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.

If CH_3COOH+OH^-=CH_3COO^-+H_2O+q_1 and H^+ OH^-=H_2O+q_2 then the enthalpy change for the reaction, CH_3COOH=CH_3COO^(-)+H^+, is equal to:

Answer»

`q_1+q_2`
`q_1-q_2`
`q_2-q_1`
`-q_1-q_2`

ANSWER :B
Discussion
2.

If certain atom has atomic number of 22. Then what is the maximum oxidation state of It?

Answer»

1
2
3
4

Answer :D
Discussion
3.

If C_(aq) is in equivalent dm^(-3) and K is in S cm^(-1), then which one among the following is correct?

Answer»

`wedgeaq = (1000K)/C_(AQ)`
`wedgeaq = C_(eq)/(1000 K)`
`wedgeaq = (1000 K)/(C_("molar")`
none of these

Answer :A
Discussion
4.

If carbon is present in cast iron in the form of cementite' then it is known as

Answer»

WHITE CAST iron
Grey cast iron
Wrought iron
None of these

Answer :A
Discussion
5.

If CO_2 is passed in excess into lime water, the milkiness first formed disappears due to:

Answer»

REVERSAL of the ORIGINAL reaction
Formation of volatile calcium derivatives
Formation of water soluble calcium bicarbonate
The solution GETTING heated by exothermic reaction

Answer :C
Discussion
6.

If C_(a_(3))(PO_(4))_(2) and H_(3)PO_(3) contain same number of 'P' atom then the ratio of oxygen atoms in these compounds respectively is

Answer»

`8//3`
`2//3`
3
`4//3`

ANSWER :D
Discussion
7.

If C+O_(2)rarrCO_(2)+94.2 kcalH_(2)+(1)/(2)O_(2)rarrH_(2)O+68.3 kcalCH_(4)+2O_(2)rarrCO_(2)+2H_(2)O+210.8 kcal

Answer»


Solution :`C+O_(2)rarrCO_(2)+94.2 " KCAL...(i)"`
`H_(2)+(1)/(2)O_(2)rarrH_(2)O+68.3 " kcal...(ii)"`
On multiplication of eq. (ii) by 2 and than ADDING in eq. (i)
`C+2H_(2)+2O_(2)rarrCO_(2)+2H_(2)O+230.8 " kcal...(iii)"`
On SUBTRACTING eq. (iii) by following eq.
`CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O+210.8" kcal. we get,"`
`C+2H_(2)rarrCH_(4)DeltaH=20 kcal`.
Discussion
8.

If C(s)+O_(2)(g)to CO_(2)(g)+94.2 KcalH_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l)+68.3 KcalCH_(4)(g)+2O_(2)(g)to CO_(2)(g)+2H_(2)O+210.8 Kcal then the heat of formation of methane will be

Answer»

47.3 KCAL
`-20.0`kcal
45.9 kcal
`-47. 3` kcal

Answer :B
Discussion
9.

If C & D are the third & fourth virial coefficients. If (D)/(C ) = (V_(c))/(x). Then find the value of x.

Answer»


SOLUTION :N//A
Discussion
10.

If by mistake some radioactive substance gets inside the human body, then form the point of view of radiation damage, the most harmful will be one which emits

Answer»

`GAMMA`-RAYS
Neutrons
`beta-`PARTICLES
`ALPHA-`particle

Solution :`gamma-`rays have MAXIMUM penetrating power
Discussion
11.

If butane on combustion gives carbon monoxide. Find the number of O_(2) molecule required

Answer»

6
5.5
4.5
4

Answer :C
Discussion
12.

If by mistake radioactive substance gets into human body, then from the point of view of radiation damage, the most harmful will be one that emits.

Answer»

`GAMMA`-RAYS
neutrons
`BETA`-rays
`ALPHA`-rays

Answer :a
Discussion
13.

If Br^(-) ions having conductivity =78.1" S "cm^(-1)mol^(-1) then calculate lamda_(m)^(@)(K_(+))

Answer»


ANSWER :`73.5" S "CM^(2)MOL^(-1)`
Discussion
14.

If both adsorption and absorption take place simulataneously, the process is called ………..

Answer»

SOLUTION :SORPTION
Discussion
15.

