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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 91351. |
(A), (B) and (C) are three dicarboxylic acids such that: (A)overset(Delta)tocarboxylic acid+CO_(2)uarr (B)overset(Delta)toacid anhydride1+H_(2)Ouarr (C )overset(Delta)tocyclic ketone+H_(2)O+CO_(2)uarr then |
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Answer»
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| 91352. |
A, B and C are three complexes of chromium(III) with empirical formula H_(12)O_(6)CI_(3)Cr. All the three complexes have water and chloride ions as ligands. Complex A does not react with conc. H_(2)SO_(4), whereas, complexes B and Close 6.7% and 13.5% of their original weight respectively on treatment with concentrated H_(2)SO_(4). Identity A, B and |
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Answer» SOLUTION :`A=[Cr(H_(2)O)_(6)]Cl_(3)overset(H_(2)SO_(4))to NO reaction` (therefour All `H_(2)O` molecules are present in co-ordination sphere) `B = [CrCl(H_(2)O)]CI_(2)H_(2)Ooverset(H_(2)SO_(4))to` one mol e of `H_(2)O` is lost Molecular WEIGHT of complex = 266.5` % loss = `(18)/(266.5)xx100=6.75%``C=[CrCI_(2)(H_(2)O)_(4)]Cl_(2)2H_(2)O_(4)]Cl_(2)2H_(2)Ooverset(H_(2))to 2mol ES` of `H_(2)O` is removedTherefour% loss in weight `= (2xx18)/(266.5)XX(100)=13.50%` |
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| 91353. |
A, B and C are three complexes of chromium with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have Cl and H_(2)O as ligands. Complex A does not react with conc. H_(2)SO_(4). Complexes B and C lose 6.75% and 13.5% of their original weight respectively on heating with conc. H_(2)SO_(4). Identify A, B and C. |
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Answer» Solution :Remembering that coordination NUMBER of Cr is 6, different conditions may be discussed as under : The formula `H_(12)O_(6)Cl_(3)Cr` suggests that there are 6 `H_(2)O` molecules, 3Cland 1 Cr. As Cl and `H_(2)O` are present as ligands andcomplex A does not LOSE `H_(2)O` on adding conc. `H_(2)SO_(4)`, this means, no `H_(2)O` is present outside the coordination sphere. Hence, the formula of A is `[Cr(H_(2)O)_(6)]Cl_(3)` Molecular mass of the complex `H_(12)O_(6)Cl_(3)Cr=12+96+106.5+52=266.5` amu As B loses 6.75% of `H_(2)O, H_(2)O` lost from 1 molecule `=(6.75)/(100)xx266.5=18` amu= `1 H_(2)O` molecule Thus, in B, one `H_(2)O` molecule is present outside the coordination sphere. Hence, complex B is `[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`. As C loses 13.5% of `H_(2)O, H_(2)O` lost from 1 molecue `=(13.5)/(100)xx266.5=36` amu = `2 H_(2)O` molecules Thus, in C. two `H_(2)O` molecules are present outside the coordination sphere. Hence, complex C is `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2 H_(2)O`. |
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| 91354. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(2)O_(6)Cl_(3)Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H_(2)SO_(4), whereas complexes B and C lose 6.75 % and 13.5 % of their original mass, respectively, on treatment with concentrated H_(2)SO_(4). Identify A, B and C |
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| 91355. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H_(2)SO_(4), whereas complexes B and C lose 6.75% and 13.5% of their original weight, respectively, on treatment with concentrated H_(2)SO_(4). Which type of isomerism is possible in complex C? |
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Answer» Geometrical |
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| 91356. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H_(2)SO_(4), whereas complexes B and C lose 6.75% and 13.5% of their original weight, respectively, on treatment with concentrated H_(2)SO_(4). Complexes A, B and C are not distinguished by: |
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Answer» molar conductivity data |
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| 91357. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H_(2)SO_(4), whereas complexes B and C lose 6.75% and 13.5% of their original weight, respectively, on treatment with concentrated H_(2)SO_(4). According to the Werner theory, the structure of complex A is represented by : |
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Answer»
