94420 + Interview Questions in Chemistry in Class 12

1.	Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes : Ag^(+)""_((aq)) ""^(+e^(-))to Ag""_((s)),E^(@)=0.80V H^(+)""_((aq)) ""^(+e^(-))to 1/2H_(2(g)),E^(@)=0.00V On the basis of theri standars electrode protential values, which reaction is feasible at cathode and why ?
Answer» Solution :`Ag^(+)+E^(-)TOAG(s).` This reaction is feasible because `E^(@)` value for `Ag^(+)//Ag` is more than that of `H^(+)//H_(2).` Higher the value of `E^(@),` more easy it will be to converty ION inot its FREE STATE.

Discussion

2.	Following reaction(s) is/are involved in thermite process :
Answer» `3Mn_(3)O_(4) + 8Al RARR 9Mn + 4Al_(2)O_(3)` `Cr_(2)O_(3) +2AL rarr Al_(2)O_(3) + 2Cr.` `2FE + Al_(2)O_(3) rarr 3Al + Fe_(2)O_(3)` `B_(2)O_(3) + 2Al rarr 2B + Al_(2)O_(3)` ANSWER :A::B

Discussion

3.	Following reaction takes place in one step : 2" NO "(g)+O_(2)(g)iff2" NO"_(2)(g) How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one-third of its original volume ? Will there be any change in the order of the reaction with reduced volume ?
Answer» SOLUTION :RATE `=k[NO]^(2)[O_(2)]` Suppose initially, moles of NO = a, moles of `O_(2)=B`, volume of the vessel = V L. Then `[NO]=(a)/(V)M,""[O_(2)]=(b)/(V)M:."Rate"(r_(1))=k((a)/(V))^(2)((b)/(V))=k(a^(2)b)/(V^(3))""...(i)` New volume = `V//3`. `:." New concentrations ":[NO]=(a)/(V//3)=(3a)/(V),[O_(2)]=(b)/(V//3)=(3b)/(V)` `:." New rate"(r_(2))=k((3a)/(V))^(2)((3b)/(V))=(27ka^(2)b)/(V^(3))""...(ii)` `:.(r_(2))/(r_(1))=27" or "r_(2)=27r_(1)," i.e., rate becomes 27 times"` There is no effect on the order of REACTION.

Discussion

4.	Following reaction is occurred in button cell. (i) Ag_(2)O+H_(2)O +2e^(-)to2Ag+2OH^(-) (ii) Zn to Zn^(2+)+2e^(-) then calculate DeltaG^(Theta) [E_(Zn^(2+)\|Zn)^(Theta)=-0.76V,E_((Ag^(+)\|Ag))^(Theta)=0.34V
Answer» ANSWER :`DELTAG^(THETA)=-2.123xx10^(5)J`

Discussion

5.	Following reaction occurring in an automobile .^(2)C_(8)H_(18)(g)+25O_(2)(g)rarr16CO_(2)(g)+18H_(2)O(g). The sign of DeltaH, DeltaS and DeltaG would be
Answer» `+, -, +` `-, +, -` `-, +, +` `+, +, -` SOLUTION :For Combustion REACTION, `DeltaH` is negative, `Deltan=(16+18)-(25+2)=+7", so " DeltaS` is +ve, reaction is SPONTANEOUS, HENCE `DeltaG` is -ve.

Discussion

6.	Following reaction is of the type R-O-R' + HBr overset"Cold"to R-Br +R' -OH If R' is 3^@ alkyl group and R is 1^@ alkyl group , then
Answer» `S_N` 1 with TERTIARY alkyl group `S_N` 2 with tertiary alkyl group both of the above TYPES NONE of the above types ANSWER :C

