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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.
| 851. |
A redox reaction consists of oxidation and reduction half reactions. There is a loss of electrons in oxidation and the species which loses elecrons is sreducing agnet.Its oxidation number increases during oxidation. Similarly there is a gain of electrons during reduciton and the species wlhih gains electrons is an odidsing agent. thhe species which gains electrons is an aoxidising agent its oxidstion number decreases during reduction .The number of electrons released during oxidation is equal to number of electrons gained dunring reduciton Which of the following involves transfer of five electrons?A. `(MnO_(4))^(-)rarrMn^(2+)`B. `(CrO_(4))_(4)^(2-)rarr(Cr)^(3+)`C. `(MnO_(4))^(2-)rarrMnO_(2)`D. `(Cr_(2)O_(7))^(2-)rarr2Cr.` |
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Answer» Correct Answer - A `(overset(+7)MnO_(4))^(-)rarr(overset(+2)Mn)^(2+) "(5 electrons)"` `(overset(+6)MnO_(4))^(2-)rarr(overset(+3)Cr)^(3+) "(3 electrons)"` `(overset(+6)MnO_(4))^(2-)rarr(overset(+4)MnO_(2)) "(2 electrons)"` `(overset(+6)Cr_(2)O_(7))^(2-)rarr2overset(+3)(2Cr) "(3 electrons)"` |
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| 852. |
A redox reaction consists of oxidation and reduction half reactions. There is a loss of electrons in oxidation and the species which loses elecrons is sreducing agnet.Its oxidation number increases during oxidation. Similarly there is a gain of electrons during reduciton and the species wlhih gains electrons is an odidsing agent. thhe species which gains electrons is an aoxidising agent its oxidstion number decreases during reduction .The number of electrons released during oxidation is equal to number of electrons gained dunring reduciton When `SO_(2)` is passed through `KIO_(3)` the O.N of iodine changes fromA. `+5 to 0`B. `5 to -1`C. `-5 to 0`D. `-7 to -1` |
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Answer» Correct Answer - B The redox reaction is : `3SO_(2)+3overset(+5)KIO_(3)rarr3overset(-1)KI+SO_(3)` |
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| 853. |
The number of electrons lost in the following change is `Fe+H_(2)OrarrFe_(3)O_(4)+H_(2)` |
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Answer» Correct Answer - 8 `{:(3Fe^(0)rarr(Fe^(+(8//3)))_(3)+8e),("["(H^(+))_(2)+2erarr(H^(0))_(2)"]"xx4),(bar(3Fe+4H_(2)OrarrFe_(3)O_(4)+4H_(2))):}` `:.` Lost electron `=8` |
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| 854. |
A redox reaction consists of oxidation and reduction half reactions. There is a loss of electrons in oxidation and the species which loses elecrons is sreducing agnet.Its oxidation number increases during oxidation. Similarly there is a gain of electrons during reduciton and the species wlhih gains electrons is an odidsing agent. thhe species which gains electrons is an aoxidising agent its oxidstion number decreases during reduction .The number of electrons released during oxidation is equal to number of electrons gained dunring reduciton The reation: `2H_(2)O(L)rarr4H^(+)(aq)+O_(2)+4e^(-)` isA. an oxidation reactionB. a reduction reactionC. a redox reactionD. a hydrolysis reaction |
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Answer» Correct Answer - A It is an oxidation reaction as there is loss of electrons |
