94420 + Interview Questions in Chemistry in Class 12

1.	Conc. HNO_3 stains wood yellow because.
Answer» The wood is BURNED by acid Nitrocellulose is formed The PROTEINS are CONVERTED into xanthoproteins. The water is removed by acid. Answer :B

Discussion

2.	Conc. HNO_(3) oxidizes phosphorus :
Answer» `H_(3)PO_(4)` `P_(2)O_(5)`. `H_(3)PO_(3)` `H_(4)P_(2)O_(7)` ANSWER :A

Discussion

3.	Conc. HNO_3 stains and wool yellow because.
Answer» The SKIN and wool is burned by acid Nitro cellulose is formed The PROTEINS are converted into xanthoproteins. The WATER is REMOVED by acid. Answer :C

Discussion

4.	Conc. HNO_3 reacts with iron to:
Answer» RENDER iron passive Give FERROUS nitrate and NITRIC oxide Give ferric nitrate and AMMONIUM nitrate Give ferric nitrate and nitrogen dioxide Answer :A

Discussion

5.	Conc. HNO_3 oxidizes phosphorus to
Answer» `HPO_3` `H_3PO_2` `H_3PO_4` `H_3PO_4` ANSWER :D

Discussion

6.	Conc. HNO_(3) oxidises cane sugar to :
Answer» `CO_(2)and H_(2)O` `CO and H_(2)O` `CO, CO_(2) and H_(2)O` OXALIC ACID and water. Answer :D

Discussion

7.	Conc. HNO_(3) is treated with iron. The metal is passive because
Answer» it is a TRANSITION metal it forms PROTECTIVE oxide film it is REDUCED it liberates laughing gas ANSWER :2

Discussion

8.	Conc. HNO_3 is heated with P_2O_5 to form:
Answer» `N_2O` NO `NO_2` `N_2O_5` ANSWER :D

Discussion

9.	Conc. HNO_3 can be stored in container of
Answer» AL SN CU ZN ANSWER :A

Discussion

10.	Conc. HNO_(3), acid react with copper to give,
Answer» NO `N_2O` `NO_(2)` `N_(2)O_(5)` ANSWER :C

Discussion

11.	conc. H_2SO_4 reacts with C_2H_5OH at 443K to form
Answer» `CH_3COCH_3` `CH_3COOH` `CH_3CHO` `C_2H_4` ANSWER :D

Discussion

12.	Conc. H_(2)SO_(4) reacts with C_(2)H_(5)OH at 170^(@)C to form
Answer» `CH_(3)COCH_(3)` `CH_(3)COOH` `CH_(3)CHO` `C_(2)H_(4)` Solution :`underset("Ethanol")(C_(2)H_(5)OH)underset(170^(@)C)OVERSET("conc. " H_(2)SO_(4))to underset("ETHANE")(C_(2)H_(4))+H_(2)O`

Discussion

13.	Conc. H_2SO_4 is similar to conc. H_3PO_4, in that
Answer» both will oxidize `I^(-)" to "I_(2)` if neutralized by alkali, both will FORMS a PRECIPITATE with `BaCl_2` SOLUTION, which is soluble in dil. HCI both can be USED to dry `NH_3` gas both will displace volatile acids from their salts Answer :D

Discussion

14.	Conc. H_2SO_4 is diluted
Answer» by adding WATER in `H_(2)SO_(4)` by adding `H_(2)SO_(4)` in water by adding GLACIAL acetic acid in `H_(2)SO_(4)` NONE of these Solution :Concentrated `H_2SO_4`, is diluted by adding the conc. `H_2SO_4` in the water drop by drop with constant stirring because it is an exofhermic REACTION and by doing so heat is generated slowly and dissipiated in the atmosphere.

