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.

In an equilibrium A+BhArrC+D,A and B are mixed in a vessel at temperature T.The initial concentration of A was twice the initial concentration of B. After the attainment of equilibrium, concentration of C was thrice concentration of B, calculate K_(c).

Answer»

`K_c=1.8`
`[A]_(eqn)=[d]_(eqn) xx 1.66`
`K_c=2.8`
`[A]_(eqn) = [C]_(eqn) xx 2.5`

Solution :
given `[c]_(eqn)=3[B]_(eqn)`
`X=3(a-x), x=(3A)/(4)`
`[A]_(eqn)=(5a)/(4), [B]_(eqn)=(a)/(4), [C]_(eqn)=(3a)/(4), [D]_(eqn)=(3a)/(4)`
2.

In an empty cylinder piston arrangemnet, NO_2(g) at2 atm and N_(2)O_(4)(g)at 4 atm is taken and the constant pressure of 6atm and temperature, 27^circC, is maintained. N_2O_4(g)hArr2NO_2(g),K_(P)=20 atm at 300K Which of the following property(ies) of system will change correctly (as given ) with time?

Answer»

Density of sample will decrease.
AVERAGE molar mass of sample will increase.
The colour of solution becomes more and more DEEPER.
Reaction will not MOVE in any direction.

ANSWER :A::C
3.

In an electrophilic substition reaction of nitrobenzene , the presence of nitro group ______

Answer»

deactivates the ring by inductive EFFECT
activates the ring by inductive effect
decreases the charge density at ortho and para position of the ring relative to META position by resonance
increases the charge density at meta position relative to the ortho and para positions of the ring by resonance .

Solution :STATEMENTS (a) and (c ) are correct but statements (b) and (d) are WRONG .
4.

In an electrophilic substitution reaction to nitrobenzene, the presence of nitro group.

Answer»

Deactivates the ring by inductive EFFECT
Activities the ring by inductive effect
Decreases the charge density at ortho and para POSITION of the rinf relative to META position by resonance.
Increases the charge density at meta position relative to the ortho and para positions of the ring by resonance.

Solution :NITRO GROUP deactivates the benzene ring due to -I effect and decreases electron density at ortho and para positions than meta position. An electrophilic attacks on meta position.
5.

In an electron transition of hydrogen atom orbital momentum may change by :

Answer»

Hund.s RULE
`h/pi`
`h/(2pi)`
`h/(4pi)`

SOLUTION :Change in ANGULAR MOMENTUM `=(n_2- n_1) = h/(2pi)`
POSSIBLE values will be `h/(pi) , h/(2pi) , 3/(2pi) ,….`
(h and `h/(pi)` are not possible )
6.

In an electrolytic cell, the flow of electron mass.

Answer»

CATHODE to anode in SOLUTION
cathode to anode through EXTERNAL supply
cathode to anode through INTERNAL supply
anode to cathode through internal supply

Solution :is the correct answer
7.

In an electrochemical cell consisting of zinc eelctrode and normal hydrogen electrode zinc electrode acts as ……………………

Answer»


ANSWER :an ANODE
8.

In an electrochemical cell Delta G= Delta H

Answer»

<P>`T rarr 0`
`T rarr oo`
`((d E_("cell"))/(dT))_(P) = 0`
`dE_("cell")= 0`

Solution :`Delta G = Delta H rArr Delta S = 0 (or) T rarr 0`
`(Delta S)/(nF) = ((del E_("cell"))/(dT))_(P) rArr ((del E_("cell"))/(dT)) = 0 rArr del E_("cell") = 0`
9.

In an electrochemical cell ………………………….acts as the negatice pole while ………………..acts as the positive pole

Answer»


ANSWER :ANODE CATHODE
10.

