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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.
| 90551. |
A compound A on heating with caustic soda solution liberates a gas B which gives white fumes on exposure to HCl gas. Heating is continued to expel the gas completely. The resultant alkaline solution again liberates the same gas B when heated with Zn powder. However, the compound A when heated alone does not evolve any gas. Identify A and B. |
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Answer» `{:( A , B ) , ( AgNO_(3) , NO_(2)):}` |
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| 90552. |
A compound A of molecular formula C_(3)H_(7)O_(2)N on reaction with Fe and cone. HCl gives a compound B of molecular formula C_(3)H_(9)N. Compound B on treatment with NaNO_(2) HCl gives another compound C of molecular formula C_(3)H_(8)O. The compound C has molecular formula C_(3)H_(8)O. The compound C gives effervescence with Na. On oxidation with CrO_(3), the compound C gives a saturated aldehyde containing three carbon atoms. Deduce the structures of A,B and C and write the equations for the reaction involved. |
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Answer» Solution :`UNDERSET("1-Nitropropane"(A))(CH_(3)CH_(2)CH_(2)NO_(2))underset(HCl)overset(FE)(to) underset(1-"Propanamine"(B))(CH_(3)CH_(2)CH_(2)NH_(2))underset(HCl)overset(NaNO_(2))(to) underset("Propan-1-ol"(C))(CH_(3)CH_(2)CH_(2)OH)underset("Sodium propoxide")(CH_(3)CH_(2)CH_(2)ONa)+H_(2)` `underset(1-"PROPANOL"(C))(CH_(3)CH_(2)CH_(2)OH)overset(CrO_(3))(to) underset("PROPANAL")(CH(3)CH_(3)CHO)` |
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| 90553. |
A compound (A) of molecular formula C_(14)H_10O_2 is formed from C_14H_14O_2 by oxidation Cr_(2)O_(7)^(-2).(A) upon treatment with OH^(-) gives (B).(B) on treatment with conc. H_2SO_4 and heat gives compound ( C) of molecular formula C_28H_20O_4(B) responds to NaHCO_3 test and effervescence comes out.What should be the molecular wt. of B ? |
Answer» 
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| 90554. |
A compound A of formula C_3H_6Cl_2 on reaction with alkali can give B of formula C_3H_6O or C of formula C_3H_4.B on oxidation gave a compound of the formula C_3H_5O_2.C with dilute H_2SO_4 containing Hg^(2+)ion gave D of formula C_3H_6O which with bromine and alkali gave the sodium salt of C_2H_4O_2. A is: |
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Answer» `CH_3CH_2CHCl_2` |
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| 90555. |
A compound A of formula C_3H_6Cl_2 on reaction with alkali can give B of formula C_3H_6O or C of formula C_3H_4 B on oxidation gave a compound of the formula C_3H_5O_2 C with dilute H_2SO_4 containing Hg^(2+)ion gave D formula C_3H_6O which with bromine and alkali gave the sodium salt of C_2H_5O_2. T |
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Answer» `CH_3CH_2CHCl_2` |
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| 90556. |
A compound 'A' formula of C_(3)H_(6)CI_(2) on reaction with alkali can give 'B' of formula C_(3)H_(6)O or 'C' of formula C_(3)H_(4). 'B' on oxidatrtion gave a compound of the formula C_(3)H_(6)O_(2).'C' with dilute H_(2)SO_(4) containing Hg^(2+) ion gave 'D' of formula C_(3)H_(6)O which with bromine and alkali gave the sodium salt of C_(2)H_(4)O_(2). Then 'A' is |
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Answer» `CH_3CH_2CHCl_2` |
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| 90557. |
A compound (A) is greenish crystalline salt, which gave the following results. i) Addition of BaCl_(2) solution to the solution of (A) results in the formation of white ppt. (B). which is insoluble in dil HCL ii) On heating (A), water vapours and two oxides of sulphur, (C) and (D) are liberated leaving a red brown residue (E) iii) (E) dissolves in warm conc. HCI to give a yellow solution (F). iv) With HS the solution (F) yields a pale yellow ppt. (G) which when filtered, leaves a greenish filtrate (H). v) Solution (F) on treatment with thiocynaate ions gives blood red coloured compound (I). The yellow solution is of |
