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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.
| 751. |
Oxidation number of carbon in `CH_(3)-Cl` isA. `-3`B. `-2`C. `-1`D. `0` |
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Answer» Correct Answer - B `overset(*)(C )H_(3)-Cl` `x+3(+1)+(-1)xx1=0` `x+3-1=0`, `x+2=0` `x=-2` |
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| 752. |
Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?A. `3`B. `6`C. `2`D. `7` |
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Answer» Correct Answer - A `overset(+6)(C)r_(2)O_(7)^(2-)+6e^(-) rarr 2Cr^(3+)` According to the reduction half-reaction, `1` mol of ` K_Cr_2O_7` gains ` 6` mol of electrons. ` 2I^(-) rarr I_2 = 2e^(-)` To produce `1` mol of `I_2` , 2 mol of electrons are lost. Thus ,by gaining ` 6` mol of electrons , 1 mol of ` K-2Cr_2 O_7` will liberate `3` mol of ` I_2` ` Cr_2O_7^(2-) + 14H^(+) = 6I^(-) rarr 2Cr^(3+) + 7H_2O + 3I_2`. |
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| 753. |
The oxidation state of `S` in `H_(2)S_(2)O_(8)` isA. `+6`B. `+7`C. `+8`D. 0 |
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Answer» Correct Answer - A `H_(2)S_(2)O_(8)` `H-O-underset(O)underset(||)overset(O)overset(||)S-O-O-underset(O)underset(||)overset(O)overset(||)S-O-H` Two oxygen atoms are involved in peroxide linkage, hence oxidation number will be -1 (each). `H_(2)S_(2)O_(8)=underset(H)(2(+1))+underset(S)(2(x))+underset(O-O)(2(-1))+underset(O)(6(-2))=0` `x=+6` |
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| 754. |
Using stock natation, represent the following compunds: (i) `HAuCI_(4)" (ii) TI_(2)O" "(iii)FeO" (iv)Fe_(2)O_(3) (v) Bi(NO_(3))_(3)" (vi) HgCI_(2)` |
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Answer» The stock notation can be represented by kinowing the oxidation of the metal atom in each compund. `{:((i) , HAuCl_(4) : , + 1 + x + 4 (-1) = 0 , , x = + 3 , , HAu(III)Cl_(4)) , ((ii) , Tl_(2)O : , 2x + (-2)= 0 , , x = +1 , , Tl(I) O), ((iii), FeO : ,x+(-2)=0, ,x=+2, ,Fe(II)O),((iv),Fe_(2)O_(3): ,2x+2(-3)=0, ,x=+3, , Fe(III)O_(3)),((v), Cul: ,x+(-12)=0, ,x=+1, , CU(I)I),((vi),CUO: ,x+(-2)=0, ,x=+2, ,Cu(II)O),((vii),MnO: ,x+(-2)=0, ,x=+2, ,Mn(II)O),((viii), MnO_(2): , x+2(-2)=0 , x=+4 , Mn(IV)O_(2)):}` |
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| 755. |
Assertion : Conversion of potassium ferrocyanide to potassium ferricyanide is an oxidation process. Reason : Oxidation is the addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 756. |
A salt bridge contains `"_____"` and agar-agar. (i) a statured solution fo ` HCl` (ii) A saturated solution fo `KNO_3` (iii) a saturated solution of ` NH_4NO_3`.A. (i), (ii), (iii)B. (i), (ii)C. (ii), (iii)D. (iii) only |
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Answer» Correct Answer - A To complete the electric circuit of a galvanic cell, the solutions must be connected by a conducting medium through the cations and anions can move. This requirement is satisfied by a salt bridge, which in its simplest form is an inverted U tube containing an inert electrolyte such as ` KCl, KNO_3,` or `NH_4NO_3`, whose ions will not react with the other ions in solution or with the electrodes. During the course of the overall redox reaction, electrons flow extrenally from the anode through the wire and voltmeter to the cathode . In the solution, the cation move toward the cathode, while the anions move in the opposite dirction, toward the anode . |
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| 757. |
The function of a salt bridge is to .A. eliminate the impurities present in the electrolyteB. eliminate liquid-junction potential where the ions are present in excess at the junctionC. decrease the cell potential at the negative electrodeD. increase the cell potential at the positive electrode |