Discussion

7.	Following passage describes charcterstics of colloids. Answer the questions at the end of it. Lyophilic colloidal sols are much more stable than lyophobic colloidal sols. This is due to the extensive solvation of lyophilic colloidal sols, which forms a protective layer outside it and thus prevents it from forming associated colloids. Lyophillic colloidal sols also protect lyophobic colloidal sols from precipition by the action of electrolytes. This is due to formation of a protective layer by lyophilic sols outside lyophobic sols. Lyophilic colloidal sols are called protective sols. Gelatin (lyophilic) protects gold sol (lyophobic) from coagulaion on the addition of sodium chloride solution. Protective powers of different colloidal sols are measured in terms of 'gold number' (Zigmody). It is defined as the amount of protective sol in milligrams that prevents the coagulation of 10 mL of a given gold sol on adding 1 mL of 10 percent sodium chloride. Thus smaller the gold number of a lyophillic sol, the greater is the protective power. [AgI]I^(-) collodial sol can be coagulated by the addition of a suitable cation. 1 mol of [Agl]I^(-) requires mol of AgNO_3, Pb(NO_3)_2 and Fe(NO_3)_3 as
Answer» `1, 1, 1` `1, 2, 3` `1, 1/2, 1/3` `6, 3, 2` Solution :`{:(overset(+)(Ag)NO_3 = "1 MOLE"),(Pb^(+2)(NO_3)_(2) = 1/2 "mole"),(Fe^(+3) (NO_3)_3 = 1/3 "mole"):}} "to ppt 1 mole" [AGI]I^(-)`

Discussion

8.	Following problem is based on thermodynamic law. Answer the questions given at the end of it. The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transition and chemical equilibrium. Three moles of CO_(2) gas expands isothermally (in thermal contact with the surroundings, temperature =15^(@) C) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are 10.0 L and 30.0 L, respectively. Change in entropy of the universe is
Answer» `34.34 JK^(-1)` `-34.34 JK^(-1)` `20.46 JK^(-1)` `-20.46 JK^(-1)` ANSWER :C

Discussion

9.	Following passage describes charcterstics of colloids. Answer the questions at the end of it. Lyophilic colloidal sols are much more stable than lyophobic colloidal sols. This is due to the extensive solvation of lyophilic colloidal sols, which forms a protective layer outside it and thus prevents it from forming associated colloids. Lyophillic colloidal sols also protect lyophobic colloidal sols from precipition by the action of electrolytes. This is due to formation of a protective layer by lyophilic sols outside lyophobic sols. Lyophilic colloidal sols are called protective sols. Gelatin (lyophilic) protects gold sol (lyophobic) from coagulaion on the addition of sodium chloride solution. Protective powers of different colloidal sols are measured in terms of 'gold number' (Zigmody). It is defined as the amount of protective sol in milligrams that prevents the coagulation of 10 mL of a given gold sol on adding 1 mL of 10 percent sodium chloride. Thus smaller the gold number of a lyophillic sol, the greater is the protective power. 0.025g of starch sol is required to prevent coagulation of 10ml gold sol when ImL of 10% Nacl solution is present. What is gold number of starch sol
Answer» 0.025 `2.5 xx 10^(-5)` `0.25` `25` Solution :GOLD sol is expressed in m.GR - 0.025 gr = 25 m.gr

Discussion

10.	Following options contain the pairs of the name of reaction and the name of final product, which pair is incorrect ?
Answer» Kolbe-Smitt reaction 2-Hydroxy BENZOIC acid Cumene PROCESS -PHENOL and acetone Reimer-Tiemann reaction - 2-Hydroxy benzaldehyde FRIES REARRANGEMENT- Methoxy benzoic acid Solution :Fries rearrangement- Methoxy benzoic acid