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| 91358. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chlorie ions as ligands. Complex A does not react with conc. H_(2)SO_(4) whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated H_(2)SO_(4). The complex C is |
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Answer» `[Cr(H_(2)O)_(4)Cl_(2)].2H_(2)O` `% " loss "= (2xx18)/(266.5)xx10=13.50%` |
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| 91359. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chlorie ions as ligands. Complex A does not react with conc. H_(2)SO_(4) whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated H_(2)SO_(4). The complex B is |
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Answer» `[Cr(H_(2)O)_(4)Cl_(2)]CL.2H_(2)O` Molecular WEIGHT of the complex = 266.5 `%" loss "=(18)/(266.5)xx100=6.75%` |
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| 91360. |
A, B and C are three complexes of chromium (III) with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have water and chlorie ions as ligands. Complex A does not react with conc. H_(2)SO_(4) whereas complexes B and C lose 6.75% and 13.5% of their original weight respectively, on treatment with concentrated H_(2)SO_(4). The complex A is |
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Answer» `[Cr(H_(2)O)_(4)(H)_(4)(O)_(2)]Cl_(3)` |
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| 91361. |
(A), (B) and (C) are thrce non-cyclic fuactional isomsers of a carbonyl compound with molecular formula C4H8O Isomery (A) and (C) give positive. Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive Iodoform test. Isomers (A) and (B) on reduction withZn/Hg – Conc. HCl give the same product (D). Write the structuress of (A), (B), (C) and (D). |
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Answer» Solution :`A=CH_3CH_2 CH_2 CHO` `B=CH_2 CO CH_2 CH` `C=(CH_3)_2 CHCHO` `D=CH_3CH_2CH_2CH_2` |
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| 91362. |
A,B and C are isodiapher, while C,D and E are isobars. Calculate the difference of protons between A and E""_(82)^(206)AtoBtoCtoDtoE |
Answer» 
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| 91363. |
A, B and C are hydroxy-compounds of the elements X, Y and Z respectively. X, Y and Z are in the same period of the periodic table. A Gives an aqueous solution of pH less than seven. B reacts with both strong acid and strong alkalis. C gives an aqueous solution which is srongly alkaline. which of the following statements is/are true? I: The three elements are metals. II: The electronegativities decreases from X to Y to Z. III: The atomic radius decreases in the order X, Y and Z. IV: X, Y and Z could be phosphorus, aluminium and sodium respectively. |
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Answer» I, II, III only correct |
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| 91364. |
A auto-catalyst is: |
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Answer» CATALYST for catalyst |
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| 91365. |
(A) Atomic weight=Specific heat (cal/mol)xx64. (R) The formula is valid for metals only. |
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Answer» If both (A) and (R) are CORRECT and (R) is the correct EXPLANATION of (A). |
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| 91366. |
(A)Atomosphere contain 78% by volume of nitrogen (R ) Oxygen is the most abundant element in the earth crust |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 91367. |
(A) Atomicity of argon is unity(R) Ratio of molar heat capacities of argon is 1.67 |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 91368. |
(A): At boiling point of the liquid vapour pressure equal to atmospheric pressure (R): Volume of a solution increases by increasing the temperature. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 91369. |
(A) Atomic size of silver is almost equal to that of gold. (R) d subshell has low penetration power and produce poor shielding effect. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91370. |
(A) At all conditions Mg can reduce Al_(2) O_(3) (R) Below 135^(@)CMg can reduce Al_(2) O_(3) |
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Answer» Both (A) and (R) are TRUE and (R) is the correct explanation of (A) |
