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.

For a reversible process at equilibrium, the change in entropy may be expressed as:

Answer»

`DeltaS = Tq_(REV)`
`DeltaS = (DELTAH)/(T)`
`DeltaS = (q_(rev))/(T)`
`DeltaS = T DeltaH`

ANSWER :C
2.

For a real gas 'X'the Boyle point is 240^@C. If the van der Waals constant 'b'is 0.08 dm^3 mol^(-1), calculate the value of constant 'a'for 'X'.

Answer»

Solution :The BOYLE POINT, `T_(b)` is given as `T_(b) = (a)/(Rb)`
van der WAALS. CONSTANT .a. for the gas .X. ` = T_(b) xx R xx b = 513 xx 0.0821 xx 0.08 = 3.37` atm `L^(2)MOL^(-2)`.
3.

For a real gas , the compressibility factor Z has different values at different temperatures and pressures . Which of the following is not correct under the given conditions ?

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Z lt 1 at very low PRESSURE .
Z gt 1 at HIGH pressure .
Z = 1 under all CONDITIONS .
Z = 1 at intermediate pressure

Solution :Z = 1 under all conditions for an IDEAL gas.
4.

For a real gas, following equation (P+(a)/(TV_(m)^(2)))(V_(m)-beta)=RT, where alpha and beta are positive constants. Select the correct option(s):

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`T_(C)=(8alpha)/(27Rbeta)`
`V_(cm)=3beta`
Second virial coefficient `=beta-(ALPHA)/(RT)`
THIRD virial coefficient `=beta^(2)`

ANSWER :BD
5.

For a reaction value of K_c is 7.105 xx 10^(-5), what will be the change in free energy ?

Answer»

`Delta G GT 0`
`Delta G lt 0`
`Delta G = 0`
`Delta G =1`

SOLUTION :If `K_(C) lt 1` than , `Delta G gt 0`
6.

For a reaction whose standrad enthalpy change is -100 kJ, what final temperature is needed to double the equilibrium constant from its value at 298 K?

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Solution :`"log"K_(2)/K_(1)=(DeltaH^(@))/(2.303R)[(T_(2)-T_(1))/(T_(1)*T_(2))]`
`LOG2=(-100000)/(2.3xx8.3)[(T_(2)-298)/(298T_(2))]`
`0.301=(-100000)/(19.09)((T_(2)-298))/((298T_(2)))`
Solving for `T_(2)`, we get `T_(2)~~293K`
7.

Which of the following reaction is spontaneous at room temperature

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`DELTA H and Delta G` should be negative
Only `Delta H`should be negative
`Delta H and Delta G` should B e positive
Only `Delta G` should be positive

ANSWER :B
8.

For a reaction the Delta S value is -20 J mol^(-1) k^(-1) . If the temperature is increased from 0^(@) C at 25^(@)C , the increases in the value of Delta Gis (in J mol^(-1))

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500
`-500`
0
`14900`

ANSWER :A
9.

For a reaction taking place in a container in equilibriumwith its surroundings, the effect of temperature on its equilibrium constant K in terms of entropy is described by

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With increase in temperature, the VALUE of K for endothermic reaction increase because unfavourable changein entropy of the surroundings DECREASES
With increase temperature , the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases.
With increase in temperature, the value of K for endothermic reaction increases because the entropychange of the SYSTEM is positive.
With increase in temperature , the value of K for exothermic reaction decreasesbecause the entropy change of the system is positive.

Solution :`DeltaS_("SURR") = - (DeltaH)/( T_("surr"))`
For endothermica reaction , if`T_(surr)` increases, `DeltaS _(surr)`will increase `(:'DeltaH = +ve)`
For exothermicreactions K increases with increase of temperature `(K= K_(f) //K_(b))` and in endothermicreaction, `k_(f)` increases with increase of temperature) WHEREAS for exothermic reaction,K decreases with increase of temperature ( because in these reactions, `k_(b)` increases with increase oftemperature) .
10.