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Answer» `FeCl_(3)` |
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| 90558. |
A compound (A) is greenish crystalline salt, which gave the following results. i) Addition of BaCl_(2) solution to the solution of (A) results in the formation of white ppt. (B). which is insoluble in dil HCL ii) On heating (A), water vapours and two oxides of sulphur, (C) and (D) are liberated leaving a red brown residue (E) iii) (E) dissolves in warm conc. HCI to give a yellow solution (F). iv) With HS the solution (F) yields a pale yellow ppt. (G) which when filtered, leaves a greenish filtrate (H). v) Solution (F) on treatment with thiocynaate ions gives blood red coloured compound (I). White ppt (B) is of |
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Answer» `K_(2)SO_(4)` |
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| 90559. |
A compound (A) is greenish crystalline salt, which gave the following results. i) Addition of BaCl_(2) solution to the solution of (A) results in the formation of white ppt. (B). which is insoluble in dil HCL ii) On heating (A), water vapours and two oxides of sulphur, (C) and (D) are liberated leaving a red brown residue (E) iii) (E) dissolves in warm conc. HCI to give a yellow solution (F). iv) With H2S the solution (F) yields a pale yellow ppt. (G) which when filtered, leaves a greenish filtrate (H). v) Solution (F) on treatment with thiocynaate ions gives blood red coloured compound (I). Compound (A) is |
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Answer» `CuSO_(4)` |
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| 90560. |
A compound (A) is greenish crystalline salt, which gave the following reactions. (i)Addition of BaCl_(2) solution to the solution of (A) results in the formation of white precipitate (B) which is insoluble in dilute HCl. (ii)On heating (A), water vapours and two oxides of sulphur (C ) and (D)are liberated leaving a red brown residue (E). (iii)(E) dissolves in warm concentrated HCl to give a yellow solution (F). (iv)Solution (F) on treatment with thiocyanate ions gives blood red coloured compound (G). Identify the compounds from (A) to (G). |
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Answer» |
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| 90561. |
A compound 'A' having the molecularformula C_(5)H_(12)O, on oxidation gives a compound 'B' withmolecularformula C_(5)H_(10)O. Compound 'B' gave a 2, 4- dinitrophenylhydrazinederivative but did not answer haloform test or silver mirror test. Thestructure of compound 'A' is |
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Answer» `CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2) -OH`  Since (B) on reaction with 2, 4-DNP forms a derivative, it implies that (B) has  group. (B) gives -ve Tollens' test, hence it is not an aldehyde, but it is a KETONE. (B) gives -ve haloform test, thus it is not a methyl ketone. (B) is formed from the OXIDATION of (A), thus (A) is a`2^(@)` alcohol, and among the given options, (A) is : `CH_(3)-CH_(2)-underset(OH)underset(|)(CH)-CH_(2)-CH_(3)` and `therefore (B)` is `CH_(3)-CH_(2)-underset(O)underset("||")C-CH_(2)-CH_(3)` |
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| 90562. |
(a) Compound (A) having the empirical formula gives the cannizaro reaction to form (B) and (C). Compound (B) has a neutralisation equivalent of 112.08 acetone participates with (A) in claisen condensation like reaction to form (D) which has the formula C_(8)H_(8)O_(2). oxidation of (C) gives (B). when (B) is heated at 200-225^(@)C gives (E) hich has the formula C_(4)H_(4)O is formed. hydrogen under pressure in the presence of Ni reduces (E) to (F). vigorous treatment of (F) with HCl gives (G) which is a dichloroalkane. treatment of (G) with KCN followed by hydrolysis and acidification gives adipic acid. give structures of (A) to (G). (b) Give stereochemistry of the products obtained by the hydroboration-oxidation of cis and trans-2-phenyl-2-butene? |
Answer» SOLUTION : (B) Hydroboration-oxidation is a method of syn. Hydration and follows ANTI markownikov.s RULE.  
