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.

Standard entropies of X_(2), Y_(2) and XY_(3) are 60,40 and 50 kJ"mol"^(-1) respectively. For the reaction: 1/2 X_(2) + 3/2Y_(2)iff XY_(3), Delta H = -30 kJ to be at equilibrium, the temperature should be

Answer»

500 K
750 K
1000 K
1250 K

Solution :`DeltaS^(@) = S^(@)(XY_(3)) -1/2S^(@)(X_(2)) - 3/2S^(@)(Y_(2))`
` = 50 -1/2(60)-3/2(40) = -40J K^(-1) "mol"^(-1)`
`T = (DeltaH^(@))/(DeltaS^(@)) = (-30 xx 10^(3) J"mol"^(-1))/(-40 J K^(-1)"mol"^(-1))` = 750 K
Discussion
2.

Standard entropies of X_(2), Y_(2) andXY_(3) are 60,40 and 50JK^(-1)mol^(-1) respectively. For the reaction(1)/(2)X_(2) + (3)/(2)Y_(2) hArr XY_(3) , DeltaH= - 30 kJto be at equilibrium, the temperature should be

Answer»

500K
750 K
1000K
1250 K

Solution :`(1)/(2) X_(2) + ( 3)/(2)Y_(2) HARR XY_(3)`
`DeltaS^(@) = SigmaS_(P)^(@) - Sigma _(R)^(@) = 50-( 30+60) = - 40 JK^(-1) MOL^(-1)`
`DeltaG^(@) =DeltaH^(@) - T DeltaS^(@) ` . At equilibrium , `DeltaG^(@) = 0`. HENCE,`TDELTAS^(@) = DeltaH^(@)`
or `T= (DeltaH^(@))/( DeltaS^(@)) = ( -30 xx 10^(3) J mol^(-1))/( -40JK^(-1) mol^(-1))= 750 K`
Discussion
3.

Standar entropies of x_(2), y_(2) and xy_(3) are 60,40 and 50JK^(-1) mol^(-1) respectively for the reaction to be at equilibrium, the temperature should be (1)/(2) x_(2) +(3)/(2) y_(2) hArr xy_(3) Delta H= - 30kJ

Answer»

750K
1000K
1250K
500K

Solution :`(1)/(2) y_(2) + (3)/(2) y_(2) hArr xy_(3)`
`Delta S = Delta S_(xy) - ((1)/(2) sy_(2) + (3)/(2) sy_(2))`
`50 - ((1)/(2) xx 60 + (3)/(2) xx 40) = -40JK^(-1) mol^(-1)`
`Delta G = Delta H - T Delta S, Delta G = 0 = T = (Delta H)/(Delta S) = 750k`
Discussion
4.

Standard entropies of x_(2) , y_(2) and xy_(3) are 60,40 and 50 "J/K mol"^(-1) respectively for the reaction (1)/(2) x_(2) + (3)/(2) y_(2) to xy_(3) , Delta H = - 30 "kJ" to be at equilibrium the temperature should be.

Answer»

500 K
750 K
1000 K
1250 K

Solution :`Delta S = 50 - ((60)/(2) + (3)/(2) XX 40)`
`= 50 (30+ 60) = - 40 "J/K mol"`
`Delta G = 0`
`therefore Delta H = T Delta S therefore T = (-30 xx 10^(3) )/( - 40) = 750 K`
Discussion
5.

Standard entropies of X_(2) , Y_(2) and XY_(3) are 60,40 and 50 "J/K mol"^(-1) respectively for the reaction (1)/(2) X_(2) + (3)/(2) Y_(2) to XY_(3) , Delta H = - 30 "kJ" to be at equilibrium the temperature should be.

Answer»

750 K
1000 K
1250 K
500 K

Solution :`Delta S` for the reaction `(1)/(2) X_(2) + (3)/(2) Y_(2) HARR XY_(3)`
`Delta S = 50- (30 + 60)`
`= - 40 J`
For equilibrium `Delta G = 0 = Delta H - T Delta S`
`T= (Delta H )/( Delta S ) = (- 30000)/( - 40)`
`= 750 K`
Discussion
6.