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| 91371. |
(a) Assign a reason for the following statements : Alkylamines are stronger bases than arylamines. (b) How would you convert methylamine into ethylamine ? |
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Answer» Solution :(a) It is because alkyl groups are electron releasing (+I) -effect. Therefore, they are stronger bases than arylamines because ARYL group is electron withdrawing (-I) effect. The lone pair of electrons on N in arylamine is delocalised due to resonance with benzene ring. (B) `UNDERSET("Methylamine")(CH_(3)NH_(2))+HNO_(2) rarr underset("Methanol")(CH_(3)OH) overset(PCl_(5))(rarr) underset("Chloromethane")(CH_(3)Cl) overset(KCN)(rarr) underset("Ethane nitrile")(CH_(3)C equiv N) underset(4[H])overset(Na//C_(2)H_(5)OH)(rarr) underset("Ethylamine")(CH_(3)CH_(2)NH_(2))` |
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| 91372. |
(A) AS_(2)O_(3) is less acidic when compared to AS_(4)O_(10) (R )Arsenic is a metalloid. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 91373. |
(A) as important laboratory reagent, turn red litmus blue imparts golden yellow colour in flame and is a gas precipitating agent, (A) reacts with Zn or AI forming H_(2)gas(A)gives while ppt with ZnCI_(2) or AICI_(3) but ppt . Diwsolves in exces of (A), what is (A) and explain reaction |
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Answer» Solution :(A)turns blue litmus red` rArr (A)` is basic in nature (A) imparts golden yellow colour in flame `rArr` (A) has `Na^(o+)` (A) gives `H_(2)` gaswith `Zn` or `AI rArr `(A) is `NaOH`. EXPLANATION: `2NaOH +Zn rarr Na_(2) ZnO_(2) +H_(2) uarr` `2NaOH + 2H_(2)O +2AI rarr 2NaAIO_(2) +3H_(2)uarr` `ZnCI_(2) +2NaOH rarr underset("White PPT")(Zn(OH)_(2)) darr +2NaCI` `Zn(OH)_(2) +2NaOH rarr underset("SODIUM zincate")(Na_(2)[Zn(OH)_(4)]) or Na_(2)ZnO_(2)` `AICI_(3) +3NaOH rarr underset("White ppt.")(AI(OH(_(3) darr) +3NaCI` `AI(OH)_(3) +NaOH rarr underset("Sodium meta-aluminate")(Na[AI(OH)_(4)] or NaAIO_(2))` |
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| 91374. |
A, arrhenius factor or pre-exponential factor corresponds to the ________ frequency. |
| Answer» SOLUTION :COLLISION | |
| 91375. |
(a) Arrange the following inthe order of property indicated for each set : (i) F_2,Cl_2,Br_2, I_2 - Increasing bond dissociation enthalpy . (ii) HF, HCl ,HBr, HI - Decreasing acid strength (iii) NH_3 ,PH_3 , AsH_3, SbH_3, BiH_3 - Decreasing base strength . (b) Write the conditions to maximize the yield toH_2SO_4 by contact process . |
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Answer» Solution :(a) (i) Increasing BOND dissociation enthalpy `I_2 lt F_2 lt Br_2 lt Cl_2` (ii) Decreasing acid strength HI > HBR > HCl > HF (iii) Decreasing base strength `NH_3 gt PH_3 gt AsH_3 gt SbH_3 gt BiH_3` (b) Conditions for maximum yield of `H_2SO_4` bycontact process (i) Low temperature - 720 K (ii) High pressure - 2 ATM (iii) Purity of gases (iv) Presence of a CATALYST `Pt//V_2O_3` (v) An excess of `O_2` |
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| 91376. |
(a) Arrange the following in the order of property indicated for each set :(I) F_2,Cl_2 , Br_2 , I_2 - Increasing bond dissociation enthalpy. (II ) HF, HCI, HBr, HI – Decreasing acid strength. (III) NH_3, PH_3, AsH_3 , SbH_3, BiH_3 - Decreasing base strength. (b) Write the conditions to maximize the yield of H_2 SO_4 by contact process. |
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Answer» Solution :(a)(I )increasingbonddissociationenthalpy `I_2 lt F_2 lt Br_2 lt Cl_2` (ii) Decreasingacidstrength `HI GT HBr gtHCIgtHF ` `(iii) ` Decreasingbasestrength ` NH_3 gt PH_3gt AsH_3gt SbH_3gt BiH_3` (b)conditionsfor maximumyieldof `H_2 SO_4 `bycontact PROCESS (i )Low temperature- 720K (ii) HIGHPRESSURE -2atm (iii)purityof gases ( IV) Presenceof acatalyst`Pt //V_2 O_3` ` (v ) ` An excess of ` O_2 ` |
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| 91377. |
(a) Arrange the following compounds in the increasing order of their acid strength : p - cresol, p - nitrophenol, phenol (b) Write teh mechanism (using curved arrow notation) of the following reaction : CH_(2)=CH_(2)overset(H_(3)O^(+))rarr CH_(3)-CH_(2)^(+)+H_(2)O |
Answer» SOLUTION :(a)  (B) `-CH_(3)` group is electron REPELLING and nitro group is electron - attracting. 