Discussion

11.	Following passage describes charcterstics of colloids. Answer the questions at the end of it. Lyophilic colloidal sols are much more stable than lyophobic colloidal sols. This is due to the extensive solvation of lyophilic colloidal sols, which forms a protective layer outside it and thus prevents it from forming associated colloids. Lyophillic colloidal sols also protect lyophobic colloidal sols from precipition by the action of electrolytes. This is due to formation of a protective layer by lyophilic sols outside lyophobic sols. Lyophilic colloidal sols are called protective sols. Gelatin (lyophilic) protects gold sol (lyophobic) from coagulaion on the addition of sodium chloride solution. Protective powers of different colloidal sols are measured in terms of 'gold number' (Zigmody). It is defined as the amount of protective sol in milligrams that prevents the coagulation of 10 mL of a given gold sol on adding 1 mL of 10 percent sodium chloride. Thus smaller the gold number of a lyophillic sol, the greater is the protective power. Gold number of haemoglobin is 0.03. Hence, 10 mL of gold sol will require haemoglobin so that gold is not coagulated by ImL of 10% NaCl solution
Answer» 0.03 MG 30 mg 0.30 mg 3 mg SOLUTION :Gold sol is EXPRESSED in m.gr = 0.03 m.gr

Discussion

12.	Followingmechanism has been proposed for a reaction ,
Answer» `r=K[A]^2[B]` `r=K[A][B]` `r=K[A]^2` r=K[A][C] ANSWER :B

Discussion

13.	Followingmechanism has been proposed for a reaction , ,The rate law expression for the reaction
Answer» `r=K[A]^2[B]` `r=K[A][B]` `r=K[A]^2` r=K[A][C] ANSWER :B

Discussion

14.	Following is the titration curve of CH_3 COOHagainst NaOH added with phenolphthalein as the indicator. K_("in") value of phenolphthalein is 4.0 xx 10^(10) Choose the incorrect statement.
Answer» It begins to change colour from the pH 9.4. It begins to change colour from acid (COLOURLESS) at pH 8.4 to the base form (reddish PINK) at pH 10.4. Phenolphthalein is suitable indicator for `CH_3 COOH-NaOH` titration. Phenolphthalein is a weak acid. Solution :`pK_a=- logK_(In )= 10 - log4=9.4` thus, it isthe equivalencepointpH .colourchangestartsfrom`(pK_(I n) -1)` andcompleteat `(pK_(I n )+1)`

Discussion

15.	Following is the substitution reaction in which -CN replaces -Cl. R-Cl+underset("alcoholic")(KCN)underset(Delta)toR-CN+KCl To obtain propanonitrile,R-Cl should be
Answer» CHLOROETHANE 1-Chloropropane Chloromethane 2-Chloropropane Solution :Propanonitrile obtained by the reaction between chloroethane ANF KCN on heating. `CH_(3)CH_(2)Cl+underset(("Aalcoholic"))(KCN)underset(DELTA)tounderset(("propanonitrile"))(CH_(3)CH_(2)CN)+KCL`.

Discussion

16.	Following is the substitution reaction in which -CN is replaced by - Cl underset("alconolic ")(R-Cl + K - CN) overset(Delta)(to)R - CN + KClTo obtain propane nitrite, R – Cl should be
Answer» CHLORO ethane 1 - chloro PROPANE chloro methane 2 -chloro propane SOLUTION :

Discussion

17.	Following is the graphical presentation of volumes occupied by different gases at S.T.P. Which is/are not placed at correct position?
Answer» `H_(2), He` `He, NH_(3)` `NH_(3), CH_(4)` `CH_(4), H_(2)` Solution :Molar MASS of `H_(2)`, He, `NH_(3)` and `CH_(4)` are 2,4,17, and 16 G `mol^(-1)` RESPECTIVELY. 2 g of `H_(2)` at S.T.P. = 22.4 L 1 g of `H_(2)` at S.T.P. = 11.2 L 4 g of He at S.T.P. = 22.4 L 17 g `NH_(3)` at S.T.P. = 22.4 L 3g of `NH_(3)` at S.T.P. = 3.95 L 16 g of `CH_(4)` at S.T.P. = 22.4 L 4 g of `CH_(4)` at S.T.P. = 5.6 L

Discussion

18.	Following is hydrated maximum at the position At which position the following compound will be maximum hydrated?
Answer» 1 2 3 equal ANSWER :B