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| 90563. |
A compound 'A' having the molecualr formula C_(5)H_(12)O on oxidation gives a compound 'B' with molecular formula C_(5)H_(10)O compound B gave a 4-dinitrophenyl hydrazine derivative but did not answer halo form test or silver nitrate test. The structure of compound A is |
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Answer» `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)OH` 
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| 90564. |
A compound (A) having molecular formula C_(7)H_(11) Br is optically acitve. A react with HBr in the absence of peroxide to give isomeric products (B) and (C). Tretting (A)with potassium-tutoxide give (D). (D) on reductive ozonolysis gives two moles of formaldehyde and one mole of 1,3 cyclopentanedione.(A) in the presence of peroxide reacts with HBr to given (E). |
Answer»  
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| 90565. |
A compound (A) having molecular formula C_4H_9Cl, on heating with alcoholic KOH solution, gives two isomeric alkenes (B) & (C ) . The mixture of (B) & ( C) ,on ozonolysis produces three compounds : 1 HCHO, 2CH_3CHO, 3 CH_3CH_2CHO. Ascertain structures of (A) , (B) and ( C) . |
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Answer» SOLUTION :The compound (A), when heated with alcoholic KOH solution, undergoes dehydrochlorination to give two isomeric alkenes (B) and (C ). So the molecular formula of these two alkenes is `C_4H_8` . `C_4H_9Cl (A) overset("Alcoholic KOH")underset("heat") to C_4H_8(B)+C_4H_8(C )` Mixture of (B) and (C ) on, OZONOLYSIS, produces three carbonyl COMPOUNDS `(HCHO,CH_3CHO &CH_3CH_2CHO)`. Now, the TOTAL number of carbon atoms of the compound or compounds obtained by ozonolysis must be equal to the number of carbon atoms present in the compoundundergoing ozonolysis. so , let us suppose that as a result of ozonolysis of the alkene B ,1 molecule of `CH_3CH_2CHO`and 1 molecule of HCHO (total number of carbon atomsof the two compounds = 4) are obtained and the ozonolysis of ( C) produces 2 molecules of `CH_3CHO` (total number of carbon atom of the two molecules =4 ) . So, the structure of the alkene B is `CH_3CH_2CH=CH_2` (1-butene) & the structure of alkene C is `CH_3CH=CHCH_3` (2-butene). `[{:(CH_3CH_2CH=O+O=CH_2rArrCH_3CH_2CH=CH_2),(CH_3CH=O+O=CHCH_3rARrCH_3CH=CHCH_3):}]` Since the compound (A) on dehydrochlorination, produces 1-butene and 2-butene, the structure of (A ) is: `CH_3CH_2CH(Cl)CH_3` (2-chlorobutene) 
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| 90566. |
A compound A having molecular formula C_(3)H_(7)ON reacts with Br_(2) in presence of NaOH to give compound 'B'. This compound 'B' reacts with HNO_(2) to form alcohol and N_(2) gas. Identify compound 'A' and 'B' and write the reaction involved. |
| Answer» Solution :A. is `CH_(3)CH_(2)CONH_(2) OVERSET(Br_(2)) underset("NaOH") to CH_(3)CH_(2)NH_(2) overset (HNO_(2))to C_(2)H_(5)OH + N_(2)H_(2)` | |
| 90567. |
A compound A having molecular formula (C_3 H_9 N) on treatment with benzenesulfonyl chloride gives an insoluble solid. Identify A |
| Answer» SOLUTION :Only secondary amines form insoluble solids on treatment with BENZENE sulfonyl CHLORIDE. THUS, A is a secondary amine having the structure, `C_2 H_5 underset(H) underset(। )N-CH_3` (N-methylethanamine) | |
| 90568. |
A compound(A) hasthe molecularformula C_(6)H_(10)O_(4). It is a sweet - smellingliquid , sparinly souble in water .(a) When (A) is boiledunder reflux withaqueoussodium hydroxide ,a clearsolution , (B) is obtained . (b) When solution (B) is distiled , the distillate(C) turns warm acidifiedsodiumdichromatesolutiongreen . (c) Whena portionof theresidual, liquidform (B) isneutralizedwithhydrochloricacid, and calciumchloride solutionadded , a white precipitated , (D) is observed. (d)Whenthe |
