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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.
| 1101. |
Justify giving reaction that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant. |
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Answer» `F_(2)` can oxidize `Cl^(-)` to `Cl_(2) , Br^(-)` to `Br_(2)` and `l^(-)` to `I_(2)` as : `F_(2 (aq)) + 2 Cl_((s))^(-) to 2 F_((aq)) + Cl_(g)` `F_(2 (ag)) + 2Br_(aq)^(-) to 2 F_(aq)^(-) + Br_(2 (l))` `F_(2 (aq)) + 2I_(aq)^(-) to 2 F_(aq)^(-) + I_(2 (s))` On the other hand , `Cl_(2) , Br_(2)` and `I_(2)` cannot oxidize `F^(-)` to `F_(2)` . The oxidizing power of halogens increases in the order of `l_(2) lt Br_(2) lt Cl_(2) lt F_(2)` . Hence , fluorine is the best oxidant among halogens . Hl and HBr can reduce `H_(2)SO_(4)` to `SO_(2)` , but HCl and HF cannot . Therefore , Hl and HBr are stronger than HCl and HF. `2HI + H_(2)SO_(4) to I_(2) + SO_(2) + 2H_(2)O` `2HBr + H_(2)SO_(4) to Br_(2) + SO_(2) + 2H_(2)O` Again `l^(-)` can reduce `Cu^(2+)` to `Cu^(+)` , but `Br^(-)` cannot . `4I_(aq)^(-) + 2Cu_(aq)^(2+) to Cu_(2)I_(2 (s))_(2 (aq))` Hence , hydroiodic acid is the best reductant among hydrohalic compounds . Thus , the reducing power of hydrohalic acids increases in the order of `HF lt HCl lt HBr lt Hl`. |
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| 1102. |
Which of the following can act both as oxidising agents and reducing agents?A. `HNO_(2)`B. `H_(2)O_(2)`C. `H_(2)S`D. `SO_(2)` |
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Answer» Correct Answer - A::B::D These can undergo increase as well as decrease in oxidation number |
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| 1103. |
Among the following , what is the total number of speices which are very good oxidising agents/reducing agents/neither oxidising nor reducing ones ? a. `F_(2)` b. `F^(ө)` c. `Na` d. `Na^(o+)` e. `MnO_(4)^(ө)` f. `I^(ө)` `Cl^(ө)` h. `Ce^(4+)` i. `Cr_(2)O_(7)^(2-)` j. `CrO_(4)^(2-)` k. `HNO_(3)` l. `Fe^(2+)` |
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Answer» Correct Answer - C Species which are netiher oxidising agent nor reducing agent are. b. `F^(ө) , d. `Na^(o+)` , g. `Cl^(o+)` |
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| 1104. |
While sulphate dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why? |
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Answer» In sulphur dioxide `(SO_(2))` , the oxidation number (O.N) of S is +4 and the range of the O.N. that S can have is from + 6 to -2 . Therefore , `SO_(2)` can act as an oxidising as well as a reducing agent . In hydrogen peroxide `(H_(2)O_(2))`, the O.N. of O is –1 and the range of the O.N. that O can have is from 0 to –2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, `H_(2)O_(2)` can act as an oxidising as well as a reducing agent In ozone `(O_(3))`, the O.N. of O is zero and the range of the O.N. that O can have is from 0 to –2. Therefore, the O.N. of O can only decrease in this case. Hence, `O_(3)` acts only as an oxidant. In nitric acid `(HNO_(3))`, the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to –3. Therefore, the O.N. of N can only decrease in this case. Hence, `HNO_(3)` acts only as an oxidant. |
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| 1105. |
Justify that the following reaction are redox reactions: a. `CuO(s)+H_(2)(g)rarr Cu(s)+H_(2)O(g)` b. `Fe_(2)O_(3)(s)+3CO(g)rarr 2Fe(s)+3CO_(2)(g)` c. `4BCl_(3)(g)+3LiAlH_(4)(s) rarr 2B_(2)H_(6)(g)+3LiCl(s)+3AlCl_(3)(s)` d. `2K(s)+F_(2)(g)rarr 2K^(o+)F^(Θ)(s)` e. `4NH_(3)(g)+5O_(2)(g)rarr 4NO(g)+6H_(2)O(g)` |
