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.

………………element subtimes at 889 K.

Answer»

SOLUTION :ARSENIC
Discussion
2.

Element of which group form anions most readily

Answer»

OXYGEN family
Nitrogen group
HALOGENS
Alkali metals

Solution :Halogens are most ELECTRONEGATIVE ELEMENTS.
Discussion
3.

Element present in the volcanic ashes is ………………. .

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OXYGEN
SULPHUR
Selenium
Tellurium

Solution :Sulphur
Discussion
4.

Element of lanthanide series belonging to

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III A
III B
IV A
IV B

ANSWER :B
Discussion
5.

Element of atomic number 23 is placed in the periodic table in

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s-block
p-block
d-block
f-block

Answer :C
Discussion
6.

Element of atomic number 24 belongs to

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s-block
p-block
d-block
both s-and d- blocks

Solution :All PERIODS of p-block conain 6 ELEMENTS.
Discussion
7.

Element furnishing coloured ions in the aqueous medium is

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Zn
Hg
Cu
Al

Answer :C
Discussion
8.

Element found in plant systems which forms an important constituent of photosynthesis is:

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Fe
Cu
Na
Mg

Answer :D
Discussion
9.

Element easily reducible is :

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Fe
Ag
Cu
Sr

Answer :A
Discussion
10.

Element E has bcc crystal arrangement, if its unit cell has edge length of 1.469 xx 10^(-10)meter, then atomic radius of this element is .... meter.

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`2.252 XX 10^(-8)` 
`4.682 xx 10^(-9)`
`6.361 xx 10^(-11)`
`3.582 xx 10^(-10)`

ANSWER :C
Discussion
11.

Element corresponding to which of these//this atomic number belongs to p-block in Modern Periodic Table:

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19
35
53
83

Answer :B::C::D
Discussion
12.

Element 'A' having 5 electrons in its valence shell and principle quantum number value for last electron is 3. The 'ic' acid of element 'A' is tribasic acid and its corresponding salts is known as 'atc' salt. The 'ous acid' 'A' is dibasic acid and its corresponding salts is known as 'its' salts. [''For example : The salt of sulphuric acid is sulphate salt '' and the salt of sulfurous acid is sulphite salt] nxx'ic' acid 'A' -H_(2)O=(y)_(n), if n = 3 How many atoms are sp^(3) hybridised in 'y' ?

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9
10
3
12

Solution :
Discussion
13.

Element A has 3 electron in the outermost orbit andelement B has 7 electrons in the outermost orbit.The formula of the compound formed between A and B would be

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` A_2B_3`
`A_3B_2`
`A_2B`
`AB_3`

ANSWER :D
Discussion
14.

Element A burns innitrogento give an ions compound B reactswith water to give C and D .Asolution of chemes ''milky'' on bubbling carbon dioxide identify A,B,C and D

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SOLUTION :The gives DATE SUGGEST the following reactions
`underset((A))(3Ca) +N_(2) rarr underset((B))(Ca_(3)N_(2))`
`ca_(3)N_(2) + 6H_(2)O rarrunderset((C))( 3Ca(OH)_(2)) + underset((D)) (2NH_(3))`
`underset((E))(Ca(OH)_(2)) +CO_(2) rarr CaCO_(3) +H_(2)O`
It is giventhat a solution of C Become 'milk'` on bubblingcarbon dioxide .Therefore it must becalciumor bariumhydroxidewhich is morethanmagnessiumhydroxide.
Discussion
15.

Element A (atomic mass 112) and element B (atomic mass 27) form chlorides . Solutoins of these chlorides are electrolysed separately and it is found that the same quantity of electricity is passed 5.6 gm of A deposited while only 0.9 gm of B was deposited . if the valency of B is 3 , the valency of A is

Answer»

2
3
4
`-2`

SOLUTION :Equivalent mass of A = `("ATOMIC mass")/("VALENCY") = (112)/(X)`
Equivalent mass of B = `("Atomic mass")/("valency") = (27)/(3) = 9`
`("Mass of A")/("Mass of B") = ("Equivalent mass of A")/("Equivalent mass of B") = (5.6)/(0.9)`
`THEREFORE (112//x)/(9) = (5.6)/(0.9)`
`therefore x = (112 xx 0.9)/(9 xx 5.6) = 2`
Valency of A = 2
Discussion
16.