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Answer» Correct Answer - B A salt bridge can be any medium through which ions can slowly pass . A salt bridge can be made by bending a piece of glass tune into the shape of a 'U' , filling it with a hot saturated salt `//5%` agar-agar solution, and allowing it to cool. The cooled mixture 'sets' to the consistency of firm gelatin. As a result, the solution does are still able to move. A salt bridge serves three functions : (a) It allows electrical contact between the two solutions. (b) It prevents mixing of the electrode solutions. (c ) It maintains the electrical neutrality in each half-cell as ions flow into and out of the salt bridge. |
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| 758. |
Assertion: Fluorine exists only in `-1` oxidation state. Reason: Fluorine has `2s^(2)2p^(5)`configuration.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 - B It is correct that fluorine exists only in `-1` oxidation state because it has `1S^(2)2p^(5)` electronic configuration and thus shows only `-1` oxidation state in order to complete its octet. Hence, both assertion and reason are true and reason is not the correct explanation of assertion. |
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| 759. |
What volume of 0.1M `H_(2)O_(2)` solution will be required to completely reduce 1 litre of 0.1 M `KMnO_(4)` in acidic medium?A. 2500mlB. 500mlC. 1000mlD. 1200ml |
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Answer» Correct Answer - a |
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| 760. |
A substance which participates readily in both acid-base and oxidation-reduction reactions is:A. `NaCO_(3)`B. `KOH`C. `KMnO_(4)`D. `H_(2)C_(2)O_(4)` |
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Answer» Correct Answer - d |
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| 761. |
A substance which participates readily in both acid-base and oxidation-reduction reactions is:A. `Na_(2)CO_(3)`B. `KOH`C. `KMnO_(4)`D. `H_(2)C_(2)O_(4)` |
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Answer» Correct Answer - d |
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| 762. |
A 0.1 mole of a metal is burnt in air to form oxide. The same oxide is then reduced by 0.05 M, 4 litres `S_(2)O_(3)^(2-)` (acidic medium) to +3 oxidation state of metal. What is the oxidation state of metal in oxide ? |
| Answer» Correct Answer - 5 | |
| 763. |
The reaction `Cl_(2) + S_(2)O_(3)^(2-) + OH^(-) to SO_(4)^(2-) + Cl^(-) + H_(2)O` Starting with 0.15 mole `Cl_(2)` , 0.010 mole `S_(2)O_(3)^(2-)` and 0.30 mole `OH^(-)` mole of `Cl_(2)` left in solution will beA. 0.11B. 0.01C. 0.04D. 0.09 |
| Answer» Correct Answer - A | |
| 764. |
Statement `N` atom has two different oxidation states in `NH_(4)NO_(2)`. Explanation 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. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C `N` in `NH_(4)^(+)` is in `-3` oxidation state and in `NO_(2)^(-)` it is in `+3` oxidation state. |
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| 765. |
Assertion: Oxygen atom in both `O_(2)` and `O_(3)` has oxidation number zero. Reason: In `Fe_(2)O`, oxidation number of `O` is `+2`.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 - B The reason is that the sum of oxidation number of elements in a molecule is equal to zero. |
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| 766. |
Saturated solution fo `KNO_3` is used to make "salt bridge" because .A. ` KNO_3` is highly soluble in waterB. Velocity of ` NO_3^(-)` is greater than that of ` K^(+)`C. Velocities of both ` K^+` and ` NO_3^(-)` are nearly the sameD. Velocity of ` K^(+)` is greater than that of ` NO_3^(-)` |
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Answer» Correct Answer - C Both ` K^+ (aq)` and ` NO_3^(-)` (aq) have nearly the same velocity . As a result , the electrical charges in both anode and cathode compartments are nullified almost simultaneously. This helps in smooth flow of current. |
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| 767. |
