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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 901. |
Equivalent weight of `H_(3)PO_(2)` when it disproportionates into `PH_(3)` and `H_(3)PO_(3)` is (mol.wt. of `H_(3)PO_(2)=M`)A. `M`B. `(3M)/(4)`C. `(M)/(2)`D. `(M)/(4)` |
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Answer» Correct Answer - B `H_(3)PO_(2) rarr PH_(3)` `P^(+)+4e^(-) rarr P^(3-)` `:. Eq. wt. (H_(3)PO_(2))=M//4` `H_(3)PO_(2) rarr H_(3)PO_(3)` `P^(+) rarr P^(3+)+2e^(-)` `:. Eq.wt. (H_(3)PO_(2))=M//2` Hence, Eq.wt. `(H_(3)PO_(2))=(M)/(4)+(M)/(2)=(3)/(4)M` |
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| 902. |
During nitration of benzene withnitrating mixture, `HNO_(3)` acts asA. acidB. oxidising agentC. reducing agentD. Both A and B |
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Answer» Correct Answer - D `HNO_(3)` is both an acid and an oxidising agent. |
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| 903. |
A `0.56` g sample of limestones is dissolved in acid and the calcium is precipitated as calcium oxalate .The precipitate as calcium oxalate the prepcipate is filtered washed with water and dissolved in dil `H_(2)SO_(4)` The solution required `40ml` of `0.25NKmnO_(4)` solutions for titration .Calculate percentage of `CaO` in limestone sample. |
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Answer» Correct Answer - 50 |
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| 904. |
A `0.56` g sample of limestones is dissolved in acid and the calcium is precipitated as calcium oxalate .The precipitate as calcium oxalate the prepcipate is filtered washed with water and dissolved in dil `H_(2)SO_(4)` The solution required `40ml` of `0.25NKmnO_(4)` solutions for titration .Calculate percentage of `0.25N` `KMnO_(4)` solution for titration .Calculate of `0.25N KMnO_(4)` solution for titration ,Calculate percentage of `CaO` in limestone sample. |
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Answer» `CaCO_(3)+2H^(+)rarrCa^(2+)+CO_(2)+H_(2)O` `Ca^(2+)+C_(2)O_(4)^(2-)rarrCaC_(2)O_(4)` `CaC_(2)O_(4)+KMnO_(4)` in presence of `H_(2)SO_(4)` show the following redox changes: `C_(2)^(3+)rarr2C^(4+)+2e` `Mn^(7+)+5erarrMn^(2+)` The above set of reaction shows: Meq.of `CaO="Meq.of" CaC_(2)O_(4)="Meq.of" KMnO_(4)` (Valence factor is two for `Ca` throughout the changes) `(w_(CaO))/(56//2)xx1000=40xx0.25` `therefore w_(CaO)=0.28` `therefore % "of"CaO=(0.28)/(0.56)xx100=50%` |
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| 905. |
In the course of a chemical reaction an oxidant -A. Loses electronB. Gain electronC. Both loses and gain electronsD. Electron change does not occur |
| Answer» Correct Answer - B | |
| 906. |
The oxidation number of phosphorus do not involve oxidation reduction?A. `+3`B. `+2`C. `+1`D. `-1` |
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Answer» Correct Answer - C `Ba^(2+) (H_(2)PO_(2))_(2)^(-1xx2)`: `H_(2)PO_(2)^(ө): 2+x-4= -1 impliesx= +1` Oxidation state of `P= +1` |
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| 907. |
An element forms an oxide, in which the oxygen is `20%` of the oxide by weight, the equivalent weight of the given element will beA. `32`B. `40`C. `60`D. `128` |
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Answer» Correct Answer - A In metal oxide `O=20%` `M=80%` If `20 g` of oxygen associates with `80 g`, then `8 g` of oxygen associates with metal `(80)/(20)xx8=32g` of metal `:. ` Eq.wt. of metal `=32 g` |
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| 908. |
