinity of Fluorine is less than that of Chlorine because

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Electronegativity of Fluorine is more
2p SUB shell of F is smaller
Chlorine is a STRONGER oxidant
BOND dissociation energy of `F_2` is less

Answer :B
3.

Electron affinity of chlorine is more than that of fluorine explain.

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ANSWER :DUE to small SIZE fluorine REPELS the incoming ELECTRON
4.

Electronaffinityis positivewhen

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O changesinto `O^(-)`
`O^(-)`changesinto `O^(2-)`
O changesinto `O^(+)`
electronaffinityis alwaysnegative

Solution :Sinceelectronaffinityis negative of electrongain enthalpythereforeelectronaffinity is + vewhenO CHANGES to`O^(-)`i.e.,OPTION(a) is CORRECT.
5.

Electron affinity is measured in

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No UNITS
`"kcal MOL"^(-1)`
`"kJ mol"^(-1)`
Both (2) and (3)

Answer :D
6.

Electron affinity is

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Energy required to TAKE out an ELECTRON from an ISOLATED gaseous atom
The tendency of an atom to attract an electron towards itself
Energy absorbed when an electron is a~o an isolated atom in gaseous state
Energy RELEASED when an electron is ADDED to nn isolated atom in the gaseous state

Answer :D
7.

Electromnn - defcienent hydroride is/are

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`BH_(3)`
`AlH_(3)`
`BeH_(2)`
All

Answer :4
8.

Electromeric effect is a temporary effect. Assign reason.

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Solution :The ELECTROMERIC effect is noticed in the organic compounds containing multiple bond i.e., DOUBLE bond `( gt C=C lt` or `gt C=O)` or triple blood `(-C equiv N)` under the influence of the attacking reagent. The moment the attacking reagent is REMOVED, pi `(pi)` electron pair comes BACK to its original position forming multiple bond again. Therefore, electromeric effect is a temporary effect. for details, consult section 12.22.
9.

Electromagnetic radiationof wavelength242nm isjustsufficientto ionise thesodiumatom .Calculatethe ionisation energyofsodiumin Kj mol^(-1)

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Solution :where `lambda= 242nm =242 xx 10^(9) m`wavelengthper photon`E= (hc )/(lambda ) ` permoleE = `(N_(A) hc )/(lambda )`
`n=6.626 xx 10^(34)Js`
`c=3.0xx 10^(8) m`
`N_(A)= 6.022xx 10^(23)` permole
`=(6.022 xx 10^(23) mol^(1)xx 6.626 xx 10^(34)js xx3.0 xx 10^(8) MS^(-1)J)/( (242 xx 10^(9) m))`
`=494.7 xx KJ mol^(-1)495kJ mol^(1)`
Ionisationenthalpy=495kJ `mol^(1)`
10.

Electromeric effect involves the complete transfer of

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`sigma `-ELECTRON
`PI`-electron
proton
both ` sigma" and " pi ` electrons

Solution :N/A
11.

Electromeric effect can be relayed in

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CARBON chain joined by SINGLE bond
carbon chain joined by double bond
carbon chain joined by triple bond
conjugated SYSTEM

ANSWER :D
12.

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJ mol^(-1)

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SOLUTION :`E = N hv = Nh (C)/(lamda) = ((6.02 xx 10^(23) mol^(-1)) xx (6.626 xx 10^(-34) JS) xx (3 xx 10^(8) ms^(-1)))/(242 xx 10^(-9) m)`
`= 4.945 xx 10^(5) J mol^(-1) = 494.5 kJ mol^(-1)`
13.

Electromagnetic radiation with minimum wavelength is

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ULTRA violet
Radio wave 
X-ray 
Infra red 

Answer :C
14.

Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out

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In the PRESENCE of `NaCl`
In the presence of fluorite
In the presence of cryolite which forms a MELT lower melting temperatures
In the presence of cryolite which forms a melt with high melting temperature.

Solution :Electrolytic reduction iof ALUMINA `(Al_(2)O_(3))` to ALUMINIUM by Hall-Heroult, cryolite which forms a melt lower melting temperature.
15.

Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_(2)O_(2) The number of peroxybonds present in X and Y. Then x + y is?

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SOLUTION :`X = H_2SO_4, Y = H_2S_2O_8`
16.

Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_(2)O_(2). The number of peroxy (O-O_ bonds present in X and Y respectively, are

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1,1
1,2
0,1
0,0

Answer :C
17.

Electrolysis of potassium salt of 3-methyl butanoic acid gives an alkane ‘A’. The name of alkylIodide which gives same alkane on treatment with sodium metal in dry ether solvent is

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1-Iodo-3-methyl butane
Isobutyl IODIDE
Neopentyl iodide
It can't be PREPARED by wurtz reaction

Solution :Alkanes properties
18.

Electrolysis of molten saline hydride generally gives

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HYDROGEN at CATHODE
hydrogen at ANODE
OXYGEN at cathode
oxygen at anode

Answer :B
19.

Electrolysis of KCl. MgCl_2 . 6H_2O gives: potassium only magnesium only magnesium and chlorine potassium and magnesium.