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| 91378. |
(a) Arrange alkyl halides water and alkane in order of decreasing density . (b) Arrange chloromethanes and water in order of decreasingdensity. (c)Arrange MeX inorder of decreasing bond length. (d) Arrange MeX in order of decreasing bond strength. (e) Arrange MeX in order of activity. (f)Arrange alkyl halides (1^(@),2^(@),3^(@)) in order of decreasing S_(N^(2)) reactivity. (g) Arrange the order of reactivity of alcohols towards HX. (h) Arrange the decreasing order of reactivity in haloforms. (i) Arrange the following isomeric bromides in order of decreasing reactivity in S_(N^(2)) displacement: underset((I))("(b) 2-Bromo-2- methylbutane") ,underset((II))(1-"Bromopentane"), underset((III))(2-"Bromopentane") (C) underset((I))(1-"Bromo-3-methylbutane") ,underset((II))(2-"Bromo-2-methylbutane") underset((III))(2-"Bromo-3-methylbutane") (d) underset((I))(1-"Bromobutane") , underset((II))(1-"Bromo-2- methylbutane") underset(III)(1-"Bromo -3- methylbutane") underset((IV))(1-"Bromo-2,2-dimethylpropane") |
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Answer» SOLUTION :`(a) RI gt RBr gt H_(2)O gt RCI gt RF gt RH` `(b) C CI_(4) gt CHCI_(3) gt CH_(2)CI_(2) gt H_(2)O gt CH_(3)CI` `(C) MeI gt MeBr gt MECI gt MeF` `(d) MeF gt MeCI gt MeBr gt MeI` `(E) MeI gt MeBr gt MeCI gt MeF` `(F) MeX gt RCH_(2) -X gt R_(2)CH -X gt R_(3)C -X` `(g) Allyl, BENZYL gt 3^(@) gt 2^(@) gt 1^(@)` `(h) CHI_(3) gt CHBr_(3) gt CHCI_(3) gt CH_(3) F` `(i) (a) I gt IV gt III gt II ""(b)II gt III gt I` (C) `I gt III gt II""(d) I gt III gt II gt IV` |
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| 91379. |
(a) Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenol (b) Write the mechanism (using curved arrow notation) of the following reaction: CH_(2)=CH_(2)overset(H_(3)O^(+))toCH_(3)-CH_(2)^(+)+H_(2)O or Write the structures of the products when butan-2-ol reacts with following: (a) CrO_(3) (b) SOCl_(2) |
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Answer» Solution :(a) p-cresol `LT` PHENOL `lt` p-nitrophenol (b)  (a) `CH_(3)CH_(2)underset("Butan-2-ol")(CHCH_(3))underset(O)OVERSET(CrO_(2))toCH_(3)CH_(2)underset("Butan-2-one")underset(O)underset(||)C-CH_(3)` (b) `CH_(3)CH_(2)underset("Butan-2-ol")underset(OH)underset(|)(CH)CHCH_(3)overset(SOCl_(2))toCH_(3)CH_(2)underset("2-Chlorobutane")underset(|)(CHCH_(3))` |
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| 91380. |
a. Arrange AgF, AgCI, AgBr and AgI is the increasing order of solubility in water b. NO_(2)^(Theta) interferes to the icingTest of NO_(3)^(Theta)suggest a chemicalmethodof removed burned with a blue flame initial but isputoff uinstandly even as gas appear coming Example. c. While testingoxalate, gas obtained burns with a blue flame intially but is put off instantly even as gas appears coming. Explain. d. I^(Theta) also interfere in the 'Ring Test' of NO_(3)^(Theta) suggest a chemicalreagent that removed I^(Theta) e. Colourless of (A)overset(Delta) to(B) gas + (C ) gas aquens solution of (c ) turn restlitrman blue , aquerous solutionof (A) and (B) also give while ppt , with AgNO_(3) solutionsouble in aqneous solutionof (C ) identify (A) ,(B) and (C ) f. Can youy defect Br^(Theta) and I^(Theta) by layer if present together ? |
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Answer» SOLUTION :`AgI lt AgBr lt AgCI lt AgF` b. Boll the mixture `NH_(4)O_(2)^(Theta)` is decomposed as `N_(2)` `NaNO_(2) + NH_(4)CI OVERSET(Delta) to NaCI + N_(2) uarr + 2H_(2)O` c. `H_(2) C_(2)O_(4) overset(Delta) toH_(2)O_((1)) + CO_(2(g))` d. `HgCI_(2)` removes `I^(Theta)`as `HgI_(2)` `HgCI_(2) + 2I^(Theta) rarr underset("ORANGE ppt")(HgI_(2) uarr + 2CI^(Theta)) ` e. (A) `NH_(4)CI (B): HCI (C ): NH_(3)` f. Reducin g power `I^(Theta) gt Br^(Theta) gt CI^(Theta) gt F^(Theta)` ADD `CI_(2)` and `CHCI_(3)` into mixture of `I^(Theta)` and `Br^(Theta)` `2I^(Theta) CI_(2) rarr underset("Violet")(I_(2) +) 2CI^(Theta)` `2Br^(Theta) + CI_(2) rarr underset("Orange yellow")(Br_(2)) + 2CI^(Theta)` `Br^(Theta)` will be also oxides `I^(Theta)` to `I^(Theta)` .Thus first there will be violetcolour in `CGCI_(3)` layer Extract aq layer (containing `Br^(Theta))` and repeat adding `CI_(2)` water and `CHCI_(3)` orange -yellow colour in `CHO_(3)` layer condifirms `Br^(Theta)` |
