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.

Consider the following first order reaction, 2A(g) rarr 2B(g) + C(g) Starting with pure A having pressure 2 atm initially, the total pressure is 3 atm after 2 hrs. Then total pressure after 4 hrs will be :

Answer»

4 ATM
`1.5` atm
5 atm
`3.5` atm

Answer :D
Discussion
2.

Consider the following first order reaction taking place at 308 K in 0.5 N HCl Sucrose to Glucose + Fructose At t = 0, the initial total rotation of the mixture is 32.4^(@). After 10 minutes, the total rotation of sucrose is 28.8^(@). If the rotation of sucrose per mole is 85^(@), that of glucose is 74^(@) and fructose is -86.04^(@), what will be the half-life of the reaction ?

Answer»

Solution :Suppose rotation per mole of sucrose, GLUCOSE and FRUCTOSE are `r_(1),r_(2)` and `r_(3)` respectively
Sucrose `to` Glucose + Fructose
`{:("At "t=0,,,a,,,0,,,0),("At time "t,,,a-x,,,x,,,x):}`
`ar_(1)=32.4" or "a=(32.4)/(r_(1))=(32.4)/(85)=0.38`
`(a-x)r_(1)+xr_(2)-xr_(3)=28.8`
`ar_(1)+x(r_(2)+r_(3)-r_(1))=28.8`
or `x(r_(2)+r_(3)-r_(1))=28.8-32.4=-3.6`
`x(74-86.04-85)=-3.6" or "x=0.037`
For a first order reaction
`k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(10)log""(0.38)/(0.38-0.037)=9.21xx10^(-3)min^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(9.21xx10^(-3))=75.2" min"`
Discussion
3.

Consider the following figure at 500 K.Assuming ideal gas behaviour, calculate the total pressure if the barriers are removed from the compartment.Assume that the volume of barriers is negligible.

Answer»


SOLUTION :Number of moles of A,B,C and D are GIVEN as :
`n_A=(PV)/(RT)=(1xx1)/(RT),n_B=(1.5xx2)/(RT)=(3)/(RT)`
`n_C=(2.5xx4)/(RT)=(10)/(RT),n_D=(2XX3)/(RT)=(6)/(RT)`
`n=n_A+n_B+n_C+n_D=(1+3+10+6)/(RT)=20/(RT)`
Total PRESSURE is given as
`P_("Total")=(nRT)/Vimplies P_("Total")=20/(RT)xx(RT)/7=2.85`
Discussion
4.

Consider the following experimental facts. I. When Cl_2 gas is passed into KI solution containingCHCl_3, violet colour appears in CHCl_3 layer. II. When Cl_2 gas is passed into KBr solution containing CHCl_3, orange colour appears in CHCI_3 layer. III. When Brą gas is passed into KI solution containing CHCI_3, violet colour appears in CHCl_3 layer. Select the correct experimental facts.

Answer»

I and II
I and III
II and III
I, II and III

Solution :Oxidisingpower`:F_2gt CI_2gt Br_2gt I_2`
thus `2KI +CI_2to2KCI+I_2"(violetin " CHCI_3)`
`CI_2 `oxidisesKIto `I_2`
thus(I )ISTRUE
`2KBr+CI_2` OXIDISES KBrto `Br_2 `
thus(II )is true

` 2KIBr_2to2KBr+I_2 `
thus(III)is alsotrue
( violetin ` CHCI_3)`
Discussion
5.

Consider the following equilibrium in a closed container N_(2)O_(4)(g)hArr2NO_(2)(g) At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements hold true regarding the equilibrium constant (K_(p)) and degree of dissociation (alpha)?

Answer»

Neither `K_(p)` nor `alpha` changes
Both `K_(p)` and `alpha` change
`K_(p)` changes but `alpha` does not change
`K_(p)` does not change, but `alpha` changes

Solution :For the equlibria `:N_(2)O_(4)(G)hArr2NO_(2)(g),K_(p)=K_(C)`because here `Deltan=1[K_(p)=K_(C)xx(RT)^(Deltan)].` will remain constant. Further SINCE volume is halved, the pressure will be double so `alpha` will decrease so as to MAINTAIN the CONSTANCY of `K_(C) or K_(p).`
`N_(2)O_(4)hArr2NO_(2)`
`{:(a,0),(a-X,2x):}`
Let total pressure =P
`thereforep_(NO_(2))=(2x)/(a+x)xxP,p_(N_(2)O_(4))=(a-x)/(a+x)xxP`
`impliesK_(p)=((P_(NO_(2)))^(2))/(P_(N_(2)O_(4)))=(4x^(2)P^(2))/((a+x)^(2))xx((a+x))/(P(a-x))=(4x^(2)P)/((x^(2)-x^(2)))`
Since `K_(p)=` constants, so `x prop(1)/(sqrt(P)).` So when volume is halved, pressure gets doubled and thus x `( or alpha)` will decrease.
Discussion
6.