Discussion

15.	Conc. H_2SO_4 is not
Answer» HYGROSCOPIC DEHYDRATING AGENT Sulphonating agent EFFLORESCENT ANSWER :D

Discussion

16.	Conc, H_2SO_4 is added to the following compounds, Mark the compound which will give CO_2.
Answer» FORMIC acid Sugar Oxalic acid Ethyl alcohol Answer :C

Discussion

17.	Conc. H_(2)SO_(4) is added followed by heating to each of the following test tubes labelled (I) to (IV). Identify in which of the above test tube the following change will be observed. Support your answer with the help of chemical equation: (a) formation of black substance (b) evolution of brown gas (c ) evolution of colourless gas (d) formation of a brown substance which on dilution becomes blue. (e) disappearance of yellow powder along with evolution of colourless gas.
Answer» Solution :(a) `underset("White")(C_(12)H_(22)O_(11))rarr underset("Black substance")(12C)+11H_(2)O` (b) `2NaBr+2H_(2)SO_(4)rarr underset("Brown gas")(Br_(2)uarr)+Na_(2)SO_(4)+SO_(2)+2H_(2)O` (c ) `2KCl+H_(2)SO_(4)""underset("(Colourless gas")(2HCluarr+K_(2)SO_(4))` (d) `Cu+2H_(2)SO_(4)rarr CuSO_(4)+SO_(2)+H_(2)O` (e ) `3S+2H_(2)SO_(4)rarr underset("(Colourless gas)")(3SO_(2)uarr+2H_(2)O)`

Discussion

18.	Conc. H_(2)SO_(4) heated with excess of C_(2)H_(5)OH at 1406^(@)C to form
Answer» `CH_3CH_2-O-CH_3` `CH_3CH_2-O-CH_2CH_3` `CH_3-O-CH_2-CH_2-CH_3` `CH_2=CH_2 ` ANSWER :B

Discussion

19.	Conc. H_(2)SO_(4) heated with excess of C_(2)H_(5)OH at 140^(@) C to form
Answer» `CH_(3)CH_(2)-O-CH_(3)` `CH_(3)CH_(2)-O-CH_(2)CH_(3)` `CH_(3)-O-CH_(2)-CH_(2)-CH_(3)` `CH_(2)=CH_(2)` Solution :`CH_(3)CH_(2)-OH+HO-CH_(2)-CH_(3) UNDERSET(140^(@)C)overset("conc." H_(2)SO_(4))to underset("Diethyl ETHER")(CH_(3)CH_(2)-O-CH_(2)-CH_(3)+H_(2)O)`

Discussion

20.	Conc. H_2SO_4 displaces HCl from chloride because:
Answer» Conc. `H_2SO_4` is STRONGER than HCl HCl is a gas whereas `H_2SO_4` is a liquid Sulphates are more SOLUBLE in WATER than chlorides Sulphates are less soluble in water than chlorides Answer :A

Discussion

21.	Conc. H_2SO_4 can be diluted by
Answer» adding boiling water to conc. `H_2SO_4` adding COLD water to conc. `H_2SO_4` passing steam through conc. `H_2SO_4` None of these. Solution :Conc. `H_2SO_4` can only be DILUTED by adding conc. `H_2SO_4` to cold water TAKEN in excess.

Discussion

22.	Conc. H_(2)SO_(4) chars paper, wood and sugar because it removes...............from them.
Answer» SOLUTION :WATER MOLECULES

Discussion

23.	Conc. H_2 SO_4 is diluted
Answer» By adding water in `H_(2)SO_(4)` By adding `H_(2)SO_(4)` in water By adding glacial acid in `H_(2)SO_(4)` None of the above Solution :CONCENTRATED `H_(2)SO_(4)` is diluted by adding the conc. `H_(2)SO_(4)`in the water drop by with constant stirring because it is an exothermic reaction and by doing so HEAT is generated SLOWLY and dissipiated in the ATMOSPHERE.