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Answer» Solution :(A) ` = underset(CO_(2)C_(2)H_(5))underset(|)(CO_(2)C_(2)H_(5))`or `underset(CO_(2)C_(3)H_(7))underset(|)(CO_(2)CH_(3))` (B) Mixture of `underset(CO_(2)Na^(+))underset(|)(CO_(2)^(-)Na^(+))` andeither `C_(2)H_(5)OH` or `CH_(3)OH` and `C_(3)H_(7)OH` (D) `underset(CO_(2))underset(|)(CO_(2)^(-)) Ca^(+ +)""` (E) = `underset(CONH_(2))underset(|)(CONH_(2))` `C_(3)H_(7)` is EITHER `CH_(3)CH_(2)CH_92) ` - or `(CH_(3))_(2)CH`. If the iodoform test is positive A is . 
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| 90569. |
A compound A has the molecular formula C_(5)H_(9)Cl. It does not react with bromine in carbon tetrachloride. On treatement with a stron base it produces a single compound B. B has a molecular formula C_(5)H_(6) and reacts with bromine in carbond tetrachloride. Ozonolysis of B produces a compound C which has a molecular formula C_(5)H_(6)O_(2). Which of the following structure is that of A? |
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| 90570. |
A compound(A)hasmolecularformulaC_(5)H_(9)CI Itdoes notreactwithbrominein C CI_(4) On treatmentwitha strongbaseit producesa singlecompound(B)(B) has amolecular formulaC_(5)H_(8) andreactswithBaeyer'sreagent . Reductiveozonotysis of(B)produces acompound(C )whichhasa molecularformulaC_(5)H_(8) O_(2)Whatwouldbe theoxidativeozonolysisproductof B ? |
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| 90571. |
A compound A has molecular formula C_2Cl_3OH. It reduces Fehling.s solution and on oxidation gives a monocarboxylic acid B, A is obtained by action of Cl_2 on ethyl alcohol. A is: |
| Answer» Answer :A | |
| 90572. |
A compound A has a molecular formula C_(7)H_(7)NO. On treatment with Br_(2) " and " KOH, A gives an amine B which gives carbylamine test. B upon diazotisation and coupling with phenol gives as azody. A can be |
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Answer» `C_(6)H_(5)CONHCH_(3)`  
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| 90573. |
A compound A has molecular formula C_(2)Cl_(3)OH. It reduces Fehling's solution and on oxidation gives a monocarboxylic acid B. A is obtained by the action of Cl_(2) on ethyl alcohol. A is |
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Answer» chloral `C_(2)H_(5)OH overset(Cl_(2))to underset("Chloral (A)")(C Cl_(3)CHO) overset([O])to underset("Trichloroacetic acid (B)")(C Cl_(3)COOH)` |
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| 90574. |
A compound 'A' has a molecular formula C_2Cl_3OH. It reduces Fehling solution and on oxidation gives a monocarboxylic acid (B). A is obtained by action of chlorine on ethyl alcohol.Ais : |
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Answer» Chloral |
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| 90575. |
A compound A has a molecular formula C_(2) CI_(3),OHIt reduces Fehling's solution and on oxidation, gives a monocarboxylic acid B. A can be obtained by the action of chlorine on ethyl alcohol. A is |
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Answer» chloroform As (A) reduces Fehling.s solution and on OXIDATION gives a monocarboxylic acid (B). It means (A) must be an ALDEHYDE. `underset((A))(CCI_(3)CHO)` This is further confirmed by the reaction `C_(2)H_(5)OH+CI_(2)overset([O])toCH_(3)underset(CI_(2))(CHO)tounderset("CC"I_(3)CHO)` `A = chloral ("CC"I_(3)CHO)` |
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| 90576. |
A compound [A] gives positive iodoform test in 5 minutes. When 6.0 g of [A] is treated with sodium metal, 1120 mL of hydrogen gas is evolved at NTP. It is assumed that [A] contains one atom of oxygen per molecule. Further when [A] reacts with PBr_(3), a compound [B] is formed which on reacting with benzene in the presence of anhydrous AlCl_(3) gives a compound [C]. the compound [C] is a well known industrial compound and is used in the commercial preparation of phenol. The compound [A] is a |