Consider the following equilibrium, Which of the following best describes the equilibrium constant K of this interconversion ?

Answer»

`K LT 1`
`K GT 1`
K = 1
K cannot be DEDUCED from this information

Answer :B
Discussion
7.

Consider the following equilibrium in a closed container: N_(2)O_(4(g))iff2NO_(2(g)) At a fixed temperature, the volume of the reactioncontainer is halved. For this change, which of the following statements, holds true regarding the equilibrium constant (K_(p)) and degree of dissociation (alpha)?

Answer»

<P>NEITHER `K_(p)` nor `ALPHA` changes
Both `K_(p) and alpha` change
`K_(p)` changes but `alpha` not change
`K_(p)` does not change, but `alpha` changes

Solution :`K_(p)` is constant at constant temperature. As VOLUME is halved, PRESSURE will be doubled. Hence equilibrium will shift in the backward direction i.e., degree of dissociation decreases.
Discussion
8.

Consider the following equilibrium in a closed container :N_(2)O_(4)(g) hArr 2NO_(2)(g)At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements, holds true regarding the equilibrium constant (K_(p)) and the degree of dissociation (alpha) ?

Answer»

<P>NEITHER `K_(p)`nor `ALPHA` changes.
both `K_(p)` and CHANGE
`K_(p)` changes but `alpha` does not changes
`K_(p)` does not change but `alpha` changes.

Answer :C
Discussion
9.

Consider the following equilibrium in a closed container N_2O_4(g) hArr 2NO_2(g) At a fixed temperature , the volume of the reaction container is halved. For the this change, which of the following statement holds true regarding the equilibrium constant (K) and degree of dissociation (alpha) ?

Answer»

NEITHER `K_p "nor" ALPHA` CHANGES
both `K_p and alpha` CHANGE
`K_p` changes, but `alpha` does not change
`K_p` does not change ,but `alpha` changes

Answer :D
Discussion
10.

Consider the following equilibrium AgCldarr+2NH_(3)hArr[Ag(NH_(3))_(2)]^(+)+Cl^(-) White precipitate of AgCl appears on adding which of the following?

Answer»

`NH_(3)`
aqueous NaCl
aqueous `HNO_(3)`
aqueous `NH_(4)CL`

SOLUTION :`2HNO_(3)(aq)+[Ag(NH_(3))_(2)]^(+)+Cl^(-)toAgCl(s)darr+2NH_(4)^(+)+2NO_(3)^(-)`
Discussion
11.

Consider the following equation H_(4)P_(2)O_(7) + 2NaOHto Na_(2)H_(2)P_(2)O_(7) + 2H_(2)O If 534 gm of H_(4)P_(2)O_(7) is reacted with 3.0 xx 10^(24) formula units of NaOH ,then total number of moles of H_(2)O is produced is (N_(A) = 6 xx 10^(23) )

Answer»


ANSWER :5
Discussion
12.

Consider the following equations for a cell reaction {:(A+Biff C + D,E^(o)=x" volt, K"_(eq)="k"_(1)),(2A+2Biff2C+2D,E^(o)=y" volt, K"_(eq)="k"_(2)):} then which of these is the correct relation ?

Answer»

`X = y , K_(1) = K_(2)`
`x = 2Y, K_(1) = 2K_(2)`
`x=y, K_(1)^(2)= K_(2)`
`x^(2)=y, k_(1)^(2)=K_(2)`

ANSWER :C
Discussion
13.

Consider the following E^(o) values E_(Fe^(3+)//Fe^(2+))^(o)=+0.77V E_(Sn^(2+)//Sn)^(o)=-0.14V Under standard conditions the potential for the reactio Sn_((s))+2Fe_((aq))^(3+)to2Fe_((aq))^(2+)+Sn_((aq))^(2+) is

Answer»

0.91 V
1.40V
1.68V
0.63V

Solution :For `Sn_((s))+2Fe_((aq))^(3+)to2Fe_((aq))^(2+)+Sn_((aq))^(2+)`
`E_(cell)^(o)=E_(SN//Sn^(2+))^(o)+E_(Fe^(3+)//Fe^(2+))^(o)=(0.14)+(0.77)`
`E_(cell)=0.91` VOLTS
Discussion
14.

