94420 + Interview Questions in Chemistry in Class 12

1.	Pb\|PbSO_(4)\|H_(2)SO_(4)(aq)\|\|PbCl_(2) saturated solution \|Cl_(2)\|Pt As the cell discharged:
Answer» Conductivity of ANODE solution decreases Concentration of `PB^(2+)` increases at anode solution and decreases at cathode `E_("cell")^(@)` depends on `K_(sp)` of `PbSO_(4)` and `K_(sp)` of `PbCl_(2)` `E_("cell")^(@)` is zero Solution :Anode: `Pb(s)+SO_(4)^(2-)toPbSO_(4)(s)+2e^(-)` Cathode: `Cl_(2)+2e^(-)to2Cl^(-)` `2Cl^(-)+Pb^(2+)toPbCl_(2)(s)` At anode, concentration of `SO_(4)^(2-)` FALLS, so conductivity of anode solution decreases. As `[SO_(4)^(2)]` decreases `[Pb^(2+)]` increase at anode to satisfy `K_(sp)`. At cathode concentration of `Pb^(2+)` decreases. `E_("cell")^(@)=E_(Pb^((2+))/(Pb))^(@)+(0.059)/(2)logK_(sp)(PbCl_(2))-` `(E_((Pb^(2+))/(Pb))+(0.059)/(2)logK_(sp)(PbSO_(4)))` `=(0.059)/(2)log(K_(sp)(PbCl_(2)))/(K_(sp)(PbSO_(4))): E_("cell")^(@)ne0`

Discussion

2.	PbOhArrPb+1/2O_2, K=2xx10^(-1) ZnOhArr +1/2 O_2 K=2xx10^2 CuOhArr +1/2 O_2: K=2xx10^(2) Ag_2OhArr2Ag+1/2 O_2 ,K=2xx10^4 On the basis of equilibrium constant, which of the following is most stable oxide ?
Answer» ZnO CuO PBO `Ag_2O` ANSWER :C

Discussion

3.	PbO_(2) is a stronger oxidising agent than SnO_(2). Or PbO_(2) can act an oxidiside agent.
Answer» ANSWER :In `PbO_(2)` and `SnO_(2)` both lead and tin present in `+4` oxidation state.But due to STRONGER INERT pair effect,`Pb^(2+)` ion is more STABLE than `Sn^(2+)` ion. In other words, `Pb^(4+)` IONS i.e. `PbO_(2)`, is more easily reduced to `Pb^(2+)` ions than `Sn^(4+)` ions are reduced to `Sn^(2+)` ions.Thus, `PbO_(2)` acts as a stronger oxidising agent than `SnO_(2)`.

Discussion

4.	PbO_(2) on heating evolves
Answer» `NO_(2)` `O_(2)` `N_(2)` `N_(2)O` Solution :`PbO_2` on heating EVOLVE `O_2` GAS

Discussion

5.	PbO_2 is :
Answer» Acidic Basic Neutral Amphoteric Answer :D

Discussion

6.	PbF_(4), PbCL_(4) exists PbBr_(4) and PbI_(4) do not exist because of
Answer» large size of `Br^(-) and I^(-)` STRONG oxidising character of `Pb^(4+)` Strong REDUCING character of `Pb^(4+)` Low electronegativity of `B^(-) and I^(-)` SOLUTION :F and Cl are more oxidising in nature and can achieve Pb in (IV) O. S but `Br_(2) and I_(2)` can not achieve Pb in (IV) O.S secondly `Pb^(4+)` is strong in oxidising nature and in its presence, `Br^(-) and i^(-)`can not exists.