Which statement is wrong?A. Oxidation number of oxygen is `+1` in preoxidesB. Oxidation number of oxygen is `+2` is oxygen difluorideC. Oxidation number of oxygen is `-(1)/(2)` in superoxidesD. Oxidation number of oxygen is `-2` is most of its compound |
| Answer» Correct Answer - A | |
| 909. |
How many mL. of aqueous solution of `KMnO_(4)` containing `158g/L` must be used to complete the conversation of `75.0g` of `KI` to iodine by the reaction `KMnO_(4)+KI+H_(2)SO_(4)rarrK_(2)SO_(4)+MnSO_(4)+I_(2)+6H_(2)O` |
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Answer» `N_(KMnO_(4))=(158)/(31.6xx1)=5.0` M `(Mn^(7+)+5erarrMn^(2+)` `therefore E=(M)/(5)=(158)/(5)=31.6)` Meq.of `KI=(75xx1000)/(166)=451.8` `(2I^(-)rarrI_(2)+2e therfore E=(M)/(1)=(166)/(1))` Now Meq.of `KMnO_(4)="Meq.of" KI` `5xxV=45.451.8` `therefore V=90.36 mL` |
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| 910. |
Balance the chemical equation by the oxidation number method. (vii) `H_(2)O_(2) + PbS to PbSO_(4) + H_(2)O` |
| Answer» `4H_(2)O_(2) + PbS to PbSO_(4) +4H_(2)O` | |
| 911. |
Which of the following reactions do not involve oxidation reduction ? I. `2Cs+2H_(2)Orarr2CsOH+H_(2)` II. `2CuI_(2)rarr2CuI+I_(2)` III. `NH_(4)Br+KOHrarrKBr+NH_(3)+H_(2)O` IV. `4KCN+Fe(CN)_(2)rarrK_(4)[Fe(CN)_(6)]`A. `I,II`B. `I,III`C. `I,III,IV`D. `III,IV` |
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Answer» Correct Answer - D i. `{:(,RbrarrRb^(1)+e^(-)("Oxidation"),,,),(H_(2)O+,e^(-)rarroverset(ө)OH+(1)/(2)e^(-)("Oxidation"),,,):}]` Hence, redox reaction. ii. `{:(e^(-)+,Cu^(2+)rarrCu^(1+)("Reduction"),,,),(,2I^(ө)rarrI_(2)+2e^(-)("Reduction"),,,):}]` Hence, redox reaction iii. `underset(x=-3)overset(o+)(NH_(4))rarrunderset(x=-3)(NH_(3))`(No change in oxidation reaction). Hence, reaction is not redox. iv. No change in the oxidation number of either `Fe^(2+)` or `CN^(ө)` in both reactant and product, hence, not work reaction. |
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| 912. |
The reaction `2K_(2)MnO_(4)+Cl_(2)rarr 2KMnO_(4)+2KCl` is an example ofA. RedoxB. Reduction onlyC. NeutralizationD. Disproportionation |
| Answer» Correct Answer - A | |
| 913. |
The weight of `MnO_(2)` and the volume of HCl of specific gravity 1.2 g `mL^(-1)` and `4%` nature by weight, needed to produce 1.78 L of `Cl_(2)` at STP. The reaction involved is: `MnO_(2)+4"HCl"toMnCl_(2)+2H_(2)O+Cl_(2)` |
| Answer» Weight of `MnO_(2)` =6.913 g, Volume of HCl =241.66 ml | |
| 914. |
Given one expample of disprop[ortionation reaction `2H_(3)overset(+1)PO_(2)overset(heat)rarroverset(-3)PH_(3)+H_(3)overset(+5)PO_(4)` |
| Answer» Since P undergoes decrease as well as increase in oxidation state it is an example of disproportionation reaction. | |
| 915. |
What is the maximum weight of `Cl_(2)` obtained by the action of `1g` `HCl` on `1g MnO_(2)`? |
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Answer» `{:(,MnO_(2)+,4HClrarr,MnCl_(2)+,Cl_(2)+,2H_(2)O,,,),("Initial mole",(1)/(87),(1)/(36.5),-,-,-,,,),(,=0.011,=0.027,,,,,,),("Final mole",[0.011-(0.027)/(4)],,(0.027)/(4),(0.027)/(4),,,,):}` `therefore` Mole of `Cl_(2) "formed" =(0.027)/(4)` `therefore wt."of"Cl_(2) "formed" =(0.027)/(4)xx71=0.4792g` |
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| 916. |
Balance the chemical equation by the oxidation number method. (viii) `MnO_(2) + HCl to MnCl_(2) + Cl_(2) + H_(2)O` |
| Answer» `MnO_(2) + 4HCl to MnCl_(2) + 2H_(2)O + Cl_(2)` | |
| 917. |
Preparation of `Cl_(2)` from `HCl` and `MnO_(2)`, involves the process of:A. Oxidation of `MnO_(2)`B. Reduction of `MnO_(2)`C. DehydrationD. Oxidation of chloride ion |
| Answer» Correct Answer - B::D | |
| 918. |
The oxidation number of `C` in `HNC` isA. `+2`B. `-3`C. `+3`D. zero |
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Answer» Correct Answer - A `1+(-3)+a=0` `:. a=+2` |