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Answer» (a) `CuO_((s)) + H_(2 (g)) to Cu_(s) + H_(2)O_(g)` Let us write the oxidation number of each element involved in the given reaction as : `overset(+2)(Cu)overset(-2)(O)_(s) + overset(0)(H_(2 (g))) to overset(0)(C)u_(s) + overset(+1)(H_(2)) overset(-2)(O)_(g)` Here , the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also , the oxidation number of H increases from 0 in `H_(2)` to +1 in `H_(2)O` i.e., `H_(2)` is oxidized to `H_(2)O` . Hence , this reaction is a redox reaction . (b) `Fe_(2)O_(3(s)) + 3 CO_((g)) to 2 Fe_((s)) + 3CO_(2 (g))` Let us write the oxidation number of each element in the given reaction as : `overset(+3)(F)e_(2)overset(-2)(O)_(3 (s)) + 3overset(+2)(C)overset(-2)(O)_((g)) to 2 overset(0)(Fe)_((s)) + 3 overset(+4)(C)overset(-2)(O_(2)(g))` Here , the oxidation number of Fe decreases from +3 in `Fe_(2)O_(3)` to 0 in Fe i.e., `Fe_(2)O_(3)` is reduced to Fe . On the other hand , the oxidation number of C increases from +2 in CO to +4 in `CO_(2)` i.e., CO is oxidized to `CO_(2)` . Hence , the given reaction is a redox reaction . (c) The oxidation number of each element in the given reaction can be represented as : `4 overset(+3)(B) overset(-1)(C)l_(3 (g)) + 3 overset(+1)(Li)overset(+3)(Al)overset(-1)(H)_(4(s)) to 2 overset(-3)(B_(2))overset(+1)(H)_(6 (g)) + 3 overset(+1)(Li)overset(-1)(C)l_((s)) + 3 overset(+3)(A)l overset(-1)(C)l_(3 (s))` In this reaction, the oxidation number of B decreases from +3 in `BCl_(3)` to `-3` in `B_(2)H_(6)` i.e., `BCl_(3)` is reduced to `B_(2)H_(6)` . Also , the oxidation number of H increases from `-1` in `LiAlH_(4)` to +1 in `B_(2)H_(6)` i.e., `LiAlH_(4)` is oxidized to `B_(2)H_(6)`. Hence, the given reaction is a redox reaction. (d) `2K_((s)) + F_(2 (g)) to 2K^(+) F_((s))^(-)` The oxidation number of each element in the given reaction can be represented as: `2 overset(0)(K)_((s)) + overset(0)(F)_(2 (g)) to 2 overset(+1)(K^(+)) overset(-1)(F^(-))_((s))` In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in `F_(2)` to `–1` in KF i.e., `F_(2)` is reduced to KF . Hence , the above reaction is a redox reaction (e) ` 4NH_(3 (g)) + 5 O_(2 (g)) to 4 NO_((g)) + 6 H_(2)O_((g))` The oxidation number of each element in the given reaction can be represented as: `4 overset(-3)(N) overset(+1)(H)_(3 (g)) + 5 overset(0)(O)_(2 (g)) to 4 overset(+2)(N) overset(-2)(O)_((g)) + 6 overset(+1)(H_(2)) overset(-2)(O)_((g))` Here, the oxidation number of N increases from –3 in `NH_(3)` to +2 in NO. On the other hand, the oxidation number of `O_(2)` decreases from 0 in `O_(2)` to` –2` in NO and `H_(2)O` i.e., `O_(2)` is reduced. Hence, the given reaction is a redox reaction. |
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| 1106. |
Oxidizing agents are spectes thatA. oxidize other substancesB. contain atoms that are reducedC. gain (or appear to gain ) electronsD. exhibit any one of the above characteristics |
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Answer» Correct Answer - D Oxidizing agents oxidize the others and get reduced by gaining electron by decreasing oxidation number. |
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| 1107. |
Which fo the follwing compounds fo oxygen has fractional oxdation number ?A. `CaO`B. `OF_2`C. `RbO_2`D. `Na_2O_2` |