The standard molar engthalpy of vaporisation of benzene Delta_(vap)H^(@) at 353 K is 30.8 kJ mol^(-1). If the benzene vapours behave as an ideal gas, the change in internal energy of vaporisation of 78 g of benzene at 353 K in kJ mol^(-1) is

Answer»

`+37.56`
`-43.76`
`+43.76`
`+40.66`

ANSWER :A
Discussion
7.

Standard enthalpy of vaporisation Delta_("vap") H^(@) for water at100^(@)C is40.66 kJ mol^(-1) . The internal energy change of vaporisation of water at100^(@)C ( in kJmol^(-1)) is:

Answer»

`+37.56`
`-43.76`
`+ 43.76`
`+40.66`

Solution :`underset(1 "mole")(H_(2)O(l)) rarrunderset("1 mole")(H_(2)O(g)), Deltan_(g)=1`
`DeltaU =DeltaH - Deltan_(g)RT`
`= 40.66xx 1000 J mol^(-1)-1 XX 8.314 JK^(-1) xx 373 K`
`= 40660 - 3101 J mol^(-1)`
`=37559 J mol^(-1)`
`=37. 56 kJ mol^(-1)`
Discussion
8.

Standard enthalpy of formation of H_(2(g)) , CI_(2(g)) and HCI_((g)) are 218, 121.68 and -92.31 KJ/mole respectively. Calculate the change in standard enthalpy for the reaction (1)/(2) H_(2(g)) + (1)/(2) CI_((g)) to HCI_((g))

Answer»

`+431.99` KJ
`-262.14` kJ
`-431.99` kJ
`+ 247.37` kJ

Answer :B
Discussion
9.

Standard enthalpy of ........... is not zero.

Answer»

graphite
`Na_((s))`
`N_(2(L))`
Rhombic sulphur

Answer :C
Discussion
10.

Standard enthalpy of formation is zero for ........... from following

Answer»

`F_(2(g)) `
`F_((g))`
`HF_((AQ))`
`F_((q))^(-)`

ANSWER :A
Discussion
11.

Standard enthalpies of C_(6)H_(6)(l),H_(2)O(g) " and "CO_(2)(g) are respectively 11.7, -68.3 and -94 kCal. Calculate the calorific value of benzene.

Answer»


ANSWER :10.008 KCAL
Discussion
12.

Standard electrode potential values Fe^(2+)//Fe,E^(@)=-0.44 V , Fe^(3+)//Fe^(2),E^(@)=0.77 v If Fe^(2+),Fe^(3+) and Fe blocks are kept together it will lead to

Answer»

increase in `Fe^(3+)`
DECREASE in `Fe^(3+)`
`Fe^(2+)//Fe^(3+)` REMAINS unchanged
`Fe^(2+)` decreases

Solution :Higher is the `E^(@)` stronger is the OXIDISING agent THUS `Fe^(3+)` with `E^(@)=0.77V` is the stronger oxidising agent
`Fe^(3+)+e^(-)rarrFe^(2+)`
Thus the amount of `Fe^(3+)` will decrease
Discussion
13.

Standard electrode potential values, (E^ɵ) for Al^(3+)//Alis -1.66 V and that of Tl^(3+) //Tlis +1.26 V. Predict about the formation of M^(3+) ion in solution and compare the electropositive character of the two metals.

Answer»

Solution :Standard electrode potential values for two half cell reactions suggest that aluminium has high tendency to make `Al_((AQ))^(3+)`ions, whereas `Tl^(3+)`isnot only unstable in solution but is a powerful oxidising agent also. Thus `Tl^+`is more stable in solution than `Tl^(3+)` . Aluminium being able to FORM +3 ions easily, is more electropositive than thallium.
Discussion
14.

Standard electrodepotential values, E^(@) at Al^(3+)//Al is -1.66 V andthat of Tl^(3+)//Tl is +1.26 V. Predict about theformation of M^(3+)ions in solution and compare the electropositive characterof the twometals.