For a reaction, SO_(2) (g) + 1//2O_(2)(g) to SO_(3)(g) if K_(p)=1.7 xx10^(12) at 20^(@)C and 1 atm pressure , then value of K_(c) is

Answer»

<P>`1.7xx10^(12)`
`0.7xx10^(12)`
`8.33xx10^(12)`
`1.2xx10^(12)`

Solution :`K_(p)=K_(C)(RT)^(DELTAN)`
or `K_(c)=(K_(p))/((RT)^(Deltan))=(1.7xx10^(12))/((0.082xx293)^(-1//2))`
`=1.7xx10^(12)xxsqrt(0.082xx293)`
`=1.7xx10^(12)xx4.9=8.33xx10^(12)`
11.

For a reaction R_(1), Delta G = x KJ mol^(-1). For a reaction R_(2), Delta G = y KJ mol^(-1). Reaction R_(1) is non-spontaneous but along with R_(2) it is spontaneous. This means that

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X is -ve, y is +ve but in magnitude `x GT y`
x is +ve, y is -ve but in magnitude `y gtx`
Both x and y are -ve but not equal
Both x and y are +ve but not equal.

ANSWER :B
12.

For a reaction N_(2)(g)+3H_(2)(g)to2NH_(3)(g),DeltaH=-24 kcal at 700 K and 10 atm pressure, calculate magnitude of change in internal energy if 1.68 kg of N_(2)(g) and 0.3 kg of H_(2) are mixed and reaction undergoes 60% completion:

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21.2 kcal
636 kcal
1200 kcal
1090 kcal

Answer :D
13.

For a reaction Delta H= 3 KJ and Delta S = 10 J/Kelvin at what temperature the reaction will be spontaneous ?

Answer»

`300K`
`200K`
`273K`
`373 K`

Answer :D
14.

For a reaction AX_(5)hArr AX_(2), 1% of AX_(5) is dissociated at a total pressure of 1 atm, the equilibrium constant K_(P) is approximately equal to

Answer»

`10^(-3)`
`10^(-4)`
`0.1xx10^(-4)`
1

Answer :B
15.

For a reaction at equilibriumH_(2)(g) + Cl_(2)(g) hArr 2HCl(g) K = 4, " the calue of " (K_(b)[HCl]^(2))/(K_(f)[H_(2)][Cl_(2)])is 1.

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ANSWER :TRUE
16.

For a reaction at 25^(circ)C,DeltaG=12.7kJ when the reaction quotient Q=10.0. What is the value of DeltaG^(circ) for this reaction?

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`-12.1kJ`
`7.0kJ`
`18.4kJ`
`37.5kJ`

ANSWER :B
17.

For a reaction aA + bB hArr cC + dD at equilibrium G^(theta) = 0 always .

Answer»


ANSWER :FALSE
18.

For a reaction A (g)hArrB(g)at equilibrium, the partical pressure of B is found to be one fourth of the paritcal pressure of A. The value of DeltaG^(circ) of the reaction AhArrB is :

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RT In 4
`-RT` In 4
RT LOG 4
`-RT log 4`

ANSWER :A
19.

For a reaction , A (g) to A(l) , Delta H = -3RT . The correct statement for the reaction is

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`DELTAH = Delta U ne O`
`|Delta H| gt |Delta U|`
`|Delta H| LT |DeltaU|`
`Delta H = Delta U = O`

ANSWER :B
20.

For a reaction A+BhArr2C the equilibrium concentration of (A) and (B) are 20 mole//Lwhen volume is doubled the new equilibrium concentration of (A) was found to be 15 mol//L then

Answer»

Ratio of CONCENTRATION of `A "and" B` at new EQUILIBRIUM is 3//4
Value of equilibrium constant for both cases are remain same
Concentration of `C` at new equilibrium become half
Equilibrium concentration of `C` at new equilibrium `(10sqrt20)/(sqrt20-sqrt15)`