If both (dc)/(dt) & specific rate have same units then rate law is

Answer»

`R=K[A]^(2)`
`R=K[A]^(1//2)`
`R=K[a]^(-2)`
`R=K`

ANSWER :D
Discussion
16.

If bond lengths by x, y and z respectively, then, estimate the order of bond lengths.

Answer»

X GT y gtz
y gt x gt Z
y gt z gt x
z gt y gt x

Solution :
Discussion
17.

If Boiling point of water is 100^(@)C. How much gram of NaCl is added in 500 g of water to increase its boiling point of water by approx 1^(@)C.(K_(b))_(H_(2)O)="0.52 K x kh/mole".

Answer»

2.812 g
28.12 g
14.06 g
7.03 g

Solution :`DeltaT_(b)="I K"_(b)m`
`1=2xx0.52xx(w)/(58.5)xx(1000)/(500)`
`w=28.125g`
Discussion
18.

If BOD of river is high, this implies that the river is

Answer»

NON POLLUTED at all
Highly polluted with inorganic CHEMICALS
Highly polluted with organic chemicals which are decomposed by microorganims
Polluted with pesticides

Answer :C
Discussion
19.

If benzene. Toluene and para-xylene are mixed in molar ratio 1:2:3. Then answer the following question carefully. Estimate the dew point pressure (At the dew point pressure all liquid vaporiaed ) for the above mixture (approximately)

Answer»

180 torr
152 torr
138 torr
156 torr

Solution :At the DEW pointpressure all LIQUID vaporized
`(1)/(P_(("dew point")))=[(Y_(B))/(P_(B)^(@))+(y_(T))/(P_(T)^(@))+(Y_(X))/(P_(X)^(@))]"" (Y_(B)=(1)/(6), Y_(T)=(2)/(6), Y_(X)=(3)/(6))`
Discussion
20.

If benzene. Toluene and para-xylene are mixed in molar ratio 1:2:3. Then answer the following question carefully. Estimate the bubble point pressure (At the bubble point pressure all liquid exist. No trace of vapour inmixture) for the above mixture.

Answer»

600 torr
170 torr
`(500)/(3)` torr
160 torr

Solution :At the bubble point PRESSURE all liquid. Exist. No trace of vaoure in mixture
`P_("bubble") [(1)/(6)xx300+(2)/(6)xx180+(3)/(6)xx120)]=170` torr
Discussion
21.

If benzene toluene and para-xylene are mixed in molar ratio 1 : 2: 3. Estimate the dew point pressure (At the dew point pressure all liquid vaporized) for the above mixture (approximately)

Answer»

<P>180 torr
152 torr
138 torr
156 torr

Solution :At the dew point PRESSURE all liquid VAPORIZED
`(1)/(P_("(dew point)")) = [(y_(B))/(P_(B)^(@)) + (y_(T))/(P_(T)^(@)) + (y_(X))/(P_(X)^(@))] (y_(B) = (1)/(6), y_(T)= (2)/(6), y_(x) = (3)/(6))`
Discussion
22.

If benzene toluene and para-xylene are mixed in molar ratio 1 : 2: 3. Estimate the bubble point pressure (At the bubble point pressure all liquid exist. No trace of vapour in mixture) for the above mixture.

Answer»

600 torr
170 torr
`(500)/(3)` torr
160 torr

Solution :At the bubble point PRESSURE all liquid EXIST. No trace of VAPOUR in mixture
`P_("bubble") = [(1)/(6) xx 300 + (2)/(6) xx 180 + (3)/(6) xx 120] = 170` torr
Discussion
23.

If B.E. per nucleon of B 11 ,(the atomic mass of B-11 is 11.00931 u). , is (0.1 xx y) approximately , then what is the value of 'y'

Answer»


SOLUTION :`Q = (DELTA m xx 931.5)/(11) Me V , Q = ((11.00931 - 11)/(11)) xx 931.5 = (0.00931)/(11) xx 931.5 = 0.788= 0.800`
Discussion
24.

If BaCl_(2) ionizes to an extent of 80% in aqueous solution,the value of van't Hoff factor is

Answer»

2.6
0.4
0.8
2.4

Answer :A
Discussion
25.