Answer»

Solution :The negative valueof `E^(@)` for `Al^(3+)//Al` SUGGESTS that Al has a strongtendency to form`Al^(3+)`(aq) IONS. On the other hand,the positivevalue of`E^(@)`for `Tl^(3+)//Tl` suggests that Tl does not have HIGH tendency to form `Tl^(3+)` ions. Since Al can form `Al^(3+)` ions more easily than `Tl` does to form `Tl^(3+)` ions, THEREFORE Al ismore ELECTRONEGATIVITY than Tl.
Discussion
15.

Standard electrode potential data are used for understanding the stability of an oxidant in a redox titration . Some half reactions and their standard potentials are given below: MnO_4^(-) (aq) +8H^(+)(aq)+5e^(-)rarrMn^(2+)(aq)+4H_2O(l) ""E^@=1.51V Cr_2O_7^(2-) (aq)+14H^(+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_2O(l) ""E^@=1.38V Fe^(3+)(aq)+e^(-) rarr Fe^(2+)(aq) ""E^@=0.77V Cl_2(g)+2e^(-) rarr2Cl^(-) (aq)""E^@=1.40V Identify the only incorrect statement regarding the quantitative estirnation of aqueous Fe(NO_3)_2.

Answer»

`MnO_4^(-)` can be used inaqueous HCL
`Cr_2O_7^(2-)` can be used in aqueous HCl
`MnO_4^(-)` can be used inaqueous `H_2SO_4`
`Cr_2O_7^(2-)` can be used in aqueous `H_2SO_4`

SOLUTION :`MnO_4^(-)` will positive , the reaction is spontaneous .
`2KMnO_4+16H^(+)+16Cl^(-)rarr 2Mn^(2+)+8H_2O+5Cl_2`
The corresponding CELL is
`Pt|Cl_2(g)|Cl^(-)(aq)||MnO_(4)^(-)(aq), Mn^(2+),H^(+)|Pt`
`E_("cell")^@=E^@` (cathode) `-E^@` (anode)
= 1.51 - 1.40 = 0.11 V
Since , `E_("cell")^@` is +ve , the above reaction is feasible . Therefore , `MnO_4^(-)` will oxidise both `Fe^(2+) and Cl^(-)` IONS in aqueous medium . THUS, quantitative estimation of `Fe(NO_3)_2` is not suitable.
Discussion
16.

Standar enthalpy of formation of threecombustible isomers compound A,B,C are -20Kcal/mol ,30kcal /mol and 40kcal /mol respectively then what willbe order of their enthalpy of combustion ?

Answer»

`DeltaH_(C)^(@)[A]ltDeltaH_(c)^(@)[B]lt DeltaH_(c)^(@)[C]`
`DeltaH_(c)^(@)[B]gtDeltaH_(c)^(@)[A]GT DeltaH_(c)^(@)[C]`
`DeltaH_(c)^(@)[A]gtDeltaH_(c)^(@)[B]gt DeltaH_(c)^(@)[C]`
`DeltaH_(c)^(@)[C]gtDeltaH_(c)^(@)[A]gt DeltaH_(c)^(@)[B]`

Answer :c
Discussion
17.

Stand electrode potential of three metals X,Y and Z are -1.2 V + 0.5 V and -3.0 V respecitvelyThe reducing power of these metals will be

Answer»

`XgtYgtZ`
`YgtZgtX`
`YgtXgtZ`
`ZgtXgtY`

Solution :Lower the `E^(@)` stronger is the REDUCING AGENT thus the reducing power decrease in the ORDER
`Z(-30 V)gtX(-1.2 V)GTY(+0.54V)`
Discussion
18.

Stacking of hexagonal close packed layers give rise to

Answer»

HCP structure
CCP structure
Simple cubic structure
FCC structure

Solution :Stacking of hexagonal CLOSE PACKED LAYER gives rise to HCP and FCC structure.
Discussion
19.

Stacking of square close packed layers give rise to

Answer»

BCC STRUCTURE
FCC structure
Simple cubic structure
HCP structure

Solution :Stacking of square CLOSE packed layer gives RISE to BCC and simple cubicstructure.
Discussion
20.

Stable from of C may be represented by the formula :

Answer»

C
`C_(2)`
`C_(3)`
`C_(4)`

SOLUTION :Is the stable FORM of element C. It is actually `Cl_(2).`
Discussion
21.