Solution :`{:(,A,+,2B,hArr,2C),("at vol=v LIT",20,,20,,a),("at vol=2v lit",(20)/(2),,(20)/(2),,(a)/(2)):}`
at new equilibrium
`((20)/(2)+(X)/(2))((20)/(2)+x)((a)/(2)-x)`
given that `(A)_(new)=15=((20)/(2)+(x)/(2)rArrx=(15-10)xxL`
`x=101`
for IST equilibrium `K_(C)=((a)/(2))^(2)/(((20)/(2)(20)/(2))^(2))=((a)/(2))^(2)/(10+100)=(((a)/(2))^(2)/(1000))`
for IInd equilibrium `K_(C)=((a)/(2)-10)^(2)/((20/(2)+(x)/(2))((20)/(2)-x)^(2))=((a)/(2)-10)^(2)/(15xx25)=((a)/(2)-10)^(2)/(275)`
`(K_(C) "for" I=K_(C) "for II")`
`((a)/(2))^(2)/(1000)=((a)/(2)-10)^(2)/(375)`
`((a)/(2))/(31.62)=((a)/(2)-10)/(19.365)`
`19.365xx(a)/(2)=31.62xx(a)/(2)-31.62xx10`
`9.6825xx0=15.81xx0-316.2`
`6.1275xxa=316.2`
`a=51.60M`
`K_(C)=(C)^(2)/((20)(20)^(2))=(51.6xx51.6)/(20xx20xx20)=332.82xx10^(-3)`
`K_(C)=0.333` approxly.
21.

For a reaction A + B rarr C + D, the entropy change is +68.4 JK^(-1) and the enthalpy change is -124.6 kJ. Calculate the Gibb's energy change at 298 K.

Answer»

Solution :`Delta G = Delta H - T Delta S`
Given, `Delta H = -1246000 J, T = 298 K` and `Delta S = +68.4 JK^(-1)`
Substituting in the formula, we get
`Delta G = -124600 - 298 xx 68.4`
`= -144983 J = -144.983 KJ`.
22.

Fora reaction:2A(g) toB(g), DeltaH =-40 Kcal. Ifrate constantfor disapperance of A is 10^(-2) M^(-1) sec^(-1) then identify the opationswhich will be correct [Assumegases to behaveideallyand reactionto be occuring at 300 K][Take : R=2 cal/mole K]

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THEREACTION must be ELEMENTARY reaction.
Change in internal energyof the reactionwillbe `-19.7 kcal//"mole"`ofA consumed.
The rateat which HEATWILL beliberatedinitiallywhen 2 moles of A is takenin 3 LITRE rigid vessel will beapprox `0.26 kcal//sec`.
Time requriedfor concentrationof A

Answer :b,d
23.

For a reaction 2A(g) hArr B (g) Q_(c) gt K " if "'A' is added maintainingQ_(c) gt K,the reaction will move in backward direction.

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ANSWER :TRUE
24.

For a reaction 2 Cl_((g))rarrCl_(2)What are the signs of DeltaH and DeltaS ?

Answer»

Solution :The GIVEN reaction represents the formation of bonds. Hence energy is released , ie., `DELTAH` is - ve
Further two moles of atom will have GREATER randomness than ONE mole of molecules. Hence randomness increases so `DeltaS=+ve`
25.

For a process change in enthalpy is 36 kJ.mol and change in entropy is 120 J "mole "^(-1) K ^(-1). The temperature at which the sywstem attains eequibrium is

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`50 ^(@)C`
`47^(@)C`
`27^(@)C`
`12^(@)C`

ANSWER :C
26.

For a polyprotic acid, H_(3)PO_(4) its three dissociation constanst K_(1),K_(2) and K_(3) are in the order

Answer»

`K_(1) = K_(2) gt K_(3)`
`K_(1) = K_(2) = K_(3)`
`K_(1) gt K_(2) gt K_(3)`
`K_(1) lt K_(2) lt K_(3)`

SOLUTION :For any polyprotic acid `H_(3)A`, the successive dissociation constants reduces as `K_(1) gt gt K_(2) gt gt K_(3)`.
27.

For a physical equilibrium, H_(2) O (Ice) hArr H_(2) O (Water ) which of the following is the true statement:

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The pressure CHANGE do not affect the equilibrium
More of ice melts, if pressure on the SYSTEM is increased
More of LIQUID freezes, if pressure on the system is increased
More of ice melts, if TEMPERATURE of the system is increased

Solution :Le-chatelier.s principle
28.