Discussion

7.	PbCl_(2) is soluble in (a) ____. AgCl is soluble in (b) _____ white Hg_(2)CI_(2) is (c ) ___by NH_(3)
Answer» ANSWER :a.hot WATER B. AQ ,`NH_(3)` c. Blackneid

Discussion

8.	PbCl_(2)darr+"hot water"toPb^(2)(aq.)+2Cl^(-)(aq.)
Answer» For PRECIPITATE formation formation reaction For precipitate dissolution reaction For precipitate exchange reaction For no reaction Answer :B

Discussion

9.	PbCl_2 is insoluble in cold water. Addition of HCl increases its solubility due to
Answer» FORMATION of soluble complex ANIONS like `[PbCl_3]^(-)` oxidation of `Pb(II) ` to `Pb(IV)` formation of `[Pb(H_2O)_(6)]^(2+)` formation of polymeric lead COMPLEXES. SOLUTION :Addition of chloride ions to a suspension of `PbCl_2` gives rise to soluble complex ions as shown : `PbCl_(2(s)) + Cl^(-) to [PbCl_(3)]_((aq))^(-)` `PbCl_(2(s)) + 2Cl^(-) to [PbCl_(4)]_((aq))^(2-)`

Discussion

10.	PbCI_(2) exists white PbBr and PbI do not exist.
Answer» Solution : `Pb^(4+)` is an oxidizing agent and readily changes into `Pb^(2+)` The Br’ and I ions are reducing agents. The redox REACTION occurs indicating that `PbBr_(4)` and `Pbl_(4)` are UNSTABLE compounds `Pb^(4+)+4BrtoPbBr_(2)Br_(2)`

Discussion

11.	Pb(CH_(3)COO)_(2) gives ...... colour with H_(2)S
Answer» Black white red orange ANSWER :A

Discussion

12.	Pb(CH_(3)COO)_(2) gives . . . Colour with H_(2)S
Answer» Orange Red BLACK White Solution :`(CH_(3)COO)_(2),Pb+H_(2)S to 2CH_(3)COOH+underset("black ppt.")(PBS) darr`

Discussion

13.	Pb_(3)O_(4)+4HNO_(3)to2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)OWhich is true about the lead ions in Pb_(3)O_(4) ?
Answer» `PB^(2+)` reacts with `HNO_(3)` to GIVE lead nitrate `Pb^(4+)` reacts with `HNO_(3)` to give lead nitrate `Pb^(2+)` reacts with `HNO_(3)` to give `PbO_(2)` `Pb^(4+)` reacts with `HNO_(3)` to give `PbO_(2)`. Answer :A

Discussion

14.	Pb^(2+) belongs to :
Answer» GROUP I Group II Group IV Both group II and I. ANSWER :D

Discussion

15.	Pb//PbSO_(4)_((s)), SO_(4(aq))^(-2) Pb_((s))+SO_(4)^(-2) hArr PbSO_(4(s))+ ?
Answer» `E^(-)` `3E^(-)` `2E^(-)` no electron Solution :`(Pb_((s)) to PB^(++)+2e^(-))+SO_(4)^(--) to PbSO_4+2e^(-)`

Discussion

16.	Pb shows oxidation state of
Answer» `+3,+4` `+1,+2` `+2,+4` Only +4 Answer :C

Discussion

17.	Pb hasbeen placedin groups I and IIbecause
Answer» It shows the valency of ONE and two It is partlysoluble in `H_(2)O` It forms INSOLUBLE `PnCI_(2)` It from lead sulphide Answer :b

Discussion

18.	Pb and Sn are extracted from their chieforeby
Answer» carbon reduction and self reduction. self reduction and carbon reduction electrolysis and self reduction self reduction and electrolysis. Solution :`PbO and PbSO_(4)` GET reduced by PbS itself which is already present in MIXTUE so because the reduction took PLACE by itself, hence is KNOWN as self reduction. `2PbO+PbS overset(Delta)to 3Pb+SO_(2)uarr` `PbSO_(4)+PbS overset(Delta)to 2Pb+2SO_(2) uarr`

Discussion

19.	Pb+Dil.HNO_(3) overset(Warm)toP+Quarr+H_(2)O Incorrect statement for Q is:
Answer» Paramagnetic COLOURLESS GAS It is oxidized to paramagnetic coloured gas by air It combines with `Fe_(2)(SO_(4))_(3)` It is also obtained by disproportionation of `HNO_(2)` ANSWER :C