Stable from of A may be represented by the formula :

Answer»

A
`A_(2)`
`A_(3)`
`A_(4)`

SOLUTION :is CORRECT answer.
Discussion
22.

Stable form of C may be represented by the formula :

Answer»

C
`C_(2)`
`C_(3)`
`C_(4)`

Solution :`C_(2)`
In the ELECTRONIC configuration of C INDICATE chlorine. Its stable FORM is dichlorine`(Cl_(2))`i.e. `C_(2)`
Discussion
23.

Stable form of A may be represented by the formula :

Answer»

A
`A_(2)`
`A_(3)`
`A_(4)`

Solution :A
In the given ELECTRONIC configuration A represents NOBLE gas As the octet is COMPLETE. A is neon which has 10 ATOMIC number.
Discussion
24.

Stability order of the following species ?

Answer»

`IgtIIgtIII`
`IIIgtIgtII`
`IIIgtIIgtI`
`IgtIIIgtII`

ANSWER :A
Discussion
25.

Stability of the species Li_(2), Li_(2)^(-) and Li_(2)^(+) increases in the order of :

Answer»

`Li_(2) lt Li_(2) lt Li_(2)^(+)`
`Li_(2) lt Li_(2)^(+) lt Li_(2)^(-)`
`Li_(2) lt Li_(2)^(+) lt Li_(2)`
`Li_(2) lt Li_(2)^(-) lt Li_(2)^(+)`

Solution :`Li_(2) sigma (1S)^(2) sigma^(**) (1s)^(2) sigma (2s)^(2)` ,
BOND ORDER ` = (1)/(2) (4-2) = 1 `
`Li_(2) ^(+) sigma (1s)^(2) sigma^(**) (1s)^(2) sigma (2s)^(1)` ,
Bond order`= (1)/(2) (3-2) = (1)/(2)`
`Li_(2)^(-) sigma (1s)^(2) sigma^(**) (1s)^(2) sigma (2s)^(2)(2s)^(1) `,
Bond order = `(1)/(2) (4-3) = (1)/(2)`
`LI_(2)^(-)` is less STABLE than ` Li_(2)^(+)` because ` Li_(2)^(-)` has more
number of electrons in the antibonding molecular
orbitals . Otherwisegreater the bond order , greater
is the stability . HENCE , the correct order is
`Li_(2)^(-) lt Li_(2)^(+) lt Li_(2) ` .
Discussion
26.

Stability of monovalent and trivalent cations of Ga. In lie the following sequence

Answer»

`Ga^(3+) LT In^(3+) gt TL^(3+)`
`Ga^(3+) gt In^(3+) gt Tl^(3+)`
`Tl^(o+) gt In^(o+) gt Ga^(o+)`
`Ga^(o+) gt "In"^(o+) gt Tl^(o+)`

Solution :`{:(Ga^(+3)gtIn^(+3)gt Tl^(+3)),(Tl^(+)gt In^(+) gtGa^(+)):}}`DUE to inert pair effect
Discussion
27.

Stability of iso-butylene can be best explaned by

Answer»

INDUCTIVE effect
mesomeric effect
hyperconjugative effect
steric effect

Answer :C
Discussion
28.

"Stability of carbocation depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom, involvement of neighbouring groups in hyperconjugation and resonance". Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange the carbocations in order of increasing stability.

Answer»

SOLUTION :2-Methylbutane has four DIFFERENT sets of equivalent hydrogen atoms marked as a, b, c and d.
`{:(overset(a)(C)H_(3)-overset(b)(C)H-overset(c)(C)H_(2)-overset(d)(C)H_(3)),("|"),(""underset(a)(C)H_(3)):}`
Removal of one hydrogen form any of these equivalent sets of hydrogens gives four different carbocations I, II, III and IV as shown below :

Now stability of carbocations decreases in the order : `3^(@) gt 2^(@) gt 1^(@)`. Since III being the `3^(@)` carbocation, it is the most STABLE. This is followed by carbocation (II) which is `2^(@)`. Out of the other two `1^(@)` carbocations, carbocation (IV) has an electron-donating `CH_(3)` group at `alpha`-carbon while carbocation (I) has a `CH_(3)` group at `beta`-carbon. Since +I effect decreases with distance, therefore, carbocation (IV) is more stable than carbocation (I). Thus, the overall stability of these four carbocations increases in the order : `I lt IV lt II lt III`.
Discussion
29.