The reaction `P_4 + 3 NaOH + 3 H_2 O rarr 3 NaH_2 PO + PH_3` is an example of.A. disproportionation reactionB. desplacement reactionC. combination reactionD. decomposition reaction |
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Answer» Correct Answer - A `overset(0)(P)_(4)+3NaOH+3H_(2)O rarr 3NaH_(2)overset(+1)(P)O_(2)+overset(-3)(P)H_(2)` Disproprtionation reactions are a special tupe of redox reactions. In a disproportionation reaction, an element in one oxidation state is simultaneouly oxidized and reduced. |
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| 768. |
How many gm of `K_(2)Cr_(2)O_(7)` is present in 1 L of its N/10 solution in acid medium ?A. 4.9B. 49C. 0.49D. 3.9 |
| Answer» Correct Answer - A | |
| 769. |
Which of the following changes involve reduction ?A. The conversion of ferrous sulphate to ferric sulphateB. The conversion of `H_(2)S` to SC. The conversion of `Cl_(2)` to `NaCl`D. The conversion of Zn to `ZnSO_(4)` |
| Answer» Correct Answer - C | |
| 770. |
When `Cu_(2)S` is converted into `Cu^(2+)` & `SO_(2)` then equivalent weight of `Cu_(2)S` will be (M=mol. Wt. of `Cu_(2)S`)A. MB. `M/2`C. `M/4`D. `M/8` |
| Answer» Correct Answer - D | |
| 771. |
In `XeO_(3)` and `XeF_(6)` the oxidation state of `Xe` isA. `+4`B. `+6`C. `+1`D. `+3` |
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Answer» Correct Answer - B The oxidation state of `Xe` in both `XeO_(3)` and `XeF_(6)` is `+6` `{:(overset(**)(X)eO_(3),,overset(**)(X)eF_(6)),(x-2xx3=0,,x-6=0),(x=+6,,x=+6):}` |
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| 772. |
The oxidation number of Cr in `CrO(5)` which has the following structure is A. `+4`B. `+5`C. `+6`D. `+3` |
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Answer» Correct Answer - C It has four O atoms as peroxide with oxidation number = - 1 and one O atom with oxidation number = - 2 . Hence, `x+4(-1)+1(-2)=0` or `x=+6` |
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| 773. |
Depict the galvanic in whiCHM the reaction `:` `Zn(s)+2Ag^(o+)(aq) rarr Zn^(2+)(aq)+2Ag(s)` takes place. Further show `:`` a.` WhiCHM of the electrode is negatively CHMarged ? `b.` The carriers of the current in the cell. `c.` Individual reaction at eaCHM electrode. |
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Answer» The galvanic cell corresponding to the given redox reaction can be represented as : `Zn|Zn_((aq))^(2+)||Ag_((aq))^(+) | Ag` (i) Zn electrodes is negatively charged because at this electrode , Zn oxidizes to `Zn^(2+)` and the leaving electrons accumulate on this electrode . (ii) Ions are the carriers of current in the cell. (iii) The reaction taking place at Zn electrode can be represented as : `Zn_((s)) to Zn_((aq))^(2+) + 2e^(-)` And the reaction taking place at Zn electrode can be represented as : `Ag_((aq))^(2+) + e^(-) to Ag_((s))` (iv) In aqueous solutions, `CuCl_(2)` ionizes to give `Cu^(2+)` and `Cl^(-)` ions as : `CuCl_(2 (aq)) to Cu_((aq))^(2+) + 2Cl_((aq))^(-)` On electrolysis , either of `Cu^(2+)` ions or `H_(2)O` molecules can get reduced at the cathode . But the reduction potential of `Cu^(2+)` is more than that of `H_(2)O` molecules . `Cu_((aq))^(2+) + 2 e^(-) to Cu_((aq)) , E^(@) = + 0.34V` `H_(2)O_((l)) + 2e^(-) to H_(2 (g)) + 2OH^(-) , E^(@) = - 0.83V` Hence , `Cu^(2+)` ions are reduced at the cathode and get deposited . Similarly , at the anode , either of `Cl^(-)` or `H_(2)O`is oxidized . The oxidation potential of `H_(2)O` is higher than that of `Cl^(-)` . `2Cl_((aq)) to Cl_(2(g)) + 2e^(-) , E^(@) = -1.36 V` `2H_(2)O_((l)) to O_(2 (g)) + 4 H_((aq))^(+) + 4e^(-) , E^(@) = -1.23 V` But oxidation of `H_(2)O` molecules occurs at a lower electrode potential than that of `Cl^(-)` ions because of over -voltage (extra voltage required to liberate gas ) . As a result , `Cl^(-)` ions are oxidized at the anode to liberate `Cl_(2)` gas . |
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| 774. |