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Answer» Correct Answer - C Oxygen has an oxidation number of `-1//2` is superoxides (such as `RbO_2`) which contain the superoxide ion `(O_2^(-))`. Oxidation number is a formal concept adopted for our convenience. The numbers are determined by relying on rules. These rules can result in a fractional oxidation number . This does not mean that electronic charges are split . |
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| 1108. |
Oxidation state of Fe in `FeO_(4)` is:A. `3//2`B. `4//5`C. `5//4`D. `8//3` |
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Answer» `overset(x)Fe_(3)O_(4)^(-2)` `3x+4(-2)=0 or x=8//3` |
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| 1109. |
Consider the following unbalanced redox reaction: `H_(2)O+AX+BYrarrHA+OY+X_(2)B` The oxidation number of `X` is `-2` and niether `X` nor water is involved in the redox process. The positive oxidation states of `B` and `Y` in `BY` are respectively,A. `+1, -1`B. `+2,-2`C. `+3,-3`D. All of these |
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Answer» Correct Answer - D In `BY`, the oxidation state of `B=lt+4` Oxidation state of `y=lt+2`. If the oxidation state of `B is +1` and that of `Y is -1` then both will be oxidised. If the oxidation state of `B is +2` and that of `Y is -2`, then both will be oxidised. If the oxidation state of `B is +3` and that of `Y is -3`, then both will be oxidised. |
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| 1110. |
`IO_(3)^(-)+I^(-)+H^(+) to`A. 5,1,6B. 1,5,6C. 6,1,5D. 5,6,1 |
| Answer» Correct Answer - A | |
| 1111. |
Calculate the oxidation number of the underlined elements in the following compounds: a. `Kunderline(Mn)O_(4)` , b. `underline(Cr)O_(2)Cl_(2)` , c. `Naunderline(I)O_(3)` |
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Answer» Correct Answer - A a. `+7, b. +6`, c. `+5` |
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| 1112. |
Assertion (A): The two `Fe` atoms in `FeO_(3)O_(4)` hace different oxidation numbers. Reason (R ): `Fe^(2+)` ions decolourise `KMnO_(4)` solution.A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct but `(R )` is incorrect.D. If `(A)` and `(R )` are incorrect. |
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Answer» Correct Answer - B Assertion: `Fe_(3)O_(4)-=(overset(+2)(FeO).overset(+3)(Fe_(2)O_(3)))` The oxidation states of `Fe` in `FeO` and `Fe_(2)O_(3)` are `+2` and `+3`. Reason : `Fe^(2+)+underset(("Pink"))(MnO_(4)^(ө))rarrFe^(3+)underset(("Colourless"))(Mn^(2+))` Both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation for `(A)`. |
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| 1113. |
Which one of the following compounds does not decolourise an acidified aqueous solution of `KMnO_(4)`A. `FeCl_(3)`B. `FeSO_(4)`C. `SO_(2)`D. `H_(2)O_(2)` |
| Answer» Correct Answer - A | |
| 1114. |
Calculate the oxidation number of the underlines elements: a. `underline(P)H_(3)` , b. `underline(M)gO` , c. `Hunderline(N)O_(3)` , d. `H_(3)underline(P)O_(4)` |
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Answer» Correct Answer - A a. `-3`, b. `+4`, c. `+5`, d. `+5` |
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| 1115. |
Assign oxidation numbers to the elements in the following ionic compounds. a. `NaBr` , b. `MgO` , c. `AlF_(3)` |
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Answer» Correct Answer - A a. +`, -1` , b. `+2, -2` , c. `+3, -1` |
