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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.
| 601. |
Which of the following can act as an oxidising agent ?A. `H_(2)SO_(4)`B. `HNO_(3)`C. `KMnO_(4)`D. `K_(2)Cr_(2)O_(7)` |
| Answer» Correct Answer - A::B::C::D | |
| 602. |
150 ml `(M)/(10) Ba(MnO_(4))_(2)` in acidic medium can oxidise completelyA. 150 ml 1M `Fe^(2+)`B. 50 ml 1M `FeC_(2)O_(4)`C. 100 ml 1M `C_(2)O_(4)^(2-)`D. 75 ml 1M `K_(2)Cr_(2)O_(7)` |
| Answer» Correct Answer - A::B | |
| 603. |
In the reaction `xHI+yHNO_(3) rarr NO+I_(2)+H_(2)O`A. `x=3`, `y=2`B. `x=2`, `y=3`C. `x=6`, `y=2`D. `x=6`, `y=1` |
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Answer» Correct Answer - C `xHI+yHNO_(3) rarr NO+I_(2)+H_(2)O` `[((2I^(-)toI_(2)+2e)xx3),((overset(+5)(N)O_(3)^(-)+3Ctooverset(+2)(N)O)xx2)]overset(....(1))(.....(2))` Adding (`1`) and (`2`) `implies 6I^(-)+2NO_(3)^(-) rarr 2NO+3I_(2)` ` 6HI+2HNO_(3) rarr 2NO+3I_(2)+4H_(2)O` `implies x=6`, `y=2` |
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| 604. |
In which of the following, oxidation number of chloride is `+5` ?A. `Cl_(2)O_(7)`B. `ClO_(3)^(-)`C. `ClO^(-)`D. `ClO_(4)^(-)` |
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Answer» Correct Answer - B `ClO_(3)^(-)` `x + 3(-2) = -1` `x - 6 = -1` `x = 6 - 1` `x = +5` |
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| 605. |
The oxidation number of carbon is zero inA. `HCHO`B. `CH_(2)Cl_(2)`C. `C_(12)H_(22)O_(11)`D. `CHCl_(3)` |
| Answer» Correct Answer - A::B::C | |
| 606. |
In a balanced equation `H_(2)SO_(4)+xHI rarr H_(2)S+YI_(2)+zH_(2)O`, the value of `x`, `y`, `z` areA. `x=3`, `y=5`, `z=2`B. `x=4`, `y=8`, `z=5`C. `x=8`, `y=4`, `z=4`D. `x=5`, `y=3`, `z=4` |
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Answer» Correct Answer - C The values of `x`, `y`, `z` are `8`, `4`, `4` respectively hence the reaction is `H_(2)SO_(4)+8HI rarr H_(2)S+4I_(2)+4H_(2)O` |
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| 607. |
`H_(2)O_(2)` reduces `K_(4)Fe(CN)_(6)`A. In neutral solutionB. In acidic solutionC. In non-polar solutionD. In alkaline solution |
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Answer» Correct Answer - B When `H_(2)O_(2)` reduces with `K_(4)[Fe(CN)_(6)]`. It is present in acidic solution. `2K_(4)[Fe(CN)_(6)+H_(2)SO_(4)+H_(2)O_(2) rarr 2K_(3)[Fe(CN)_(6)]+K_(2)SO_(4)+2H_(2)O` |
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| 608. |
Carbon has zero oxidation number inA. `CH_(4)`B. `CH_(3)Cl`C. `C Cl_(4)`D. `CH_(2)Cl_(2)` |
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Answer» Correct Answer - D `CH_(2)Cl_(2)`, `x + 2(+1) + 2(-1) = 0` `x + 2 -2 = 0` `x = 0` |
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| 609. |
`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate toA. `2//3` mole of `MnO_(4)^(-)` and `1//3` mole of `MnO_(2)`B. `1//3` mole of `MnO_(4)^(-)` and `2//3` mole of `MnO_(2)`C. `1//3` mole of `Mn_(2)O_(7)` and `1//3` mole of `MnO_(2)`D. `2//3` mole of `Mn_(2)O_(7)` and `1//3` mole of `MnO_(2)` |
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Answer» Correct Answer - A `MnO_(4)^(2-)` in neutral aqueous medium is disproportionate to `(2)/(3)` mole of `MnO_(4)^(-)` and `(1)/(3)` mole of `MnO_(2)`. |
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| 610. |
Phosphorus has the oxidation state `+3` inA. Ortho phosphoric acidB. Phosphorus acidC. Meta phosphoric acidD. Pyrophosphoric acid |
| Answer» Correct Answer - B | |