For a perfectly cyrstalline solid C_(p,m)= a T^(3) + bT, where a and b are constants. If C_(p,m) is 0.40 J/K mol at 10K and 0.92 J/K mol at 20K, then molar entropy at 20K is 0.2x xx R joules. Then the value of x is

Answer»

0.92 J/K mol
8.66 J/K mol
0.813 J/K mol
none of these

Solution :`S_(20)- S_(0) = underset(0)OVERSET(20)int (C_(P))/(T) dT = underset(0)overset(20)int (aT^(3) + bT)/(T) dT`
`=[(aT^(3))/(3) + bT]_(0)^(20)`
`S_(20) = (a)/(3) (20)^(3) + b(20)`
`C_(10) = a(10)^(3) + b(10) =0.4`
`C_(20) = a(20)^(3) + b(20) = 0.92`
`rArr a = 2 xx 10^(-5) , b= 0.038`
`rArr S_(20) = (2)/(3) xx 10^(-5) xx (20)^(3) + 0.038 xx 20`
= 0.813 J/mol -K
29.

For a perfectly crystalline solid C_(p.m) = aT^(3), where a is constant. If C_(p.m) is 0.42 J/K - mol at 10K, molar entropy at 10K is

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<P>0.42J/K -mol
0.14 J/K -mol
4.2 J/K-mol
zero

Solution :`DELTA S = underset(0)overset(10)int (C_(P))/(T) dT` 3))/(T) dT = [(aT^(3))/(3)]_(0)^(10) = a xx (1000)/(3)`
`C_(P) = 0.42 = a xx 10^(3) rArr a= (0.42)/(10^(3))`
`rArr S_(10) - S_(0) = (0.42)/(1000) xx (1000)/(3) = (0.42)/(3)`
`S_(10) = 0.14` J/mol-K
30.

For a particular value of azimuthal quantum number (l), the total number of magnetic quantum number values (m) is given by

Answer»

`L = (m+1)/(2)`
`l = (m-1)/(2)`
`l = (2m+1)/(2)`
`N = (2l+1)/(2)`

ANSWER :B
31.

For a particular reversible reaction at temperature T, DeltaH and DeltaS were found to be both +ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when

Answer»


SOLUTION :`DeltaG = DeltaH -TDELTAS`
At EQUILIBRIUM, `DeltaG =0`
` 0 = DeltaH -TDeltaS`
or `T_(e)DeltaS = DeltaH`
`T_(e) = (DeltaH)/(DeltaS)`
SINCE it is an endothermic reaction, it is favoured by high temperature so that `T gt T_(e)`.
32.

For a particular reverisble reaction at temperature T,DeltaH and DeltaSwere found to be both =ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when

Answer»

`T_(e)` is 5times T
`T = T_(e)`
`T_(e) gt T`
`T gt T_(e)`

SOLUTION :`DELTAG =DeltaH - T DELTAS`. At equilibrium, `DeltaG = 0`. Hence,`T_(e)DeltaS = DeltaH`. As `DeltaH `and `DeltaS` are`+ve` , for REACTION to be spontaneous `DeltaG` should be `-ve`. This can be so only if `T DELTA S gt DeltaH `, i.e., `T Delta S gt T_(e) DeltaS ` or `T gt T_(e)`
33.

For a of orbital, the values of m are

Answer»

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

ANSWER :D
34.

For a non-ideal gas, the compressibility factor (Z) is defined as Z=(PV_(m))/(RT),V_(m)= molar volume Compressibility of an unknown gas at 600 K and 1.0 atm was found to be 1.2 Also, this gas was found to effuse 1.58 times slower than the pure methane gas under identical conditions Take R=0.0821 L-atm-mol^(-1)k^(-1) Molar volume of the gas in the given experimental condition is:

Answer»

`41.0` L
`39.4` L
`59.1` L
`27.3` L

Answer :C
35.

For a non-ideal gas, the compressibility factor (Z) is defined as Z=(PV_(m))/(RT),V_(m)= molar volume Compressibility of an unknown gas at 600 K and 1.0 atm was found to be 1.2 Also, this gas was found to effuse 1.58 times slower than the pure methane gas under identical conditions Take R=0.0821 L-atm-mol^(-1)k^(-1) Density of the gas in the above mentioned experimental condition is:

Answer»

`0.98 GL^(-1)`
`0.68 gL^(-1)`
`1.02 gL^(-1)`
`1.47 gL^(-1)`

ANSWER :B
36.

For a non-electrolytic solution

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` i=+ve`
` i= - ve`
` i=0`
` i=1`

ANSWER :D
37.