In an atom with 2K, 6L, 1M and 2N eletrons the number of electrons with m = 0, s = +(1)/(2) are

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2
7
8
16

Solution :Each subshell i.e., s, p, d, f have an orbital for which m = 0. Thus, for the above CONFIGURATION,
`(1s^(2))/(K), (2s^(2)2p^(6))/(L), (3s^(2)3p^(3)3d^(3))/(M), (4s)/(N)`
The no. of electrons for which m = 0 is `s = + (1)/(2)` equal to = 1+1+1+1+1+1+1 = 7.
11.

In an atom when an electron jumps from K-shell to M-shell

Answer»

ENERGY is ABSORBED
Energy is emitted
Energy is NEITHER absorbed nor emitted
Sometimes energy is absorbed and some TIMES emitted

Answer :A
12.

In an atom two electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is

Answer»

`1:4`
`4:1`
`1:8`
`8:7`

ANSWER :C
13.

In an atom, the total number of electron having quantum number, n = 4(m_(1)) =1 and m_(s) = -1//2 is

Answer»


Solution :When `n = 4, L = 0, 1, 2, 3`
When `l = 0, m = 0`
When `l = 1, m = -1, 0, + 1`
(Two p-orbitals have `|m| = 1`)
When `l = 1, m = -2, -1, 0 + 1, +2`
(Two d-orbitals have `|m| = 1`)
when `l = 3, m = -3,-2,-1, 0 , + 1, + 2, +3`
(Two f-orbitals have `|m| =1`)
Thus, `|m_(1)| =1` for two p, two d and two f-orbitals.
Each of these orbitals has only one electron with `m_(s) = - (1)/(2)`. HENCE, total number of ELECTRONS = 6
14.

In an atom __________ charged nucleus there.

Answer»

LARGE positively
TINY positively
larger NEGATIVELY
tiny negatively

Answer :B
15.

In an atom an electron is moving with a speed of 600 ms^(-1) with an accuracy of 0.005 % certain ty w ith w hich the p osition of the electron can be located is....

Answer»

`5.10 xx 10^(-3)` m
`3.84 x 10^(-3)` m
`1.92 xx 10^(-3)` m
`1.52 xx 10^(-4)` m

Solution :speed v=600 `ms^(-1) xx0.005 xx10^(-2)`
accuracy =0.005 %
MASS of electron=`9.1xx10^(-3)` jkg
`=(6.26xx10^(-34))/(4xx3.141xx9.11xx10^(-31)xx6xx0.005)`
`=1.929xx10^(-3)` m
`=1.93xx10^(-3)` m
16.

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity with which the position of the electron can be located is (h = 6.6 xx 10^(-34) kg m^(2) s^(-1), " mass of electron, " m_(e) = 9.1 xx 10^(-31) kg)

Answer»

`1.52 XX 10^(-4) m`
`5.10 xx 10^(-3) m`
`1.92 xx 10^(-3)m`
`3.84 xx 10^(-3) m`

SOLUTION :`Delta v = (0.005)/(100) xx 600 MS^(-1) = 3 xx 10^(-2) ms^(-1)`
`Delta x xx m Delta v = (h)/(4pi)`
`:. Delta x = (h)/(4pi m Delta v)`
`= (6.6 xx 10^(-34) KG m^(2) s^(-1))/(4 xx 3.14 xx 9.1 xx 10^(-31) kg xx 3 xx 10^(-2) ms^(-1))`
`= 1.92 xx 10^(-3) m`
17.

In an arrangement of type ABABA... Identical atoms of I layer A and III layer A are joined by a line passing through their centers. Suggest the correct statement

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No void is found on the line
Only TETRAHEDRAL VOIDS are found on the line
Only OCTAHEDRAL voids are found on the line
Equal number of tetrahedral and octahedral voids are found on the line

Solution :Only tetrahedral, since there is one tetrahedral voids just above the atom & one just below the atom
18.