| 611. |
`MnO_(4)^(2-)` in neutral aqueous medium is disproportionate toA. `2//3` mole of `MnO_(4)^(-)` and `1//3` mole of `MnO_(2)`B. `1//3` mole of `MnO_(4)^(-)` and `2//3` mole of `MnO_(2)`C. `1//3` mole of `Mn_(2)O_(7)` and `1//3` mole of `MnO_(2)`D. `2//3` mole of `Mn_(2)O_(7)` and `1//3` mole of `MnO_(2)` |
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Answer» Correct Answer - A `MnO_(4)^(2-)` in neutral aqueous medium is disproportionate to `(2)/(3)` mole of `MnO_(4)^(-)` and `(1)/(3)` mole of `MnO_(2)`. |
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| 612. |
Calculate the oxidation number of suplhur in the folowing molecules and ions: `(i) H_(2)SO_(4) , (ii) Na_(2)SO_(4), (iii) SO_(4)^(2-), (iv)S_(2)O_(4)^(2-)` |
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Answer» (i) S in `H_(2)SO_(4)` Let the oxidation number of s in `H_(2)SO_(4)` be x Write the oxidation number of each atom above its symbol `H_(2)^(+1) overset(x)S O_(4)^(-2)` Calculate the sum of the oxidation numbers of all the atoms Since `H_(2)SO_(4)` is a neutral molecule, the sum of oxidation numbers must be zero. (ii) S in `Na_(2)SO_(4)` Let the oxidation number of S in `Na_(2)SO_(4)` be x Write the oxidation number of each atom above its symbol `overset(+1)Na_(2) overset(x)S overset(-2)O_(4)` Calculate the sum of the loxidation numbers of all the atoms 2(+1) +x+4(-2)=2+x-8=x-6 since `Na_(2)SO_(4)` is a neutral molecule, the sum of the oxidation numbers must be zero x-6=0 or x=+6 (iii) S in `SO_(4)^(2)` ion Let the oxidation number of S in `SO_(4)^(2)` ion be x write the oxidatioin numbers of both the atoms above their symbols calculate the sum of the oxidation numbers of both the atoms `x+4(-2)=x-8` Since `SO_(4)^(2)` ion is an anion the sum of the oxidation numbers must be -2 x-8=-2 or x= -2+=+6 (iv) S in `S_(2)O_(4)^(2)`ion "Let the oxidation number of S in `S_(2)O_(4)^(2)` ion be x. Write the oxidation numbers of both atoms above their symbols" `overset(x)S_(2) overset(-2)O_(4)` "Calculate the sum of oxidation number of both the atoms" `2(x)+4(-2)=2x-8` Since `S_(2)O_(4)^(2)` "is an anion the sum of the oxidation numbers must be" -2 `2(x) -8=-2 or 2 x =- 2 + 8 = 6 or x =6//2 =+3` |
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| 613. |
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_(4)^(2-)ltSO_(3)^(2-)ltS_(2)O_(6)^(2-)`B. `SO_(3)^(2-)ltS_(2)O_(4)^(2-)ltS_(2)O_(6)^(2-)`C. `S_(2)O_(4)^(2-)ltS_(2)O_(6)^(2-)ltSO_(3)^(2-)`D. `S_(2)O_(6)^(2-)ltS_(2)O_(4)^(2-)ltSO_(3)^(2-)` |
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Answer» Correct Answer - A `S` in `S_(2)O_(4)^(2-)`, `SO_(3)^(2-)` and `S_(2)O_(6)^(2-)` has `+3`, `+4` and `+5` oxidation numbers. |
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| 614. |
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_(4)^(2-)gtS_(2)O_(6)^(2-)gtSO_(4)^(2-)gtSO_(3)^(2-)`B. `S_(2)O_(6)^(2-)gtSO_(3)^(2-)gtS_(2)O_(4)^(2-)gtSO_(4)^(2-)`C. `SO_(4)^(2-) gt S_(2)O_(6)^(2-)gtSO_(3)^(2-)gtS_(2)O_(4)^(2-)`D. `SO_(3)^(2-) gtSO_(4)^(2-)gtS_(2)O_(4)^(2-)gtS_(2)O_(6)^(2-)` |
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Answer» Correct Answer - C `SO_(3)^(2-) rarr x + (-6)=-2 rArr x= +4` `SO_(4)^(-) rarr x + (-8)=-2rArr x=+6` `S_(2)O_(4)^(2-) rarr 2x+(-8)=-2rArr x=+3` `S_(2)O_(6)^(2-) rarr 2x+(-12)=-2 rArr x= +5` Hence `SO_(4)^(2-) gt S_(2)O_(6)^(2-) gt SO_(3)^(2-) gt S_(2)O_(4)^(2-)` |
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| 615. |