Discussion

20.	Pb + Conc.HNO_(3) gives
Answer» `Pb(NO_(3))_(2)+NO_(2)` `PbNO_(3)+NO` `Pb(NO_(3))_(4)+NO_(3)` `Pb(NO_(3))_(3)+N_(2)O` SOLUTION :`HNO_(3)` is the BEST solvent for lead. `Pb+4HNO_(3)("conc.")toPb(NO_(3))_(2)+2NO_(2) +2H_(2)O`

Discussion

21.	Pb and Sn are extracted from their chief ores by
Answer» Carbon REDUCTION and self reduction. Self reduction and carbon reduction. ELECTROLYSIS and self reduction. Self reduction and electrolysis. Solution :`{:(PBS + 2PbO rarr 3Pb + SO_(2) ("self reduction")),(SnO_(2) + C rarr Sn + CO_(2) ("Carbon reduction")):}`

Discussion

22.	Pb and Sn are extracted from their chief ore respectively by
Answer» CARBON reduction and self reduction Self reduction and carbon reduction ELECTROLYSIS and self reduction Self reduction and electrolysis. ANSWER :B

Discussion

23.	Pb and Sn are extracted from their chief ores by :
Answer» CARBON REDUCTION and SELF reduction RESPECTIVELY. ELECTROLYSIS and self-reduction respectively. selfreduction and electrolysis respectively self reduction and carbon reduction respectively. Answer :D

Discussion

24.	Pb and Sn are extracted from their chief ore by :
Answer» CARBON REDUCTION and self reduction self reduction and carbon reduction electrolysis and self reduction self reduction and electrolysis Solution :The chief ore of lead is galena (PbS_ and is extracted with the help of self reduction. At the same time the chief ore of tin is tin STONE `(SnO_(2))` and is ISOLATED from that by carbon reduction (SMELTING).

Discussion

25.	Pauling's equation for determining the electronegativity of an element is
Answer» `X_(A)-X_(B)=0.208sqrt(Delta)` `X_(A)+X_(B)=0.208sqrt(Delta)` `X_(A)-X_(B)=0.208 Delta^(2)` `X_(A)-X_(B)= sqrt(Delta)` Solution :ACCORDING to PAULING, `X_(A)-X_(B)=0.208sqrt(Delta)` `or, X_(A)-X_(B)=0.208[E_(A-B)-(E_(A-A)xxE_(B-B))^(1//2)]^(1//2)`

Discussion

26.	Pauling.s electronegativity scale is based on experimental value of:-
Answer» ATOMIC radii bond energies bond lengths electron affinity Answer :B

Discussion

27.	Paul Walden, the brilliant son of a farmer succeeded in finding a concept that is now widely known as umbrella effect in S_N2 reactions. In an S_N2 reaction, the configuration at the reaction centre is inverted as a umbrella turns inside out in a strong wind. An ideal S_N1 reaction, on the other hand, yield nearly racemised product as it proceeds through planar carbocation. S_N2 reactions proceed smoothly on primary alkyl halide whereas tertiary halides are fit for S_N1 mechanism. However, evidences indicate that all the molecules of particular substances do not necessarily react by a single mechanism.After all, the intention of the molecule is very clear, they want to follow the low energy path.If both paths have nearly the same energy barrier, then follow a mixed mechanism.For example, secondary alkyl halides follow a mixed mechanism, say 60% molecule react by S_N2 and rest 40%by S_N1 Stereochemical features, as mentioned above, are noticeable only when the reaction is carried out on a particular enantiomer.Even the reaction centres, which are paths of suitably substituted rings, also exemplify the said features. There are two basic mechanisms S_N1 and S_N2 All except one of the following substances give at least one of these two reactions enthusiastically in the absence of catalyst Identify the odd out that reacts sluggish under both the mechanisms.
Answer» `R-CH_3-x` `R_3C_X` `R_3C-CH_2-X` `R-CO-CH_2-X` Solution :The susbstrate through is a primary halide but it is excessively branced at the `beta`-position.Such SUBSTRATE fails miserable in `S_N2` reactions even though it's a primary halide. Being a primary halide it does not react actively in `S_N1` mechanism as well.However, it reacts quite FLUENTLY by `S_N1` mechanism in PRESENCE of `Ag^+` CATALYST.