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| 919. |
What is the correct representation of reaction occurring when HCl is heated with `MnO_(2)` ?A. `MnO_(4)^(-)+5Cl^(-)+8H^(+)rarr Mn^(2+)+5Cl^(-)+5H_(2)O`B. `MnO_(2)+2Cl^(-)+4H^(+)rarr Mn^(2+)+Cl_(2)+2H_(2)O`C. `2MnO_(2)+4Cl^(-)+8H^(+)rarr 2 Mn^(2+)+2Cl_(2)+4H_(2)O`D. `MnO_(2)+4HCl rarr MnCl_(4)+Cl_(2)+H_(2)O` |
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Answer» Correct Answer - B `MnO_(2)+4HCl rarr MnCl_(2)+Cl_(2)+2H_(2)O` or in ionic form `MnO_(2)+2Cl^(-)+4H^(+)rarr Mn^(2+) + Cl_(2) + 2H_(2)O` |
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| 920. |
If ` 10.0 mL` of hypo solution `(Na_2S_2O_3. 5H_2 )` is decolorized by `15 mL` of `M//40` iodine solution , then the concentration of hypo solution is `"____" g dm^(-3)`.A. ` 24.6`B. ` 8.6`C. ` 18.6`D. ` 31.6` |
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Answer» Correct Answer - C The balanced equation for the redox reaction is ` 2S_2O_3^(2-) + I_2 rarr 2I^(-) +S_4O_6^(2-)` Applying molarity equation , we have ` (M_1V_1) /n_1 (Na_2S_2O_3) = (M_2V_2)/n_2 (I_2)` `(M_1(10 mL))/2 = ((1//40) (15 mL))/1` or ` M_1 = (1xx 15 xx 2)/(10xx 40) = 3/(40) "mol L"^(-1)` Strenght of hypo solution ` = ("Molarity" ) xx ("Molar Mass")` ` =(3/(10) "mol L"^(-1)) (248 "gmol"^(-1))` ` =18.6 "g L"^(-1)` ` =18.6 "g dm"^(-3)`. |
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| 921. |
Oxidation number of C in HNC is :A. `+2`B. `-3`C. `+3`D. Zero |
| Answer» Correct Answer - A | |
| 922. |
What is the oxidation state of sodium in sodium amalgam `(Na//Hg)`? |
| Answer» Sodium amalgam is simply a homogeneous mixture of the two metals and no chemical reaction has rtaken place.Both the metals are in elemental state and their oxidation states are zero. | |
| 923. |
Oxidation number of silver in silver amalgam isA. `+1`B. 0C. `-1`D. `+2` |
| Answer» Correct Answer - B | |
| 924. |
`0.56 g` of lime stone was treated with oxalic acid to give `CaC_(2)O_(4)`. The precipitate decolorized `45 ml` of `0.2N KMnO_(4)` in acid medium. Calculate `%` of `CaO` in lime stone. |
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Answer» `underset(CaCO_(3))("Limestone")overset("Oxalic acid")rarrCaC_(2)O_(4)overset(KMnO_(4))rarr"decolorizes"` `therefore` Redox changes: For `CaC_(2)O_(4), C_(2)^(3+)rarr2C^(4+)+2e` For `KMnO_(4), 5e+Mn^(7+)rarrMn^(2+)` `therefore` Meq.of `CaCO_(3)=` Meq.of `CaC_(2)O_(4)=` meq.of `KMnO_(4)` `because` Meq.of `CaCO_(3)=` Meq.of `CaO` ( since `CaO` is present in `CaCO_(3)`) `therefore` Meq.of `CaO=` Meq.of `KMnO_(4)` `(w)/(56//2)xx1000=45xx0.2` Wt.of `CaO=0.252g` `therefore %` of `CaO` in limestone `=(0.252)/(0.56)xx100=45%` |
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| 925. |
Oxidation number of silver in silver amalgam isA. `+1`B. zeroC. `-1`D. none of these |
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Answer» Correct Answer - B Oxidation number of an element in its amalgam is zero. |
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| 926. |
Oxidation state of nitrogen is incorrectly given for:A. `{:("Compound","Oxidation State"),([Co(NH_(3))_(5)Cl]Cl_(2),-3):}`B. `{:("Compound","Oxidation State"),(NH_(2)OH,-1):}`C. `{:("Compound","Oxidation State"),((N_(2)H_(5))_(2)SO_(4),+2):}`D. `{:("Compound","Oxidation State"),(Mg_(3)N_(2),-3):}` |
| Answer» Correct Answer - C | |
| 927. |
In which of the following cases is the oxidation state of `N` atom wrongly calculated?A. Compound= `NH_(4)Cl`, Oxidation state= `-3`B. Compound= `(N_(2)H_(5))_(2)SO_(4)`, Oxidation state= `+2`C. Compound= `Mg_(3)N_(2)`, Oxidation state= `-3`D. Compound= `NH_(2)OH`, Oxidation state= `-1` |