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| 855. |
The number of electrons lost in the following change is `Fe+H_(2)OrarrFe_(3)O_(4)+H_(2)`A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - D `Fe_(3)O_(4)-=FeO+Fe_(2)O_(3)` Therefore, the change can be represented as `FerarrFe^(2+)+2e^(-)` `2Fe rarr2Fe^(3+)+6e^(-)` `:.` Number of electrons transferred = 8. |
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| 856. |
In the following reaction: `3Fe+4H_(2)OrarrFe_(3)O_(4)+4H_(2)`, if the atomic weight of iron is `56`, then its equivalent weight will beA. 42B. 21C. 63D. 84 |
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Answer» Correct Answer - b |
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| 857. |
For the following balanced redox reaction, `2MnO_(4)^(-) + 4H^(+) + Br_(2) hArr 2Mn^(2+) + 2BrO_(3)^(-) + 2H_(2)O`. If the molecular weight of `MnO_(4)^(-)` and `Br_(2)` are x & y respectively thenA. Equivalent weight of `MnO_(4)^(-)` is `(x)/(5)`B. Equivalent weight of `Br_(2) ` is `(y)/(5)`C. Equivalent weight of `Br_(2)` is `(y)/(10)`D. n-factor ratio of `MnO_(4)^(-)` and `Br_(2)` is 2:1 |
| Answer» Correct Answer - A::C | |
| 858. |
Given below are few statements regarding electrode potentials. Mark the correct statements.A. (i) and (ii)B. (i) and (iii)C. (ii) and (iii)D. (i), (ii) and (iii) |
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Answer» Correct Answer - D All the statements are correct. |
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| 859. |
Phosphorus acid can act both as oxidising agent as well as reducing agent while phosphoric acid is only an oxidising agent. Explain. " " Or Phosphorus acid undergoes disproportionation reaction but phosphoric acid does not. Explain. |
| Answer» The behaviour is linked with the oxidation state of phosphorus atom in the two acids.In phosphorus acid `(H_(3)OPO_(3))`. the oxidation state of P is +3.It is in a position to increase its oxidation state (acts as redcuiing agent) as well as sdecrease it (acts as oxidising agent).The oxidation state of phosphorus in phosphoric acid `(H_(3)PO_(4))` is +5 .Since it cannot increase it vbeyond 5, it cannot act as reducing agent .Hower it can behave as oxidising agent bty decreasing its oxidation state. | |
| 860. |
3.65gm equimole mixture of `NaOH` and `Na_(2)CO_(3)` is titrated against `0.1M HCl` using phenolphathalein as an indicator, `V_(1)mL` of acid was required to reach end point. In another experiment `3.65gm` of same mixture is titrated against 0.2M HCl using methyl orange as an indicator, `V_(2)mL` of acid was required to reach end point. `V_(1)+V_(2)` is :A. 875mLB. 750mLC. 500mLD. 1000mL |
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Answer» Correct Answer - a |
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| 861. |
Among the following , identify the species with an atom in +6 oxidaiton state of metalA. `MnO_(4)^(-)`B. `Cr(CN)_(6)^(3-)`C. `NiF_(6)^(2-)`D. `CrO_(2)CI_(2)` |
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Answer» Correct Answer - D oxidation of stte of Cr is +6 `(Cr_(2))O_(7))^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O` Equivalent mass of `K_(2)Cr_(2)O_(7)`=(Molecular)/(6)mass` |