For a molecule to be stable

Answer»

`N_(B) LT N_(a)`
`N_(b) = N_(a)`
`N_(b) GT N_(a)`
`N_(b) ge N_(a)`

Answer :c
38.

For a mutielectron atom, the maximum of 2p-orbital in radial probability distribution graph is nearer the nucleus than that of 2s-orbital . Therefore, 2p-orbital should be closer to the nucleus and lower in energy than 2s-orbital. But 2s-orbital has lower energy than 2p-orbital. Explain.

Answer»

Solution :The radial probability DISTRIBUTION graphs for 2s - and 2p- orbitals are shown below:

It is clear from the figure that in case of 2s - orbital, there is a small additional PEAK or lamp. This indicates that a 2s-electron spends some of its time NEAR the nucleus. In other words, 2s-electron penetrates the `1s^2` - core (or K-shell, shown shaded in the figure). Due to penetration, a 2s- electron gets less shielding from other electrons and therefore, feels more nuclear charge. As a result, a 2s-electron is ATTRACTED more strongly by the nucleus than a 2p - electron. THUS, 2s has lower energy than a 2p - electron.
39.

For a molecule given select the correct statement (s). CH_(3)-CH=CH-CH=C=C=CH-CH=C=CH-CH_(3)

Answer»

It has only one optically inactive from, is its trans geometrical isomer.
It can SHOW geometical and as well as optical isomerism.
Total stereoisomers possible is 4.
Two PAIRS of ENANTIOMERS are possible.

Solution :
40.

For a mixture of I mole He and 1 mole Ne, select the correct statements(s)

Answer»

Molecules of the two gases strike the wall of the container with same frequency
Molecules of the HELIUM strike the wall more frequently
`u_(av_(He)) gt u_(av_(He))`
`P_(He) gt P_(N e)`

Solution :`(U_(AV))_(He) > (U_(av))_(N e) as M_(He)< M_(N e)`
Same moles `IMPLIES` P same `implies` same frequency .
41.

For a liquid the vapour pressure is given by log_(10) P = (-400)/(T) + 10 Vapour pressure of the liquid is 10^(x) mm Hg at 400K. The value of x will be _______

Answer»

<P>

Solution :`log P = (-400)/(T) + 10`
at 400K, log `P = (-400)/(400) + 10 = - 9 RARR P = 10^(-9)`
42.

For a liquid, enthalpy of fusion is 1.435 xx 10^(3) cal mol^(-1) and molar entropy change is 5.26 cal mol^(-1). Calculate the melting point of the liquid.

Answer»


Solution :`Delta H_(f) = 1.435 xx 10^(3), Delta S= 5.26`
`Delta S= (Delta H)/(T_(MP))`
`rArr M.P. = (1.435 xx 10^(3))/(5.26) = 273 = 0^(@)C`
43.

For a liquid, enthalpy of fusion is 1.435 kCal mol^(-1) and molar entropy change is 5.26 cal mol^(-1) K^(-1). The freezing point of liquid in celcius will be ________

Answer»


SOLUTION :Tmp `= (DELTA H "fusion")/(Delta S "fusion") = (1.435 xx 10^(3))/(5.26) = 272.81 K`
`= 0^(@)C`
44.

For a hypothetical reaction , X rarr Y , the enthalpy and entropy changes are 46.3 kJ mol^(-1) and 108.80 JK^(-1) mol^(-1) respectively. Find the temperature. Find the temperature above which this reaction is spontaneous

Answer»


ANSWER :425.6 K
45.

For a hypothetical reaction, the following kinetic data suggested that the overall order of this reaction is {:("[A]",[B],[C]),(mol" "dm^(-3),mol" "dm^(-3),mol" " dm^(-3) s^(-1)),(2.0,1.0,1.0),(2.0,4.0,2.0),(1.0,4.0,1.0):}

Answer»

3
1.5
1
0.5

Solution :LET rate of reaction , `R=[A]^(a) [B]^(b)`
`1=(2)^(a)(1)^(b)""…(i)`
`2=(2)^(a)(1)^(b)""…(ii)`
`1=(1)^(a)(4)^(b)""…(iii)`
Dividing (ii) by (i)
`((2)^(a)(4)^(b))/((2)^(a)(1)^(b))=(2)/(1)`
`4=2 to b=1//2`
Dividing (ii) by (iii)
`((2)^(a)(4)^(b))/((1)^(a)(4)^(b))=(2)/(1)`
`2^(a) = 2 IMPLIES a=1`
`therefore` Rate law EXPRESSION is , `r=[A][B]^(1//2)`
Order of reaction `=1+(1)/(2)=(3)/(2)=1.5`
46.