Discussion

24.	Compute the heat of reaction at 1000^(@) C foir 1/2H_(2)(g) + 1/2 Cl_(2)(g) to HCl(g), DeltaH_(298 K)^(@) = 92.236 kJ mol^(-1) C_(P)^(@)(H_(2),g) = 29.0284 - 0.8355 xx 10^(-3)T + 2.0097 xx 10^(-8) T^(2) C_(P)^(@) (Cl_(2),g) = 31.6555 + 1.0134 xx 10^(-2) T - 4.0337 xx 10^(-8) T^(2) C_(P)^(@)(HCl, g) = 28.1359 + 1.8078 xx 10^(-3) T + 1.5453 xx 10^(-8) T^(2)
Answer» <P> Solution :`DeltaH_(1273 K)^(@) = DeltaH_(298 K)^(@) + int_(298 K)^(1273 K) {(DeltaC_(p))DT}` where, `DeltaCO_(P) = C_(P)^(@)(HCl,g) -1/2C_(P)^(@)(H_(2),g) -1/2C_(P)^(@)(Cl_(2),g)` `=(28.1359 - 1/2 xx 29.0284 - 1/2 xx 31.6556) + (1.8078 + 1/2 xx 0.8355 - 1/2 xx 10.134) xx 10^(-3) T + (1.5453 - 1/2 xx 2.0097 + 1/2 xx 4.0337) xx 10^(-6) T^(2)` Hence, `DeltaH_(1273 K)^(@) =[-92.236 xx 10^(2) + (-2.2061) -(2.8415 xx 10^(-3))(1273^(3)/3 - 298^(3)/3)] J"mol"^(-1)` `=-92236 - 2151.0 + 2176.2 + 1736.0 = -94827 J "mol"^(-1) = -94.827 kJ "mol"^(-1)`

Discussion

25.	Finkelstein reaction for the preparation of alkyl iodide is based upon the fact that:
Answer» Sodium iodide is soluble in METHANOL, while sodium chloride is insoluble in methanol. Sodium iodide is solublc in methanol, while NaCl and NaBr are insoluble in mcthanol. Sodium iodide is insolublc in methanol, while NaCl and NaBr are soluble. The three halogens differ CONSIDERABLY in their electronegativity Solution : `R-X+underset(("Solube" in CH_3OH,Me_2CO))NaIoverset("ACETONE")rarrR-I+underset(("Insolube" in CH_3OH,Me_2CO))NaXdarr`(where X = Cl or Br)

Discussion

26.	Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure.
Answer» SOLUTION :Since, anions OCCUPY fcc POSITIONS and half of the tetrahedral HOLES are occupied by cations. Since, there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres/unit volume of the unit cell is `=4 xx (4/3 pi r_(a)^(3) +1/2 xx 8 xx (4/3 pi r_(c )^(3)))/(16sqrt(2)r^(3)) = pi/(3sqrt(2))[1+(t_(c)/t_(a))^(3)]` `therefore` For tetrahedral holes. `r_(0)/r_(a) = 0.225 = pi/(3sqrt(2))(1 + (0.225)^(3)) = 0.7496` `therefore` Void volume =1-0.7496 = 0.2504 / unit volume of unit cell. % Void space = 25.04 %

Discussion

27.	Compute the heat of formation of liquid methyl alcohol in kJ mol^(-1), using the following data. Heat of vaporisation of liquid CH_(3)OH = 38 kJ/mol. Heat of formation of gaseous atoms from the elements in their standard states , H, 218 KJ/mol, C, 715 KJ/mol , O, 249 KJ/molAverage Bond energies C-H , 415 KJ/mol , C – O, 356 KJ/mol , O – H, 463 KJ/mol
Answer» SOLUTION :`-266 KJ MOL^(-1)`

Discussion

28.	Conant Finkelstein reaction for the preparation of alkyl iodide is based upon the fact that
Answer» SODIUM IODIDE is soluble in METHANOL, while sodium chloride is insoluble in methanol Sodium iodide is soluble in methanol, while NaCl and NaBr are insoluble in methanol Sodium iodide is insoluble in methanol, whileNaCl and NaBr are solubl The three halogens differ CONSIDERABLY in their electronegativity Solution :`R-X+underset("soluble in" (CH_(3)OH, Me_(2)CO))(NaI) overset("acetone")(to)R-I+underset("Insoluble in " (CH_(2)OH, Me_(2)CO))(NaXdarr)`