Which of the following is a disproporationation reaction ?A. `Cu_(2)O+2H^(o+)rarrCu+Cu^(2+)+H_(2)O`B. `2CrO_(4)^(2-)+2H^(o+)rarrCr_(2)O_(7)^(2-)+H_(2)O`C. `CaCO_(3)+2H^(o+)rarrCa^(2)+H_(2)O+CO_(2)`D. `Cr_(2)O_(7)^(2-)rarr2overset(ө)OHrarr2CrO_(4)^(2-)+H_(2)O` |
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Answer» Correct Answer - A Statement is self explanatory. |
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| 616. |
In the reaction given in euaqution 8, mention: I. Which reactant is oxidised? To what? II. Which reactant is the oxidiser? III.Which reactant is reduced? To what? IV. What reactant is the reducer? |
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Answer» Correct Answer - A I. Zince metal is oxidised to (II) `Zn^(2+), Pb^(2+)` is the oxidiser. (III) `Pb^(2+)` is reduced to `Pb`,(IV) `Zn` is the reducer. b. `I^(ө)` is oxidised to `I_(2),Fe^(3+)` is oxidiser `Fe^(3+)` is reduced to `Fe^(2+), I^(-)` is the reducer. c. `Na` is oxidised to `na^(+), Cl_(2)` is the oxidiser. `Cl_(2)` is reduced to `Cl^(ө),Na` is the reducer. d. `Mg` is oxidised to `Mg^(+2), Cl_(2)` is the oxidiser. `Cl_(2)` is reduced to `Cl^(ө), Mg` is the reducer. e. `Zn` is oxidised to `Zn^(2+), H^(o+)` is the reducer. `H^(o+)` is reduced to `H_(2), Zn` is the reducer. |
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| 617. |
Which of the following is not a disproportionation reaction ?A. `2PhCHO overset(Al(OEt)_(3))rarrPhCOOCH_(2)Ph`B. `overset(CHO)underset(COOH)"|"+overset(ө)OHrarrunderset(COO^(ө))overset(CH_(2)OH)|+underset(COO^(ө))overset(COO^(ө))|`C. `NaH+H_(2)OrarrNaOH+H_(2)`D. All |
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Answer» Correct Answer - C `(a) and (b)` are disproportionation reactions. `(c )` is an oxidation reaction. `2H^(ө)rarrH_(2)+2e^(-)` |
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| 618. |
Oxidation number of sulphur in `H_(2)SO_(5)` isA. `+2`B. `+4`C. `+8`D. `+6` |
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Answer» Correct Answer - D `H-O-O-overset(O)overset(||)underset(O)underset(||)(S)-O-H` `2xx(+1)+x+3xx(-2)+2xx(-1)` (for `H`) (for `S`) (for `O`) (for `O-O`) `implies 2+x-6-2=0` ` implies x=6` |
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| 619. |
The oxidation state of `Fe` in `Fe_(3)O_(8)` isA. `3//2`B. `4//5`C. `5//4`D. `16//3 |
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Answer» Correct Answer - D `Fe_(3)O_(8)` `3x-16=0implies x=(16)/(3)` |
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| 620. |
The oxidation state of `S` in `H_(2)S_(2)O_(8)` isA. `+2`B. `+4`C. `+6`D. `+7` |
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Answer» Correct Answer - C `H_(2)S_(2)O_(8)`: It is peroxodisulphuric acid. So oxidation state of `S= +6` |
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| 621. |
In which of the following compounds, the oxidation state of transition metal is zero ?A. `CrO_(5)`B. `4//5`C. `FeSO_(4)`D. `Fe(CO)_(5)` |
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Answer» Correct Answer - D `overset(0)(Fe)(overset(0)(CO))_(5)` Oxidation state of `Fe=0` |
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| 622. |
What volume of `3` molar `HNO_(3)` is needed to oxidise `8 g` of `Fe^(3+)`, `HNO_(3)` gets converted to `NO` ?A. `8 mL`B. `16 mL`C. `32 mL`D. `64 mL` |
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Answer» Correct Answer - B Meq.of `HNO_(3)="Meq.of" Fe^(3+)` `["Eq.of" HNO_(3)=M//3]` or `3xx3xxV=(8)/(56)xx1000` `:. V=15.87 mL` |