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Answer» Correct Answer - B Oxidation state of `N` in (b) is `-2` |
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| 928. |
Oxidation number of sodium is sodium amalgam is :A. `+2`B. `+1`C. `-3`D. Zero |
| Answer» Correct Answer - D | |
| 929. |
In which of the following cases is the oxidation state of `N` atom wrongly calculated?A. (Compound `=NH_(4)Cl`, Oxidation state `= -3`)B. (Compound `=(N_(2)H_(5))_(2)SO_(4)`, Oxidation state `= +2`)C. (Compound `=Mg_(3)N_(2)`, Oxidation state `= -3`)D. (Compound `=NH_(2)OH`, Oxidation state `= -1`) |
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Answer» Correct Answer - B Oxidation state of `N` in (`b`) is `-2`. |
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| 930. |
`xMnO_4^(-) +yH_2O_2 rarr 2Mn^(2+) + 5 H_2O + 9O_2 + Ze^(-)` In this reaction, the values of ` x,y`, and `z`, respectively, are .A. ` 2,5,6`B. ` 5,2, 9`C. ` 3,5,5`D. ` 2,6,6` |
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Answer» Correct Answer - A Redox reaction should be balanced with respect to the number of atoms and ionic charges . To balance Mn atome on both sides, we multiply ` MnO_4^(-)` by `2`. To balance H and O atoms on both sides, we multiply `H_(2)O_(2)` by 5. To balance ionic charge, we add 6 electrons on the right-hand side. `2MnO_(4)^(-)+5H_(2)O_(2) rarr 2Mn^(2+)+5H_(2)O+9O_(2)+6e^(-)` |
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| 931. |
Which of the following shows highest oxidation number of combined state:A. OsB. RuC. Both (1) and (2)D. None |
| Answer» Correct Answer - C | |
| 932. |
In which one of the following changes there are transfer of five electrons?A. `MnO_(4)^(-) rarr Mn^(2+)`B. `CrO_(4)^(2) rarr Cr^(3+)`C. `MnO_(4)^(2-) rarr Mn^(2+)`D. `Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)` |
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Answer» Correct Answer - A `MnO_(4)^(-) rarr Mn^(2+)+5e^(-)`. |
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| 933. |
Identify the element which can have highest oxidation numbersA. `N`B. `O`C. `Cl`D. `C` |
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Answer» Correct Answer - C Chlorine has oxidation state `-1` to `+7`. |
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| 934. |
A metal ion `M^(3+)` loses three electrons , its oxidation number will beA. `+3`B. `+6`C. 0D. `-3` |
| Answer» Oxidation number changes from +3 to +6 | |
| 935. |
The oxidation states of sulphur in the anions `SO_(3)^(2-), S_(2)O_(4)^(2-)`, and `S_(2)O_(6)^(2-)` follow the orderA. ` S_2 O_6^(2-) lt S_2 O_4^(2-) lt SO_3^(2-)`B. ` S_2O_4^(2-) lt SO_3^(2-) lt S_2O_6^(2-)`C. `SO_3^(2-) lt S_(2)O_(6)^(2-) lt SO_(3)^(2-)`D. ` S_2O_4^(2-) lt S_2O_6^(2-) lt SO_3^(2-)` |
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Answer» Correct Answer - B We can work out the oxidation number of S by using the fact that in polyatomic ions, the algebraic sum of all the oxidation numbers must equal the charge on the ion. Let the oxidation number of S be x. (a) `SO_3^(2-) rArr + 3 (-2) =- 2` `x- 6=-2` ` x= +4` (b) `S_2O_4^(2-) rArr 2x +4 (-2) =-2` ` 2 x - 8 =- 2` ` x= +3` (c) ` S_2O_6^(2-) rArr 2x + 6(-2) =-2` ` 2x- 12 =-2` `x= +5` Thus, the order of oxidation number is `overset(+3)(S_(2))O_(4)^(2-) lt overset(+4)(S)O_(3)^(2-) lt overset(+5)(S_(2))O_(6)^(2-)` |
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| 936. |
A metal ion `M^(3+)` loses three electrons , its oxidation number will beA. ZeroB. `+6`C. `+2`D. `+4` |
| Answer» Correct Answer - B | |
| 937. |
The oxidation number of `S` in `Na_(2)S_(4)O_(6)` isA. `(2)/(3)`B. `(3)/(2)`C. `(3)/(5)`D. `(5)/(2)` |