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| 1116. |
To an acidic solution of an anion, a few drops of `Kmno_(4)` solution are added. Which of the following, if present, will not decolourise the `KMnO_(4)` solution?A. `CO_(3)^(2-)`B. `NO_(2)^(ө)`C. `S^(2-)`D. `Cl^(ө)` |
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Answer» Correct Answer - A Oxidation state of `C` in `CO_(3)^(2-)` is `+4`, which is maximum. So, it will not be oxidised. |
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| 1117. |
Which of the following is not a disproprotionation reaction? I. `NH_(4)NO_(3)overset(Delta)rarrN_(2)O+H_(2)O` II. `P_(4)overset(Delta)rarrPH_(3)+HPO_(2)^(ө)` III. `PCl_(5)overset(Delta)rarrPCl_(3)+Cl_(2)` IV. `IO_(3)^(ө)+I^(ө)rarrI_(2)`A. `I,II`B. `I,III,IV`C. `II,IV`D. `I,III` |
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Answer» Correct Answer - B Reduction `II` is disproportionation, while I, III, and IV are not. |
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| 1118. |
Assertion: In the titration of `Na_(2)CO_(3)` with `HCl` using methyl orange indicator, the volume of acid required is twice that of the acid required using phenolphthalein as indicaton. Reason: Two moles of `HCl` are required for the complete neutralisation of one mole of `Na_(2)CO_(3)`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-8B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-8C. Statement-1 Is True, Statement-2 is False .D. Statement-1 is True False, Statement-2 is True. |
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Answer» Correct Answer - b |
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| 1119. |
Which of the following reaction are example of redox reactions?A. `KMnO_(4)+H_(2)SO_(4)+COto`B. `K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+HCl to`C. `CuSO_(4)+KCN("excess") to`D. `CuSO_(4)+KI("excess") to` |
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Answer» Correct Answer - a,b,c,d |
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| 1120. |
The equivalent weight fo `Na_2S_2O_3` in the reaction `2Na_2S_2O_3 + I_2 rarr Na_2 S_4O_6 + 2 NaI` will be .A. `M//2`B. `M`C. `M//0.5`D. `M//8` |
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Answer» Correct Answer - B `overset(+2)(S_(2))O_(3)^(2-) rarr overset(+2.5)(S_(4))O_(6)^(2-)` Sincec there are two S atoms in `S_(2)O_(3)^(2-)`, we must calculate the change in oxidation number for two S atoms only: ` 2 (+2) rarr 2 ( +2.5)` or ` +4 rarr +5` `:.` Equivalent weight `= (M)/1`. |
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| 1121. |
In this reaction: `S_(2)O_(8)^(2-)+2I^(-) to 2SO_(4)^(2-)+I_(2)`A. Oxidation of iodide into iodine takes placeB. Reduction of iodine into iodide takes placeC. Both oxidation and reduction of iodine takes placeD. All of these |
| Answer» Correct Answer - A | |
| 1122. |
Oxidation state of `P` in `H_(4)P_(2)O_(5), H_(4)P_(2)O_(6), H_(4)P_(2)O_(7)` are respectivelyA. `+3, +5,+4`B. `+5,+3,+4`C. `+5,+4,+3`D. `+3,+4,+5` |
| Answer» Correct Answer - D | |
| 1123. |
Oxidation state of `P` in `H_(4)P_(2)O_(5), H_(4)P_(2)O_(6), H_(4)P_(2)O_(7)` are respectivelyA. `+3,+5,+4`B. `+5,+3+4`C. `+5,+4,+3`D. `+3,+4,+5` |
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Answer» Correct Answer - D Let O.N of P in `H_(4) P_(2)O_(5)=x` `4(+1)+2x +5 (-2)=0` `:. x=+3` O.N of P in `H_(4)P_(2)O_(6)=x` `4(+1)+2x+6(-2)=0` `:. x= + 4` O.N of P in `H_(4)P_(2)O_(7)=x` `4(+1)+2x+7(-2)=0` `:. x=+5`. |