Consider the following eneryg profile for the reaction. X + Y = R +S. Which of the following deductions about the reaction is not correct ?

Answer»

the energy of activation for the backward REACTION is 80kJ
the FORWARD reaction is endothermic.
`Delta H` for the forward reaction is 20kJ
the energy of activation for the forward reaction is 60kJ.

Answer :A
Discussion
15.

Consider the following elements with their period number and valence electrons. {:("Elements","Period number","Total valence"e^(-)),(P,2,4),(Q,2,6),(R,3,7),(S,3,3),(T,3,6),(U,3,4):} According to the given informations, answer the following questions : Choose the incorrect statement :

Answer»

`SR_(3)` is a hypovalent COMPOUND
`UR_(4)` can ACT as a LEWIS acid
`PQ_(2)` can not acts as Lewis acid
`UR_(4) gt SR_(3)` : Lewis acidic CHARACTER

Solution :`SR_(3) : AlCl_(3) "" UR_(4) : SiCl_(4) "" PQ_(2) : CO_(2)` can act as Lewis acid.
Lewis acidic character of `SiCl_(4) gt AlCl_(3)`
Discussion
16.

Consider the following Ellingham diagram for carbon Which of thefollowing statements is incorrect for the above Ellingham diagram

Answer»

Upto `710^(@)C`, the reaction of formation of `CO_(2)` is energetically more favourable but above `710^(@)C,` theformation of `CO` is preferred
Carbon can be used to reduce any metal oxide at a sufficiently high TEMPERATURE
Carbon reduces many oxides at elevated temperature because `DeltaG^(@)` VS temperature line has a negative slope
`DeltaS^(@)"["C(s)+1/2O_(2)(g)toCO(g)"]"`
`ltDeltaS^(@)[C(s)+O_(2)(g)toCO(g)]`

Solution :(d) Since, `DeltaG^(@) = DeltaH^(@)-TDELTAS^(@)`
where ,`DeltaG^(@)=`standard Gibbs' free energy of the RACTION
`DeltaS^(@)=` standard ENTROPY ofthe reaction
`DeltaH^(@)=` standard enthalpy of the reaction
T= temperature
from the above equation it is clear that `DeltaG^(@)` will be more negative when `DeltaS^(@)` is less negative (or `DeltaS^(@)` is high).
Discussion
17.

Consider the following elements with their period number and valence electrons. {:("Elements","Period number","Total valence"e^(-)),(P,2,4),(Q,2,6),(R,3,7),(S,3,3),(T,3,6),(U,3,4):} According to the given informations, answer the following questions : Choose the correct statement :

Answer»

Q has maximum value of electron affinity
R has maximum value of ELECTRONEGATIVITY
S has maximum ATOMIC size
T and U are same GROUP elements

Solution :Maximum electron affinity is of R. `"" ` Maximum electronegativity is of Q
T and U belong to DIFFERENT GROUPS.
Discussion
18.

Consider the following elements with their electronegativity value: {:("Element",A,B,C,D),("Electronegativity (Pauling scale)",3.77,1.12,2.25,3.10):} Q. Select correct statement:

Answer»

OXIDE of element D is more acidic than that of A
Oxides of elements C and D are basic in nature
Oxide of element B is acidic in nature
BOH is more basic than `H_(2)O`

Solution :Due to more DIFFERENCE of ELECTRONEGATIVITY in B-O bond, BOH is more basic than water.
Discussion
19.

Consider the following elements with their electronegativity value: {:("Element",A,B,C,D),("Electronegativity (Pauling scale)",3.77,1.12,2.25,3.10):} Q. Incorrect statements is:

Answer»

AOH is more ACIDIC than DOH
BOH is more basic than COH
AB' molecule is predominantly ionic
D-OH' bond is more weaker than 'B-OH' bond in polar SOLVENT

Solution :Bont energy is INVERSELY PROPORTIONAL to differnce of ELECTRONEGATIVITY in polar solvent. Hence, `'B-OH'` is more basic 'D-OH'.
Discussion
20.