Element A( atomic mass 112) and element B (atomic mass 27) from chlorides. Solutions of these chlorides are electrolysed seperately and it is found that when the same quantity of electricity is passed 5.6 g of A was deposited while only 0.9 g if B was deposited. If the valency of B is 3, then the valency of A is _________.

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2
3
4
`-2`

ANSWER :A
Discussion
17.

Elemantary unimolecular reactions have first order rate laws, elementary bimolecular reactions have second order rate laws. A rate law is often derived from a proposed mechanism by imposing the state approximation or by assuming that there is a pre -equilibrium. A proposed mechanism must be consistent with the experiment rate law. In a gas phase reaction, a reaction takes place only if the reactant molecules get________________during the molecular collisions

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threshold energy
activation emergy
both
None of these

Solution :If gaseous reactant MOLECULES have the energy GREATER than (or) EQUAL to activation energy, then the reaction TAKES place.
Discussion
18.

Elemantary unimolecular reactions have first order rate laws, elementary bimolecular reactions have second order rate laws. A rate law is often derived from a proposed mechanism by imposing the state approximation or by assuming that there is a pre -equilibrium. A proposed mechanism must be consistent with the experiment rate law. The molecularity of the elementary reaction C_(2)N_(2)to2CN is

Answer»

zero
ONE
two
three

Solution :MOLECULARITY is one
Discussion
19.

Elemantary unimolecular reactions have first order rate laws, elementary bimolecular reactions have second order rate laws. A rate law is often derived from a proposed mechanism by imposing the state approximation or by assuming that there is a pre -equilibrium. A proposed mechanism must be consistent with the experiment rate law. The decomposition of O_(3) obeys themechanism give below Step 1: O_(3)hArrO_(2)+(O) (both forward and backward reactions are fast) Step 2: O_(3)+(O)overset("slow")(to)O_(2)+O_(2) (ignore) backward reaction), the rate of reaction is given by

Answer»

`k[O_(3)]^(2)[O_(2)]`
`k[O_(3)]^(2)[O_(2)]^(-1)`
`k[O_(3)]^(2)`
`k[O_(2)]^(2)`

Solution :`K_(E)=[(O_(2)][O])/([O_(3)])IMPLIES[O]=(K_(e).[O_(3)])/([O_(2)]),r=K_(1)[O_(3)][O]=r=K_(1).[O_(3)].K_(C).([O_(3)])/([O_(2)]),r=K.[O_(3)]^(2)[O_(2)]^(-1)`
Discussion
20.

'Elektron'' is an alloy of

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MG and Zn
Fe and Mg
Ni and Zn
Al and Zn

Solution :`Mg-95%+Zn-5%`
Discussion
21.

Electrorefining is used for

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NA
Mg
Cu
`Cl_(2)`

Solution :Cu-Electrorefining is USED for OBTAINING 100% COPPER.
Discussion
22.

Electrophilic subsitution reaction in phenol take place at :

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p-position
m-position
o-position
o-and p-position

Answer :D
Discussion
23.

Electrophilic substitution in pyridine preferably occurs at position

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`alpha`
`GAMMA`
`beta`
`alpha, beta or gamma ` DEPENDING on the CONDITIONS of the reaction

Answer :C
Discussion
24.

Electrophilic substitution in phenol takes place at

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ORTHO and para-positions
meta-position
ortho-position only
para-position only

ANSWER :A
Discussion
25.

Electrophilic substitution in case of aromatic amines takes place more readily than in benzene. Explain.

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SOLUTION :It is because `NH_2` GROUP is ELECTRON releasing and increases electron DENSITY on benzene ring at 2.4 and 6 positions.
Discussion
26.

Electrophilic substitution in phenol takes place at _______ position.

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<P>o- and p- position
m-position
o-position
p-position

Answer :A
Discussion
27.

Electrophilic substitution in case of aromatic amines takes place more readily than benzene. Why?