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| 1124. |
When `KMnO_(4)` acts as an oxidising agnet and ultimetely from `MnO_(4)^(2-), MnO_(2), Mn_(2)O_(3)`, and `Mn^(2+)`, then the number of electrons transferred in each case, respectively, areA. ` 1,3,4,5`B. ` 3,5,7,1`C. ` 1,5,3,7`D. ` 4,3,1,5` |
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Answer» Correct Answer - A Let x be the oxidaton number of Mn. ` MnO_4^(2-) rArr x + 4 (-2) =-1` ` x-8 =-1` `x = + 7` `MnO_4^(2-) rArr x +4 (-2) =-2` ` x-8= -2` ` x= +6` `MnO_2 rArr x + 2(-2) =0` ` x-4 =0` `x rArr +4` `Mn_2O_3 rArr 2x +3 (-2) =0` ` 2x -6 =0` ` x=+3` `Mn^(2+) rArr x = +2` `:. overset(+7) (M)nO_(4)^(-)+e^(-) rarr overset(+6)(M)nO_(4)^(2-)` `+3e^(-) rarr overset(+4)(M)nO_(2)` `+4e^(-)rarroverset(+3)(M)n_(2)O_(3)` `+5e^(-) rarr overset(+2)(M)n^(2+)` |
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| 1125. |
When `KMnO_(4)` acts as an oxidising agnet and ultimetely from `MnO_(4)^(2-), MnO_(2), Mn_(2)O_(3)`, and `Mn^(2+)`, then the number of electrons transferred in each case, respectively, areA. 4, 3, 1, 5B. 1, 5, 3, 7C. 1, 3, 4, 5D. 3, 5, 7, 1 |
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Answer» Correct Answer - c |
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| 1126. |
When `KMnO_(4)` acts as an oxidising agnet and ultimetely from `MnO_(4)^(2-), MnO_(2), Mn_(2)O_(3)`, and `Mn^(2+)`, then the number of electrons transferred in each case, respectively, areA. `4,3,1,5`B. `1,5,3,7`C. `1,3,4,5`D. `3,5,7,1` |
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Answer» Correct Answer - C `{:(Mn^(7+),+,e,rarr,Mn^(6+),),(mn^(7+),+,3e,rarr,Mn^(4+),),(2Mn^(7+),,,overset(+8e)rarr,(Mn^(3+))_(2)","),(Mn^(7+),+,5e,rarr,Mn^(2+)):}` |
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| 1127. |
When `KMnO_(4)` acts as an oxidising agnet and ultimetely from `MnO_(4)^(2-), MnO_(2), Mn_(2)O_(3)`, and `Mn^(2+)`, then the number of electrons transferred in each case, respectively, areA. `4`,`3`,`1`,`5`B. `1`,`5`,`3`,`7`C. `1`,`3`,`4`,`5`D. `3`,`5`,`7`,`1` |
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Answer» Correct Answer - C Number of `e^(-)` transferred in each case is `1`, `3`, `4`, `5`. |
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| 1128. |
Assertion (A): `H_(2)O_(2)` acts only as an oxidising agnet. `H_(2)O_(2)rarrH_(2)O+O` Reason (R ): All peroxides behave as oxidising agnets only.A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct but `(R )` is incorrect.D. If both `(A)` and `(R )` are incorrect. |
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Answer» `H_(2)O_(2)` acts both as an oxidising and a reducing agent. `(H_(2)O_(2)rarrO_(2)+2H^(o+)+2e^(-))` (Oxidation, acts as a reducing agent) `(2e^(-)+2H^(o+)+H_(2)O_(2)rarr2H_(2)O)`, (Reduction, acts as oxidising agent) In `H_(2)O_(2)`, the oxidation number of `O` is `-1`. `O` can have a minimum oxidation number of `-2` and a maximum oxidation number of Zero. The oxidation number can either decrease from `-1` to `-2` or can increase from `-1` to zero. Correct `(R )`: All peroxides can act either as an oxidising agent or as a reducing agent or both. |
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| 1129. |
Assertion (A): `KMnO_(4)` is a stronger oxidising agent than `K_(2)Cr_(2)O_(7)`. Reason (R ): This is due to increasing stability of the lower species to which they are reduced.A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct but `(R )` is incorrect.D. If `(A)` and `(R )` are incorrect. |
