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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.
| 351. |
Which of the following will act as cathode when connected to standard hydrogen electrode which has `E^(@)` value given as zero ? (i) `Zn^(2+)//Zn, E^(@)=-0.76V` (ii) `Cu^(2+)//Cu, E^(@)=+0.34 V` (iii) `Al^(3+)//Al, E^(@)=-1.66 V` (iv) `Hg^(2+)//Hg, E^(@)=+0.885 V`A. (i) and (ii)B. (ii) and (iv)C. (i) and (iii)D. (i), (ii), (iii) and (iv) |
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Answer» Correct Answer - B When the value of standard reduction potential is positive, the electrode undergoes reduction and acts as cathode. |
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| 352. |
Which of the following electrodes will act as anodes, which connected to Standard Hydrogen Electrode?A. `Al//Al^(3+)" "E^(Θ) =-1.66`B. `Fe//Fe^(2+)" "E^(Θ)=-0.44`C. `Cu//Cu^(2+)" "E^(Θ)=+0.34`D. `F_(2)(g)//2F^(-)(aq)" "E^(Θ)=02.87` |
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Answer» Correct Answer - a,b All electrodes which have negative electrode potentials are stronger reducing agents than `H_(2)` gas and hence acts as anodes when connected to standard hydrogen electrodes. Thus, `Al^(3+)//Al(E^(Θ)=-1.66V)` and `Fe^(2+)//Fe(E^(Θ)=-0.44V)` act as anode. |
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| 353. |
The largest oxidation number exhibited by an element depends on its outer eletronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number ?A. `3d^(1)4s^(2)`B. `3d^(3)4s^(2)`C. `3d^(5)4s^(1)`D. `3d^(5)4s^(2)` |
| Answer» The element with `3d^(5) 4s^(2)` configuration (Mn) can exhibt largest oxidation states up[ to +7] | |
| 354. |
In the balanced chemical reaction `IO_(3)^(ө)+aI^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)` `a, b,c`, and `d`, respectively, correspond toA. `5,6,3,3`B. `5,3,6,3`C. `3,5,3,6`D. `5,6,5,5` |
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Answer» Correct Answer - A The balanced equation is `{:(12H^(o+)+10e^(-)+2IO_(3)^(ө)rarrI_(2)+6H_(2)O),(10I^(ө)rarr5I_(2)+10e^(-)),(ulbar(10I^(ө)+12H^(o+)+2IO_(3)^(ө)rarr6H_(2)O+6I_(2))):}` or `underset((a))(5I^(ө))+underset((b))16H^(o+)+IO_(3)^(o+)rarrunderset((c ))(3H_(2)O)+underset((d))(3I_(2))`. |
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| 355. |
One mole of `CaOCl_(2)` is dissolved in water andexcess of `KI` added. `Hypo(Na_(2)S_(2)O_(3))` required to react with the oxidised part completely isA. `1` moleB. `2.0` molesC. `1.5` molesD. `2.5` moles |
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Answer» Correct Answer - B `underset(1 mol)(CaOCl_(2))+H_(2)O rarr Ca(OH)_(2)+underset(1 mol)(Cl_(2))` `Cl_(2)+2KI rarr 2KCl+underset(1 mol e)(I_(2))` `I_(2)+KI hArr KI_(3)` `I_(2)`(in `KI`) `+2Na_(2)S_(2)O_(3) rarr 2Nal+Na_(2)S_(4)O_(6)` |
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| 356. |
How many milliliters fo a `0.05 M KMnO_4` solution are required to oxidize `2.0 g FeSO_4` in a dilute acid solution ?A. ` 32. 56 mL`B. `62.53 mL`C. ` 25.36mL`D. ` 52.63 mL` |
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Answer» Correct Answer - D Normality `=n_("factor")xx`Molarity `overset(+7)(M)nO_(4)^(-) rarroverset(+2)(M)n^(2+), n_("factor")=5` `Fe^(2+) rarr Fe^(3+), n_("factor")=1` Normality of `KMnO_(4)=(5)(0.05)` `=0.25 N` Volume of `KMnO_(4)=V` milliliters Thus, milliequivalents of `KMnO_(4)=NxxV` `=0.25 V` Equivalents of `FeSO_(4)=("Mass"_(FeSO_(4)))/("Gram equivalent mass"_(FeSO_(4)))` `["Note that eq. wt. of "FeSO_(4)=("Formula weight")/("Change in O.N.")=152/1]` Milliequivalent of `FeSO_(4)=2/152xx1000` According to the law of equivalence, `"Milliequivalents"_(KMnO_(4))="Miliequivalents"_(FeSO_(4))` `0.25 V= 2/(152) xx 1000` ` V= (2xx 1000)/(152 xx 0. 25) ` ` = 52. 63 mL`. |
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| 357. |
Which separation technique is based on difference in the volatility of the of the substances to be separated ?A. FiltrationB. DistillationC. Solvent extractionD. Paper chromatography |
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Answer» Correct Answer - b |
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| 358. |