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Answer» Correct Answer - D `Na_(2)overset(*)(S_(4))O_(6)` `2+4x-12=0` `4x=10 x=(10)/(4) x=(5)/(2)`. |
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| 938. |
In which of the following reactions, the underlined element has decreased its oxidation number during the reaction?A. `Fe + CuSO_(4) to Cu + ul(Fe)SO_(4)`B. `ul(H_(2)) + Cl_(2) to 2ul(H)Cl`C. `ul(C) + H_(2)O to ul(CO)+ H_(2)`D. `ul(Mn)O_(2) + 4HCl to ul(Mn)Cl_(2) + Cl_(2) + 2H_(2)O` |
| Answer» Correct Answer - D | |
| 939. |
In which of the following reactions the underlined substance is oxidised?A. `3Mg+ ul(N_(2)) to Mg_(3)N_(2)`B. `2Kl + ul(Br_(2)) to 2KBr + I_(2)`C. `ul(Cu)O + H_(2)O`D. `ul(C)O + Cl_(2) to COCl_(2)` |
| Answer» Correct Answer - D | |
| 940. |
`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. ` 2//3` mol of `MnO_4^(-)` and `1//3` mol of `MnO_2`B. ` 1//3` mol of `MnO_4^(-)` and `2//3` mol of `MnO_2`C. ` 1//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2`D. ` 2//3` mol of `Mn_(2)O_(7)` and `1//3` mol of `MnO_2` |
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Answer» Correct Answer - A ` MnO_4^(2-)` is quite strongly oxdizing and is only stable in very strong alkali. In dilute alkali, water, or acidic solutions, it disproportionates : ` MnO_4^(2-) rarr MnO_4^(-) MnO_2` Oxidation half-reaction: `overset(+6)(M)nO_(4)^(2-) rarr overset(+7)(M)nO_(4)^(-) +e^(-)` Reduction half-reaction: `overset(+6)(M)nO_(4)^(2-)+2e^(-) rarr overset(+4)(M)nO_(2)` Balance ionic charges by adding `H^(+)` ions: `MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)` `MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)` Balance H and O by adding `H_(2)O` molecules `MnO_(4)^(2-) rarr MnO_(4)^(-)+e^(-)` `MnO_(4)^(2-)+2e^(-)+4H^(+) rarr MnO_(2)+2H_(2)O` To equalize electron transfer, we multiply oxidation half-reaction by 2 and add them, cancelling electrons on both sides: `3MnO_(4)^(2-) +4H^(+) rarr 3MnO_(4)^(-)+MnO_(2)+2H_(2)O` Similarly we can balance it by means of `OH^(-)` ions: `3MnO_(4)^(2-)+2H_(2)O rarr 2MnO_(4)^(-)+MnO_(2)+4OH^(-)` Thus, 3 mol of `MnO_(4)^(2-)` yield 2 mol of `MnO_(4)^(-)` and 1 mole of `MnO_(2)`. Consequently, 1 mol of `MnO_(4)^(2-)` disproportionates to yield `2//3` mol of `MnO_(4)^(2-)` and `1//3` mol of `MnO_(2)`. Note that in neutral aqueous meadium, we can balance the charge either using `H^(+)` ions or using `OH^(-)` ions. |
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| 941. |
The oxidation state of phosphorus vary fromA. `-1` to `+1`B. `-3` to `+3`C. `-3` to `+5`D. `-5 ` to `+1` |
| Answer» Correct Answer - C | |
| 942. |
Oxidation states of iodine vary fromA. `-1` to `+ 1`B. `-1 " to " +7`C. `+3 " to " +5`D. `-1 " to " 5` |
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Answer» Correct Answer - B Iodine shows `-1` oxidation state in iodides (Kl etc), `+1, +3, +5` and `+7` oxidation states in interhalogen compounds and polyhalides like `lCl, lCl_(3), IF_(5), IF_(7)` etc. |
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| 943. |
Oxidation number of S in `Na_(2)S_(4)O_(6)` isA. 1.5B. 2.5C. 3D. 2 |
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Answer» Correct Answer - B `Na_(2)S_(4)O_(6), 2(+1) + 4x + 6 (-2) = 0` `4x + 2 -12 = 0` `4x = 10` `x = 2.5` |
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| 944. |
The conductivity of a saturated solution of `BaSO_4` is ` 3. 06 xx 10^(-6) "ohm"^(-1) cm^(-1)` and its equivalent conductance is `1.53 "ohm"^(-1) cm^2 equiv^(-1)`. The `K_(sp)` for `BaSO_4` will be .A. `4xx10^(-12)`B. `2.5xx10^(-9)`C. `2.5xx10^(-13)`D. `4xx10^(-6)` |