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Answer» Both `(A)` and `(R )` are correct, and `(R )` is the correct explanation for `(A)` In `KMnO_(4)`, the oxidation state of `Mn is +7`. ( highest oxidation state) and is reduced to `Mn^(2+)`, whereas in `K_(2)Cr_(2)O_(7)`, the oxidation state of `Cr +6` ( highest oxidation state) and is reduced to `Cr^(3+)`. `Mn^(2+)` is more stable than `Cr^(3+)`. Hence, `KMnO_(4)` is a stronger oxidising agnet than `K_(2)Cr_(2)O_(7)`. |
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| 1130. |
In the disproportionation reaction, `3HCIO_(3) rarr HCIO_(4) +CI_(2)+2O_(2)+H_(2)O` What is the equivalent mass of the oxidising agnet ? |
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Answer» `overset(x)CIO_(3)rarr overset(0)CI2, x+2(-3)=-1or x=-1+6=5` change in oxidation number= +5 Equivalent mass of `CIO_(3) (Molecular mass)/(Oxidation number change")`=`(35..5+48)/(5)` `= (83.56)/(5)=16.7` |
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| 1131. |
In the determination of the molar mass of a solid acid by titrating it with a standardized base, which procedural error will yield a molar mass that is smaller than the actual value?A. Adding the standardized base to a buret containing drops of waterB. Dissolving the weighed solid acid in twic the recommended volume of waterC. Using half as many drops of indicator as suggestedD. Weighing out half of the recommended mass of solid acid |
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Answer» Correct Answer - a |
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| 1132. |
A NaOH solution is to be standarized by titrating its against a known mass of potassium hydrogen phthalate. Which procedure will give a molarity of NaOH that is too low ?A. Deliberately weighing one half the recommended amount of potassium hydrogen phtalate.B. Dissoving the potassium hydrogen phtalate in more water than is recommeded.C. Neglecting to fill the tip of the buret with NaOH solution before titrating.D. Lossing some of the potassium hydrogen phthalate solution from the flask before |
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Answer» Correct Answer - c |
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| 1133. |
Kinetics can be studied by titration usingA. `Na_(2)S_(2)O_(3)`B. `KmMnO_(4)`C. `H_(2)C_(2)O_(4)`D. |
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Answer» Correct Answer - a |
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| 1134. |
Which of the following changes requires a reducing agent ?A. `H_(3)AsO_(3)rarrHAsO_(4)^(2-)`B. `BrO_(3)^(ө)rarrBrO^(ө)`C. `CrO_(4)^(2-)rarrCr_(2)O_(7)^(2-)`D. `Al(OH)_(3)rarr[Al[OH)_(4)]^(ө)` |
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Answer» Correct Answer - B `4e^(-)+BrO_(3)^(ө)rarrBrO^(ө)` `x-6= -1, x-2=-1` `x=5, x=+1` The reduction undergoes reduction, i.e., it acts as oxidising agent, so it requires a reducing agent. |
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| 1135. |
In the following reactions: a. `3MnO_(2)+4Alrarr3Mn+2Al_(2)O_(3)` b. `2MnO_(4)^(ө)+16H^(o+)+10Cl^(ө)rarr2Mn^(2+)+5Cl_(2)+8H_(2)O` Which species is reduced and which is oxidised? |
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Answer» Correct Answer - A::B::C::D a. `MnO_(2)` is reduced and aluminium is oxidised. b. `MnO_(4)^(ө)` is reduced, and `Cl^(ө)` is oxidised. |
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