The oxidation number of `S` in `Na_(2)S_(4)O_(6)` isA. `+2.5`B. `+2 and +3` (two S have +2 and other two have +3)C. `+2 and +3` (three S have +2 and one S has +3)D. `+5 and 0` (two S have +5 and the other two have 0) |
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Answer» Correct Answer - D The structural formula of `Na_(2)S_(4)O_(6)` (sodium tetrathionate) is as follows `{:(" O O"),(" || ||"),(Na^(+)O^(-)-S-S-S-S-O^(-)Na^(+)),(" || ||"),(" O O"):}` The O.N of the two central sulphur atoms which are linked only to each other i.e., -S-S- is zero. Let the O.N of the other two sulphur atoms be x each. Here, O.N. of O is -2 and O.N. of tetrathionate ion = -2 `:. 6xx(-2)+2xx0+2xx x=-2` `-12+2x=-2` `x=5` Thus, two S atoms in `Na_(2)S_(4)O_(6)` have O.N. of zero each, while the other two S atoms have O.N of 5 each. |
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| 775. |
In a galvanic cell .A. the flow of electrons through a wire is not possibleB. the anode is the positive terminal and the cathode is the negative terminalC. chemical energy is converted into electrical energyD. electrical energy is converted into chemical energy |
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Answer» Correct Answer - C Galvanic (or ovltaic) cells are electrochemical cells in which spontaneous redox reactions produce electrical energy . The two halves of the redox reaction are separated, requiring electron transfer to occur through an external circuit. In this way, useful electrical energy is obtained. Everyone is familiar with some galvanic cells. The batteries commoly used in mobile phones, laptops,flashlights, portable radios, photographite equipment, and many toys and appliances are galvanic cells. Automobile batteries consist of voltaic cells connected in series, so their voltage add. |
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| 776. |
The oxidation number of `S` in `Na_(2)S_(4)O_(6)` isA. `+0.5`B. `2.5`C. `+4`D. `+6` |
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Answer» Correct Answer - B `Na_(2)S_(4)O_(6): 2+4x-12=0` `:. X=(10)/(4)=2.5` Oxidation number of `S= 2.5` |
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| 777. |
Equivalent weight of `N_(2)` in the change `N_(2) rarr NH_(3)` isA. `28//6`B. `28`C. `28//2`D. `28//3` |
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Answer» Correct Answer - A `6e+N_(2)^(0) rarr 2N^(3-)` `EN_(2)=(28)/(6)` |
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| 778. |
Assertion :- Equivalent weight of `NH_(3)` in the reaction `N_(2) rarrNH_(3)` is `17//3` while that of `N_(2)` is `28//6`. Reason :- `"Equivalent weight" =("Molecular weight")/("number of "e^(-)" lost or gained/mole")`A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False. |
| Answer» Correct Answer - A | |
| 779. |
Oxidation number of `Cl` in `NOClO_(4)` is:A. `+7`B. `-7`C. `+5`D. `-5` |
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Answer» Correct Answer - A Calculate ox.no. of `Cl` in `NOClO_(4)` by assuming `ClO_(4)^(-)` and `NO^(+)`. |
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| 780. |
The strongest reducing agent isA. `HNO_(3)`B. `H_(2)S`C. `H_(2)SO_(3)`D. `SnCl_(2)` |
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Answer» Correct Answer - B `H_(2)S` is the strongest reducing agent among these as the oxidation state of `S` is minimum in `H_(2)S`, i.e. `-2` |
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| 781. |
Which of the following is correct ?A. Oxidation fo a substance is followed by reduction of another substanceB. Reduction of a substance is followed by oxidation of another substance.C. Oxidation and reduction are complementary processesD. All of these |
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Answer» Correct Answer - C Oxidation and reduction always occur simultaneously, and to the same exent, in ordinary chemical reactions. For brevity, we usually call oxidation-reduction reactions as redox reactions. |
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| 782. |