Discussion

19.	Following is first order reaction: N_(2)O_(5) ("solution") to2NO_(2) ("solution")+(1)/(2)_(O_(2)(g))In which CCl_(4) is a solute .It is k=5.0xx10^(-4)S^(-1) The concentration of N_(2)O_(5) in initial is 0.25 mol L^(-1) (i)What will be the rate of reaction initially? (ii)Calculate half life (t_(1)/(2)) (iii)How much time require to compute 75 % reaction ? (iv)Calculate the concentration of N_(2)O_(5) and NO after 30 min .
Answer» SOLUTION :(i)`1.25xx10^(-4)` mol `L^(-1) s^(-1)` (ii) `t_((t)/(2))=1386s` (iii) 2773 s (IV) `[N_(2)O_(5)]=0.1 mol L^(-1)` and `[]NO_(2)=0.30 mol L^(-1)`

Discussion

20.	Following is a first order gas phase reaction: A_((g))toB_((g))+C_((g)).At t time ,total pressure =p_(t) and partial pressure of A=p_(A) atm.So,derive the integrated rate equation for this reaction .
Answer» <P> Solution :Suppose,the partial pressure of A,B and C are `p_(A),P_(B)` and `p_(c)` respectively so,`p_(t)=p_(A)+p_(B)+p_(C)` Ifx atm be the decrease in pressure of A at time t and one MOLE each of B and C is being formed,the increase in pressure of B and C will also be x atm each. These expression is in following equation So,total pressure (`p_(i)`-x+x+x)=`p_(t)` `therefore p_(i)+x=p_(t)` `therefore x=(p_(i)-p_(i)` atm) Thus at equilibrium `p_(A)=(p_(i)-x)` atm `(2p_(i)-p_(t))=[R]_(t)` At initially t=0 time `p_(i)=[R]_(0)` For FIRST order reaction `k=(2.303)/(t)` LOG `([R]_(0))/([R]_(t))` `therefore k=(2.303)/(t)` log `(p_(i))/((2p_(i)-p_(i)))`

Discussion

21.	Following ions are given : Cr^(2+), Cu^(2+), Cu^(+), Fe^(2+),Fe^(3+),Mn^(3+) Identify the ion which is (i) a strong reducing agent. (ii) unstable in aqueous solution. (iii) a strong oxidising agent. Give suitable reason in each.
Answer» Solution :(i) `Cr^(2+)` is a strong reducing agent because `E_(Cr^(3+)//Cr^(2+)=-0.41 V.)^(o)` The values with other metal IONS are positive. (ii) `Cu^(+)` ion is unstable in aqueous solution because the ENTHALPY of hydration `Cu^(+)` ion is very small. (iii) `Cu^(2+)` is a strong oxidising agent because it has a tendency to GET REDUCED with `E^(o)` value of `+0.34` V.

Discussion

22.	Following interconversion was done by Vandana and Upasana in ICL(International chemical Laboratory) New York. HC-=C-Hoverset(?)rarr DC-=C-CH_(2)-CH_(2)-OH Vandana's method : (P) CH_(3)MgBr (1 aq.) "followed by" " and " NH_(4)Cl (Q)CH_(3)MgBr(1eq.) followed by DOD. UPASANA's method (P) CH_(3)MgBr(excess)(Q)Cl-CH_(2)CH_(2)-Cl (R)Aq.KOH(S)D_(2)O Find out the correct statement(s) based on above formation.
Answer» VANDANA's method is correct and upasana's method is wrong upasana's method is correct and vandana's method is wrong Both the METHODS can give desired profuct but Upasana's method is better. Director of ICL (INTERNATIONAL chemical Laboratory) Arvind Vyas firds vandana and Upasana because both applied the wrong method. Solution :N//A

Discussion

23.	Following graphs are obtained when different gases are subjected to change in pressure at constant temperature. Identify the option which has correctly matched gas wth the graph.
Answer» `1 rarr O_(2), 2 rarr H_(2), 3 rarr He` `1 rarr CH_(4),2 rarr N_(2),3 rarr CO` `1 rarr H_(2), 2 rarr O_(2), 3 rarr SO_(2)` `1 rarr SO_(3), 2 rarr CO_(2), 3 rarr He` Answer :C