Discussion

29.	Compute Delta_(r)G for thereactionH_(2)O (l, 1, atm, 323 K ) to H_(2)O (g, 1 atm, 232K)Given that :Delta_(vap) H at 373K = 40.639 kJmol^(-1), C_(p)(H_(2)O, l)= 75.312 JK^(-1) mol^(-1), ""C_(p)(H_(2)O, g) = 33.305 J K^(-1)mol^(-1).
Answer» <P> Solution :`DeltaH_(373) = DeltaH _(323) - DeltaC_(p)DELTAT` `40639 = DeltaH_(323)-(33.305 - 75.312) XX 50` `DeltaH_(323) = 42739. 35 J//mol`

Discussion

30.	Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data. The heat of vaporisation of liquid methyl alcohol =38 "kJ/mole". The heats of formation of gaseous atoms from the elements in their standard states: H,218 "kJ/mole", C,715 "kJ/mole", O, 249 "kJ/mole". Average bond energies : C-H=415 "kJ/mole" C-O=356 "kJ/mole" O-H=463 "kJ/mole".
Answer» Solution :`C+2H_(2)+(1)/(2)O_(2) to CH_(3)OH,DELTAH=?` For reactants : HEAT of atomisation of 1 mole of `C=715 kJ` Heat of atomisation of 4moles of `H=4xx218 kJ` Heat of atomisation of 1 mole of `O=249 kJ` For products: Heat of formation of 3 MOLES of C-H bonds `=-3xx415 kJ` Heat of formation of 1 mole of C-O bonds `=-356 kJ` Heat of formation of 1 mole of O-H bonds `=-463 kJ` Heat of CONDENSATION of 1 mole of `CH_(3)OH` to liquid `=-38 kJ` On adding, we get `DeltaH` of formation of `CH_(3)OH` (l). `DeltaH=-266 "kJ mole"^(-1)`.

Discussion

31.	Compute the charge on one mole of electrons.
Answer» Solution :One mole of electrons = `6.02 XX 10^(23)` electrons Charge on one electron = `1.6021 xx 10^(-19)` `THEREFORE`Charge on one mole of electrons = `6.02 xx 10^(23) MOL^(-1)xx 1.6021 xx 10^(-19) C = 96487 C mol^(-1)`

Discussion

32.	Compund (A), C_(5)H_(11)NO is not soluble in cold alkaline or acidic solutions. When (A) is refluxed in NaOH solution, a gas (B) is evolved and salt (C) is formed. Acetyl chloride reacts with (B) to give (D), (C_(4)H_(9)NO). (B) reacts with HNO_(2) to give a yellow oil (E). Give structures of (A) to (E) with reason.
Answer» Answer :(A) `=CH_(3)CH_(2)CON(CH_(3))_(2),B=(CH_(3))_(2)NH,C=CH_(3)CH_(2)COONa,D=CH_(3)CON(CH_(3))_(2)` and `E=(CH_(3))_(2)N-N=O

Discussion

33.	Compund A reacts with PCl_(5) to give B which on treatment with KCN followed by hydrolysis gave propanoic acid. What is A and B respectively
Answer» `C_(3)H_(8) and C_(3)H_(7)Cl` `C_(2)H_(6) and C_(2)H_(5)Cl` `C_(2)H,Cl and C_(2)H_(5)Cl` `C_(2)H_(5)OH and C_(2)H_(5)Cl` Solution :`{:(C_(2)H_(5)OH+PCl_(5)to C_(2)H_(5)Cl+POCl_(3)+ HCL),(" "(A)""(B)),(C_(2)H_(5)Cl+KCNto C_(2)H_(5)CN+KCl),(" "(B)""DARR"HYDROLYSIS"),(""C_(2)H_(5)-COOH):}`