On the basis of structure, the oxidation of two Cl atoms in `CaOCl_(2)` respectively areA. `-1 and +1`B. `+2,-2`C. `-2,+2`D. `-1 and +3` |
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Answer» Correct Answer - A `CaOCl_(2)` can be represented as `Ca^(2+)(Cl^(-))(ClO^(-))` O.N of Cl in `Cl^(-1)=-1` O.N of Cl in `ClO^(-) = -1` |
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| 783. |
The oxidation number of N and Cl in `NOClO_(4)` respectively areA. `+2 " and " +7`B. `+3 " and " +7`C. `-3 " and " +5`D. `+2 " and " -7` |
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Answer» Correct Answer - B `NOClO_(4)` is actually `NO^(+)ClO_(4)^(-)` is x. `N overset(+)(O): x + (-2) = +1` `x = +1 + 2 = +3` Let the oxidation state of Cl in `ClO_(4)^(-)` is y. `ClO_(4)^(-) : y + (-2) xx 4 = -1` `y - 8 = 1, y = +7` |
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| 784. |
How many electrons should `X_(2)H_(4)` liberate so that in the new compound X shows oxidation number of `-1//2`(E.N. X gt H)A. 10B. 4C. 3D. 2 |
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Answer» Correct Answer - C `X_(2)H_(4)` or `X_(2)^(-4)rarr X_(2)^(-1)+3e^(Theta)` `2x=-1` `x=-(1)/(2)` |
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| 785. |
The oxidation number of `Cl` in `CaOCl_(2)` isA. `-1 and +1`B. `+2`C. `-2`D. None of these |
| Answer» Correct Answer - A | |
| 786. |
Assertion: `Cu` is stronger reducing agent than `H^(+)`. Reason: `E^(0)` of `Cu^(2+)//Cu` is negative. |
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Answer» Correct Answer - D `E^(0)` of `Cu^(2+)//Cu` is `+0.34V` and positive `E^(0)` means that the redox couple is a weaker reducing agent than the `H^(+)//H_(2)` couple. |
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| 787. |
Oxidation number of `Cl` in `NOClO_(4)` is:A. `+11`B. `+9`C. `+7`D. `+5` |
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Answer» Correct Answer - C The compound may be written as `NO^(+)ClO_(4)`. For `ClO_(4)^(-)`, Let oxidation number of Cl = a `a+4xx(-2)=-1` `a=+7` Hence, the oxidations number of Cl in `"NOClO"_(4)` is `+7` |
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| 788. |
Which of the following is the most powerful oxidizing agent?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `l_(2)` |
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Answer» Correct Answer - A Fluorine is a most powerful oxidizing agent because it consits of `E^(o)=+2.5 vol t`. |
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| 789. |
Which of the following can funtion as an oxidizing as well as a reducing agent ?A. `N_2O_5`B. `H_2S`C. `H_2SO_4`D. `SO_2` |
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Answer» Correct Answer - D A species can work both as an oxidizing agent and a reducing agent if the central atom is present in the intermediate oxidation state so that it can be oxidized (increase of oxidation number ), playing the role of a reducing agent, as well as reduced (decrease of oxidation number ), playing the role of an oxidizing agent. For S, the lowest oxidation number is ` -2` and the highest oxidation number is ` +6`. In ` SO_2`, the ` O.N.` of S is ` +4` which is neither highest (i.e. , +6) nor lowest (i.e., -2) . Thus, it can act both as an oxidizing and a reducing agent. In ` N_2O_5,N` atom has its highest ` O.N.`, thus, it functions only as an oxidizing agent. In ` H_2S, S` atom has its lowest ` O.N.`, thus, it functions only as a reducing agent. Similarly. ` H_2SO_4` can function only as an oxidizing agent as S gas its maximum oxidation number. The terms oxidation number and oxidation state are used interchangeably. However, the term oxidation state is preferred when the atom in a given species undergoes a change of oxidation number. |
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| 790. |
Which of the following is the most powerful oxidising agent ?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `l_(2)` |