For a hypothetical reactionP(g) + Q(g) hArr R(g) + S(g) , " a graph between log K and " T^(-1)" is a straight line as hsown in the fig. in which "theta = tan^(-1) 0*5 and OA = 10. " Assuming "Delta H^(@) " is independent of temperature , calculate the equilibrium constant of the reaction at 298 K and 798 K respectively.

Answer»

Solution :Effect of temperature on equilibrium constant K is given by van't Hoff equation, viz.
` log K = -(DeltaH^(@))/(2* 303 RT) +C`where C = constant of integration
THUS , a plot of log or `1/T`, i.e., `T^(-1) ` is a straight LINE with SLOPE = `(Delta H^(@))/(2* 303 R)` and intercept = C
`:. C = 10 and " slope " = tan theta= 0*5 = (DeltaH^(@))/(2*303 xx 8*314) or DeltaH^(@)= 9* 574 " Jmol"^(-1)`
`:. log K= 10 -(9*574)/(2*303 xx 8* 314 xx 298 ) =9* 9827`
or K= Antlog `(9 *9827)= 9* 96 xx 10^(9)`
As `DeltaH^(@)` is independent of temperature , K will also be independent of temperature . Hence , K wil be same at 298 K and 798 K.
47.

For a hypothetical hydrogen like atom, the potential energy of the system is given by U(r) = (-Ke^2)/(r^3) ,where r is the distance between the two particles, If Bohr.s model of quantization of angular momentum is applicable then velocity of particle is given by:

Answer»

`v= (n^2 h^3)/(Ke^2 8 pi^3 m^2)`
`v= (n^3 h^3)/(8Ke^2 8 pi^3 m^2)`
`v= (n^3 h^3)/(Ke^2 8 pi^3 m^2)`
`v= (n^2 h^3)/(24Ke^2 8 pi^3 m^2)`

SOLUTION :`(d[U(r)])/(DR) = (3Ke^2)/(r^4) implies ` Magnitude of the force
`therefore (3Ke^2)/(r^4) = (mv^2)/(r)` and
we KNOW mvr `=(nh)/(2PI) ` or `r= (nh)/(2pin.v)`,
`3Ke^2 XX(8pi^3m^3 v^3)/(n^3 h^3) = mv^2 , v=(n^3 h^3)/(24ke^2 pi^3 m^2)`
48.

For a hydrogen -like particle, derive the expression : v_(n) = ((Ze^(2))/(mr_(n)))^(1//2) where v_(n) is the velocity of the electron at distance r_(n) from the nucleus, Z is the atomic number of the H-like particles, m and e are the charge and mass of the electron.

Answer»

Solution :For H-like particle, FORCE of ATTRACTION between the electron and the nucleus = centrifugal force i.e., `(Ze^(2))/(r_(n)^(2)) = (mv_(n)^(2))/(r_(n)) or v_(n)^(2) = (Ze^(2))/(mr_(n)) or v_(n) = ((Ze^(2))/(mr_(n)))^(1//2)`
49.

For a homogeneous gaseous reaction, 2A(g)+3B(g)hArr4C(g)+D(g),K_(eq)=(8)/(1.5)^(3) If in a 2 litre rigid container starting with 4 moles of A and 6 moles of B equilibrium was established then identify the options which is/are correct.

Answer»

CONCENTRATION of B at equilibrium is 1.5M.
Concentration of D at equilibrium is 1M.
Concentration of A at equilibrium is 1M.
Concentration of C at equilibrium is 2 M.

ANSWER :A::C::D
50.

For a homodiatomic molecule with a total number of electrons 14, after filling sigma**2s M.O. the next electron goes to

Answer»

`SIGMA _(2PZ)`
`pi_(2px)`
`pi_(2PY)`
`pi_(2px) orpi_(2py)`

ANSWER :D