Discussion

34.	Comproportionationn occurs between:
Answer» `Cl^(-)(aq.)+ClO^(-)(aq.)+OH^(-)(aq.)` `PH_(3)(g)+H_(3)PO_(4)` acid `Na_(2)S(aq.)+Na_(2)SO_(3)(aq.)` `MNO_(4)^(2-)(aq.)+Mn^(2+)(aq.)+ZnSO_(4)(aq.)` Solution :`MnO_(4)^(-)(aq.)+Mn^(2+)(aq.) underset("Comproportionation Reaction")overset(ZN^(2+)(aq.))toMnO_(2)`

Discussion

35.	Compressibility factor (Z) for N_(2)" at "-50^(@)C and 800 atm pressure is 1.95 calculate mole of N_(2) gas required to fill a gas cylinder of 100 L capacity under the given conditions.
Answer» Solution :We have `Z=(PV)/(NRT)` `THEREFORE" mole of "N_(2) =(PV)/(ZRT)` `=(800xx100)/(1.95xx0.0821xx223)=2240.8`

Discussion

36.	Compressibility factor (Z) is plotted against pressure at different temperature for same gas
Answer» `T_(4) gt T_(3) gt T_(2) gt T_(1)` `T_(1) gt T_(2) gt T_(3) gt T_(4)` `T_(1) gt T_(2) gt T_(3) gt T_(4)` `T_(1) gt T_(2) gt T_(3) gt T_(4)` ANSWER :B

Discussion

37.	Compressibility factor (Z) for H_2 (g) at STP is
Answer» `Z = 1` `[Z = 1 +` `a`/`(V_m RT)]` `[Z = 1 -` `a`/`(V_m RT)]` `[``Z = 1` `+` `Pb`/`Rt``]` Answer :2

Discussion

38.	Compressibility factor under critical state of a gas is
Answer» `frac{5}{8]` `frac{8}{3}` `frac{3}{8R}` `frac{3}{8}` ANSWER :D

Discussion

39.	Compreshension -I The experimental values of colligative properties of many solutes in solution resembles calculated value of colligative properties. However in same cases, the experimental value of colligative property differ widely than those obtained by calculation. Such experimental values of colligative properties are known as Abnormal values of colligative properties are : (i) Dissociation of solute : It increases the colligative properties. (ii) Association of solute : It decreases the colligative properties e.g. : Dimerisation of acetic acid in benzene One mole I_(2) (solid) is added in 1 M, 1 litre Kl solution . Then
Answer» Osmostic pressure of solution increases FREEZING point of solution increases Relative lowering in vapour pressure decreases No CHANGE in BOILING point of solution Answer :4

Discussion

40.	Compreshension-II In non-ideal solutions, at one of the intermediate compostions, the total vapour pressure is highest and the boiling point is lowest. At this point, the composition of the liquid and vapour phase is same. So, if liquid mixture vapouriese at this point and vapours are condensed, teh condensate contains same compositon as present in original liquid mixture. it means at this point liquid behaves like a pure liquid and is called an Azeotropic mixture. Choose the correct answer :
Answer» Ideal solutions cannot be separated into their components by fractional distillation For ideal solutions ENTHALPY of mixing is ALWAYS greater than zero Only non-ideal solution showing POSITIVE DEVIATION cannot be separated out by FRACTION distillation Non-ideal solution showing both positive and negative deviation cannot be separated out by fractional distillation Answer :4

Discussion

41.	Compreshension -I The experimental values of colligative properties of many solutes in solution resembles calculated value of colligative properties. However in same cases, the experimental value of colligative property differ widely than those obtained by calculation. Such experimental values of colligative properties are known as Abnormal values of colligative properties are : (i) Dissociation of solute : It increases the colligative properties. (ii) Association of solute : It decreases the colligative properties e.g. : Dimerisation of acetic acid in benzene 4 different 100 ml solutions are prepared by mixing 1 mole each of NaCl (NH_(2))_(2)CO.Na_(2)SO_(4) and K_(4)[Fe(CN)_(6)] at temperature T. Correct order of osmotic pressure is
Answer» `(NH_(2))_(2)CO` solution `gt` NACL solution `gt Na_(2)SO_(4)` solution `gt K_(4)[Fe(CN)_(6)]` solution NaCl solution `gt Na_(2)SO_(4)` solution `gt (NH_(2))_(2)CO` solution `gt K_(4)[Fe(CN)_(6)]` solution `K_(4)[Fe(CN)_(6)]` solution `gt Na_(2)SO_(4)` solution `gt `NaCl solution `gt (NH_(2))_(2)CO` solution `Na_(2)SO_(4)` solution `gt (NH_(2))_(2)CO` solution `gt `NaCl solution `gt K_(4)[Fe(CN)_(6)]` solution Answer :3