`CN^(-)` is oxidised by `NO_(3)^(-)` in presence of acid : `can^(-)+bNO_(3)^(-)+cH^(+)rarr(a+b)NO+aCO_(2)+(c)/(2)H_(2)O` `What are the values of a, b, c in that order:A. 3, 7, 7B. 3, 10, 7C. 3, 10, 10D. 3, 7, 10 |
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Answer» Correct Answer - d |
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| 359. |
In the reaction `X^(-)+XO_(3)^(-)+H^(+)rarrX_(2)+H_(2)O` , the molar ratio in which `X^(-)` and `XO_(3)^(-)` react is :A. `1:5`B. `5:1`C. `2:3`D. `3:2` |
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Answer» Correct Answer - b |
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| 360. |
A student is asked to measure 12 mL of a liquid as precisely as possible. Which piece of equipment should she select for this task ?A. 25 mL beakerB. 25 mL graduated cylinderC. 25 mL conical flaskD. 25 mL volumetric flask |
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Answer» Correct Answer - b |
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| 361. |
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 Out of the following only one is redox reaction.Identify it.A. `Ca(OH)_(2)+2HCIrarrCaCI_(2)+2H_(2)O`B. `BaCI_(2)+MgsO_(4)rarrBaSO_(4)+MgCI_(2)`C. `2S_(2)O_(7)^(2-)+23H_(2)Orarr2SO_(4)^(2-)+4H^(+)`D. `Cu_(2)S+2FeOrarr2Cu+2Fe+SO_(2)` |
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Answer» Correct Answer - D There is a change in oxidation state of the species only in this reaction `overset(+1)Cu_(2)overset(-2)S+2overset(2+)Feoverset(-2)Orarr2overset(0)Fe+overset(+4)SO_(2)^(-2)` |
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| 362. |
Assertion : The oxidation numbers are artificial, they are useful as a book keeping device of elements in reactions Reason : The oxidation numbers do not usually represent real charge on atoms, they are simply conventions that indicate what the maximum charge could possibly be on an atom in a molecule.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - A | |
| 363. |
Which of the following statement is true for the electrochemical Daniell cell ?A. Current flows from zinc electrode to copper electrodeB. Electrons flow from copper electrode to zinc electrodeC. Cations move towards copper electrodeD. Cations move twoards zinc electrode |
| Answer» Cation move towards copper electrode. | |
| 364. |
The oxidation number of chlorine in HOCl isA. `-1`B. ZeroC. `+1`D. `+2` |
| Answer» Correct Answer - C | |
| 365. |
In a Daniell cell, when a Za electrode and a Cu electrode ane connected with a wireA. electrons flow from the Zn electrode to the Cu electrode through the wireB. electrons flow from the Cu electrode to the Zn electrode through the wireC. current flow from the Zn electrode to the cu electrode through the wireD. electrons flow from the Zn electrode to the cu electrode through the cell |
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Answer» Correct Answer - A Electrons are released (by oxidation) at the anode (Zn electrode ) and consumed (by reduction) at the cathode (Cu electrode). Therefore, electrons flow through thewire from anode to cathode, i.e., in all voltaic cells, the electrons flow spontaneously from the negative electrode to the positive electrode. |
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| 366. |
`-1//3` oxidation state of nitrogen will be obtained in case of :A. Ammonia `(NH_(3))`B. Hydrazoic acid `(N_(3)H)`C. Nitric oxide `(NO)`D. Nitrous oxide `(N_(2)O)` |
| Answer» Correct Answer - B | |
| 367. |
Two oxidation states for chlorine are found in the compound :A. `CaOCl_(2)`B. `KCl`C. `KClO_(3)`D. `Cl_(2)O_(7)` |
| Answer» Correct Answer - A | |
| 368. |
`{:(,"Compounds",,"O.N."),((A),KMn^(**)O_(4),(1),+4),((B),Ni^(**)(CO)_(4),(2),+7),((C),[Pt^(**)(NH_(3))Cl_(2)]Cl_(2),(3),0),((D),Na_(2)O_(2)^(**),(4),-1):}` The correct code for the O.N. of asterisked atom would be:A. `{:(A,B,C,D),(1,2,3,4):}`B. `{:(A,B,C,D),(4,3,2,1):}`C. `{:(A,B,C,D),(2,3,1,4):}`D. `{:(A,B,C,D),(4,1,2,3):}` |
| Answer» Correct Answer - C | |
| 369. |
The substatnce that will reduce ` Ag^(+)` to Ag but will not reduce `Ni^(2+) ` to Ni is.A. `Al`B. `Mg`C. `Pb`D. `Zn` |
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Answer» Correct Answer - C This implies that ` Ag^(+) //Ag` has higher reduction potential than the substance while `Ni^(2+) //Ni` has lower reduction potential than the substance : `Ag^(+) //Ag (0.80 V) gt Pb^(2+) //Pb (-0.13 V) gt Ni^(2+) //Ni (0.25 V)` . |