If at a particular tempreture, the density of 18M H_(2)SO_(4) is 1.8g Cm^(-3), calcualte: (a) Molality, (b) % concentration by weight of solution (c ) Mole fraction of water and H_(2)SO_(4), (d) Relative decrease in vapour pressure with repsect to H_(2)O solvent assuming H_(2)SO_(4) almost unionised at this high concentration.

Answer»


Answer :(a) `500` molal, (B) `98%`,(C ) `H_(2)SO_(4) = 0.9, H_(2)O = 0.1`,(d) `0.9`;
Discussion
26.

If Avogadro number (N_(A)) is changed from 6.022xx10^(23) to 6.022xx10^(20)"mol"^(-1), this would change

Answer»

the RATIO of the chemical species to each other in a balanced equation
the ratio of the elements to each other in a compound
the definition of the MASS in UNITS of gram
the mass of one mole of CARBON

Solution :Mass of 1 mole of carbon, i.e., `6.022xx10^(23)` atoms of carbon = 12 g
If Avogadro's number is CHANGED, the mass of 1 mole of carbon, i.e.,
`6.022xx10^(20)` atoms of carbon`=(12.6.022xx10^(20))/(6.022xx10^(23))`
`therefore"Massof 1 mole of carbon is changed."`
Discussion
27.

If avogado number N_(A), is changed from 6.022xx10^(23) mol^(-1) to 6.022xx10^(20) mol^(-1) this would change

Answer»

The DEFINITION of MASS in units of grams
The mass of one mole of carbon
the RATIO of chemical species to each other in a BALANCED equation
The ratio of elements to each other in a compound

Answer :B
Discussion
28.

If Aufbau rule is not followed "" _(19)K will be placed in

Answer»

<P>s BLOCK
p block
d block
F block

ANSWER :C
Discussion
29.

If atomic number of an inert gas is Z then an element with which of the following atomic number will has highest I.E.

Answer»

`Z-2`
`Z-1`
`Z+1`
`Z+2`

SOLUTION :ATOMIC NUMBER of inert gas = Z
SECOND highest IE in a period = HALOGENS (Z–1)
Discussion
30.

If at certain temperature the vapour pressure of pure water is 25 mm Hg and that of a very dilute aqueous urea solution is 24.5 mm Hg, the molarity of the solution is

Answer»

<P>`0.02`
`1.2`
`1.11`
`0.08`

Solution :`(P^(@)-P_(s))/(P^(@))=(w_(B))/(m_(B))xx(m_(A))/(w_(A)), (w_(B))/(w_(B)xx w_(A))xx1000xx(m)/(1000)`
`RARR (P^(@)-P_(s))/(P^(@))=" molality " xx(m)/(1000)`
`(25-24.5)/(25)=m xx(18)/(1000)`
`m=(0.02xx1000)/(18)=1.11`
Discussion
31.

If Aufbau principle and Hund's rule were not followed

Answer»

Ca would have been d-block element with zero dipole moment
Zn would have been s-block element
Ti would have been diamagnetic
`Fe^(3+)` ion would have 5 unpaired electrons

Solution :For no consideration of Aufbau PRINCIPLE nad HUND's rule, the electronic configuration `Ca(20),[AR]3d^(2),Zn(30):[Ar]3d^(10)4s^(2)` (filling of 4s subshell after 3d subshell), `Ti(22), [Ar]3d^(2)` (both electrons paired up)
Discussion
32.

If at 298 K the bond energies of C-H,C-C,C=CandH-H bonds are respectively 414, 347, 615 and 435 kJ mol^(-1), the value of enthalpy change for the reaction given below at 298 K will be H_(2)C=CH_(2)(g)+H_(2)(g)toH_(3)C-CH_(3)(g)

Answer»

`-250` KJ
`+125` KJ
`-125` KJ
`+250` KJ

Answer :C
Discussion
33.

If at 100^@C temperature copper possess fcc arrangements. Edge length is xÅ . Then find out approximate density ("g/cm"^3) of Cu at this temperature. (Atomic weight of Cu = 63.55 u)

Answer»

`105/(x^3)`
`211/(x^3)`
`205/(x^3)`
`422/(x^3)`

Solution :For fcc unit cell, Z = 4
`therefore D = (63.5 XX 4)/(6 xx 10^(23) xx x^3 xx 10^(-24)) "g/cm"^3`
`therefored = (63.5 xx 4 xx 10)/(6) "g/cm"^3`.
`therefore d = (423.33)/(x^3) = 422/(x^3)`.
Discussion
34.