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Answer» Correct Answer - D `lambda_(m)=(1000K)/(S)=(1000xx3.06xx10^(-6))/(S)=1.53` `S=2xx10^(-3)(mol)/(litre)` `K_(sp_((BaSO_(4)))=S_(2)=(2xx10^(-3))^(2)=4xx10^(-6)` |
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| 945. |
Assertion: `N` atom has two different oxidation states in `NH_(4)NO_(2)`. Reason: One `N` atom has `-ve` oxidation number as it is attached with less electronegative `H` atom and other has `+ve` oxidation number as it is attached with more electronegative atom.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. |
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Answer» Correct Answer - A `N` in `NH_(4)^(+)` is in `-3` oxidation state and in `NO_(2)^(-)` it is `+3` oxidation state. |
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| 946. |
Redox equations are balanced either by ion-electron method or by oxidation number method. Both methods lead to the correct from of the balanced equation. The ion electron methodd has two advantages. So some chemists prefer to use the ion-electron method for redox reactions carried out in dilute aqueous solutions, where free ions have more or less independent existance. The oxidation state method for redox reactions is mostly used for solid chemicals or for reactions in concentrated acid media. For the reaction `As_(2)S_(3)rarrAs^(5+)+SO_(4)^(2-)` the `n`-factor isA. `11`B. `28`C. `61`D. `(5)/(3)` |
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Answer» Correct Answer - B `{:(As_(2)^(3+) rarr 2 As^(5+) + 4e^(-)),(2x =62x = 10),(S_(3)^(2-)rarr 3 SO_(4)^(2-) + 24e^(-)),(3x = -63 x - 24 = -6),(3x-18)]n-"factor"=28` |
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| 947. |
In a balanced redox reaction net gain of electron (s) is equal to net loss of electrons (s).`n_("factor")` is a reaction specific parameter and for intermolecular redox reaction n-factor of oxidising reducing agent is the no. of moles of electron gained /lost by one mole of compound. 50mL 0.1 M `CuSO_(4)` are mixed with 50mL of 0.1 M KI then, number of moles of electrons involved in the reaction will be:A. 4B. 2.5C. `2.5xx10^(-3)`D. none of these |
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Answer» Correct Answer - c |
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| 948. |
`1 g` sample of `AgNO_(3)` is dissolved in `50 mL` of water, It is titrated with `50 mL` of `KI` solution. The `Agl`percipitated is filtered off. Excess of `KI` filtrate is titrated with `M//10KIO_(3)` in presence of `6 M HCl` till all `I^(-)` converted into `ICI`. It requires `50 mL` of `M//10 KIO_(3)` solution. `20 mL` of the same stock solution of `KI` requires `30 mL` of `M//10KIO_(3)` under similar conditions. Calculate `%` of `AgNO_(3)` in sample. The reaction is `KIO_(3)+2KI+6HClrarr3ICl+3KCl+3H_(2)O` |
| Answer» `%` of `AgNO_(3)=85` | |
| 949. |
A solution contains `Na_(2)CO_(3)` and `NaHCO_(3). 10 mL` of the solution required `2.5 mL "of" 0.1M H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further `2.5 mL "of" 0.2M H_(2)SO_(4)`was required. The amount of `Na_(2)CO_(3)` and `NaHCO_(3)` in `1 "litre"` of the solution is: |
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Answer» Weight of `NaCO_(3)` in 1 litre =5.3 g Weight of `NaHCO_(3)` in 1 litre =4.2 g |
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| 950. |
In `C+H_(2)O rarr CO+H_(2)`, `H_(2)O` acts asA. oxidising agentB. reducing agentC. both (`a`) and (`b`)D. none of these |
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Answer» Correct Answer - A In this reaction `H_(2)O` acts as oxidising agent. |
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