Discussion

28.	Paul Walden, the brilliant son of a farmer succeeded in finding a concept that is now widely known as umbrella effect in S_N2 reactions. In an S_N2 reaction, the configuration at the reaction centre is inverted as a umbrella turns inside out in a strong wind. An ideal S_N1 reaction, on the other hand, yield nearly racemised product as it proceeds through planar carbocation. S_N2 reactions proceed smoothly on primary alkyl halide whereas tertiary halides are fit for S_N1 mechanism. However, evidences indicate that all the molecules of particular substances do not necessarily react by a single mechanism.After all, the intention of the molecule is very clear, they want to follow the low energy path.If both paths have nearly the same energy barrier, then follow a mixed mechanism.For example, secondary alkyl halides follow a mixed mechanism, say 60% molecule react by S_N2 and rest 40%by S_N1 Stereochemical features, as mentioned above, are noticeable only when the reaction is carried out on a particular enantiomer.Even the reaction centres, which are paths of suitably substituted rings, also exemplify the said features. A reaction gives the following results. What should be the % age of back side attack assuming complete reactions ?
Answer» `65%` `30%` `35%` `85%` Solution :PERCENTAGE of BACKSIDE attack=Percentage of inversion + 1/2 (Percentage of racemisation) 70% racemisation and rest 30% inversion gives 65% backside attack.

Discussion

29.	Pauling electronegativity scale is based on experimental values of:
Answer» BOND lengths Bond energies Ionisation energies Electron affinities. Answer :B

Discussion

30.	What is the function of limestone the extraction of iron? Give equation to explain its action.
Answer» ANSWER :A

Discussion

31.	Pasteurization means____ .
Answer» preservation by anti-oxidants preservation by heating ADDITION of SWEETENING agent addition of`Cl_(2)` gas Answer :B

Discussion

32.	Passivity of iron is due to the formation of a thin layer of :
Answer» `Fe_(3)O_(4)` `Fe_(2)O_(3)` `FeS_(2)` `FeCO_(3)`. Answer :A

Discussion

33.	Passivity of iron is due to the formation of thin film of its :
Answer» Oxide Carbonate Nitride Hydroxide Answer :A

Discussion

34.	Passivity of iron is due to formation of:
Answer» `Fe_(2)O_(3)` `Fe_(3)O_(4)` `FeSO_(4)` NONE of these Answer :B

Discussion

35.	Passing H_2S gas through nitric acid produces:
Answer» ANSWER :C

Discussion

36.	Passing H_(2)S gas into a mixture of Mn^(2+) , Ni^(2+) , Cu^(2+) and Hg^(2+) ions in an acidified aqueous solution precipitatates:
Answer» `CuS` and `HgS` `MnS` and `NiS` `MnS` and `NiS` `NiS` and `HgS` SOLUTION :In presence of acidic MEDIUM, IONISATION of `H_(2)S` is supressed to less number of `S^(2-)` ions are produces.So only those sulphides are PRECIPITATED which have low solubility PRODUCT `(K_(sp))` value, For example `CuS` and `HgS`.

Discussion

37.	Passing H_(2)S gas in slightly acidic solution of a metal nitrate results in formation in formation of a black coloured precipitate (X). The precipitate is insoluble in cold-dilute acid solution as well as in ammonium sulphide solution. Boiling, concentrated HCl dissolves X, evolving a pungent smelling gas .Hot, dilute HNO_(3) dissolves the precipitate, leaving behind a white precipitate (Y). Boiling the solution for a longer time dissolves Y too. The most likely density of X and Y are
Answer» `CUS` and S `Bi_(2)S_(3)` and S `PbS` and S `Ag_(2)S` and S Answer :B