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Answer» Primary alcohol |
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| 90577. |
A compound [A] gives positive iodoform test in 5 minutes. When 6.0 g of [A] is treated with sodium metal, 1120 mL of hydrogen gas is evolved at NTP. It is assumed that [A] contains one atom of oxygen per molecule. Further when [A] reacts with PBr_(3), a compound [B] is formed which on reacting with benzene in the presence of anhydrous AlCl_(3) gives a compound [C]. the compound [C] is a well known industrial compound and is used in the commercial preparation of phenol. The molar mass of compound [A] is : |
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Answer» `60.0" g mol"^(-1)` `UNDERSET("2 moles")(2 ROH)+2Na rarr2RONa+underset(22400 mL)(H_(2))` 22400 mL of `H_(2)` is evolved from alcohol `=2` mol 1120 mL of `H_(2)` is evolved from alcohol `=1120/22400xx2=0.1` mol. Now, 0.1 mole of `ROH=` 6g `:.` 1.0 mole of `ROH =` 60 g |
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| 90578. |
A compound A dissociates by two parallel first order paths at certain temperature A(g)overset(k_(1)(min^(-1)))to2B(g)"" k_(1)=6.93xx10^(-3)min^(-1) A(g)overset(k_(1)(min^(-1)))toC(g)"" k_(2)=6.93xx10^(-3)min^(-1) The reaction is started with 1 mole of pure 'A' litre closed container with initial pressure 2 atm. What is the pressure (in atm) developed in container after 50 minutes from start of experiment? |
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Answer» `1.25` |
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| 90579. |
A compound A gives following reactions. Its structure can be |
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Answer» `CH_2=CH-(CH_2)_2-undersetunderset(O)(||)C-CH_2OH` 
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| 90580. |
A compound (A) forms a unstable pale blue colour solution in water. The solution decolourised Br_(2) water and an acidified solution of KMnO_(4). The possible compound (A) is : |
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Answer» `HNO_(2)` REACTION with `: a) Br_(2)` water b) `KMnO_(4)` |
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| 90581. |
A compound (A) called vanillin is isolated from vanilla seeds. Molecular formula of vanillin is C_(8)H_(8)O_(3). Vanillin gives violet colour on treatment with neutral FeCl_(3) . It also gives silver mirror with Tollens' reagent . Each mole of vanilin of gives one mole of AgI when subject to Zeisel's active methoxy estimation. "Vanillin"overset(conc.HBr)rarr"compound"(B), Compound (B) can alsobe obtaoned from catechol by Gattermann-Koch reaction. Compound(B) when heated with zinc dust will give : s |
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Answer»
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| 90582. |
A compound A dissociate by two parallel first order paths at certain temperature A(g)overset(k_(1)("min"^(-1))rarr 2B(g)k_(1)=6.93xx10^(-3)"min"^(-1) A(g)overset(k_(2)("min"^(-1))rarrC(g) k_(2)=6.93xx10^(-3)"min"^(-1) If reaction started with pure 'A' with 1 mole of A in 1 litre closed container with initialpressure (in atm) developed in container after 50 minutes from start of experiment? |
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Answer» 1.25 |
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| 90583. |