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| 862. |
In the standardisation of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry the equivalent mass of `K_(2)Cr_(2)O_(7)` is:A. Molecular `mass//2`B. Molecular `mass//6`C. Molecular `mass//3`D. Same as the molecular mass. |
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Answer» Correct Answer - B The redox reaction is: `(Cr_(2)O_(7))^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O` Equivalent mass of `K_(2)Cr_(2)O_(7) = "(Molecular )/(6) mass"` |
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| 863. |
How may millilitres of a `9 N H_(2)SO_(4)` solution will be required to neutralize completely `20mL` of a `3.6 N NaOH` solution ?A. `18.0mL`B. `8.0mL`C. `16.0mL`D. `80.0mL` |
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Answer» Correct Answer - b |
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| 864. |
Match the column I with column II with the type of reaction and mark the appropriate choice. A. (A) `rarr` (i), (B) `rarr` (iii), (C) `rarr` (ii), (D) `rarr` (iv)B. (A) `rarr` (iv), (B) `rarr` (iii), (C) `rarr` (ii), (D) `rarr` (i)C. (A) `rarr` (ii), (B) `rarr` (i), (C) `rarr` (iii), (D) `rarr` (iv)D. (A) `rarr` (iii), (B) `rarr` (i), (C) `rarr` (iv), (D) `rarr` (ii) |
| Answer» Correct Answer - D | |
| 865. |
Fill up the table from the given choice. A. `{:("(i)","(ii)","(iii)","(iv)",("v"),"(vi)"),(+1,+1,Cl,+1,-1,+2):}`B. `{:("(i)","(ii)","(iii)","(iv)",("v"),"(vi)"),(-1,+2,F,+1,-1,-2):}`C. `{:("(i)","(ii)","(iii)","(iv)",("v"),"(vi)"),(-1,+1,F,+1,+2,+2):}`D. `{:("(i)","(ii)","(iii)","(iv)",("v"),"(vi)"),(+2,+2,Cl,+1,+1,+6):}` |
| Answer» Correct Answer - B | |
| 866. |
Match the compounds given in column I with oxidation states of carbon given in column II and mark the appropriate choice.`{:(,"Column I",,"Column II"),("(A)",C_(6)H_(12)O_(6)," (i)"," +3"),("(B)",CHCl_(3)," (ii)"," -3"),("(C)",CH_(3)CH_(3)," (iii)"," +2"),("(D)",(CO OH)_(2)," (iv)"," 0"):}`A. (A) `rarr` (iv), (B) `rarr ` (iii), (C) `rarr` (ii), (D) `rarr` (i)B. (A) `rarr` (i), (B) `rarr ` (ii), (C) `rarr` (iii), (D) `rarr` (iv)C. (A) `rarr` (ii), (B) `rarr ` (iii), (C) `rarr` (iv), (D) `rarr` (i)D. (A) `rarr` (iii), (B) `rarr ` (ii), (C) `rarr` (i), (D) `rarr` (iv) |
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Answer» Correct Answer - A `C_(6)H_(12)O_(6):6x+12+(-12)=0rArr x=0` `CHCl_(3):x+1+(-3)=0 rArr x=+2` `CH_(3)CH_(3):x+3+x+3=0 rArr x=-3` `(CO OH)_(2):x+x+(-4xx2)+(+2)=0 rArr x=+3` |
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| 867. |
Is the reaction `BaO_(2)+H_(2)SO_(4)rarrBaSO_(4)+H_(2)O_(2)` a redox reaction? |
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Answer» `overset(+2-1)BaO_(2)+overset(+1+6-2)H_(2)SO_(4)rarroverset(+2+6-2)BaSO_(4)+overset(+1-1)H_(2)O_(2)` In the reaction, none of the reacting species undergoes change in O.N during the reaction .Therefore it is not a redox reaction. |
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| 868. |
Assertion: Equivalent weight of `FeC_(2)O_(4)` in the reaction, `FeC_(2)O_(4)+` Oxidising agent `rarr Fe^(3+)+CO_(2)` is `M//3`, where `M` is molar mass of `FeC_(2)O_(4)`. Reason: In theabove reaction, total two mole of electrons are given up by `1 mol e` of `FeC_(2)O_(4)` to the oxidising agent.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - C | |