Discussion

38.	PassingH_(2)S gasa mixtureof Mn^ (2+) ,Ni^(2+) ,Cu^(2+) and Hg^(2+) ions in an acidifiedaqaeous solution precipitate
Answer» `CuS and HGS` `MnS and CuS` `MnS and NIS` `NiS and HgS` Solution :In acid solution `[S^(2-)]` is LOW so, `CuS` and `HgS` willprecipitatemore easillymoredueto low `K_(sp)` MnS willprecipitate If `H_(2)S` is passed in alkalinesolution

Discussion

39.	Passing chlorine gas through dry slaked lime to produce …………………… .
Answer» `CAOCL` `CaOCl_(2)` `CAO` `CaCl_(2)` SOLUTION :`CaOCl_(2)`

Discussion

40.	Passage - VIII : In this passage five observation are given. Question are asked with reference to the givenobservations Observation (i) Gaseousoxygen is colourless whereas liquid and solid oxygen are coloured substance. Observation (ii) When O_(2) is cooled below a certain temperature its paramagnetic character decreases. Observation (iii) In ice H_(2)O molecules are H bonded Observation (iv) Ozone is responsible for tailing of Hg(l) Observation (v) O_(3)(g) is almost unavailable in lower atmosphere Reference to observation (v) O_(3)(g) is unavailable in lower atmosphere because -
Answer» at lower altitude`O_(3)`produced is decomposed to `O_(2)` due to the HIGHER temperature at lower temperature there is noise pollution, therefore `O_(3)` produced is decomposedto `O_(2)` `O_(3)(g)` is HEAVIER than air, hence availablein the coal mines in the lower altitude UV RADIATION of higher frequencies required for the dissociation of gaseous `O_(2)` to atomic oxygen, are unavailable SOLUTION :In the lower altitude UV radiation of higher frequencies required for the dissociation of gaseous`O_(2)` to atomic oxygen, are unavailable

Discussion

41.	Passage - VII : An aqueoussolution of a gas (x) gives the following reactions (1) It decolourises an acidified K_(2)Cr_(2)O_(7) solution (2) On boiling with H_(2)O_(2) cooling it and then adding on aqueous solution of BaCl_(2), a precipitation insoluble in dilute HCl is obtained (3) On passing H_(2)S in the solution white turbidity (y) is obtained (4) When gas 'x' is heated with concentratedHNO_(3) evolves a brown coloured gas (A) (5) When 'x' also dissolves in Na_(2)SO_(3) solution on heating a clear solution (C ) is formed. The brown coloured gas is
Answer» NO `NO_(2)` `SO_(2)` `H_(2)S` SOLUTION :`FeSO_(4)` on decomposition GIVE `SO_(3)`

Discussion

42.	Passage - VII : An aqueoussolution of a gas (x) gives the following reactions (1) It decolourises an acidified K_(2)Cr_(2)O_(7) solution (2) On boiling with H_(2)O_(2) cooling it and then adding on aqueous solution of BaCl_(2), a precipitation insoluble in dilute HCl is obtained (3) On passing H_(2)S in the solution white turbidity (y) is obtained (4) When gas 'x' is heated with concentratedHNO_(3) evolves a brown coloured gas (A) (5) When 'x' also dissolves in Na_(2)SO_(3) solution on heating a clear solution (C ) is formed. "C" is
Answer» `Na_(2)SO_(3)` `Na_(2)S_(2)O_(3)` `Na_(2)SO_(4)` `Na_(2)S_(4)O_(6)` Solution :Thermally more STABLE salts are `Al_(2)(SO_(4))_(3), CuSO_(4), FeSO_(4)`

Discussion

43.	Passage - VIII : In this passage five observation are given. Question are asked with reference to the givenobservations Observation (i) Gaseousoxygen is colourless whereas liquid and solid oxygen are coloured substance. Observation (ii) When O_(2) is cooled below a certain temperature its paramagnetic character decreases. Observation (iii) In ice H_(2)O molecules are H bonded Observation (iv) Ozone is responsible for tailing of Hg(l) Observation (v) O_(3)(g) is almost unavailable in lower atmosphere O_(3) on reaction with Hg(l) as per observation (iv) produce -
Answer» HgO `Hg_(2)O` `HgO_(2)` `Hg_(2)O_(3)` SOLUTION :`O_(3)+2HG to Hg_(2)O+O_(2)`