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Solution :Due to RESONANCE, the electrons on the N-atom in aniline are delocalized over the benzene RING. As a result, the electron DENSITY on the benzene ring increases as compared to benzene itself. In other words, aniline GETS ACTIVATED and hence electrophilic substitution occurs more readily in aniline.
Discussion
28.

Electrophilic substitution in aromic amines can be carried out by _________ the -NH_(2) group by acetylation.

Answer»


ANSWER :PROTECTING
Discussion
29.

Electrophilicbrominationofp- toluene sulphonicacid, followedby heat with50%H_(2)SO_(4) producesorhto-bromotoluenc. Whichof thefollowingintermediate lendsto thisproduct ?

Answer»




ANSWER :C
Discussion
30.

Electrophilic attack is very difficult for pyridine, but at which position it is relatively easy?

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Only 2
Only 4
Only 3
2 and 4 both

Answer :C
Discussion
31.

Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate reaction. H_(3)C-HC=CH_(2)+H^(+)to?

Answer»

`2^(@)` Carbonation
`1^(@)` Carbocation
`2^(@)` Carbocation
`1^(@)` CARBANION

ANSWER :C
Discussion
32.

Electrophilic addition reaction is easily given by

Answer»

Acetylene
Ethene
Ethane
Benzene

Solution :Inspite of the fact that the two `PI`-bonds are present in alkynes, these are less reactive than ALKENES towards addition reactions (due to symmetrical cylindrical nature of `pi`-electron cloud). Further , alkynes do not show geometrical isomerism due to their LINEAR structure.
The `pi`-electrons of alkynes are more tightly attached to the carbon atoms than `pi`-electrons in alkene and hence less easily AVAILABLE for addition reactions with electrophiles. As a result, alkynes are less reactive than alkenes towards electrophilic addition reaction . A majority of addition reactions of alkynes are catalysed by heavy metal ions such as `Hg^(2+)` and `Ba^(2+)`, etc., while no such catalysts are needed in case of electrophilic addition reactions to alkenes.
Discussion
33.

Electrophiles are usually…….

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LEWIS acid
Lewis base
Bronsted acid
Bronsted base

Solution :Lewis acid are ELECTROPHILE because they ACCEPT electron PAIR.
Discussion
34.

Electrophiles are:

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ELECTRON loving
Electron hating
Nucleus loving
Nucleus hating

Answer :A
Discussion
35.

Electrophile that participates in nitration of benzene is

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`NO^(+)`
`NO_(2)^(+)`
No
`NO_(3)^(-)`

Solution : `HNO_(3)+2H_(2)SO_(4)rightarrowNO_(2)^(+)+2HSO_(4)^(-)+H_(3)O^(-)`
Discussion
36.

Electrophile in the case of chlorination of benzene in the presence of FeCl_3 is

Answer»

`CL^+`
`Cl^-`
`Cl`
`FeCl_3`

SOLUTION :In CHLORINATION ELECTROPHILE is `Cl^+` (chloronium IONS)
Discussion
37.

Electrons of energy 10.2 and 12.09 e Vatom"^(-1) can cause radiations to be emitted from H atoms. Name the spectral line in the Balmer series.

Answer»

SOLUTION :`H_(ALPHA)` LINE
Discussion
38.

Electrons in various sub shell are filled in increasing order of their energies. Pairing of electrons in various orbitals of a subshell takes place only after each orbital is half filled. Moreover no two electrons in an atom can have the same of four quantum numbers. Which of the following have the same number of s-electrons as the d- electrons in Fe^(2+)?

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Li
Na
N
P

Answer :D
Discussion
39.

Electrons occupy the available sub-level which has lower n+l value. This is called:

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HUND's rule
Aufbau principle
Heisenberg's UNCERTAINTY principle
Pauli's principle

Answer :B
Discussion
40.