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Answer» Correct Answer - A Fluorine is the strongest oxidising agent. Its reduction potential `(E_("red")^(@))` is `+2.87 V`, hence it is easily reduced to `F^(-)` by gaining electron. |
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| 791. |
The oxidation number of `S` in `H_(2)S_(2)O_(8)` isA. `+8`B. `-8`C. `+6`D. `+4` |
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Answer» Correct Answer - C In `H_(2)S_(2)O_(8)`, two O atoms form peroxide linkage i.e. `H-O-overset(O)overset(uarr)underset(O)underset(darr)"S"-O-O-overset(O)overset(uarr)underset(O)underset(darr)"S"-O-H` `2xx1+2a+6(-2-)+2(-1)=0` `therefore" "a=+6` Thus the oxidation number of S in `H_(2)S_(2)O_(8)` is `+6` |
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| 792. |
Which of the following is the strongest oxidizing agent ?A. `HOCl`B. `HClO_(2)`C. `HClO_3`D. `HClO_4` |
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Answer» Correct Answer - A The oxidizing power of oxyacids decreases in the following order : ` HOC lt HClO_2 lt HCIO_3 lt HCIO_4` due to increase in the thermal stability which follows the order ` HOCl lt HClO_2 lt HClO_3 lt HClO_4`. |
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| 793. |
Which of the following is the most powerful reducing agent ?A. `F^(-)`B. ` CI^(-)`C. `Br^(-)`D. ` I^(-)` |
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Answer» Correct Answer - D The reducing power of halide ions follows the order `I^(-) gt Br^(-) gt Cl^(-) gt F^(-)` `I^(-)` is the strongest reducing agent because it has the highest oxidation potential . |
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| 794. |
The oxidation number of nitrogen in `NH_(2)OH` is :A. ZeroB. `+1`C. `-1`D. `-2` |
| Answer» Correct Answer - C | |
| 795. |
The oxidation number of nitrogen atoms in `NH_(4)NO_(3)` are:A. `+3,+5`B. `+3,-5`C. `-3,+5`D. `-3,-5` |
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Answer» Correct Answer - C There are two N atoms in `NH_(4)NO_(3)`, but one N atom has negative oxidation number (attached to H) and the other has positive oxidation number (attached to O). Therefore evaluation should be made separately as - `{:("oxidation number of N is NH"_(4)^(+),"Oxidation number of N in NO"_(3)^(-)),(a+4xx(+1)=+1,and a+3(-2)=1),(therefore" " a=-3,therefore" "a=+5):}` Here the two oxidation number are `-3-` and `+5` respectively. |
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| 796. |
`H_(2)O` can be oxidised toA. `H_(2)` and `O_(2)`B. `O_(2)`C. `OH^(-)`D. `O^(2-)` |
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Answer» Correct Answer - B `underset(+1)underset(uarr)(2H_(2))underset(-2)underset(uarr)(O) rarr underset(0)underset(uarr)(2H_(2))+underset(0)underset(uarr)(O_(2))` `H` is reduced to `H_(2)` and `O` is oxidised to `O_(2)`. |
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| 797. |
Oxidation number if nickel in `Ni(CO_(4))` isA. `0`B. `+4`C. `-4`D. `+2` |
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Answer» Correct Answer - A If any central metal non atom combined with carbonyl group than central metal atom shows always zero oxidation state. |
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| 798. |
Which of the following equations is a balanced one?A. `5BiO_(3)^(-)+22H^(+)+Mn^(2+)rarr5Bi^(3+)+7H_(2)O+MnO_(4)^(-)`B. `5BiO_(3)^(-)+14H^(+)+2Mn^(2+)rarr5Bi^(3+)+7H_(2)O+2MnO_(4)^(-)`C. `2BiO_(3)^(-)+4H^(+)+Mn^(2+)+Mn^(2+)rarr2Bi^(3+)+2H_(2)O+MnO_(4)^(-)`D. `6BiO_(3)^(-)+12H^(+)+3Mn^(2+)rarr6Bi^(3+)+6H_(2)O+3MnO_(4)^(-)` |
| Answer» Correct Answer - B | |
| 799. |
The oxidation number of nitrogen in `NH_(2)OH` is : |
| Answer» Correct Answer - C | |
| 800. |
The oxidation state of nickel in `K_(4)Ni(CN)_(4)` is:A. `+1`B. `+2`C. `-1`D. `0` |
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Answer» Correct Answer - D `4xx1+a+4xx(-1)=0, :. a=0` |
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