| 869. |
When `S_(2)O_(8)^(2-)` oxidise `Fe^(2+)` then product formed is :-A. `s_(2)O_(6)^(2-)`B. `SO_(4)^(2-)`C. `S_(2)O_(8)^(2-)`D. `SO_(5)^(2-)` |
| Answer» Correct Answer - A | |
| 870. |
Correct order of oxidising strength is :-A. `MnO_(4)^(-)gt VO_(2)^(+)gt Cr_(2)O_(7)^(2^(-))`B. `Cr_(2)O_(7)^(2-)gtMnO_(4)gtVO_(2)^(+)`C. `Cr_(2)O_(7)^(2-)gtVO_(2)^(+)gtMnO_(4)^(-)`D. `MnO_(4)^(-)gtCr_(2)O_(7)^(2-)gtVO_(2)^(+)` |
| Answer» Correct Answer - D | |
| 871. |
निम्नलिखित संभव अभिक्रियाओं की सहायता से `Mg,Zn,Cu` और `Ag` को उनके धत्ते हुए इलेक्ट्रोड विभव के क्रम में लिखिए ।`Cu+2Ag^(+)toCu^(2+)+2Ah` `Mg+Zn^(2+)toMg^(2+)+Zn` `Zn+Cu^(2+)toZn^(2+)+Cu`A. `Mg gt Zn gt Cu gt Ag`B. `Mg lt Zn lt Cu lt Ag`C. `Zn gt Cu gt Ag gt Mg`D. `Mg gt Cu gt Zn gt Ag` |
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Answer» Correct Answer - B `{:(," Mg",lt," Zn",lt," Cu",lt," Ag"),(E^(@)(V),-2.36,,-0.76,,+0.34,,+0.80):}` |
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| 872. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `1//3`B. `3`C. `1//6`D. `6` |
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Answer» Correct Answer - A `(6e^(-)+Cr_(2)O_(7)^(2-)rarr2Cr^(3+))` `Sn^(2+)rarrSn^(4+)+2e^(-)` Equivalent of `Cr_(2)O_(7)^(2-)= "Equivalent of" Sn^(2+)` `(n=6)` , `(n=2)` `implies1 Eq= 1Eq` `(1)/(6) mol =(1)/(2) mol` `(1)/(3) "mol of" Cr_(2)O_(7)^(2-)= 1 "mol of" Sn^(2+)`. |
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| 873. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `1//3`B. `1//6`C. `2//3`D. `1` |
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Answer» Correct Answer - A `undersetulbar(Cr(2)O_(7)^(2-)+14H^(+)2Sn^(2+) rarr 3Sn^(4+)+2Cr^(3+)+7H_(2)O)(Cr_(2)O_(7)^(2-)+14H^(+)+underset((Sn^(2+) rarr Sn^(4+)+2e^(-))xx3)(6e^(-)to2Cr^(3+)+7H_(2)O)` It is clear form this equation that `3` moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-)`, hence `1` mol. Of `Sn^(2+)` will reduce `(1)/(3)`moles of `Cr_(2)O_(7)^(2-)`. |
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| 874. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `1//3`B. `1//6`C. `2//3`D. `3//4` |
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Answer» Correct Answer - A `Cr_(2)O_(7)^(2-)+3Sn^(2+)+14H^(+)rarr 2 Cr^(3+)+3Sn^(4+)+7H_(2)O` 1 mole of `Sn^(2+)` will reduce `1/3` moles of `K_(2)Cr_(2)O_(7)`. |
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| 875. |
(a) Define oxidation. (b) Which is the O.N of sup[lhur in `H_(2)S` and of carbon in `C_(6)H_(12)O_(6)`? |
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Answer» `(i) H_(2)S,x+2=0 or x=-2` `(ii) C_(6)H_(12)O_(6),6x+12-12=0 or x=0` |
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| 876. |
Oxidation number of carbon in `H_(2)C_(2)O_(4)` isA. `+4`B. `+3`C. `+2`D. `-2` |
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Answer» Correct Answer - B `H_(2)overset(*)(C_(2))O_(4)` `2+2x-2xx4=0`, `2x=8-2=6` `x=(6)/(2)=+3`. |
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| 877. |
Which is the following reaction `(s)` is (are) not oxidation reduction?A. `H^(+)+OH^(-)rarrH_(2)O`B. `(1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl`C. `CaCO_(3) rarrCaO+CO_(2)`D. `2H_(2)O_(2)rarr2H_(2)O+O_(2)` |
| Answer» Correct Answer - A::C | |
| 878. |