Discussion

44.	Passage - VIII : In this passage five observation are given. Question are asked with reference to the givenobservations Observation (i) Gaseousoxygen is colourless whereas liquid and solid oxygen are coloured substance. Observation (ii) When O_(2) is cooled below a certain temperature its paramagnetic character decreases. Observation (iii) In ice H_(2)O molecules are H bonded Observation (iv) Ozone is responsible for tailing of Hg(l) Observation (v) O_(3)(g) is almost unavailable in lower atmosphere Which of the following explain the observation (i)?
Answer» in liquid and solid oxygen there is a TRANSITION of bonding ELECTRONS from the TRIPLET state to the singlet state in liquid and solid oxygen there is a transition of antibonding electrons from the triplet state to the singlet state in liquid and solid oxygen the two unpaired electrons of gasesous oxygen are PAIRED up (a) and (c ) SOLUTION :`KHSO_(4)` is obtained in solid state

Discussion

45.	Passage - VIII : In this passage five observation are given. Question are asked with reference to the givenobservations Observation (i) Gaseousoxygen is colourless whereas liquid and solid oxygen are coloured substance. Observation (ii) When O_(2) is cooled below a certain temperature its paramagnetic character decreases. Observation (iii) In ice H_(2)O molecules are H bonded Observation (iv) Ozone is responsible for tailing of Hg(l) Observation (v) O_(3)(g) is almost unavailable in lower atmosphere Which of the following explain the observation (ii)?
Answer» because below a CERTAIN temperature `O_(2)(g)` is PARTIALLY dimerized the unpair electrons of `O_(2)(g)` is paired up in some `O_(2)` MOLECULES `O_(2)(g)` is partialy dissociated to atomic oxygen `O_(2)(g)` is CONVERTED partially to ozone Solution :The unpair electrons of `O_(2)(g)` is paired up in some `O_(2)` molecules

Discussion

46.	Passage - VII : An aqueoussolution of a gas (x) gives the following reactions (1) It decolourises an acidified K_(2)Cr_(2)O_(7) solution (2) On boiling with H_(2)O_(2) cooling it and then adding on aqueous solution of BaCl_(2), a precipitation insoluble in dilute HCl is obtained (3) On passing H_(2)S in the solution white turbidity (y) is obtained (4) When gas 'x' is heated with concentratedHNO_(3) evolves a brown coloured gas (A) (5) When 'x' also dissolves in Na_(2)SO_(3) solution on heating a clear solution (C ) is formed. y is
Answer» `SO_(2)` `H_(2)SO_(4)` `BaCl_(2)` S SOLUTION :Cation size increase ionic nature increase SOLUBILITY increase. In `Rb_(2)SO_(4) rArr Rb^(+)` is large size cation`Rb_(2)SO_(4)` is more soluble in WATER

Discussion

47.	Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) H_(2)SO_(4)+PCl_(5) to (X) overset(H_(2)O)totwo strong acids where 'x' is
Answer» `SO_(2)Cl_(2)` SOCl `ClSO_(3)H` `POCl_(3)` SOLUTION :`SO_(2)Cl_(2)+2H_(2)O to 2HCl +H_(2)SO_(4)`

Discussion

48.	Passage - VII : An aqueoussolution of a gas (x) gives the following reactions (1) It decolourises an acidified K_(2)Cr_(2)O_(7) solution (2) On boiling with H_(2)O_(2) cooling it and then adding on aqueous solution of BaCl_(2), a precipitation insoluble in dilute HCl is obtained (3) On passing H_(2)S in the solution white turbidity (y) is obtained (4) When gas 'x' is heated with concentratedHNO_(3) evolves a brown coloured gas (A) (5) When 'x' also dissolves in Na_(2)SO_(3) solution on heating a clear solution (C ) is formed. Gas 'x' is
Answer» `SO_(3)` S `SO_(2)` `H_(2)S` Solution :CU do not reacts with dil. `H_(2)SO_(4)` and not FORM `CuSO_(4)`