"Stability of carbocation depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom, involvement of neighbouring groups in hyperconjugation and resonance". Draw the possible resonance structures for CH_(3)-underset(..)overset(..)(O)-overset(+)(C)H_(2) and predict which of the structures is more stable. Give reason for your answer.

Answer»

Solution :The given CARBOCATION has two RESONANCE structures, i.e., I and II.

STRUCTURE (II) is more stable since both the carbon ATOMS and the oxygen atom have an octet of electrons.
Discussion
30.

Stability of alkyl carbocations can be explained by

Answer»

INDUCTIVE EFFECT only
hyperconjugation only
both inductive effect and hyperconjugation
electromeric effect only

Answer :C
Discussion
31.

Stability is more for the intermediate

Answer»

`C_6H_5overset(BAR"..")"C"H_2`
`C_6H_(11)OVERSET(bar(..))"C"H_2`
`""overset(bar(..))CH_3`
`CH_3overset(bar(..))CH_2`

ANSWER :A
Discussion
32.

SRP values of lanthanides lies between

Answer»

`-2.2` to `-2.4 V`
4 to 2 V
1 to 5 V
`0.1` to `-0.2 V`

ANSWER :A
Discussion
33.

sqrtV of two particles A and B are plotted against de-Broglie wavelengths. where V is the potential onthe particles. Which of the following relation is correct about the mass of particles?

Answer»

`m_A = m_B`
`m_A GT m_B`
`m_A lt m_B`
`m_A lt m_B`

Solution :`LAMBDA = 1/(SQRT(2.qvm))`
Discussion
34.

Spodoptol, a sex attractant, produced by a female fall armyworm moth, can be prepared as follows. The structure of Spodoptol is (pK_a : terminal alkynes ~ 25, alcohols ~ 17)

Answer»

`n-C_4H_9-O-CH_2-(CH_2)_7-HC=CH_2`


`HO-CH_2-(CH_2)_12-CH_3`

ANSWER :B
Discussion
35.

Spodumene is the silicate mineral of ___________

Answer»

Beryllium
Lithium
Calcium
Potassium

Answer :B
Discussion
36.

Splitting of spectral lines in an electric field is called:

Answer»

COMPTON effect
STARK effect
Zeeman effect
Shielding effect

ANSWER :C::D
Discussion
37.

Splitting of spectral lines in an electric field is called

Answer»

Zeeman effect
Shielding effect
COMPTON effect
Stark effect

Solution :Splitting of spectral LINES in magnetic FIELD is called zeeman effect and splitting of spectal lines in electric field is called stark effect.
Discussion
38.

Splitting of spectral lines in an electric field is called…………….

Answer»

ZEEMAN EFFECT
Shielding effect
Compton effect
Staark effect

Solution :Splitting of spectral lines in magnetiic FIELD is CALLED Zeeman effect and splitting of spectral lines in electric field,is called STARK effect.
Discussion
39.

Splitting of signals is caused by

Answer»

PROTON
neutron
Positron
Electron

Solution :Splitting of SIGNALS is caused by protons attached to ADJACENT CARBON provided these are not EQUIVALENT to the absorbing proton.
Discussion
40.

Split the following redox reaction in to the oxidation and reduction half rections (a) 2K(s) +CI_(2)(g)rarrKCI(s)

Answer»

Solution :`{:((a)K(s) toK^(+)(g)+e^(-)"["xx2),(Cl_(2)(g)+2E^(-)to2Cl^(-)(g)),(UL(K^(+)(g)+Cl^(-)(g)toKCl(s)"]"xx2)),(2K(s)+Cl_(2)(g)to2KCl(s)),((b) Al(s) to Al^(3+)(aq)+3e^(-)"["xx2),(ul(Cu^(2+)(aq)+2e^(-) to Cu(s)(s)"]"xx3)),(2Al(s)+3Cu^(2+)(aq) to 2Al^(3+)(g)+3Cu(s)):}`
Discussion
41.