Carbon is in the lowest oxidation state inA. `CH_(4)`B. `C Cl_(4)`C. `CF_(4)`D. `CO_(2)` |
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Answer» Correct Answer - A In (`b`, `c`,`d`) carbon shows `+4` oxidation state while in (`a`) carbon shows `-4` oxidation state. |
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| 879. |
Which is`//`are disproportionation reaction`(s)` ?A. `2RCHOoverset(Al(OEt)_(3))rarrRCOOCH_(2)R`B. `4H_(3)PO_(3)overset(Delta)rarr3H_(3)PO_(4)+PH_(3)`C. `PCl_(5)rarrPCl_(3)+Cl_(2)`D. `RCHOoverset(KOH)rarrRCOOK+RCH_(2)OH` |
| Answer» Correct Answer - A::B::D | |
| 880. |
Thermal decomposition of `(NH_(4))_(2)Cr_(2)O_(7)` involves.A. Oxidation of `N`B. Reduction of `Cr`C. Disproportionation of compoundD. Intermolecular redox process |
| Answer» Correct Answer - A::B | |
| 881. |
Oxidation state of chlorine in perchloric acid isA. `-1`B. `0`C. `-7`D. `+7` |
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Answer» Correct Answer - D `HClO_(4)` `1+x-2xx4=0.1+x-8=0` `x=8-1=+7` oxidation state. |
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| 882. |
Oxidation number of chlorine in perchloric acid is :-A. `+1`B. `+3`C. `+5`D. `+7` |
| Answer» Correct Answer - D | |
| 883. |
Nitric acid is an oxidising agent and reacts with PbO but it does not react with `PbO_(2)`. Explain why? |
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Answer» PbO is a base. It reacts with nitric acid and forms soluble lead nitrate `PbO+2HNO_(3)tounderset("Soluble")(Pb(NO_(3))_(2))+H_(2)O" (acid base reaction)"` Nitric acid does not react with `PbO_(2)`. Both of them are strong oxidising agents. In `HNO_(3)` nitrogen is in its maximum oxidation state(+5) and in `PbO_(2)`. leads is in its maximum oxidation state (+4). Therefore, no reaction takes place. |
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| 884. |
PbO and `PbO_(2)` react with HCl according to following chemical equations `2PbO+4HCl to2PbCl_(2)+2H_(2)O` `PbO_(2)+4HCl toPbCl_(2)+Cl_(2)+2H_(2)O` Why do these compounds differ n their reactivity? |
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Answer» The difference in the reactivity of the metal oxides is because of he oxidation staes of the metal `overset(+2)2Pbo+4HCIrarroverset(+2)2PCI_(2)+2H_(2)O` `overset(+4)2PbO_(2)+4HCIrarroverset(+2)2PbCI_(2)+CI_(2)+2H_(2)O` atom in the two oxides.In the first reaction, oxidtion state of Pb does not change but it undergoes a change in the second case.Therefore `PbO_(2)` reacts at faster rate than Pbo. |
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| 885. |
In the equation `H_(2)S+2HNO_(3) rarr 2H_(2)O+2NO_(2)+S` The equivalent weight of hydrogen sulphide isA. `17`B. `68`C. `34`D. `16` |
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Answer» Correct Answer - A `H_(2)S rarr overset(0)(S)+2e` Equivalent wt. `=(Mol wt.)/(2)=(34)/(2)=17` |
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| 886. |
In the equation below, which species acts the oxidation agent? `Pb(s)+PbO_(2)(s)+H^(+)(aq)+2HSO_(4)^(-)(aq)rarr2PbSO_(4)(s)+2H_(2)O(l)`A. `Pb(s)`B. `PbO_(2)(s)`C. `H^(+)`(aq)D. `HSO_(4)^(-)`(aq) |
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Answer» Correct Answer - b |
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| 887. |