Discussion

49.	Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7))Ionically the reaction between H_(2)SO_(4) & Mg may be presented as, Mg+2H_(3)O^(+) to Mg^(2+)+H_(2) uarr +2H_(2)O Therefore, in the given reaction -
Answer» oxidising AGENT is `H^(+)(H_(2)O)` reducingagent is `H^(+)(H_(2)O)` oxidising agent is `SO_(4)^(2-)` oxidising agent are `SO_(4)^(2-)` & `H_(3)O^(+)` both Solution :`H_(2)SO_(4)` act as oxidizing agent. `H_(2)SO_(4)+2PCl_(5) to UNDERSET((X))(SO_(2)Cl_(2))+2POCl_(3)+2HCl`

Discussion

50.	Passage - VI : Industrially sulphuric acid is produced by the following steps: Stage I : S+O_(2) overset(Delta)to SO_(2) Stage II : 2SO_(2) +O_(2) overset(V_(2)O_(5))to 2SO_(3) Stage III : SO_(3)+H_(2)O to H_(2)SO_(4) Since the reaction between SO_(3) and H_(2)O is violent, SO_(3) is passed into 98% H_(2)SO_(4) to produce oleum (H_(2)S_(2)O_(7)) Pure H_(2)SO_(4) does not react with metal because -
Answer» PURE `H_(2)SO_(4)` does not contain any water Pure `H_(2)SO_(4)` is not OXIDISING agent Pure `H_(2)SO_(4)` is STRONGLY H - bonded Pure `H_(2)SO_(4)` does not contain any `SO_(3)` Solution :Pure `H_(2)SO_(4)` is strongly H-bonded. Hence it does not react with METAL.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Pb|PbSO_(4)|H_(2)SO_(4)(aq)||PbCl_(2) saturated solution |Cl_(2)|Pt As the cell discharged:

PbOhArrPb+1/2O_2, K=2xx10^(-1) ZnOhArr +1/2 O_2 K=2xx10^2 CuOhArr +1/2 O_2: K=2xx10^(2) Ag_2OhArr2Ag+1/2 O_2 ,K=2xx10^4 On the basis of equilibrium constant, which of the following is most stable oxide ?

PbO_(2) is a stronger oxidising agent than SnO_(2). Or PbO_(2) can act an oxidiside agent.

PbO_(2) on heating evolves

PbO_2 is :

PbF_(4), PbCL_(4) exists PbBr_(4) and PbI_(4) do not exist because of

PbCl_(2) is soluble in (a) . AgCl is soluble in (b) _ white Hg_(2)CI_(2) is (c ) ___by NH_(3)

PbCl_(2)darr+"hot water"toPb^(2)(aq.)+2Cl^(-)(aq.)

PbCl_2 is insoluble in cold water. Addition of HCl increases its solubility due to

PbCI_(2) exists white PbBr and PbI do not exist.

Pb(CH_(3)COO)_(2) gives ...... colour with H_(2)S

Pb(CH_(3)COO)_(2) gives . . . Colour with H_(2)S

Pb_(3)O_(4)+4HNO_(3)to2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)OWhich is true about the lead ions in Pb_(3)O_(4) ?

Pb^(2+) belongs to :

Pb//PbSO_(4)_((s)), SO_(4(aq))^(-2) Pb_((s))+SO_(4)^(-2) hArr PbSO_(4(s))+ ?

Pb shows oxidation state of

Pb hasbeen placedin groups I and IIbecause

Pb and Sn are extracted from their chieforeby

Pb+Dil.HNO_(3) overset(Warm)toP+Quarr+H_(2)O Incorrect statement for Q is:

Pb + Conc.HNO_(3) gives

Pb and Sn are extracted from their chief ores by

Pb and Sn are extracted from their chief ore respectively by

Pb and Sn are extracted from their chief ores by :

Pb and Sn are extracted from their chief ore by :

Pauling's equation for determining the electronegativity of an element is

Pauling.s electronegativity scale is based on experimental value of:-

Pauling electronegativity scale is based on experimental values of:

What is the function of limestone the extraction of iron? Give equation to explain its action.