Discussion

42.	Compreshension -I The experimental values of colligative properties of many solutes in solution resembles calculated value of colligative properties. However in same cases, the experimental value of colligative property differ widely than those obtained by calculation. Such experimental values of colligative properties are known as Abnormal values of colligative properties are : (i) Dissociation of solute : It increases the colligative properties. (ii) Association of solute : It decreases the colligative properties e.g. : Dimerisation of acetic acid in benzene If degree of dissociation of an electrolyte A_(2)B_(3) is 25% in a solvent, then
Answer» Normal BOILING point =EXPERIMENTAL boiling point Normal freezing point `gt` Experimental freezing point Normal osmotic pressure`=(1)/(2)` Experimental osmotic pressure Normal MOLECULAR weight `=(1)/(4)` Experimental molecular weight Answer :3

Discussion

43.	Comprehension # 2 The clouds consist of charged particles of water dispersed in air. Some of them are +vely charged, others are -vely charged. When +vely charged clouds come closer they cause lightening and thundering whereas when +ve and -ve charged colloids come closer they cause heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulating the mist hanging in air. Smoke screen is a cloud of smoke used to hide military, naval police etc. it consists of fine particles of TiO_(2). Smoke screens consist of
Answer» fine PARTICLES of `TiO_(2)` dispersed in air by AEROPLANES fine particles of AGL dispersed in air by aeroplanes fine particles of `Al_(2)O_(3)` dispersed in air by aeroplanes None of these Answer :A

Discussion

44.	Comprehension # 2 The clouds consist of charged particles of water dispersed in air. Some of them are +vely charged, others are -vely charged. When +vely charged clouds come closer they cause lightening and thundering whereas when +ve and -ve charged colloids come closer they cause heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing electrified sand or silver iodide from an aeroplane and thus coagulating the mist hanging in air. Smoke screen is a cloud of smoke used to hide military, naval police etc. it consists of fine particles of TiO_(2). When excess of AgNO_(3) is treated with Kl solution, Agl forms
Answer» `+ve " CHARGED SOL"` `-ve " charged sol"` NEUTRAL sol True solution Answer :A

Discussion

45.	Comprehension # 1 Many lyophilic sols and few lyophobic sols when coagulated under some special conditions changes into semi rigid mass, enclosing whole amount of liquid within itself, it is called gel and the process is called gelation, Gelatin Agar-agar, gum-Arabic can be converted into gels by cooling them under moderate concentration conditions. Hydrophobic sols like silicic acid. Al(OH)_(3) are prepared by double decomposition and exchange of solvent method. Types of Gel : (i) Elastic gel : Those gel which have elastic properties. Ex : Gelatin, Strach, Agar-agar etc. (ii) Non-elastic gel : Those gel which are rigid. Ex : Silica gel. Properties of Gel : 1. Syneresis/weeping of gel : The spontaneous liberation of liquid from a gel is called syneresis or weeping of gels. It is reverse of swelling. Ex : Gelatin, Agar-Agar show syneresis at low concentration while sillicic acid shows it at high concentration. 2. Imbibition or swelling of gel : When gel is kept in a suitable liquid (water) it absorb large volume of liquid. The phenomenon is called imbibition or swelling of gel. 3. Thixotropic : Some gels when shaken to form a sol, on keeping changes into gel are termed as thixotropic gel and phenomenon is called thixotropy. Ex : Gelatin and silica liquify on shaking changing into corresponding sol and sol on keeping changes back into gel. Some types of gels like gelatin and silica liquify on shaking thereby changing into sols. The sols on standing change back into gels. This process is know as
Answer» synersesis thixotropy double decompostion peptization Answer :B