Consider the following elements A, B, C and D and their outer electronic configurtions are ns^(2)np^(1), ns^(2)np^(3), ns^(2)np^(4) and ns^(2)np^(5) respectively. Element E also has same outer electronic configuration like D but shows only single oxidation state(-1). If element A, B , C and D belong to same period as that of sodium. Consider the following compounds. (i)CE_(4)(ii)BD_(2)E_(3)(iii)DE_(3)(iv)DE_(2)(v)BD_(3)E_(2)(vi)C_(2)E_(2)(vii)DE(viii)A_(2)D_(6). Then calculate the value of "x+y", (where x and y are total number of polar and non-polar compounds).

Answer»


SOLUTION :`(6)/(2)=3`, ELEMENT `A=Al, B=P, C=S, D=Cl, E=F`
Discussion
21.

Consider the following electronic configuration of an element(P): [Xe]4f^(14)5d^(1)6s^(2) Then correct statement about element 'P' is

Answer»

It BELONGS to 6th PERIOD and 1st group
It belongs to 6th period an 2ND group
it belongs to 6th period and 3RD group
none of these

Solution :`[Xe]4f^(14)5d^(1)6s^(2)`
Atomoic number =71
Period number=6th
Group number=3rd
Discussion
22.

Consider the following electrode potentials{:(,Mg^(+2)+2e^(-)toMg,,,E^(0)=-2.37V),(,V^(+2)+2e^(-)toV,,,E^(0)=-1.18V),(,Cu^(+2)e^(-)toCu^(+),,,E^(0)=0.15V):} Which of the following reactions will proceed from left to right spontaneously ?

Answer»

`MG^(+2)+VtoMg+V^(2+)`
`Mg^(+2)+2CU^(+)toMg+2Cu^(+2)`
`V^(+2)+2Cu^(+)toV+2Cu^(+2)`
`V+2Cu^(+2)toV^(2)+2Cu^(+)`

ANSWER :D
Discussion
23.

Consider the following electrochemical cell. (a). Write a balanced net ionic equation for the spontaneous reaction that take place in the cell. (b). Calculate the standard cell potential E^(0) for the cell reaction. (c). If the cell emf is 1.6V what is the concentration of Zn^(2+)? (d). How will the cell potential be affected if Kl is added to Ag^(+) half-cell?

Answer»

Solution :(a) The SPONTANEOUS reaction taking place in the cell is :
`Zn(s)+2AG^(+)(aq) to Zn^(2+)(aq)+2Ag(s)`
(b) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.80-(-0.76)=1.56V`
(c ) ACCORDING to Nernst equation :
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Ag^(+)]^(2))`
`1.60=1.56-(0.0591)/(2)"log"([Zn^(2+)])/((0.1)^(2))`
`0.04=-0.02955" log"([Zn^(2+)])/((0.01))`
`"log"([Zn^(2+)])/((0.01))=(0.04)/((-0.02955))=-1.356`
`([Zn^(2+)])/(0.01)="Antilog"(-1.356)="Antilog"(overset(-)(2).644)4.4xx10^(-2)`
`[Zn^(2+)]=4.4xx10^(-4)M`
(d) On adding KI solution to `Ag^(+)//Ag` half-cell (cathode half cell) `Ag^(2+)` ions will be precipiated as Agl. Thus, the `[Ag^(+)]` will decrease and `E_(cell)` will reduce according to the Nernst equation.
`Ag^(+)+I to Agl`.
Discussion
24.

Consider the following E^(@) values E_(Fe)^(@).^(3+)._(//Fe^(2+))=+0.77V,E_(Sn^(2+)._(Sn))^(@)=-0.14V Under standard conditions, the potential for the reaction Sn(s)+2Fe^(3+)(aq)rarr2Fe^(2+)(aq)+Sn^(2+)(aq) is

Answer»

`1.68V`
`0.63V`
`0.91V`
`1.40V`

SOLUTION :`E_("cell")^(@)=E_(Fe^(3+)//Fe^(3+))-E_(SN^(2+) //Sn)^(@)`
`=0.77-(-0.14)=0.91V`
Discussion
25.

Consider the following E^(@) values E^(@)(Fe^(3+)|Fe^(2+))=+0.77V,E^(@)(Sn^(2+)|Sn)=-0.14Vunder standard conditions , the potemtial for the reaction : Sn(s)+2Fe^(3+)(aq)to2Fe^(2+)(aq)+Sn^(2+)(aq)is

Answer»

0.19 V
0.140 V
1.68 V
0.63 V

Solution :`E_("CELL")^(@)=E^(@)(FE^(3+)|Fe^(2+))-E^(@)(SN^(2+)|Sn)`
`=0.77-(-0.14)=+0.91V`
Discussion
26.