Discussion

24.	Following gases have equal masses at the same temperature, pressure and volume. The maximum work is done by ________________.
Answer» oxygen nitrogen hydrogen fluorine Answer :C

Discussion

25.	Following flow diagram represents the extraction of aluminium from bauxite, Coke powder is spreaded over the molten electrolyte to:
Answer» PREVENT the loss of HEAT by rediation from the surface. prevent the CORROSION of GRAPHITE anode. prevent oxidation of molten ALUMINIUM by air. both a and c Answer :d

Discussion

26.	Following flow diagram represent the extraction of aluminium form bauxite. The purpose of adding cryolite is :
Answer» To increase the electrical conductivity of pure aluminium To lower the melting point of `Al_2O_3` To remove the IMPURITIES as slag To increase the aluminium PERCENTAGE in the yield Solution :TRUE STATEMENT : Melting point of the compound `(Al_2O_3)` decreases when some other substance (cryolite) is added as impurity.

Discussion

27.	Following flow diagram represent the extraction of aluminium form bauxite. Select the incorrect statement.
Answer» Bauxite is purified by Hall's , Serpeck's and Baeyer's processes In electrochemical process for aluminium extraction , a molten mixture of `Al_2O_3,Na_3AIF_6 and CaF_2` is USED as electrolyte `AL(OH)_3` is converted into anhydrous alumina by calcination process None Solution :(A),(B) and (C )are correct statements. (A)Red bauxite CONTAINING the impurity of oxide of iron is removed by using Hall's and Baeyers methods while white bauxite containing the impurity of silica is removed by using serpeck's method. (B)`Na_3AlF_6and CaF_2` lowers th melting point and increase the electrical conductivity of molten `Al_2O_3` (C )`2Al(OH)_3underset("In absence of air")overset(Delta)to Al_2O_3+3H_2O(g)darr`. It is CALLED as calcination process.

Discussion

28.	FollowingequationillustratesC_6 H_5 Cl+ 2 NaOHunderset( 200a t m ) overset( 200-250^@ C ) toC_6 H_5 Ona + H_2 O
Answer» DOW's PROCESS Kolbe'sprocess Carbylaminetest HALOFORM reaction Answer :A

Discussion

29.	Following data is known about melting of a compound AB. DeltaH=9.2 kJ mol^(-1), DeltaS=0.008 kJ K^(-1) mol^(-1). Its melting point is
Answer» 736 K 1050 K 1150 K `1150^(@)C` Solution :`T_(m)=(DeltaH_("FUSION"))/(DeltaS_("fusion"))=(9.2)/(0.008)=1150 K`.

Discussion

30.	Following data is obtained for the reaction, N_(2)O_(5) to 2NO + 1/2O_(2) a) Show that the reaction is of first order b) Calculate the half life period.
Answer» Solution :a) For first ORDER reaction `k=2.303/t log a/(a-x)` `k_(1) = 2.303/(300s) (log) ([1.6 xx 10^(-12)])/([0.8 xx 10^(-2)])= (2.303)/(300s) xx LOG2 = 2.303/(300s)xx (0.3010) = 2.3 xx 10^(-3)`s `k_(2)= (2.303)/(600s)(0.6021) = 2.3 xx 10^(-3)`s Since the magnitude of the RATE constant reamains the same, the reaction is of first order. b) `t_(1//2) = 0.693/k = 0.693/(2.3 xx 10^(-3)s) = 301.30`S

Discussion

31.	Following compounds are given to you: 2-bromopentane, 2-bromo-2-methylbutane, 1-bromopentane (iii) Write the compound which is most reactive towards beta-elimination reaction.
Answer» SOLUTION :2-bromo-2-methylbutane is most reactive towards `BETA` elimination REACTION. This is because tertiary ALKYL halides on dehydrohalogenation form most SUBSTITUTED alkene.