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| 623. |
What volume of `3` molar `HNO_(3)` is needed to oxidise `8 g` of `Fe^(3+)`, `HNO_(3)` gets converted to `NO` ?A. 8 mlB. 16 mlC. 32 mlD. 64 ml |
| Answer» Correct Answer - B | |
| 624. |
The stoichiometric coefficient of S in the following reaction `H_(2)S + HNO_(3) to NO + S+ H_(2)O` is balanced (in acidic medium):A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - C | |
| 625. |
In which of the following pair oxidation number of Fe is same :-A. `K_(3)Fe(CN)_(6),Fe_(2)O_(3)`B. `Fe(CO)_(5),Fe_(2)O_(3)`C. `Fe_(2)O_(3),FeO`D. `Fe_(2)(SO_(4))_(3),K_(4)Fe(CN)_(6)` |
| Answer» Correct Answer - A | |
| 626. |
Choose the corret statement regarding following reaction `HNO_(2) to HNO_(3) + NO uarr`A. It is an example of disproportionation reactionB. Equivalent weight of `HNO_(3) =(3M)/(2)`C. Equivalent weight of `HNO_(3)=M`D. It is an example of intramolecular redox reaction |
| Answer» Correct Answer - A::B | |
| 627. |
`Cr_(2)O_(7)^(2-) + 14H^(+) to I_(2) + 2Cr^(3+) + 7H_(2)O` Which ions are not in balanced position in above reaction?A. `H^(+)` and `H_(2)O`B. `Cr_(2)O_(7)^(2-)` and `Cr^(3+)`C. `I^(-)` and `I_(2)`D. All are balanced |
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Answer» Correct Answer - D Since charge is not balanced for both the sides, on solving we get: `14H^(+) + Cr_(2)O_(7)^(2-) + 6I^(-) to 3I_(2) + 2Cr_(2)^(3+) + 7H_(2)O` Now, charge on LHS = charge on RHS Hence, `I^(-)` and `I_(2)` are not balanced in the given equation. |
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| 628. |
When `Na_(2)S_(2)O_(3)` is reacted with `I_(2)` to form `Na_(2)S_(4)O_(6)` and Nal then which statement is correct ?A. n-factor for `Na_(2)S_(2)O_(3)` is oneB. n-factor for `I_(2)` is twoC. 2 moles of `Na_(2)S_(2)O_(3)` is reacted with one mole of `I_(2)`D. n-factor for `Na_(2)S_(4)O_(6)` is one |
| Answer» Correct Answer - A::B::C | |
| 629. |
In the balanced equation - `[Zn+H^(+)+NO_(3)^(-)rarr NH_(4)^(-) rarr NH_(4)^(+) +Zn^(+2)+H_(2)O]` coefficient of `NH_(4)^(+)` is :-A. 4B. 3C. 2D. 1 |
| Answer» Correct Answer - D | |
| 630. |
The oxidation number of sulphur in `H_(2)S_(2)O_(8)` is:A. `+2`B. `+6`C. `+7`D. `+12` |
| Answer» Correct Answer - D | |
| 631. |
In the conversion fo ` Br_2` to `BrO_3^(-)` , the oxidation state of `Br` changes from.A. `+1` to `+5`B. 0 to `-3`C. `+2` to `+5`D. 0 to `+5` |
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Answer» Correct Answer - D `Br_(2)rarr BrO_(3)^(-)` For `Br_(2)` oxidation number `=0` For `BrO_(3)^(-)` oxidation number, `x+(-6)=-1 rArr x=+5` |
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| 632. |
The standard reduction potential of three metallic cations x,y and z are +0.52,-3.03 and -1.18V respectively .The order of reducing power is:A. `YgtZgtX`B. `XgtYgtX`C. `ZgtYgtX`D. `ZgtXgtY` |
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Answer» Correct Answer - A is the correct answer |
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| 633. |
`I_(2)+KIrarrKI_(3)` In the above reaction :-A. Only oxidation taken placeB. Only reduction takes placeC. Both the aboveD. None of the above |
| Answer» Correct Answer - C | |
| 634. |
Match List - I (compound) with list - II (Oxidation state of N) and select the correct answer using the codes given below the list :- `{:(" List - I"," List - II"),((A)KNO_(3),(a)-1//3),((B)HNO_(2),(b)-3),((C)NH_(4)Cl,(c)0),((D)NaN_(3),(d)+3),(,(e) +5):}`A. `{:(A,B,C,D),(e,d,b,a):}`B. `{:(A,B,C,D),(e,b,d,a):}`C. `{:(A,B,C,D),(d,e,a,c):}`D. `{:(A,B,C,D),(b,c,d,e):}` |