A compound (A) called vanillin is isolated from vanilla seeds. Molecular formula of vanillin is C_(8)H_(8)O_(3). Vanillin gives violet colour on treatment with neutral FeCl_(3) . It also gives silver mirror with Tollens' reagent . Each mole of vanilin of gives one mole of AgI when subject to Zeisel's active methoxy estimation. "Vanillin"overset(conc.HBr)rarr"compound"(B), Compound (B) can alsobe obtaoned from catechol by Gattermann-Koch reaction. Structure of vanillin should be : |
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Answer»
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| 90584. |
A compound (A) called vanillin is isolated from vanilla seeds. Molecular formula of vanillin is C_(8)H_(8)O_(3). Vanillin gives violet colour on treatment with neutral FeCl_(3) . It also gives silver mirror with Tollens' reagent . Each mole of vanilin of gives one mole of AgI when subject to Zeisel's active methoxy estimation. "Vanillin"overset(conc.HBr)rarr"compound"(B), Compound (B) can alsobe obtaoned from catechol by Gattermann-Koch reaction. Which of the following groups is not present in vanillin ? |
| Answer» ANSWER :A | |
| 90585. |
a) Compound (A) C_(8)H_(12) on oxidation gives an acid (B) C_(4)H_(6)O_(2). One molecule of compound (A) reacts with 3 mole of H_(2) in presence of platinum catalyst to give octane. Identify compound (A). |
Answer» Solution :The number of carbon atoms `(C_(8))` in (A) is twice the number of carbon atoms `(C_(4))` in acid (B). So 2 mole of acid `C_(4)H_(6)O_(4)` join with each other to give the compound (A).  Since, the CYCLOPROPANE ring is highly strained ring, so on catalyst hydrogenation, 3 moleof `H_(2)` are required i.e., one mol for each of cyclopropane ring and one mol for reducing the `(C=C)` bond. b)  So,2 mole of each of the products `C_(3)` are obtained by oxidation of (A). TWO (C=O) groups of two acetone are joned with two (C=O) groups of dicarboxyclic acid and then removing the oxygen atoms and joining the carbon atoms with double BONDS gives the structure of compound (A) 
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| 90586. |
A compound (A) called vanillin is isolated from vanilla seeds. Molecular formula of vanillin is C_(8)H_(8)O_(3). Vanillin gives violet colour on treatment with neutral FeCl_(3) . It also gives silver mirror with Tollens' reagent . Each mole of vanilin of gives one mole of AgI when subject to Zeisel's active methoxy estimation. "Vanillin"overset(conc.HBr)rarr"compound"(B), Compound (B) can alsobe obtaoned from catechol by Gattermann-Koch reaction. Vanillin contains : |
| Answer» Answer :C | |
| 90587. |
A compound A C_(6)H_(12)O reacts with NH_(2)OH but does not reduce Fehling's solution. On reduction with H_(2)|Pt, a gives an alcohol which on dehydration is converted into an alkene. The alkene on ozonolysis gives a liquid product which reduces tollen's reagent but does not give yellow precipitate with I_(2) with NaOH and (b) a liquid product which gives an yellow precipitate with I_(2) and NaOH but does not reduce Tollen's reagent. deduce the structure of A. |
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Answer» SOLUTION :(a) `PhCH_(2)OH+PhCOO^(-)` (b) `Me_(3)C CH_(2)OH+Me_(2)C COO^(-)` (c) 
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| 90588. |
A compound (A) C_(5)H_(9)Br on treatment with Mg in dry ether gives a compound (B) which on reaction with acetaldehyde gives an alcohol (C), on mild acidification prolonged treatment of alcohol with equivalent amount of HBr gives a compound (D) which on reaction with alcoholic KOH gives compound (E). compound (E) on ozonolysis followed by treatment with H_(2)O//Zn gives (F). the compound (F) on treatment with base gives 1 acetyl cyclopentene (G). identify (A) to (G). |
Answer» SOLUTION :
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| 90589. |
A compound (A) C_(5)H_(10) Cl_(2) on hydrolysis gives C_(5)H_(10)O which reacts with NH_(2)OH, forms iodoform but does not give Fehling test (A) is : |