Electrons are revolving around the nucleus in n_(1^(th)) orbit of an atom, have atomic number Z_1, and in the n_2 orbit of other atom, have atomic number Z_2, then [Where P= Linear momentum, L=Angular momentum, f=frequency of revolution and K.E. =kinetic energy]

Answer»

`L_1/L_2=n_1/n_2`
`P_1/P_2=(Z_1n_2)/(Z_2n_1)`
`f_1/f_2=(Z_2/Z_1)^2 (n_1/n_2)^3`
`((K.E)_1)/((K.E)_2)=(Z_1/Z_2 . n_2/n_1)^2`

SOLUTION :`L_1/L_2=(mv_1r_1)/(mv_2r_2)=(Z_1/n_1xxn_1^2/Z_1)/(Z_2/n_2xxn_2^2/Z_2)=n_1/n_2implies P_1/P_2(mv_1)/(mv_2)=(Z_1/n_1)/(Z_2/n_2)=(Z_1n_2)/(Z_2n_1)`
`(f_1)/(f_2) =(v_1/(2pir_1))/(v_2/(2pir_2))=(Z_1/n_1xxZ_1/n_1^2)/(Z_2/n_2xxZ_2/n_2^2)=(Z_1/Z_2)^2.(n_2/n_1)^3implies (K.E._1)/(K.E._2)=(1/2mv_1^2)/(1/2mv_2^2)=(Z_1/Z_2)^2XX(n_2/n_1)^2=((Z_1n_2)/(Z_2n_1))^2`
`(K.E_1)/(K.E_2)=(1/2(KZ_1e^2)/r_1)/(1/2(KZ_2e^2)/r_2)=(Z_1/n_1^2.Z_1)/(Z_2.Z_2/n_2^2)=(Z_1/Z_2.n_2/n_1)^2`
Discussion
41.

Electronic structure of… is 3d^(3)4s^(0)

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`Cr^(2+)`
`Mn^(4+)`
`Mn^(3+)`
`Fe^(3+)`

Answer :B
Discussion
42.

Electronic configuration of vanadium is

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`3d^(4)4s^(1)`
`3d^(3)4s^(2)`
`3d^(4)4s^(2)`
`3d^(2)4s^(2)`

Answer :B
Discussion
43.

Electronic configuration of noble gases is …………………….

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`ns^(2)`
`ns^(2)NP^(5)`
`ns^(1)np^(6)`
`ns^(2)np^(6)`

ANSWER :D
Discussion
44.

Electronic configuration of Pu^(7+)

Answer»

`7S^(0), 5F^(14), 6d^(0)`
`7s^(0), 5f^(3), 6d^(0)`
`7s^(0) , 5f^(0) ,6d^(0)`
` 7s^(0), 5f^(1), 6d^(0)`

Answer :B
Discussion
45.

Electronic configuration of most electronegative elements is

Answer»

`1s^2 2s^2 2p^6 3s^1`
`1s^2 2s^2 2p^6 3s^2 3p^5`
`1s^2 2s^2 2p^5 3s^1`
`1s^2 2s^2 2p^6 3s^2 3p^6`

SOLUTION :`F(1s^2 2s^2 2p^5)` is the most ELECTRONEGATIVE ELEMENT.
Discussion
46.

Electronic configuration of Ferrous ion is

Answer»

`1s^(2)2s^(2)2P^(6)3s^(2)3P^(6)3d^(6)4s^(0)`
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(2)`
`1^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)`
`1s^(2)2s^(2)2p^(6)2p^(6)3s^(2)3p^(6)3d^(5)4s^(0)`

Answer :A
Discussion
47.

Write the the electronic configuration of Fe^(2+) ion.

Answer»

`1s^(2)2s^(2)2P^(6)3S^(2)3p^(6)3d^(6)4s^(0)`
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(2)`
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(1)`
`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(5)4s^(0)`

ANSWER :A
Discussion
48.

Electronic configuration of divalent cation of Eu is

Answer»

`4f^(6)`
`4f^(7)`
`5F^(7)`
`5f^(8)`

ANSWER :B
Discussion
49.

Electronic configuration of d-block elements is......

Answer»

SOLUTION :`(n-1)d^(1-10)NS^(1-2)`
Discussion
50.

Electronic configuration of copper is 3d^(10)4s^(1) instead of 3d^(9) 4s^(1).Why?

Answer»

Solution :The configuration of COPPER is `3d^(10) 4S^(1)` due to the extra stbility of `d^(10)`SYSTEM in which all 10 electrons are completely paired and 3d-subshell in completely filled.
Discussion