Spinel is the mineral with the formula _________

Answer»


ANSWER :`MgAl_2O_4`
Discussion
42.

Spin multiplicity of p-subshell with configurationis ________ .

Answer»


ANSWER :4
Discussion
43.

Spin isomerism is shown by

Answer»

dichlorobenzene
hydrogen
dibasic acid
n-butane

Solution :ortho and PARA hydrogens are spin isomers because they DIFFER from each other in the direction of spin of their NUCLEI.
Discussion
44.

Spin m ultiplicity value of oxygen element is

Answer»

2
1.5
1
4

Solution :f ORBITAL STARTS with `4^(th)`ORBIT
Discussion
45.

Spin angular momentum of an electron has no analog in classical mechanics. However, it turns out that the treatment of spin angular momentum is closely analogous to the treatment of orbital angular momentum. Spin angular momentum = sqrt(s(s+1)h) Orbital angular momentum = sqrt(l(l+1)h) Total spin of an atom or ion is a multiple of (1)/(2). Spin multiplicity is a factor to confirm the electronic configuration of an atom or ion. Spin multiplicity (2 sum s +1) The orbital angular momentum for a 2p-electron is :

Answer»




ANSWER :A
Discussion
46.

Spin angular momentum of an electron has no analog in classical mechanics. However, it turns out that the treatment of spin angular momentum is closely analogous to the treatment of orbital angular momentum. Spin angular momentum = sqrt(s(s+1)h) Orbital angular momentum = sqrt(l(l+1)h) Total spin of an atom or ion is a multiple of (1)/(2). Spin multiplicity is a factor to confirm the electronic configuration of an atom or ion. Spin multiplicity (2 sum s +1) Total spin of Mn^(2+) = (Z = 25) ion will be :

Answer»

`(3)/(2)`
`(1)/(2)`
`(5)/(2)`
`(7)/(2)`

Answer :C
Discussion
47.

Specify the coordinate geometry around and hyridilization present about the central atom in BF_(3) is:

Answer»

`N`: Tetrahedral, `sp^(3)`, `B`: Tetrahedral, `sp^(3)`
`N:` pyramidal, `sp^(3),B:` pyramidal, `sp^(3)`
`N:`pyramidal,`sp^(3),B:` Planar, `sp^(3)`
`N:`Pyramidal, `sp^(3)`, `B:` Tetrahedral `sp^(3)`

SOLUTION :In complex `[H_(3)NtoBF_(3)]`, both `N` and `B` attains `sp^(3)`-HYBRIDISATION and acquires tetrahadral geometry.
Discussion
48.

Specific test for the detection of H_(2)O_(2) is

Answer»

Reaction with acidified `TiO_2`
Starch Iodide paper into blue
Conversion of `PbO_2` into yellow colour
Conversion of BLACK PbS into `PbSO_4`

Solution :`H_2O_2` gives orange colour (pertitanic acid) on reaction with acidified titanium DIOXIDE.
`TiO_2 + 2H_2SO_4 to TI(SO_4)_2`
` + 2H_2O overset(H_2O_2) to underset("PE ti tan ic acid")(H_2TiO_4 + 2H_2SO_4)`
Discussion
49.

Specific reagent for dehydrohalogenation of alkyl halides to get alkenes is

Answer»

AQUEOUS KOH
Anhydrous `ZnCl_(2)`
ALCOHOLIC KOH
Conc. `H_(2)SO_(4)` at `140^(@)C`

Answer :C
Discussion
50.

Specific heat ofan elementary gas is found to be 0.313 J at constant volume. If the molar mass of the gasis 40 g mol^(-1) , what is the atomicity of the gas ?

Answer»


Solution :`C_(v) = 0.313 xx 40 = 12.52J MOL^(-1) , C_(p) = C_(v)+R = 12.52 + 8.31= 20.83Jmol^(-1)`
`:. C_(p) //C_(v) = 20.83 // 12.52 = 1.66`
Discussion