In an aqueous solution of volume 500 ml, the the reaction of 2Ag^(+) + CuhArrCu^(2+) + 2Ag reached equilibrium the [Cu^(2+)] was x M. When 500 ml of water is further added, at the equilibrium [Cu^(2+)] will be

Answer»

`2x M`
`X M`
between x M and `x//2 M`
less than `x//2 M`

Solution :`{:(2Ag^(+)+,Cu,hArr,Cu^(2+),+,2Ag),(aM,,,xM,,if V is 500ml):}`
If volume is dobled, then for `K_(C) = (x)/(a^(2))` to have same value `[Cu^(2+)]` will be less than `x//2`
19.

In an antifluorite structure, cationsoccupy

Answer»

OCTAHEDRAL voids
centre of the cube
tetrahedral voids
corners of the cube

Solution :In antifluorite structure, oxide ions are presentin fcc PACKING and METAL ions OCCUPY all thetetrahedral voids.
20.

In an antifluorite structure, cations occupy

Answer»

octahedral voids
centre of the cube
tetrahedral void
corners of the cubic

Solution :In antifluorite STRUCTURE ANION FORM F.C.C structure and cation occupy all tetrahedral void
21.

In an anti bonding molecular orbital, electron density is minimum

Answer»

around one ATOM of the MOLECULE
between two nuclei
at a point AWAY from nuclei of the molecule
at no place

Answer :B
22.

Inan adiabatic process, the work involed during expansionor compressionof an idealgasis givenby :

Answer»

`nC_(V)DELTAT`
`(NR)/(gamma-1)(T_(2)-T_(1))`
`-nRP_(EXT)[(T_(2)P_(1)-T_(1)P_(2)]/(P_(1)P_(2))]`
`-2.303RT log ((V^(1))/(V_(2)))`

Answer :a,b,c
23.

In an adiabatic process, which of the following is true ?

Answer»

q=w
q=0
`DeltaE=q`
`PDeltaV=0`

ANSWER :B
24.

In an adiabatic process ,no transfer of heattakes placebetween system and surroundings.Choose the correct option for free expansion of and ideal gas under adiabatic condition from the following :

Answer»

`q=0 , Delta cancel(=) 0,w=0`
`q cancel(=) 0, DeltaT =0,w=0`
`q =0,DeltaT=0, w=0`
`q=0,DeltaT=0,w cancel (=) 0`

Solution :For FREE expansion , `P_(ext) = 0,w =- P _(ext) DeltaV=0`
For ADIABATICPROCESS , `q=0`
`:. DeltaU =q+w= 0`. This means that internal energy remains CONSTANT which is so at constant TEMPERATURE. Hence `DeltaT =0`
25.

In an adiabatic process, no transfer of heat takes place between system and surroundings.Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Answer»

`q = 0, DeltaT ne 0, w = 0`
`q ne 0 , Delta T = 0 , w = 0`
`q = 0, DeltaT = 0 , w = 0`
`q = 0, DeltaT LT 0, w ne 0`

Solution :Work done in FREE expansion of an ideal gas is ZERO.
`THEREFORE w = 0`
Also, in an adiabatic process, q = 0
`therefore DeltaU = q + w = 0`. This MEANS that during the process, `DeltaU` remains constant.Therefore, during expansion no change in temperature takes place.
`therefore DeltaT = 0`
26.

In an adiabatic process, no transfer of heat takes place between the system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Answer»

`q=0 , Delta T ne 0 , W=0`
`qne 0, Delta T =0, W=0`
`q=0, Delta T = 0, W= 0`
`q=0, Delta T lt 0 , W ne 0`

Solution :Free EXPANSION, W = 0
For ADIABATIC process, q = 0
`Delta U = q+W= 0`, this means that internal energy remains constant, Therefore, `Delta T = 0`.
In IDEAL gas there is no intermolecular attraction. Hence, when such a gas expands under adiabatic conditions into a VACUUM, no heat is absorbed or evolved since no EXTERNAL work is done to separate the molecules.
27.

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

Answer»

`Q=0, DELTAT ne 0, W=0`
`q ne0, DeltaT=0, W=0`
`q=0, DeltaT=0, W=0`
`q=0, DeltaT lt 0, W ne 0`

Solution :For free expansion (i.e. in vacuum), `P_("EXT")=0`.
Thus, `W-P_("EX")DeltaV=0`
`therefore""DeltaU=q+W=0` which is true for isothermal PROCESS where T is constant i.e., `DeltaT=0.`
28.