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POTASSIUM only
MAGNESIUM only
magnesium and CHLORINE
potassium and magnesium.

Solution :magnesium and chlorine
20.

Electrolysis of aqueous solution of sodium salt of Maleic acid and Fumaric acid gives

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Ethylene and acetylene respectively
Acetylene and ethylene respectively
Ethylene and Ethane respectively
Ethyne in both the CASES

Solution :In both the cases ethyne is FORMED. Ethylene is formed by the ELECTROLYSIS of sodium salt of Succinic ACID, while ethane is formed by the electrolysis of sodium ACETATE.
21.

Electrolysis of aqueous solution of CH_(3)COOK gives

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`CH_(4)`
`C_(2)H_(4)`
`C_(2)H_(6)`
`C_(2)H_(2)`

Solution :`2CH_(3)COOK + 2H_(2)O overset("Electrolysis")RARR H_(3)C - CH_(3) + 2NaOH + 2CO_(2) + H_(2)O`
22.

Electrolysis of aqueous solution of ammonium sulphate and sulphuric acid at anode gives

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`H_2S_2O_8`
`(NH_4)_2S_2O_8`
`H_2O_2`
`H_2`

ANSWER :B
23.

Electrolysis of an aqueous solution of potassium acetate, the hydrocarbon obtained

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Methane
ETHANE
Propane
A MIXTURE of the above three ALKANES

Solution :Kolbe.s REACTION.
24.

Electrolysis of a concentrated aqueous solution of a compound gave C_2H_6 on anode.The compound is

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`CH_3COOK`
`CH_3CH_2COOK`
`CH_3COOC_2H_5`
`CH_3OCH_3`

SOLUTION :`2CH_3COOK+2H_2Ooverset"ELECTROLYSIS"to ubrace(CH_3-CH_3+2CO_2)_"ANODE"+ubrace(2KOH+H_2)_"CATHODE"`
25.

Electrolysis of 50% H_(2)SO_(4) produces

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`H_(2)S_(2)O_(8)` at ANODE
`H_(2)SO_(4)` at anode
`H_(2)SO_(5)` at cathode
`H_(2)O_(2)` at anode

Answer :A
26.

Electrolysis of 50% H.SO, produces

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`H_2S_2O_8` at ANODE
`H_2SO_5` at CATHODE
`H_2` at anode
`H_2O_2` at anode and cathode

ANSWER :A
27.

Electrolysis of 50%" "H_(2)SO_(4) produces

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`H_(2)S_(2)O_(8)` at ANODE
`H_(2)SO_(5)` at cathode
`H_(2)` at anode
`H_(2)O_(2)` at anode & Cathode

Answer :A
28.

Electro-osmosis was discovered by

Answer»

Dorn
Reuss and PORRET
Freundlich
Langmuir

Solution :Reuss and Porret.
29.

Electrinoic configuration of species M^(2+) is 1s^(2) 2s^22p^63s^23p^63d^6 and its atomic weight is 56.The number of neutrons in the nuclcus of species M Is…………..

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26
22
30
24

Solution :`M^(2+)`:`1s^2``2s^2``2p^6``3s^2``3p^6``3d^6`
M:`Is^2``2s^2``2p^6``3s^2``3p^6``3d^8`
ATOMIC number=26
Mass number=56
No.of netutrons=56-26=30
30.

Electriccookers have a coating of thatprotectes them against fire.

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HEAVY lead
Magnesium OXIDE
ZINC oxide
SODIUM sulphate

Answer :2
31.

Electric cookers have a coating that protects them against firer. The coating is made of

Answer»

Heavy LEAD
MAGNESIUM oxide
ZINC oxide
Sodium SULPHATE

ANSWER :B
32.

Elements with high electronegativity are generally

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GOOD REDUCTANTS
HARD solids
Good oxidants
Soft solids

Answer :C
33.

Elaborate the term compound atoms as used by Dalton

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Solution :The TERM compound atoms was used by Dalton to EXPRESS the substances obtained by the combination of atoms of different elements in a FIXED, simple and WHOLE number RATIO.
34.

Eka silicon is now known as

Answer»

SCANDIUM 
Gallium
Germanium 
BORON 

ANSWER :C
35.

Eka-aluminium andeka- silicon were the namesgivenby Mendeleev for thethenunknowns elements. What isitsatomic number? Write its group number, electronic configuration, IUPAC and official names.

Answer»

Solution :The element whichcomesafternercury in thesame group of theperiodictableis calledeka- mercury . Itsvariousparameters are : Z=80 + 32 = 112 IUPAC name : Uub
OFFICIAL name : Cn ( coperniclum) E.C.=[RN]`5f^(14) 6d^(10) 7s^(2)`
36.

Eka-aluminium and Eka-siliconare knownas

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Galliumand GERMANIUM
ALUMINIUMAND SILICON
Ironandsulphur
Protonand silicon

Solution :NA
37.

Einstein was awarded Noble Prize for

Answer»

General THEORY of relativity 
The equation, `E = mc^(2)`
ENUNCIATION of QUANTUM theory 
Explanation of photoelectric EFFECT 

Answer :D
38.