If aniline is treated with conc. H_(2)SO_(4) and heated at 200^(@)C, the product is

Answer»

ANILINIUM SULPHATE
benzenesulphonic acid
m-aminobenzenesulphonic acid
sulphanilic acid.

Solution :
Discussion
35.

If Aniline is treated with 1:1 mixture of con HNO_(3) and con. H_(2)SO_(4),p-nitroaniline and mnitroaniline are formed nearly in equal amounts.This is due to

Answer»

m-directing PROPERTY of —`NH_(2)` GROUP
protonation of -`NH_(2)` which causes deactivation of benzene ring
m- and p-directing property of —`NH_(2)` group
isomerization of some p-nitroaniline into m-nitroaniline.

Solution :(b): `-NH_(2)` group in aniline is o, p-directing and activating while in presence of acids, most of the aniline GETS protonated to form anilinium ion i.e., `-ovrset(+)(N)H_(3)` which is m-directing and deactivating.
Discussion
36.

If an oil soluble dye is mixed with an emulsion and theemulsion remains colourless, then it is _______type emulsion.

Answer»

O-in-W
W-in-O
O-in-O
W-in-W

Solution :When water is the dispersion MEDIUM, it will be lyophobic for OIL SOLUBLE dye, hence will be COLOURLESS in O-in-W.
Discussion
37.

If an iron rod is dipped in CuSO_(4) solution

Answer»

Blue colour of the solution turns green
Brown LAYER is deposited on IRON ROD
No CHANGE occurs in the colour of the solution
Blue colour of the solution vanishes

Solution :Brown layer is deposited o iron rod because Cu has greater reduction POTENTIAL than that of `Fe^(2+)`.
Discussion
38.

If an ionic solid XY (X & Y are monovalent ions) is doped with 10^-2moles % of another ionic solid AY_3 , then the concentration of the cation vacancies created is

Answer»

`6.023xx10^19 mol^-1`
`60.23xx10^18 mol^-1`
`12.05xx10^21 mol^-1`
`1.205xx10^21 mol^-1`

ANSWER :D
Discussion
39.

If an iron rod is deppedin CuSO_4 solution:

Answer»

Blue COLOUR of the solution turns red
Brown LAYER is DEPOSITED on IROD rod
No change occous in the colour of the solution
None

Answer :B
Discussion
40.

If an external opposing E.M.F. is slightly greater than E.M.F. of Daniel cell the reaction will

Answer»

be reversed
remain same
move in the FORWARD direction
all are right

Solution :CONDITION of REVERSIBILITY of CELL.
Discussion
41.

If an endothermic reaction occurs spontaneously at constant temperature T and Pressure P, then whicch of the following is true

Answer»

`DeltaG gt 0`
`DeltaH lt 0`
`DeltaS gt 0`
`DeltaS lt 0`

Solution :The thermodynamic QUANTITY of the system, the decrease in whose value during a process is equal to useful work done by the system, i.e. `DeltaG=DeltaH-TDeltaS`, where H = enthalpy, S= ENTROPY and T= absolute temperature. If an ENDOTHERMIC REACTION occurs spontaneously i.e. `DeltaH=+ve` then for reaction to be spontaneous `DeltaG` should be NEGATIVE that is only possible when `DeltaS gt 0` at constant temperature and pressure.
Discussion
42.

If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point then

Answer»

`DeltaH` is -ve, `DELTAS` is + ve
`DeltaH` and `DeltaS` both are + ve
`DeltaH` and `DeltaS` both are -ve
`DeltaH` is +ve, `DeltaS` is -ve

Solution :For a reaction to be SPONTANEOUS, `DELTAG` must be negative, According to the equation -
`DeltaG=DeltaH-T.DeltaS`
If `DeltaH` and `DeltaS` both are positive, than term T. `DeltaS` will begreater than `DeltaH` at high temperature and consequently `DeltaG` will be negative at high temperature.
Discussion
43.