In an adiabatic expansion of an ideal gas

Answer»

`w=-Deltau`
`w=Deltau+DeltaH`
`Deltau=0`
`w=0`

ANSWER :A
29.

In an acidic buffer solution (pH = 4.4) , the concentration ratio of acid and salt is 2 :1 . The value of dissociation constant of weak acid may be

Answer»

`1.8 XX 10^(-4)`
`2 xx 10^(7)`
`4 xx 10^(-5)`
`2 xx 10^(-5)`

ANSWER :D
30.

In ammionium ion the covalency of nitrogen is

Answer»

3
4
2
5

Answer :B
31.

In allene (C_(3)H_(4)), the type(s) of hybridization of the carbon atoms is (are)

Answer»

sp and `sp^(3)`
sp and `sp^(2)`
only `sp^(2)`
`sp^(2)` and `sp^(3)`

Solution :`overset(sp^(2))(C)H_(2)0 = overset(sp)(CH)=overset(cp^(2))(CH_(2))`.
32.

In alkenes, pi -electrons forming carbon-carbon pi-bond are

Answer»

LOCALISED
delocalised over the ENTIRE molecule
MAY or may not be delocalised
none of the above

Answer :A
33.

In alkane compound with same number of carbon atoms straight chain isomers have higherboiling point as compared to brached chain isomers justify statement

Answer»

SOLUTION :(i) THEBOILINGPOINT ofcontinouschainalkanes increasewithincreasein thelengthofto thechain .
(II)Alkeneschainrouglyabout `30^(@)` everyaddedcarbonatomto thechain .
(III)The boilingpointdecreasewithinbranchingon the moleculei.e., as itbecomescarbonatomsstraight chainisomershavehigherboilingpointas compared tobranchedchainisomers.
34.

In alkaline medium, H_(2)O_(2) reacts with Fe^(3+) and Mn^(2+) respectively to give

Answer»

`Fe^(4+) , and Mn^(4+)`
`Fe^(2+) and Mn^(2+)`
`Fe^(2+) and Mn^(4+)`
`Fe^(4+) and Mn^(2+)`

SOLUTION :The maximum O.S. of Fe is +3 , THEREFORE, `Fe^(3+)` can only be reduced by `H_(2)O_(2)`. In contrast , the minimum stable O.S. of Mn is +2 , therefore can be further oxidised by `H_(2)O_(2)` to + 4 oxidation state. In other words, in alkaline medium, `H_(2)O_(2)` . reduces `Fe^(3+)` to `Fe^(2+)` but is oxidises `Mn^(2+)` to `Mn^(4+)` THUS, OPTION (c) is correct.
35.

In alkaline medium, H_(2)O_(2) reacts with Fe^(3+) and Mn^(2+) respectively to give ?

Answer»

`Fe^(4+)` and`Mn^(4+)`
`Fe^(2+)` and `Mn^(2+)`
`Fe^(2+)` and `Mn^(4+)`
`Fe^(4+)` and `Mn^(2+)`

ANSWER :C
36.

In alkaline earth metals, the electrons are more firmly held to the nucleus and hence

Answer»

Ionization enthalpy, of alkaline EARTH metals is greater than that of alkali metals
Alkaline EARTHS are less ABUNDANT in nature
Reactivity of alkaline earth metals is greater than that of alkali metals
Atoms of alkaline earth metals are bigger in size than alkali metals

Solution :Ionization enthalpies of alkaline earth METAL are greater than those of alkali metals due to increased nuclear charge.
37.

In alkaline conditions, KMnO_(4) reacts as follows, 2KMnO_(4) + 2KOH to 2K_(2)MnO_(4)+H_(2)O + [O] Therefore, its equivalent mass will be :

Answer»

31.6
52.7
72
`158.0`

SOLUTION :N//A
38.