The conversion of `PbO_(2)` to `Pb(NO_(3))_(2)` isA. OxidationB. ReductionC. Neither oxidation nor reductionD. Both oxidation and reaction |
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Answer» Correct Answer - B `overset(+4)(PbO_(2)) rarr overset(+2)(Pb(NO_(3))_(2))`. In this reaction reduction occurs. |
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| 888. |
Why following two reaction proceed differently? `Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O` and `Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O` |
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Answer» `Pb_(3)O_(4)` is actually a stoichiometric mixture of 2 mol of PbO and 1 mol of `PbO_(2)`. In `PbO_(2)`, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. `PbO_(2)` thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl– ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction `Pb_(3)O_(4) + 8HCl to 3 PbCl_(2) + Cl_(2) + 4 H_(2)O` can be splitted into two reactions namely : `2PbO + 4 HCl to 2 PbCl_(2) + 2H_(2)O` (acid -base reaction) `overset(4+)(Pb)O_(2) + 4 Hoverset(-1)(Cl) to overset(+2)(Pb)Cl_(2) + overset(0)(Cl_(2)) underset("redox reaction")(2H_(2)O)` Since `HNO_(3)` itself is an oxidising agent therefore, it is unlikely that the reaction may occur between `PbO_(2)` and `HNO_(3)`. However, the acid-base reaction occurs between PbO and `HNO_(3)` as : `2PbO + 4 HNO_(3) to 2 Pb (NO_(3))_(2) + 2H_(2)O` It is the passive nature of `PbO_(2)` against `HNO_(3)` that makes the reaction different from the one that follows with HCl. |
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| 889. |
Why following two reaction proceed differently? `Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O` and `Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O`A. three numbers of `Pb^(2+)` ions get oxidised to `Pb^(4+)` stateB. one number `Pb^(4+)` ion gets reduced to `Pb^(2+)` and two numbers of `Pb^(2+)` ions remain unchanged in their oxidation stateC. one number `Pb^(2+)` ion gets oxidised to `Pb^(4+)` and two numbers of `Pb^(4+)` ions remain unchanged in their oxidation statesD. three numbers of `Pb^(4+)` ions get reduced to `Pb^(2+)` state. |
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Answer» Correct Answer - B `Pb_(3)O_(4)`is a mixture of PbO and `PbO_(2)`. `Pn_(3)O_(4)-=2PnO*PnO_(2)+8HCl rarr 3PnCl_(2)+Cl_(2)+4H_(2)O` In the reaction one `Pn^(4+)` ion is reduced to `Pb^(2+)` and two `Pb^(2+)` ions remain unchanged. |
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| 890. |
In the reaction :- `C+4HNO_(3)rarrCO_(2)+2H_(2)O+4NO_(2)` `HNO_(3)` acts as :-A. An oxidising agentB. An acidC. A reducing agentD. None of them |
| Answer» Correct Answer - A | |
| 891. |
Why following two reaction proced differently? `Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O` and `Pb_(3)O_(4)+4HNO_(3)rarr2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O` |