If an elementemitsphotonin its groundstatedue toelectron transition , thenits .

Answer»

atomicnumberwill increase
ATOMICNUMBER will remainsunchaged
atomicnumberwill decrease
MASS numberwill decrease

Solution :In the ground state electronictransition will occur only when an electron from K-shell i.e., FIRSTSHELL is captured by the nucleus (K - electron capture).Electron so capturedconverts proton into neutron resultingin the decrease of atomic NUMBER.
Discussion
44.

If an element can exist in several oxidation states, it is convernient to display the reduction potentials correspondingg to the various half reactions in diagrammatic form, know as latimer diagram the latimer diagram for chlorine in acid solutio is CIO_(4)^(-)overset(+1.20V)toCiO_(3)^(-)overset(+1.18V)toHClO_(2)overset(+1.60V)toHClOoverset(1.67V)toCl_(2)overset(1.36V)toCl^(-) in basic solution. ClO_(4)^(-)overset(0.37V)toClO_(3)^(-)overset(0.30V)toClO_(2)^(-)overset(0.68V)toClO^(-)overset(0.42V)toCl_(2)overset(1.36V)toCl^(-) The standard potentials for two nonadjacent species can also be calculateed by using the concept that triangleG^(@) as an additive property but potential is not an additive property and triangleG^(@)=-nFx^(0). if a given oxidation stateis a stronger oxidising agent than the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of triangleG^(@)//F against oxidation state, known as frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram, disproportionation is spontaneous if the species lies above a straight line joining its two product species. Q. Which of the following statement is correct?

Answer»

`Cl_(2)` UNDERGOES disproportionation into `CL^(-)` and `CLO^(-)` both at `pH=0` and `pH=14`.
`Cl_(2)` undergoes disproportionation into `Cl^(-)` at `pH=14` but not at `pH=0`
`Cl_(2)` undergoes disproportionation into `Cl^(-)` and `ClO^(-)` at `pH=0` but not at `pH=14`
NONE of these

Solution :
Discussion
45.

If an element can exist in several oxidation states, it is convernient to display the reduction potentials correspondingg to the various half reactions in diagrammatic form, know as latimer diagram the latimer diagram for chlorine in acid solutio is CIO_(4)^(-)overset(+1.20V)toCiO_(3)^(-)overset(+1.18V)toHClO_(2)overset(+1.60V)toHClOoverset(1.67V)toCl_(2)overset(1.36V)toCl^(-) in basic solution. ClO_(4)^(-)overset(0.37V)toClO_(3)^(-)overset(0.30V)toClO_(2)^(-)overset(0.68V)toClO^(-)overset(0.42V)toCl_(2)overset(1.36V)toCl^(-) The standard potentials for two nonadjacent species can also be calculateed by using the concept that triangleG^(@) as an additive property but potential is not an additive property and triangleG^(@)=-nFx^(0). if a given oxidation stateis a stronger oxidising agent than the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of triangleG^(@)//F against oxidation state, known as frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram, disproportionation is spontaneous if the species lies above a straight line joining its two product species. Q. For a hypothetical element, the frost diagram is shown in figure.? which of the following oxidation state is least stable?

Answer»

`-1`
`0`
`+2`
`+3`

ANSWER :D
Discussion
46.

If an element emits an alpha-particle, then the position of daughter element will be displaced in the periodic table by :

Answer»

TWO POSITIONS on left
one POSITION on left
one position on right
two positions on right.

Answer :A
Discussion
47.

If an element can have four values of spin quantum numbers and rest of the quantum no. are same, which of the following statement would be incorrect :-

Answer»

Be would have been in the first period of P.T.
The TOTAL no of ELEMENTS in 2nd period would have been 12
There would have been 36 elements in 4th period
NUMBER of periods would have been less than the no. of periods present in the modern P.T.

Solution :
If element can have four VALUE of spin quantum number so one orbital can CONTAIN 4 electrons.
Discussion
48.