In alkaline earth metals ....... is radio active.

Answer»

Beryllium
Calcium
Barium
Radium

Answer :D
39.

In Al_2 Cl_6 the covalency of aluminium is

Answer»

6
4
3
2

Answer :B
40.

In Al_(2)Cl_(6), the covalency of aluminium is

Answer»

6
4
3
2

Solution :In `Al_(2)Cl_(6)`, the covalency of ALUMINIUM is 4
41.

In Al_(2) Cl_(6) , each Al atom is linked to how many Cl-atoms ?

Answer»


Solution :
THUS , each Al atom is LINKED to 4 CL atoms.
42.

In AgBr, there can occur

Answer»

only Schottky defect
only FRENKEL defect
Both (A) and (B)
None of these

Solution :AGBR has both Schottky and Frenkel defects.
43.

In [Ag (CN)_(2)]^(-), the number of pi bonds in

Answer»

2
3
4
6

Solution :`[N-=C-Ag-C-=N]^(-)`
THUS , NUMBER of `PI` BONDS = 4 .
44.

In afuel cell, methanol is used as fuel and oxygen is used as an oxidiser.The reaction is : CH_(3)OH(l) + 3/2O_(2)(g) to CO_(2) (g) + 2H_(2)O(l) At 298 K, standard Gibbs energies of formation for CH_(3)OH(l), H_(2)O(l) and CO_(2)(g) are -166.2, -237.2 and -394.4 kJ/mol respectively.If standard enthalpy of combustion of methanol is -726 (kJ)/(mol), efficiency of the fuel cell will be

Answer»

0.8
0.87
0.9
0.97

Solution :`CH_(3)OH(l) + 3/2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l) Delta_(F)H = -726 KJ"mol"^(-1)`
`Delta_(r)G = Delta_(f)G_("products") - SigmaDelta_(f)G_("REACTANTS")`
`= [ Delta_(f)G_(co_(2)) + 2 xx Delta_(f)G_(H_(2)O)] - [ Delta_(f)G(CH_(3)OH) + 3/2Delta_(f)G_(O_(2))]`
` = [-394.4 + 2 xx (-237.2)] - [-166.2 + 3/2(0)]`
` = -702.6 kJ"mol"^(-1)`
% Efficiency = `(Delta_(f)G)/(Delta_(f)H) xx 100 = (-702.6)/(-723) xx 100`
96.77 % aaprox 97%
45.

In adsorption chromatography mobile phase will be

Answer»

Only SOLID
Only LIQUID
Only GAS
Liquid as WEIL as gas

ANSWER :D
46.

In acidic medium, the reaction of H_(2)O_(2) with potassium permanganate produces a compound in which the oxidation state of Mn is not?

Answer»

`0`
`+2`
`+3`
`+4`

SOLUTION :N//A
47.

In acidic medium , H_(3)O_(2) changes Cr_(2)O_(7)^(-2)"to"CrO_(5) "Which has two (-O-O-) bonds.Oxidation state of Cr in" CrO_(5) is

Answer»

`+5`
`+3`
`+6`
`-10`

SOLUTION :O.N of CR may be calculated as:
`x+4(-1)+(-2)= or x=6 `
48.

In acidic medium H_(2)O_(2) changes Cr_(2)O_(7)^(2-) to CrO_(5) which has two (-O-O-) bonds. Oxidation state of Cr in CrO_(5) is

Answer»

`+5`
`+3`
`+6`
`-10`

ANSWER :C
49.

In acidic medium H_(2)O_(2) changes Cr_(2)O_(7)^(2-) to CrO_(5) which has (-O-O-) bonds oxidation state of Cr in CrO_(5) is

Answer»

`+5`
`+3`
`+6`
`-10`

Answer :C
50.

In acidic medium dichromate ion oxidises ferrous ion to ferric ion. If the grammolecular weight of potassium dichromate is 294 gm, its equivalent weight is

Answer»

294
147
49
24.5

Answer :C