Consider the following E^(0) values E_(Fe^(3+) // Fe^(2+))^(0) = + 0.77 V , E_(Sn^(2+) // Sn)^(0)=-0.14 VUnder standard condition the potential for the reaction Sn_((s)) + 2Fe_((aq))^(3+) to 2Fe_((aq))^(2+) + Sn_((aq))^(2+) is

Answer»

1.68V
0.63V
0.91V
1.40V

Answer :C
Discussion
27.

Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes 'A' and 'B' in the electrolytic cell ?

Answer»

SOLUTION :A will have NEGATIVE POLARITY.
B will have POSITIVE polarity.
Discussion
28.

Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes A and B in the electrolytic cell ?

Answer»

Solution :* ELECTRODE (A) has negative electric charge and electrode (B) has positive electric charge, because electrochemical cell of Zn-Cu is attached with electrolytic cell.
* In electrochemical cell Zn is OXIDIZED `(ZntoZn^(2+)+2e^(-))` such electrons are MOVED on the Zn METAL and hence, it is negative, so the electrode (A) attached with it become positive.
* In cell, copper ion undergoes reduction reaction and hence amount of electron get reduced and so copper become positive and the electrode attached with it become negative.
Discussion
29.

Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes 'A' and 'B' in the electrolytic cell ?

Answer»

Solution :In the electrolytic CELLS COUPLED with an electrochemical cell, electrode 'A' will have negative polarity SINCE it is connected to copper plate in the electrochemical cell which has positive polarity, At the same TIME electrode 'B' have positive polarity.
Discussion
30.

Consider the following decay ._(Z)^(A)Xrarr_(Z+1)^(A)Y+_(-1)^(0)e,X is unstable because:

Answer»

its nucleus has EXCESS ENERGY
`(n)/(p)` ratio is HIGH
`(n)/(p)` ratio is low
none of these

Answer :B
Discussion
31.

Consider the following data for the reaction: A+B toProducts Determine the order of reaction with respect to A and with respect to B and the overall order of the reaction.

Answer»

Solution :The rate law may be expressed as:
Rate = `K[A]^(p)[B]^(q)`
Comparing experiments 2 and 3,
`(Rate)_(2) = (k[0.2]^(p)[2.0]^(q))/(K[0.2]^(p)[1.0]^(q))= (8.4 xx 10^(-3))/(8.4 xx 10^(-3))=1`
`[2]^(q) = [2]^(0),q=0`
Comparing experiments 1 and 2,
`(Rate)_(2) = k[0.20]^(p)[1.0]^(q) = 8.4 xx 10^(-3)`
`(Rate)_(1) = k[0.10]^(p)[1.0]^(q) = 2.1 xx 10^(-3)`
Dividing eqn. (III) by (iv),
`(Rate)_(2)/(Rate_(1)) = (k[0.20]^(p)[1.0]^(q))/(k[0.10]^(p)[1.0]^(q)) = (8.4 xx 10^(-3))/(2.1 xx 10^(-3))=4`
ORDER w.r. to A = 2, B=0, Overall order = 2.
Discussion
32.

Consider the following decay ._(Z)^(A)Xrarr_(Z-1)^(A)Y+_(+1)^(0)e,(beta^(+))X is unstable because:

Answer»

its NUCLEUS has EXCESS ENERGY
`(n)/(p)` ratio is high
`(n)/(p)` ratio is low
none of these

Answer :C
Discussion
33.

Consider the following decay series: A to B to C to D Where, A,B and C are radioactive elements with half lives of 4,5 sec, 15 days and 1 sec respectively and D is non-radioactive element, Starting with 1 mole of A, the numbe of moles B,C andleft after 30 days are:

Answer»

One mole of D and none of A, B or C
3/4 mole of B, 1/4 mole of D and none of A or C
1/4 MOL of B, 3/4 mol of D and none of A or C
1/2 MOE of B, 1/4 mol of C, 1/4 mol of D and noen of A

Answer :d
Discussion
34.