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Answer» `Pb_(3)O_(4)` is a mixed oxide of `(PbO_(2)+2PbO)`. In `PbO_(2), Pb` is in `+4` oxidation state and the stable oxidation state of `Pb` in `PbO is +2`. `PbO_(2)` thus acts as an oxidising agent (oxidant) and therefore can oxidise `Cl^(ө)` ion of `HCl` into `Cl_(2)`. Moreover, `PbO` is a basic oxide. Thus, the reaction `Pb_(3)O_(4)+8HClrarr3PbCl_(2)+Cl_(2)+4H_(2)O` is split into two reactions as follows: `2PbO+4HClrarr2PbCl_(2)+2H_(2)O` (acid base reaction) `overset(+)PbO_(2)+4Hoverset(-1)Clrarroverset(+2)PbCl_(2)+Cl_(2)+2H_(2)O` (redox reaction) Reaction between `PbO_(2)` and `HNO_(3)` does not occur since `HNO_(3)` is itself an oxidising agent but the acid base reaction between `PbO` and `HNO_(3)` occurs as follows: `2PbO+4HNO_(3)rarr2Pb(NO_(3))_(2)+2H_(2)O` `PbO_(2)` is passive against `HNO_(3)`. That is why the reaction proceeds differently with `HCl` |
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| 892. |
`CrO_(4)` has structure as shown The oxidation number of chromium in the above compound isA. `+4`B. `+5`C. `+6`D. `+10` |
| Answer» Correct Answer - C | |
| 893. |
The reaction,`P_(4)+3NaOH+3H_(2)O to 3NaH_(2)PO_(2)+PH_(3)` is an example ofA. Disproportionation reactionB. Neutralization reactionC. Double decomposition reactionD. Displacement reaction |
| Answer» Correct Answer - A | |
| 894. |
In the convsersion of `K_(2)Cr_(2)O_(7)` to `K_(2)CrO_(4)`, the oxidation number of chromium.A. Remains sameB. IncreaseC. DecreasesD. None |
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Answer» Correct Answer - A In `K_(2)Cr_(2)O_(7) and K_(2)CrO_(4)` O.N. of Cr is +6. |
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| 895. |
When `K_(2)Cr_(2)O_(7)` is converted to `K_(2)CrO_(4)`, the change in the oxidation state of chromium isA. `0`B. `6`C. `4`D. `3` |
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Answer» Correct Answer - A `overset(+6)(K_(2)Cr_(2)O_(7)) rarr overset(+6)(K_(2)CrO_(4))`. In this reaction no change in oxidation state of chromium. |
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| 896. |
`Cr_(2)O_(5)` has structure as shown The oxidation number of chromium in the above compound isA. 4B. 5C. 6D. 10 |
| Answer» Correct Answer - C | |
| 897. |
`P_(4)+NaOH+H_(2)O rarr NaH_(2)PO_(3)+PH_(3)` isA. oxidation reactionB. reduction reactionC. both oxidation abnd reduction reactionD. none fo these |
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Answer» Correct Answer - C In this reaction, the oxidation number of P is incresasing from 0 (in `P_4`) to +3 (in `NaH_2PO_3`) as well decreasing from `0` (in `P_4`) to `-3` (in `PH_3`) Thus, it is a redox reaction. In fact, it is a disproportionation reaction. |
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| 898. |
`P_(4)+NaOH+H_(2)O rarr NaH_(2)PO_(3)+PH_(3)` isA. Oxidation reactionB. Reduction reactionC. Both oxidation and reductionD. None |
| Answer» Correct Answer - C | |
| 899. |
When P reacts with caustic soda the products are `PH_(3)` and `NaH_(2)PO_(2)`.the reaction is an example of:A. OxidationB. reductionC. both oxidation and reductionD. neutralisation |
| Answer» Correct Answer - C | |
| 900. |
What is the equivalent weight of `P` in the following reaction? `P_(4)+NaOH rarr NaH_(2)PO_(2)+PH_(3)`A. `(31)/(4)`B. `(31)/(3)`C. `(31)/(2)`D. `31xx4//3` |
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Answer» Correct Answer - D `P_(4) rarr NaH_(2)PO_(2)` `(P_(4))^(@) rarr 4P^(+)+4e^(-)` `:. ` Eq. wt. (Reductant) `=(P_(4))/(4)` `(P_(4))^(@)+12e rarr 4p^(3-)` `:.` Eq.wt.(Oxidant)`=(P_(4))/(12)` `:.` Eq. wt. `(P_(4))=(P_(4))/(12)+(P_(4))/(12)` `=P_(4)xx(16)/(48)=(4xx31)/(3)=31xx(4)/(3)` |
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