Discussion

32.	Following compounds are given to you:2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane (i) Write the compound which is most reactive towards S_(N)2reaction. (ii) Write the compound which is optically active. (iii) Write the compound which is most reactive towards beta-elimination reaction.
Answer» Solution :`underset("2-Bromopentane") (CH_(3)- CH_(2) - CH_(2) - underset(H) underset(\|)overset(Br) overset(\|) C -CH_(3)) underset("2 -Bromo-2-methylbutane") (CH_(3) - CH_(2) - underset(CH_(3))underset(\|)overset(Br)overset(\|)C - CH_(3)) underset("1-Bromopentane") (CH_(3) - CH_(2) - CH_(2) - CH_(2) - CH_(2) - Br) ` (i) 1-Bromopentane is most REACTIVE towards SN2 reaction because it is a primary ALKYL halide. (II) Compound 2-Bromopentane is optically active because it contains an asymmetric carbon atom. `CH_(3) - CH_(2) - CH_(2) - underset(H) underset(\|) overset(Br) overset(\|)(C^(***)) - CH_(3) ` (iii)2-Bromo-2-methylbutane is most reactive towards b-elimination because it gives the alkene with greatest number of alkyl GROUPS attached to the double bond. `CH_(3) - CH_(2) - underset(CH_(3))underset(\|) overset(Br) overset(\|)C - CH_(3) overset("Alc.KOH") to CH_(3) - CH = underset(CH_(3)) underset(\|) C - CH_(3) `

Discussion

33.	Following compounds are given (i) CH_(3)CH_(2)OH (ii) CH_(3)COCH_(3) (iii) CH_(3)-underset(CH_(3))underset(\|)(C)HOH (iv) CH_(3)OH
Answer» (i), (iiii) and (iv) only (II) (i), (ii) and (iii) (i) and (ii) Solution :Compounds containingg `CH_(3)CO-` or `CH_(3)CHOH-` GROUP on WARMING with `I_(2)` and NaOH FORM iodoform. Hence, compounds (i), (ii) and (iii) will give iodoform while compound (iv), i.e., `CH_(3)OH` will not.

Discussion

34.	Following compounds are given 1.CH_(3)CH_(2)OH 2. CH_(3)COCH_(3) 3.CH_(3)-underset(CH_(3))underset(\|)CHOH 4.CH_(3)OHWhich of the above compound (s) on,beingwarmed with iodine solution and NaOH, will give iodoform
Answer» 1,3 and 4 only2 1, 2 and 3 1 and 2 Solution :Terminal `CH_(3)-OVERSET(O)overset(\|\|)C-` or `CH_(3)-underset(H)underset(\|)overset(OH)overset(\|)C-` SHOW positive IODOFORM TEST.

Discussion

35.	Following compounds A and B have similar structure with delocalization of pi-electrons system. (A)(PNCl_(2))_("x/2")(B)(CH)_(x) If value of x is 6, then calculate value of "P+Q", where 'P' is total no. of sigma-bonds in compounds A and B and 'Q' is total no. of pi bonds in compounds A and B.
Answer» SOLUTION :

Discussion

36.	Write the common and IUPAC names of (i) CH_3-CH=CH-CHO (ii) CH_3-CH(OH)-CH_2-CHO (iii) (CH_3)_2C(OH)CH_2CH_3.
Answer» i and II i, III and iv only ii i, ii and iii ANSWER :D

Discussion

37.	Following compound is treated with NBS: Compound formed A is
Answer» SOLUTION :

Discussion

38.	Following compound can be named as CH_3-O-(CH_2)_4CH_3
Answer» 3-methoxypentane 2-methoxypentane 4-methoxypentane 1-methoxypentane Answer :D