Consider the following data: {:("Element","Atomic weight"),(A,12.01),(B,35.5):} A and B combine to form a new substance X. if 4 moles of B combine with 1 mole of A to give 1 mole of X, then the weight of 1 mole of X is:

Answer»

154g
74g
47.5g
160g

Answer :A
Discussion
35.

Consider the following conversions: (i) O_((g))+e^(-) to O_((g))^(-),DeltaH_(1) (ii) F_((g))+e^(-) to F_((g))^(-),DeltaH_(2) (iii) Cl_((g))+e^(-) to Cl_((g))^(-),DeltaH_(3) (iv) O_((g))^(-) +e^(-) to O_((g))^(2-),DeltaH_(4) the according to given information the incorrect statement is :

Answer»

`DeltaH_(3)` is more negative than `DeltaH_(1) and DeltaH_(2)`
`DeltaH_(1)` is LESS negative than `DeltaH_(2)`
`DeltaH_(1),DeltaH_(2) and DeltaH_(3)` are negative whereas `DeltaH_(4)` is POSITIVE
`DeltaH_(1) and DeltaH_(3)` are negative whereas `DeltaH_(2) and DeltaH_(4)` positive

Solution :order of ELECTRON gain enthalpy: `Cl gt F gt O`
Second electron gain enthalpy for an element is always positive.
Discussion
36.

Consider the following conversions : I. CH_(3) underset(CN)underset(|)("CHCHO") overset(H_(3) O^(+))(rarr) CH_(3) underset("COOH")underset(|)(CH_(3)"CHCHO") II. underset(OH)underset(|)(CH_(3)"CHCN")overset(H_(3)O^(+))(rarr) CH_(3) underset(OH) underset(|)("CHCOOH") III. overset(O)overset(||)(CH_(3)"CCN")overset(H_(3)O^(+))(rarr ) overset(O)overset(||)(CH_(3) "CCOOH") In which case(s), preference group of nomenclature changes ?

Answer»

I
I, II
I, II, III
I, III

ANSWER :A
Discussion
37.

Consider the following covalent compounds in their solid state and find the value of expression (X+Y+Z). N_(2)O_(5),Cl_(2)O_(6),PC l_(),I_(2)Cl_(6),XeF_(6),PB r_(5) Where X=total number of compounds in which central atom of cationic or anionic part is sp^(3) hybridized. Y=Total number of compounds having 90^(@) bond angle either in cationic or anionic part. Z=Total number of compounds having 109^(@)28' bond angle either in cationic or anionic part.

Answer»


SOLUTION :
x=1+1+1+1=4
y=1+1=2
z=1+1+1=3
x+y+z=4+2+3=9
Discussion
38.

Consider the following conversion : Amongst the following, how many methods can be used in any one of steps involved in the mentioned conversion? i. Hoffmann bromamide degradation ii. Gabriel phthalimide synthesis iii. Sandmeyer reaction iv. Clemmensen reduction v. Mendius reduction

Answer»


ANSWER :1
Discussion
39.

Consider the following concentration cell : M(s)|M^(+)(aq, 0.04M)||M^(+)(aq, 0.08M)|M(s).|E_("cell")| for the cell is 60 mV. In the cell, if the concentration of the solution of M^(+) ions is 0.04 M instead of 0.02M, then the cell potential will be -

Answer»

60 mV
200 mV
319.3 mV
256.3 mV

Answer :C
Discussion
40.

Consider the following configuration (A),(B) and (C ). (A) Which of the following statements is correct?

Answer»

`(A)` and `R-`configuration
`(C )` has `R-` configuration
`(B)` has `R-` configuration
All `(A),(B)` and `(C )` have `(S)-` configuraiton

Solution :In `(C )`, the PRIORITY order is `H_(2)NGT COOH gt CHO gtH`. Therefore the compound is `R`. In `(A)`, the priority order is `CH_(2)CI gt CH(CH_(3))_(2) gtCH_(3) gtH`, therefore the compound si `S`. In `(B)`, the priority order is `CH_(2)OH gt CN gt C -=CH gt CH_(3)`, therefore the compound si `S`. Note that `-CH_(2)OH` has priority over `C-=N` because one oxygen atom counts before three NITROGEN atoms.
Discussion
41.

Consider the following configurations (A),(B),(C ) and (D) (A) which of the following statements is not correct?