Discussion

46.	Compoune (A) C_5H_(10)O_5, give a tetra-acetate with Ac_2O and oxidation of (A) with Br_2 - H_2O gives an acid, C_5H_(10)O_6. Reduction of (A) with HI and red phosphorous gives 3-methylbutane. What is structure of (A)?
Answer» Solution :The formation of tetracetate INDICATES of 4OH group and OXIDATION with bromine water indicates presence of CHO group. REDUCTION with red phosphorous and HI indicates presence of one carbon in the side chain. Thus, the STRUCTURE of (A) WOULD be

Discussion

47.	{:("Compound",,,,"Total stereoisomer"),((A)[Co(en)_(2)Cl]^(+),,,,(P)2),((B)[Co(en)_(3)]Cl_(3),,,,(Q)3),((C)[Pt(NH_(3))_(2)Cl_(2)],,,,(R)4),((D)[Pt(NH_(3))_(3)Cl_(3)]^(+),,,,(S)5):} Correct code :-
Answer» A=R, B=P, C=R, D=R A=Q, B=P, C=P, D=P A=P, B=R, C=P, D=P A=Q, B=Q, C=P, D=P ANSWER :B

Discussion

48.	compounds X istreatedwith NH_(2) OH and followedby reductiongives : (CH_(3))_(2) CHCH_(2)overset(NH_(2))overset(\|)(CHCH_(2) (CH_(3))_(3) Thecompound X is
Answer» `(CH_(3))_(2)CHCH_(2) overset(NOH)overset(\|\|)(C CH_(2)) C(CH_(3))_(3)` `(CH_(3))_(2) CHCI_(2) overset( CN)overset(\|)(CHCH_(2) C(CH_(3))_(3)` `(CH_(3))_(2) CHCH_(2) overset(NO_(2))overset(\|)(CHCH_(2)) C(CH_(3))_(3)` `(CH_(3))_(2) CHCH_(2) overset(O)overset(\|\|) (C C H_(2))C(CH_(3))_(3)` Solution :`(CH_(3))_(2)CHCH_(2) overset(O) overset(\|\|)( C C H_(2)) C (CH_(3))_(3) +NH_(2) OH to ` `(CH_(3))_(2)CH- CH_(2)- overset( NOH) overset(\|\|)( C )CH_(2)C (CH_(3))_(2)overset( " Reduction ") (to) ` `(CH_(3))_(2)CH- CH_(2)-overset( NH_(2) )overset(\|)( C H) -CH_(2). C(CH_(3)) _(3)+ H_(2) O`

Discussion

49.	Compounds with spinonly magnetic moment equivalent to five unpaired electrons are
Answer» `K_(4)[Mn(CN)_(6)]` `[Fe(H_(2)O)_(6)]Cl_(3)` `K_(3)[FeF(6)]` `K_(4)[MnF_(6)]` Solution :a. `K_(4)[Mn(CN)_(6)],Mn^(2+)=[Ar]3d^(5)4s^(0)` `CN^(-)` is a strong field ligand, so pairing will TAKE place i.e. `t_(2g)^(2,2,1) e_(g)^(0,0)` B. `[Fe(H_(2)O)_(6)]Cl_(3),Fe^(3+)=[Ar]3d^(5)4s^(0)` `H_(2)O` is a weak field ligand so pairing will not occur. i.e. `t_(2g)^(1,1,1), e_(g)^(1,1)` c. `K_(3)[FeF_(6)],Fe^(3+)=[Ar]3d^(5) 4s^(0)` `F^(-)` is a weak field ligand, so, pairing will not occur. i.e `t_(2g)^(1,1,1) e_(g)^(1,1)` d. `K_(4)[MnF_(6)], Mn^(2+)=[Ar]3d^(5)4s^(0)` `F^(-)` is a weak field ligand so, pairing will not occur. i.e. `t_(2g)^(1,1,1)e_(g)^(1,1)` HENCE b, c and d complexes will SHOW spin only magnetic moment equivalent to five unpaired electrons.