Discussion

39.	Following cell has EMF 0.7995 V . Pt\| H_(2) (1) atm) \|HNO_(3) (1M)\|\| AgNO_(3) (1 M)\| Ag If we add enough KCl to Ag cell so that the final Cl^(-) is 1 M . Now the measured emf of the cell is 0.222 V . The K_(sp) of AgCl would be -
Answer» `1 xx 10^(-9.8)` `1 xx 10^(-19.6)` `2 xx 10^(-10)` `2.64 xx 10^(-14)` Solution :`2 Ag^(+) + H_(2) to 2 H^(+) + 2 Ag` `E = E^(@) - (0.0591)/(2) "log" ([H^+]^(2))/(P_(H_(2)) xx [Ag^(+)]^(2))` `0.222 = 0.7995 - (0.0591)/(2) "log" (1)/([Ag^(+)]^(2))` `[Ag^(+)] = 10^(-9.8)` `K_(SP) = [Ag^(+)] [CL^(-)] = (10^(-9.8)) xx (1) = 10^(-9.8)`

Discussion

40.	Following are two first order reactions with their half times given at 25^(@)C A overset(t_(1//2) = 30 "min")(to) Products B overset(t_(1//2) = 40 "min") (to)Products The temperature coefficients of their reaction rates 3 and 2 , respectively , between 25^(@)C and 35^(@)C . if the above two reactions are carried out taking 0.4M of each reactant but at different temperatures 25^(@)C for the first reaction and 35^(@)C for the second reaction , find the ratio of the concentration of A and B after an hour
Answer» SOLUTION :0.4M of `A overset(30 "min")(to) 0.2 M overset(30 "min")(to) 0.1 M ` (for A) 0.4 M of `B overset(20 "min")(to) 0.2M overset(20 "min") (to) 0.1 M overset(20 "min")(to) 0.05` (for B) `"("(-d[B])/(dt)` will be DOUBLED and hence `t_(1//2)` will be halved) `THEREFORE ([A])/([B]) = 2`

Discussion

41.	Following beta ketoacid will not undergo decarboxylation on heating Give reasons.
Answer» Solution :The carbon at bridge HEAD position cannot be `SP^(3)` hybridizedaccoding to Bredt.s RULE. This rule is violated if the give `beta-` ketoacid undergoes decarboxylation on heating.

Discussion

42.	Following are the values of E_(a) and DeltaH for three reactions carried out at the same temperature : I : E_(a) =20 kJ mol^(-1), DeltaH =-60 k J mol^(-1) II : E_(a)=10 kJ mol^(-1), DeltaH =-20 kJ mol^(-1) III: E_(a) =40 kJ mol^(-1) , DeltaH =+15 kJ mol^(-1) If all the three reaction have same frequency factor then fastest and slowest reations are :
Answer» `{:("FASTEST",,"Slowest"),(I,,II):}` `{:("Fastest",,"Slowest"),(II,,III):}` `{:("Fastest",,"Slowest"),(I,,III):}` can't be predicted Answer :B

Discussion

43.	Following are the transition metal ions of3d series : Ti^(4+),V^(2+), Mn^(3+), Cr^(3+) (Atomci number : Ti = 22, V = 23 , Mn = 25 , Cr = 24) Anser the following : (i) Which ion is most stable in aqueous solution and why ? (ii) Whichion is a strong oxidizing agent and why ? (iii) Which ion is colourless and why ?
Answer» Solution :(i) Their electronic configuration are `:` `Ti^(4+)= 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)` `V^(2+) =1s^(1)2s^(2)2p^(6) 3s^(2) 3p^(6) 3d^(3)` `Cr^(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)` `Mn^(3+) =1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)` Thus , `Ti ^(4+)` is the most STABLE because it has noble gas configuration. (ii) Oxidizing agent is the substance which itself is reduce most easily, i.e., can gain ELECTRON easily. `V^(2+)` and `Cr^(3+)` are also stablebecause they have filled `t_(2g)` level (i.e., `t_(2g)^(3))` discussed in unit 9. Thus, `Mn^(3+)` can gain electron easily. Moreover `Mn^(2+)` is more stable than `Mn^(3+)` . Hence, `Mn^(3+)` is the strongest oxidizing agent. (iii) Ions are COLOURED if they have incompletely filled d-orbitals. Those with fully - filled or empty d-orbitals are coloureless . As` Ti^(4+)` has empty d-orbitals, hence it is colourless.