Answer»

`(A)` and `(C )` are enantiomers
`(A)` and `(B)` are enantiomers
`(A)` and `(D)` are enantiomers
`(B)` and `(D)` are identical

Solution :The priority order is `OH gtCH =CH_(2) gtCH_(3) gtH`. Let's consider the configuration `(A)`, since the LOWEST priority `(H)` is one horizontal, `H` is moved to the vertical by making any two switches preserving the configuration. The configuration is then found to be `S`.

Alternatively, the priority order in the ORIGINAL `(A)` with `H` on the horizontal is `R`, but this is wrong and the molecules is actually `S`. similarly, configuration `B` is `R` (the enantiomers), `C` is `S` (identical) and `D` is `R` (the enantiomer). THUS `A` and `C` are not enantiomers but identical compounds.
Discussion
42.

Consider the following compounds. (P) XeF_(2)(Q) XeF_(4) (R ) XeF_(6) The correct order of tendancy of accept F^(-) ion is :

Answer»

<P>`P GT Q gt R`
`Q gt P gt R`
`P gt R gt Q`
`R gt Q gt P`

ANSWER :D
Discussion
43.

Consider the following compounds with their regular geometries. Calculate value of (x+y-z)^(2), where x is total number of axial d-orbitals having zero nodal plane, y and z are total no. of non-axial and axial d-orbitals respectively each one having two nodal planes used in hybridisation of central atoms of compounds A to E.

Answer»


Solution :
`x=d_(x^(2))=3`
`y=d_(xy)=1,(x+y-z)^(2)=(3+1-4)^(2)=0`
`z=d_(x^(2)-y^(2))+4`
Discussion
44.

Consider the following compounds.I) DDTII) Gammexane III) Carbon tetrachloride IV) ChlorobenzeneThe correct sequence of these compounds in the increasing order of percentage of chlorine in them is

Answer»

III, II, I, IV
IV, II, I, III
III, I, II, IV
IV, I, II, III

Solution :
Discussion
45.

Consider the following compounds, which of these will release CO_(2) with 5% NaHCO_(3)?

Answer»

(i), (II) and (III)
(i) and (iii)
(ii) and (iii)
(i) and (ii)

ANSWER :C
Discussion
46.

Consider the following compounds, I, 1,2-hydroxybenzene II 1,3-dihydroxybenzene III 1,4-dihydroxybenzene IV. Hydroxybenzene

Answer»

`I lt II lt III lt IV`
`IV lt I lt II lt III`
`IV lt II lt I lt III`
`I lt II lt IV lt III`

Solution :
1,4-dihydroxybenzne, adn isomer of dihydroxybenzene shows highest BOILING point due to intermolecular H-bonding followed by META ORTHO isomer. Hence, correct option is (b)
Discussion
47.

Consider the following compounds How many of the above compounds are less reactive than benzene in aromatic electophilic substitution reaction?

Answer»


ANSWER :6
Discussion
48.

Consider the following compounds (I)CH_(3)-underset(underset(CH_(3))(|))overset(overset(CH_(3))(|))C-Br(II)CH_(3)-underset(underset(CH_(3))(|))C-Br (III) CH_(3)-CH_(2)-CH_(2)BrTheir reactivity toward E1 is

Answer»

I GT II gt III
II gt I gt III
II gt III gt I
III gt II gt I

Answer :A
Discussion
49.

Consider the following compounds (i) C_(6)H_(5)COCl "" (ii) (##ERRL_CHE_V02_C20_E01_142_Q01.png" width="80%"> (iii) (##ERRL_CHE_V02_C20_E01_142_Q02##) "" (iv) (##ERRL_CHE_V02_C20_E01_142_Q03.png" width="80%"> The correct decreasing order of their reactivity towards hydrolysis is

Answer»

`(II) GT (iv) gt (III) gt (i)`
`(i) gt (ii) gt (iii) gt (iv)`
`(iv) gt (ii) gt (i) gt (iii)`
`(ii) gt (iv) gt (i) gt (iii)`

Solution :Electron withdrawing group increases REACTIVITY.
Discussion
50.

Consider the following compounds and select the incorrect statement from the following : NH_(3),PH_(3),H_(2)S, SO_(2), SO_(3), BF_(3), PCl_(3), IF_(7), P_(4), H_(2)

Answer»

3 MOLECULES out of given compounds involve `SP^(2)` hybridisation
Three molecules are hyprevalent compounds
Six molecules out of above compounds are non-planar in structure
Two molecules out of given compounds INVOLVES `(dpi-p pi)` bonding as well as also involves `(p pi-p pi)` bonding.

Answer :C
Discussion