Pasteurization means____ .

Passivity of iron is due to the formation of a thin layer of :

Passivity of iron is due to the formation of thin film of its :

Passivity of iron is due to formation of:

Passing H_2S gas through nitric acid produces:

Passing H_(2)S gas into a mixture of Mn^(2+) , Ni^(2+) , Cu^(2+) and Hg^(2+) ions in an acidified aqueous solution precipitatates:

PassingH_(2)S gasa mixtureof Mn^ (2+) ,Ni^(2+) ,Cu^(2+) and Hg^(2+) ions in an acidifiedaqaeous solution precipitate

Passing chlorine gas through dry slaked lime to produce …………………… .

Explore topic-wise InterviewSolutions in Current Affairs.

Pb|PbSO_(4)|H_(2)SO_(4)(aq)||PbCl_(2) saturated solution |Cl_(2)|Pt As the cell discharged:

PbOhArrPb+1/2O_2, K=2xx10^(-1) ZnOhArr +1/2 O_2 K=2xx10^2 CuOhArr +1/2 O_2: K=2xx10^(2) Ag_2OhArr2Ag+1/2 O_2 ,K=2xx10^4 On the basis of equilibrium constant, which of the following is most stable oxide ?

PbO_(2) is a stronger oxidising agent than SnO_(2). Or PbO_(2) can act an oxidiside agent.

PbO_(2) on heating evolves

PbO_2 is :

PbF_(4), PbCL_(4) exists PbBr_(4) and PbI_(4) do not exist because of

PbCl_(2) is soluble in (a) ____. AgCl is soluble in (b) _____ white Hg_(2)CI_(2) is (c ) ___by NH_(3)

PbCl_(2)darr+"hot water"toPb^(2)(aq.)+2Cl^(-)(aq.)

PbCl_2 is insoluble in cold water. Addition of HCl increases its solubility due to

PbCI_(2) exists white PbBr and PbI do not exist.

Pb(CH_(3)COO)_(2) gives ...... colour with H_(2)S

Pb(CH_(3)COO)_(2) gives . . . Colour with H_(2)S

Pb_(3)O_(4)+4HNO_(3)to2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)OWhich is true about the lead ions in Pb_(3)O_(4) ?

Pb^(2+) belongs to :

Pb//PbSO_(4)_((s)), SO_(4(aq))^(-2) Pb_((s))+SO_(4)^(-2) hArr PbSO_(4(s))+ ?

Pb shows oxidation state of

Pb hasbeen placedin groups I and IIbecause

Pb and Sn are extracted from their chieforeby

Pb+Dil.HNO_(3) overset(Warm)toP+Quarr+H_(2)O Incorrect statement for Q is:

Pb + Conc.HNO_(3) gives

Pb and Sn are extracted from their chief ores by

Pb and Sn are extracted from their chief ore respectively by

Pb and Sn are extracted from their chief ores by :

Pb and Sn are extracted from their chief ore by :

Pauling's equation for determining the electronegativity of an element is

Pauling.s electronegativity scale is based on experimental value of:-

Pauling electronegativity scale is based on experimental values of:

What is the function of limestone the extraction of iron? Give equation to explain its action.

Pasteurization means____ .

Passivity of iron is due to the formation of a thin layer of :

Passivity of iron is due to the formation of thin film of its :

Passivity of iron is due to formation of:

Passing H_2S gas through nitric acid produces:

Passing H_(2)S gas into a mixture of Mn^(2+) , Ni^(2+) , Cu^(2+) and Hg^(2+) ions in an acidified aqueous solution precipitatates:

PassingH_(2)S gasa mixtureof Mn^ (2+) ,Ni^(2+) ,Cu^(2+) and Hg^(2+) ions in an acidifiedaqaeous solution precipitate

Passing chlorine gas through dry slaked lime to produce …………………… .

PbCl_(2) is soluble in (a) . AgCl is soluble in (b) _ white Hg_(2)CI_(2) is (c ) ___by NH_(3)