| Answer» Correct Answer - A | |
| 635. |
X gm of metal gave Y gm of its oxide, so equivalent mass of metal is :A. `((X)/(Y-X))xx8`B. `((Y-X)/(X))xx8`C. `((Y+X)/(X))xx8`D. `(X)/(Y)xx8` |
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Answer» Correct Answer - a |
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| 636. |
In the conversion fo ` Br_2` to `BrO_3^(-)` , the oxidation state of `Br` changes from.A. 0 to 5B. 1 to 5C. 0 to `-3`D. 2 to 5 |
| Answer» Correct Answer - A | |
| 637. |
Which fo the following will release a gaseous product from `Pb_3O_4 `?A. `HCl`B. `HNO_3`C. Bitg (1) and (2)D. Neither (1) nor (2) |
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Answer» Correct Answer - A ` Pb_3O_4 + 8 HCl rarr 3 PbCl_2+ Cl_(2) +4 H_2O` `Pb_3O_4` is actually a stoichiometric mixture fo 2 mol fo `PbO` and ` 1 ` mol of ` PbO_2`. In `PbO_2`, lead is present in ` +4` oxidation state (an unstable state) while in ` PbO`, lead is present in state (an unstable state) while in `PbO`, lead is present in `+2` oxidation state (a stable state). Since lead hds a strong tendency to change from ` +4` state to` +2` state,`PbO_(2)` can act as an oxidizing agent (oxidant). Therefore, it oxidizes `Cl^(-)` ion fo ` HCl` into chlorine. On the other hand, `PbO` is a basic oxide. Therefore, it can react with hydrochloric acid `(HCl)` to form salt `(PbCl_2)` and water. Hence, the above reaction can be split into two reactions namely : (a) Redox reaction `overset(+4)(Pb) O_(2)+4Hoverset(-1)(Cl) rarr overset(+2)(Pb)Cl_(2)+overset(0)(C)l_(2)+2H_(2)O` (b) Acid-base reaction `2PbO + 4HCl rarr 2PbCl_2 +2H_2O` Since ` HNO_3` ( N is present in the highest oxidation state, +5) itself is an oxidizing agent, there is no reaction between ` PbO_2` and ` HNO_3`, However, an acid-base reaction takes place between ` PbO` and ` HNO_3` : ` 2PbO+4HNO_3 rarr 2Pb (NO_3)_2 +2 H_2O` Thus, 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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| 638. |
`HNO_(2)` acts as an oxidant with which one of the following reagent :-A. `KMnO_(4)`B. `H_(2)S`C. `K_(2)Cr_(2)O_(7)`D. `Br_(2)` |
| Answer» Correct Answer - B | |
| 639. |
The incorrect order of decreasing oxidation number of S in compound is :A. `H_(2)S_(2)O_(7)gtNa_(2)S_(4)O_(6)gtNa_(2)S_(2)O_(3)gtS_(8)`B. `H_(2)SO_(5)gtH_(2)SO_(3)gtSCl_(2)gtH_(2)S`C. `SO_(3)gtSO_(2)gtS_(8)gtH_(2)S`D. `H_(2)SO_(4)gtSO_(2)gtH_(2)SgtH_(2)S_(2)O_(8)` |
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Answer» Correct Answer - d |
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| 640. |
Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate?A. `25mL` of `0.1M KMnO_(4)`B. `25mL` of `0.2M KMnO_(4)`C. `25mL` of `0.6M KMnO_(4)`D. `15mL` of `0.1M KMnO_(4)` |
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Answer» Correct Answer - d |
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| 641. |
Which statement is wrong?A. Oxidation number of oxygen is o`+1` peroxidesB. Oxidation number of oxygen is `+2` in oxygen difluorideC. Oxidation number of oxygen is `-(1)/(2)` in superoxidesD. Oxidation number of oxygen is `(-2)` in most of its compounds |
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Answer» Correct Answer - a |
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| 642. |