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| 91381. |
A aromatic hydrocarbon (A) of molecular formula C_(6)H_(6) reacts with Con HNO_(3) and Conc. H_(2)SO_(4) gives (B) of formula C_(6)H_(5)O_(2)N. (B) on reaction with Sn/HCl gives (C) formula C_(6)H_(7)N which answers carbylamine reaction. (C) on treatment with chloroform and alkalli gives (D) of formula C_(7)H_(5)N. Identify A,B,C,D and explain the reactions involved. |
Answer» SOLUTION : 
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| 91382. |
(A) : Aromatic carboxylic acids do not undergo Friedel - Crafts reactions (R) : In aromatic carboxylic acids carboxyl group is ring deactivating group. |
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Answer» Both A & R are TRUE , R is the correct EXPLANATION of A |
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| 91383. |
(A) Aromatic aldehydes are more reactive than aliphatic aldehydes. (R) Polarity of carbonyl group is more in benzaldehyde. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91384. |
(A) 'Ar' is used for arc welding of metals or alloys.(R) Ar provides an inert atmosphere for metal (or) alloys during arc welding |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91385. |
A: Arene diazonium salts are more stable than alkane diazonlum salts. R: Positive charge is dispersed on ring in arene diazonium salts. |
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Answer» If both Assertion & Reason are TRUE and the reason is the correct explanation of the assertion, then mark (1). |
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| 91386. |
(A) : Aqueous solution of Mohr's salt exhibits the test for NH_4^(+), Fe^(2+) and SO_4^(2-) ions. (R ): Mohr's salt is a double salt . |
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Answer» Both A & R are TRUE, R is the correct EXPLANATION of A |
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| 91387. |
(A) Aqueous solution of chlorine is a bleaching agent (R ) In presence of moisture ,chlorine liberates nascent oxygen which remove the colour of organic matter |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 91388. |
(A) Aqueous solution of FeCl_3 is acidic due to cationic hydrolysis (R) Ferric chloride is a covalent compound and exists as a dimer. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91389. |
A aqueous solution containing 1.248 g of barium chloride (molar mass ="208.34 g mol"^(-1)) in 100 g of water boids at 100.0832^(@)C. Calculate the degree of dissociation of BaCl_(2)(K_(b)" for water= 0.52 K kg mol"^(-1)). |
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Answer» `""{:(BaCl_(2), rarr,BA^(2+),+,2Cl^(-),),("1 mol",,,,,),(1-alpha,,alpha,,2alpha,"Total"=1+2alpha):}` `i=1+2 alpha "or"alpha=(i-1)/(2)=(1.67)/(2)=0.835.` |
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| 91390. |
(A) Aqueous fluoride on electrolysis at inertanode liberates oxygen (R ) Fluorine reacts vigorously with water to liberate oxygen |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT explanation of (A) |
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| 91391. |
(A) Aqueous chlorine turns blue litmus to red (R ) Aquous chlorine is acidic |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 91392. |
(A) : Any part of the solution has identical chemical and physical properties (R): Solution is always a homogenous mixture |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91393. |
A antifreeze solution is prepared from 222.6 g of ethylene glycol, C_(2)H_(4)(OH)_(2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is "1.072 g mL"^(-1), then what shall be the molarity of the solution? |