Discussion

50.	Compound(s) without carboxylic acid group in it are :
Answer» BARBITURIC acid Carbolic acid Picric acid Cyanic acid Answer :A::B::C::D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Conc. HNO_3 stains wood yellow because.

Conc. HNO_(3) oxidizes phosphorus :

Conc. HNO_3 stains and wool yellow because.

Conc. HNO_3 reacts with iron to:

Conc. HNO_3 oxidizes phosphorus to

Conc. HNO_(3) oxidises cane sugar to :

Conc. HNO_(3) is treated with iron. The metal is passive because

Conc. HNO_3 is heated with P_2O_5 to form:

Conc. HNO_3 can be stored in container of

Conc. HNO_(3), acid react with copper to give,

conc. H_2SO_4 reacts with C_2H_5OH at 443K to form

Conc. H_(2)SO_(4) reacts with C_(2)H_(5)OH at 170^(@)C to form

Conc. H_2SO_4 is similar to conc. H_3PO_4, in that

Conc. H_2SO_4 is diluted

Conc. H_2SO_4 is not

Conc, H_2SO_4 is added to the following compounds, Mark the compound which will give CO_2.

Conc. H_(2)SO_(4) heated with excess of C_(2)H_(5)OH at 1406^(@)C to form

Conc. H_(2)SO_(4) heated with excess of C_(2)H_(5)OH at 140^(@) C to form

Conc. H_2SO_4 displaces HCl from chloride because:

Conc. H_2SO_4 can be diluted by

Conc. H_(2)SO_(4) chars paper, wood and sugar because it removes...............from them.

Conc. H_2 SO_4 is diluted

Finkelstein reaction for the preparation of alkyl iodide is based upon the fact that:

Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure.

Conant Finkelstein reaction for the preparation of alkyl iodide is based upon the fact that

Compute Delta_(r)G for thereactionH_(2)O (l, 1, atm, 323 K ) to H_(2)O (g, 1 atm, 232K)Given that :Delta_(vap) H at 373K = 40.639 kJmol^(-1), C_(p)(H_(2)O, l)= 75.312 JK^(-1) mol^(-1), ""C_(p)(H_(2)O, g) = 33.305 J K^(-1)mol^(-1).

Compute the charge on one mole of electrons.

Compund A reacts with PCl_(5) to give B which on treatment with KCN followed by hydrolysis gave propanoic acid. What is A and B respectively

Comproportionationn occurs between:

Compressibility factor (Z) for N_(2)" at "-50^(@)C and 800 atm pressure is 1.95 calculate mole of N_(2) gas required to fill a gas cylinder of 100 L capacity under the given conditions.

Compressibility factor (Z) is plotted against pressure at different temperature for same gas

Compressibility factor (Z) for H_2 (g) at STP is

Compressibility factor under critical state of a gas is

Compoune (A) C_5H_(10)O_5, give a tetra-acetate with Ac_2O and oxidation of (A) with Br_2 - H_2O gives an acid, C_5H_(10)O_6. Reduction of (A) with HI and red phosphorous gives 3-methylbutane. What is structure of (A)?

{:("Compound",,,,"Total stereoisomer"),((A)[Co(en)_(2)Cl]^(+),,,,(P)2),((B)[Co(en)_(3)]Cl_(3),,,,(Q)3),((C)[Pt(NH_(3))_(2)Cl_(2)],,,,(R)4),((D)[Pt(NH_(3))_(3)Cl_(3)]^(+),,,,(S)5):} Correct code :-

compounds X istreatedwith NH_(2) OH and followedby reductiongives : (CH_(3))_(2) CHCH_(2)overset(NH_(2))overset(|)(CHCH_(2) (CH_(3))_(3) Thecompound X is

Compounds with spinonly magnetic moment equivalent to five unpaired electrons are

Compound(s) without carboxylic acid group in it are :