Efficiency of energy conversion in nuclear fission of uranium and nuclear fusion of hydrogen are respectively:

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`0.09%, 0.35%`
`0.35%, 0.09%`
`0.09%, 0.09%`
`0.35%, 0.35%`

ANSWER :A
39.

Efficiency of a Carnot's engine is 100% when

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sing is at`0^(@)C`
sink is at 0K
source is at `273^(@)C`
sourceis at`100^(@)C`

Solution :Efficiency Carnot's cycle is `ETA=(T_(2) - T_(1))/(T_(2))`
`T_(1) =` TEMPERATURE of sink,
`T_(2) =` Temperature of source
For`100%` efficiency , `eta= 1`
HENCE, `1= 1- (T_(1))/(T_(2))` or `(T_(1))/(T_(2))=0` or `T_(1) = 0 K`
40.

Effective magnetic moment of actinides with 'n' unpaired electrons can be calculated using following relation:

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`g SQRT(J(J+1))BM`
`sqrt(N(n+1))BM`
`sqrt(n(n+2))BM`
`1.9sqrt(n+1)BM`

ANSWER :A
41.

Effect of fluorosis disease is......

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irritation in stomach.
causes heart diseases.
weakness of teeth and bones.
weakness of vision

Solution :DAISES like flourocis OCCURS when the AMOUNT of flourine will be more than 10 ppm in drinking water. In which, teeth and bones getting weaked.
42.

Edge length of a cube is 400 pm , its body diagonal would be

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500 pm
600 pm
566 pm
693 pm

Solution :Body DIAGONAL =`SQRT3A` =1.732 x 400 pm =692.28 pm `approx` 693 pm
43.

Edge length of a cube is 400 pm, its body diagonal would be

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500 pm
600 pm
566 pm
693 pm

SOLUTION :BODY diagonal = `SQRT3 a`
` = 1.732 xx 400 "pm" = 692. 28"pm" = 693 "pm" `
44.

Ease of aromatic nucleophilic substitution among these compound will be in the order as :

Answer»

`(V) gt (II) gt (III) gt (I) gt (IV)`
`(IV) gt (I) gt (III) gt( II) gt (V)`
`(IV) gt (III) gt (II) gt (V) gt (I)`
`(V) gt (III) gt (II) gt (IV) gt (I)`

Solution :De activated group INCREASE , reactivity of aromatic NUCLEOPHILIC SUBSTITUTION reaction .
45.

Ease of formation of anion is favoured by

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LOWER VALUE of IONISATION energy
higher value of electronegativity
lower value of ELECTRON affinity
higher value of electron affinity

Answer :D
46.

Earth's protective umbrella existing in the stratosphere is known as ______

Answer»


ANSWER :OZONE LAYER
47.

E_(Al^(3+)//Al)^0 =- 1.66 V and E_(Tl^(3+)//Tl)^0 = 1.26 V. Then which of the following statements is correct?

Answer»

ALUMINIUM has high TENDENCY to form `Al^(3+)` ions in aqueous solution.
`Tl^(3+)` is unstable in aqueous solution
`Tl^(3+)` is a powerful OXIDISING agent
Al is a powerful reducing agent

Answer :D
48.

Each unit cell of NaCl consists of 4 chlorine ions and:

Answer»

13 NA ATOMS
4 Na atoms
6 Na atoms
8 Na atoms

Answer :B
49.

Each rubidium halide crystallising in the NaCl type lattice has a unit cell length 0.30 Å greater than for corresponding potassium salt (r_(K^+)=1.33 Å) of the same halogens . Hence, ionic radius of Rb^+ is

Answer»

1.03 Å
1.18 Å
1.48 Å
1.63 Å

Solution :As RBX crystallies in NaCl type lattice, unit cell length of RbX =`2(r_(Rb^+)+r_(X^-))`
Unit cell length of KX =`2(r_(K^+) + r_(X^-))`
`therefore 2(r_(Rb^+)+r_(X^-))-2(r_(K^+)+r_(X^-))`=0.30 Å
or `r_(Rb^+) - r_(K^+)`=0.15 Å
or `r_(Rb^+) = r_(K^+)`+ 0.15 Å =1.33 +1.15=1.48 Å
50.

Each rubidium halide crystallising in the NaCl type lattice has a unit cell length0.30 Ågreaterthan for corresponding potassium salt ( r_(K^(+))= 1.33 Å ) of the same halogen. Hence, ionic radius ofRb^(+) is

Answer»

1.03 Å
1.18 Å
1.48 Å
1.63 Å

Solution :As RBX crystalliesin NaCl type LATTICE, unit CELL length of RbX = `2 ( r_(Rb^(+)) + r_(X^(-))`
Unit cell length of KX =`2 (r_(K^(+)) + r_(X^(-))`
` 2( r_(Rb^(+)) + r_(X^(-)))-2 (r_(K^(+)) + r_(X^(-))) = 0.30 Å`
` or r_(Rb^(+)) - r_(K^(+)) = 0.15 Å`
` orr_(Rb^(+) = r_(K^(+)) + 0.15 Å = 1.33 + 1.15 = 1.48 Å`