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Answer» Solution :`"MASS of the solute, "C_(2)H_(4)(OH)_(4)="22.6 g,Molar mass of "C_(2)H_(4)(OH)_(2)="62 g mol"^(-1)` `therefore"Moles of the solute"=("222.6 g")/("62 g mol"^(-1))="3.59 ,Mass of the solvent =200 g = 0.200 KG"` `"Molality "=("3.59 moles")/("0.200 kg")="17.95 mol kg"^(-1)` `"Total mass of the solution = 422.6 g"` `"Volume of the soltuion "=("422.6 g")/("1.072 g ml"^(-1))="394.2 ml = 0.3942 L , Molarity "=("3.59 moles")/("0.3942 L")="9.11 mol L"^(-1)`. |
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| 91394. |
(A): Antiferromagnetic substances possess almost zero magnetic moment (R): There are no unpaired electrons in anti ferromagnetic substances |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 91395. |
(a) Answer the following questions : (i) Which elements of the first transition series has highest second ionisation enthalpy ? (ii) Whichelements of the first transition series has highest third ionisation enthalpy ? (iii) Which element of the first transitionseries has lowest enthalpy of atomisation? (b)Identify the metaland justify your answer. (i) CarbonylM(CO)_(5) (ii) MO_(3)F |
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Answer» Solution :(i) Cu. This is because ELECTRONIC configuration of Cuis `3d^(10)4s^(1)`. After loss of 1st electron, it acquires stableconfiguration of `3d^(10) `. Hence, removal removal of 2nd electron is very difficult. (ii) Zn.This is because`Zn= 3d^(10) 4s^(2)` and `Zn^(2+) = 3d^(10)` which is again fully filledand hence is very stable. Removal of 3rd electronrequires very high ENERGY. (iii) Zn. This is because it hascompletely filled 3d subshell and no unpaired electron is availableformetallic bonding . (b) (i) Following EAN rule (ii) `overset(x-2)(MO_(3))F^(-1), x-6-1=0, x= +7, i.e., ` M is in OXIDATION state `+7` Mn shows an oxidationstate of `+7`.Hence, the compound is `MnO_(3)F` |
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| 91396. |
(A) Anti ferromagnetic substances posses zero magnetic moment. (R) MnO is an anti-ferromagnetic substance |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 91397. |
(a) Answer the following question: (i) Which element of the first transition series has highest second ionisation enthalpy? (ii) Which element of the first transition series has highest third ionisation enthalpy? (iii) Which elements of the first transition series has lowest enthalpy of atomisation? (b) Identify the metal and justify your answer: (i) Carbonyl M(CO)_(5) (ii) MO_(3)F |
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Answer» Solution :(a) (i) Cu In Cu the second ELECTRON is to be removed from `3d^(10)` CONFIGURATION which is highly stable. (ii) Zn The third electron is to be removed from `d^(10)` configuration which is highly stable. (iii) Zn-Zinc has all electrons paired and HENCE it cannot form metallic bonds. (b) (i) `Fe(CO)_(5)` As per Effective Atomic NUMBER (EAN) rule EAN=Z- [Number of electrons lost to form metalion] + [Number of electrons gained from donor atom] `:.` EAN `=26-0 + (5 xx 2)= 36` (ii) `MnO_(3)F`: Mn shown (+7) oxidation state. |
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| 91398. |
(A) Anisole is more reactive than phenol towards electrophilic aromatic substitution reactions(R)-OCH_3group is more ring activating than -OH group present on benzene ring |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 91399. |
(A) Anisole undergoes electrophilic substitution reaction at ortho & para position(R) Anisole is less reactive than phenol towards electrophilic aromatic substitution reaction |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 91400. |
A: Aniline does not undergo Friedel Craft's reaction. R: NH_(2) group of Aniline is deactivating group for all reactions |
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Answer» If both Assertion & Reason are TRUE and the reason is the CORRECT explanation of the assertion, then mark (1). |
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