6 g of a mixture of ammonium sulphate and ammonium chloride was made up to 1000cc with water . 25cc of this solution was boiled with 50cc of `M/(10)` sodium hydroxide until no more ammonia waws evolved and it was then found that the excess of sodium hydroxided required 24.3 cc of `N/(10)` hydrochloric acid for neutrilisation. What was the mass percentage of ammonium chloride in the mixture? |
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Answer» Correct Answer - 56 |
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| 643. |
Assertion: `NH_(4)NO_(3)` on heating give `N_(2)O`. Reason: `NH_(4)NO_(3)` on heating shows disproportionation.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 - C `NH_(4)NO_(3) rarr N_(2)O+2H_(2)O` (intermolecular redox) |
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| 644. |
In the reaction ` 2K MnO_4 + 16 HCl rarr 5Cl_2 + 2MnCl_2 + 2 KCl+ 8H_2O` the reduced product is .A. `Cl_2`B. ` KCl`C. ` H_2O`D. `MnCl_2` |
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Answer» Correct Answer - D In the given redox reaction, ` KMnO_4` is reduced to `MnCl_2` while ` HCl` is oxidized to ` Cl_2`: `Koverset(+7)(M)nO_(4)+Hoverset(-1)(C)l rarr overset(0)(C)l_(2)+overset(+2)(M)nCl_(2)+KCl+H_(2)O` Thus, the reduced produce is `MnCl_2`. |
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| 645. |
Which of the following represents redox reactions?A. `Cr_(2)O_(7)^(2-)+2overset(ө)OHrarr2CrO_(4)^(2-)+H_(2)O`B. `SO_(3)^(2-)+H_(2)O+I_(2)rarrSO_(4)^(2-)+2I^(ө)+2H^(o+)`C. `Ca(OH)_(2)+Cl_(2)rarrCa(Ocl)_(2)+CaCl_(2)`D. `PCl_(5)rarrPCl_(3)+Cl_(2)` |
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Answer» Correct Answer - B::C::D `(b), (c ), and (d)` are redox reactions. In `(a)`, there in no change in the oxidation state `(+6)` of `Cr`. |
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| 646. |
Which of the following represents redox reactions?A. `Cr_(2)O_(7)^(2-)+2OH^(-)rarrCrO_(4)^(2-)+H_(2)O`B. `SO_(3)^(2-)+I^(-)rarrI_(2)+SO_(4)^(2-)`C. `Ca(OH)_(2)+Cl_(2)rarrCa(ClO)_(2)+CaCl_(2)`D. `PCl_(5)rarrPCl_(3)+Cl_(2)` |
| Answer» Correct Answer - B::C::D | |
| 647. |
In the redox reaction `xMnO+yPbO_(2)+zHNO_(3) rarr HMnO_(4)+Pb(NO_(3))_(2)+H_(2)O`A. `x=2`, `y=5`, `z=10`B. `x=2`, `y=7`, `z=8`C. `x=2`, `y=5`, `z=8`D. `x=2`, `y=5`, `z=5` |
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Answer» Correct Answer - A The balanced chemical equation is `underset(x)underset(uarr)(2MnO)+underset(y)underset(uarr)(5PbO_(2))+underset(z)underset(uarr)(10HNO_(3)) rarr 2HMnO_(4)+5Pb(NO_(3))_(2)+4H_(2)O` |
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| 648. |
For the reaction, `2KClO_(3)rarr2KCl+3O_(2)`, which statements `(s)` are (are) correct?A. It is disproprtionationB. It is intramolecular redox changeC. `Cl` atoms are reducedD. Oxygen atoms are oxidised |
| Answer» Correct Answer - B::C::D | |
| 649. |
Balance the chemical equation by the oxidation number method. (ii) `P + HNO_(3) to HPO_(3) + NO + H_(2)O` |
| Answer» `3P+ 5HNO_(3) to 3HPO_(3) + 5NO + H_(2)O` | |
| 650. |
The coefficients w, x,y,z in the reaction w `Cr_(2)O_(7)^(2-) + xFe^(2+) to y Cr^(3+) + z Fe^(3+) + H_(2)O`A. w-1, x-2, y-6, z-6B. w-6, x-1, y-2, z-4C. w-1, x-6, y-2, z-6D. w-1, x-2, y-4, z-6 |
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Answer» Correct Answer - C `Cr_(2)O_(7)^(2-) + 6Fe^(2+) + 14H^(+) to 2Cr^(3+) + 6Fe^(3+) + 7H_(2)O` |
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