Discussion

44.	Followingare the reaction and conclusions related to radioactive disintegration with initial moles of each species being 10.
Answer» SOLUTION :

Discussion

45.	Following are the events taking place to explain adsorption theory I : Desorption II : Diffusion of the reactants along the surface III : adsorption of the reactants IV : formation of the activated surface complex These events are taking place in the following order
Answer» I, II, III, IV II, III, IV, I III, IV, I, II IV, III, II, I Solution :`UNDERSET((i))overset("DIFFUSION")(rarr)underset((ii))overset("adsorption")(rarr)underset({:("COMPLEX"),((iii)):})overset("activated")(rarr)underset((iv))overset("Desorption")(rarr)`

Discussion

46.	Following are the atoms having the number of neutrons and protons as given below : Select correct conclusion (s) :
Answer» A , B ANDC, are isotopes A ANDD are ISOTONES A and E are isobars B and D are isodiaphers Solution :for isotopes .Z. is same , for isobars .A. is same for isotones .n. is same , for isodiapheres (A - 2Z) is same

Discussion

47.	Following are neutral oxides except .......
Answer» `N_2O` NO `NO_2` CO Solution :`NO_2,NO,N_(2)O,CO ` - NEUTRAL oxides , `NO_2-` ACIDIC OXIDE

Discussion

48.	Following are equimolal (=equimolar) aqueous solutions. (A) 1 m glucose (B) 1 m NaCI © 1 m BaCI_(2)(D) 1 m Na_(3)PO_(4) € 1 m benzoic acid Assume 100%ionisation in B, C, D and 100% dimer formation in E, arrange the following in increasing (i)boiling point (ii) freezing point (iii) osmotic pressure
Answer» Solution : (i) Boiling point of solution `T_(B)=T_(b)+(DeltaT_(b))=T_(0)+i m K_(b)` Greater the value of more is the boiling point Increasing order of boiling point is : `EltAltBltCltD` (ii) Freezing point of solution `T_(f)=T_(0)=im K_(f)` Greater the volume of i, LESSER is the freezing point: `DltCltBltAltE` Osmotic PRESSURE of solution `pi=i` CRT Greater the value of i more is the osmotic pressure. Increasing order of osmotic pressure is : `EltAltBltCltD`.

Discussion

49.	Following are equimolal (=f equimolar) aqueous solutions (A) 1 m glucose (B) 1m NaCl (C) 1m BaCl_(2) (D) 1m Na_(3)PO_(4) (E) 1m benzoic acid. assume 100% ionisation in B,C,D and 100% dimer formation in E arrange them in increasing (1) boiling point, (2) freezing point, (3) osmotic pressure, (4) vapour pressure.\
Answer» Solution : (1) Boiling point of solution `T=T_(0)+(DeltaT_(b))` `DeltaT_(b)=mK_(b)i` ltbRgt THUS, greater the value of `i`, greater the value of `DeltaT_(b)` and boiling point of solution. `therefore EltAltBltCltD` (2) Freezing pont of solution `T=T_(B)-(DeltaT_(f))` `DeltaT_(f)=mK_(f)i` Thus, greater the value of `i`, greater the value of `DeltaT_(f)` but smaller the value of freezing point. Thus, `EltAltBltCltD` (4)Vapour pressure of solution `P_("solution")=P_("solvent")^(0)-DELTAP` `(DeltaP)/(P^(0))=X_("solute")=(n_(1)i)/(n_(1)i+n_(2))=(n_(1)i)/(n_(2))` `therefore` The greater the value of `i`, greater the value of `Delpap` hence, smaller the value of vapour pressure of solution. Thus, `DltCltBltAltE`.

Discussion

50.	Following are a group of compounds showing acidic behaviour: Phenols are less acidic than carboxylic acids. Why?
Answer» ANSWER :This is because phenoxide `(C_6H_5O^(theta))` ion is LESS resonance stabilised than carboxylate anion `R-COO^(theta)`. In three resonance structures of phenoxide ion, the negative charge is on less ELECTRONEGATIVE CARBON. But in carboxylate ion, negative charge is one more electronegative oxygen.

Discussion

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