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Answer» `H_(3)C-overset(CI)overset(|)UNDERSET(CI)underset(|)C-CH_(2)CH_(2)CH_(3)` |
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| 90590. |
A compound (60 g) on analysis gave C = 24 g, H = 4 g and O =32 g. Its empirical formula is : |
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Answer» `C_2H_4O_2` |
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| 90591. |
A compond A has the molecular formula C_5H_9Cl.It doesnot react with bromine in carbon tetrachloride.On treatment with strong base it produces a single structure isomer B.B has a molecular formula C_6H_8 and reacts with bromine in carbon tetrachloride. Ozonolysis of B produces a compound C which has a molcular formula C_5H_8O_2.Which of the following structures is that of A ? |
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| 90592. |
(A): Complexes containing ambidentate ligands exhibit co-ordination isomerism. (R) : Co-ordination isomerism is shown by the compounds in which both cation and anions are complexes. |
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Answer» Both A & R are TRUE, R is the CORRECT EXPLANATION of A |
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| 90593. |
A complex with the composition [MA_(3)B]^(n+-) is found to have no geometrical isomers. The possible structure (s) of the complex is (where A and B are monodentate ligands) |
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Answer» tetrahedral  `implies`In this square planar complex, only one configuration is possible. Hence, geometrical ISOMERISM is not OBSERVED.  `implies`Tetrahedral complexes NEVER show geometrical isomerism. |
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| 90594. |
A complex whose IUPAC name is not correctly writtein is: |
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Answer» |
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| 90595. |
A complex shown helow can be exhibit: |
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Answer» OPTICAL ISOMERISM only |
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| 90596. |
A complex shown below can exhibit : |
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Answer» OPTICAL ISOMERISM only |
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| 90597. |
A complex PtCI_(4).5NH_(3) shows a molar conductance of 402 ohm^(-1) cm^(-2) mol^(-1) in water and precipitate3 mole of AgCl with AgNO_(3). The formula of the complex is |
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Answer» `[Pt(NH_(3))_(6)]Cl_(4)` To GIVE 3 mole of AgCl with `AgNO_(3),3CL^(-)` ions must be present in IONIZATION sphere of the complex. `:.` Formula of complex `[Pt(NH_(3))_(5)Cl]Cl_(3)`. |
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| 90598. |
A complex of the type [M("AA")_(2_X_(2)]^(n+) is known to be optically active. What does this indicate about the structure of the complex ? Give one example of such complex. |
| Answer» SOLUTION :An OPTICALLY active COMPLEX of the type `[M("AA")_(2)X_(2)]^(n+)` INDICATES CIS-octahedral structure, e.g., cis`[Pt(en)_(2)Cl_(2)]^(2+)` or cis- `[Cr(en)_(2)Cl_(2)]^(+)`. | |
| 90599. |
A complex of the type [M(A A)_(2)X_(2)]^(n+) is known to be optically active. What does this indicate about the structure of the complex ? Give one example of such complex. |
| Answer» Solution :An optically ACTIVE COMPLEX of the type `[M(A A)_(2)X_(2)]^(N+)` indicates cis-octahedral structure, e.g., cis-`[Pt(en)_(2)Cl_(2)]^(2+)` or `cis-[Cr(en)_(2)Cl_(2)]^(+)`. | |
| 90600. |
A complex of platinum, ammonia and chlorine produces four ions per molecule in the solution. The structure consistent with the observation is: |
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Answer» `[PT(NH_3)_4]CI_4` |
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