If an element can exist in several oxidation states, it is convernient to display the reduction potentials correspondingg to the various half reactions in diagrammatic form, know as latimer diagram the latimer diagram for chlorine in acid solutio is CIO_(4)^(-)overset(+1.20V)toCiO_(3)^(-)overset(+1.18V)toHClO_(2)overset(+1.60V)toHClOoverset(1.67V)toCl_(2)overset(1.36V)toCl^(-) in basic solution. ClO_(4)^(-)overset(0.37V)toClO_(3)^(-)overset(0.30V)toClO_(2)^(-)overset(0.68V)toClO^(-)overset(0.42V)toCl_(2)overset(1.36V)toCl^(-) The standard potentials for two nonadjacent species can also be calculateed by using the concept that triangleG^(@) as an additive property but potential is not an additive property and triangleG^(@)=-nFx^(0). if a given oxidation stateis a stronger oxidising agent than the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of triangleG^(@)//F against oxidation state, known as frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram, disproportionation is spontaneous if the species lies above a straight line joining its two product species. Q. What is the potential couple (ClO^(-))/(Cl^(-)) at pH=14?

Answer»

`1.78V`
`-0.94V`
`0.89V`
`-0.89V`

Solution :`triangleG^(0)=triangleG_(1)^(0)+triangleG_(2)^(0)` USING this `E^(0)=(0.42+1.36)/(2)V=0.89V`
Discussion
49.

If an element can exist in several oxidation states, it is convernient to display the reduction potentials correspondingg to the various half reactions in diagrammatic form, know as latimer diagram the latimer diagram for chlorine in acid solutio is CIO_(4)^(-)overset(+1.20V)toCiO_(3)^(-)overset(+1.18V)toHClO_(2)overset(+1.60V)toHClOoverset(1.67V)toCl_(2)overset(1.36V)toCl^(-) in basic solution. ClO_(4)^(-)overset(0.37V)toClO_(3)^(-)overset(0.30V)toClO_(2)^(-)overset(0.68V)toClO^(-)overset(0.42V)toCl_(2)overset(1.36V)toCl^(-) The standard potentials for two nonadjacent species can also be calculateed by using the concept that triangleG^(@) as an additive property but potential is not an additive property and triangleG^(@)=-nFx^(0). if a given oxidation stateis a stronger oxidising agent than the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of triangleG^(@)//F against oxidation state, known as frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram, disproportionation is spontaneous if the species lies above a straight line joining its two product species. Q. Which of the following couple have same value of potential at pH=0 and pH=14?

Answer»

`(ClO_(4)^(-))/(ClO_(3)^(-))`
`(ClO_(2)^(-))/(Cl_(2))`
`(ClO_(2)^(-))/(Cl_(2))`
`(Cl_(2))/(CL^(-))`

Answer :D
Discussion
50.

If an element can exist in several oxidation states, it is convenient to display the reduction potentials corresponding to the various half reactions in diagrammatic from, known as Latimer diagram. The Latimer diagram for chlorine in acid solutionis Cl_(4)^(-)overset(+1.20V)toClO_(3)^(-)overset(+1.60V)toClO_(2)^(-)overset(+1.60V)to ClO^(-)overset(+1.67V)toCl_(2)overset(+1.36V)toCl^(-) In basic solution is : ClO_(4)^(-)overset(+0.37V)toClO_(3)^(-)overset(+0.30V)toClO_(2)^(-)overset(+0.68)to ClO^(-)overset(+0.42V)toCl_(2)overset(+1.36)toCl^(-). The standard potentials for two nonadjacent species can also be calculated by using the concept that DeltaG^(@) is an additive property but using potential is not an assitive property and DeltaG^(@)=-nFx^(0), If a given oxidation state is a the next higher oxidation state disproportionation can occur. The reverse of relative stabilities of the oxidation state can also be understood by drawing a graph of DeltaG^(@)//F against oxidation state, known as Frost diagram, Choosing the stability of zero oxidation state arbitrery as zero. The moststable oxidation state of a apecies lies lowest in the digram Disproportionation is spontaneous if the species lies above a straight line joining its two product species. Which of the following statement is correct accordingto abovequestion ?

Answer»

`A^(+1)` undergoes disproportionation into A and `A^(2+)`
`A^(2+)` undergoes disproportionation in AAND `A^(3+)`
A undergoes COMPROPORTIONATION into `A^(+1)` and `A^(-1)`
All of the above

Answer :A
Discussion