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| 370. |
Assertion: `HClO_(4)` is a stronger acid than `HClO_(3)`. Reason: Oxidation state of `Cl` in `HClO_(4)` is `+VII` and in `HClO_(3)+V`.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 Both assertion and reason are true but reason is not the correct explanation of assertion. Greater the number of negative atoms present in the oxy-acid make the acid stronger. In general, the strengths of acids that have general formula `(HO)_(m)ZO_(n)` can be related to the value of `n`. As the value of `n` increase, acidic character also increases. The negative atoms draw electrons away from the `Z`- atom and make it more positive. The `Z`-atom, therefore, becomes more effective in withdrawing electron density away from the oxygen atom that bonded to hydrogen, in turn the electrons of `H-O` bond are drawn more strongly away from the `H`- atom. The net effect makes it easier from the proton release and increases the acid strength. |
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| 371. |
When `N_(2)` is converted into `NH_(3)`, the equivalent weight of nitrogen will be:A. 1.67B. 2.67C. 2.63D. 4.67 |
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Answer» Correct Answer - d |
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| 372. |
Of the following elements, which one has the same oxidation state in all of its compounds ?A. HydrogenB. FluorineC. CarbonD. Oxygen |
| Answer» Correct Answer - B | |
| 373. |
which of the following is a redox reaction ?A. `NaClrarrKNO_(3)rarrNaNO_(3)+KCl`B. `CaC_(2)O_(4)+2HClrarrCaCl_(2)+H_(2)C_(2)O_(4)`C. `Mg(OH)_(2)+2NH_(4)ClrarrMgCl_(2)+2NH_(4)OH`D. `Zn+2AgCNrarr2Ag+Zn(CN)_(2)` |
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Answer» Correct Answer - D `{:(Zn,rarr,Zn^(2+),+,2e),(Ag,+e,rarr,Ag,):}` |
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| 374. |
Which of the following is not a redox reaction ?A. Rusting or ionB. Evaporation of waterC. Buring og gasolineD. Human respiration |
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Answer» Correct Answer - B It is a physical change `H_2O(1) rarr H_2O(g)` In rusting of iron. Fe is oxidized to `Fe_2O_3.xH_2O` (called rust). In burning or gasoline, carbon of hydrocarbon is oxidized to `CO_(2)`. In human respiration, C of glucose is oxidized to `CO_(2)`. |
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| 375. |
Which fo the following is a redox reaction ?A. `Na_2SO_4 + BaCl_2 rarr BaSO_4 + 2NaCl`B. ` SO_2 + H_2O rarr H_2SO_3`C. `2 CuSO_4 + 4Kl rarr 2Cu l +2K_2SO_4`+I_2D. ` CuSO_4 + 4NH_3 rarr [Cu(NH_3)_4]SO_4` |
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Answer» Correct Answer - C Reaction (1) is a double displacement reaction, reaction(2) involves hydration, and reaction (4) involves the formation fo a complex compound. Reaction (3) is a redox reaction as the oxidation number of Cu is reduced while that of `I` is increased `3overset(+2)(C)uSO_(4)+4Koverset(-1)(I) rarr2 overset(+2)(C)uI+2K_(2)SO_(4)+overset(0)(I_(2))` |
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| 376. |
`2MnO_(4)^(-)+5H_(2)O_(2)+6H^(-) rarr 2Z+5O_(2)+8H_(2)O`. In this reaction `Z` isA. `Mn^(+2)`B. `Mn^(+4)`C. `MnO_(2)`D. `Mn` |
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Answer» Correct Answer - A `2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) rarr 2Mn^(2+)+5O_(2)+8H_(2)O`. |
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| 377. |
Which reaction is possible at anode?A. `2Cr^(3+)+7H_(2)OrarrCr_(2)O_(7)^(2-)+14H^(+)`B. `F_(2)rarr2F^(-)`C. `O_(2)+4H^(+)rarr2H_(2)O`D. None of the above |
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Answer» Correct Answer - A Oxidation occurs at anode. |
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| 378. |
`2MnO_(4)^(-) + 5H_(2)O_(2) + 6H^(+) rarr 2Z + 5O_(2) + 8H_(2)O` Identify Z in the above reaction.A. `Mn^(2+)`B. `Mn^(4+)`C. `Mn`D. `MnO_(2)` |
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Answer» Correct Answer - A `2MnO_(4)^(-) + 5H_(2)O_(2) + 6H^(+) rarr 2Mn^(2+) + 5O_(2) + 8H_(2)O` |
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| 379. |
Which fo the following combination reactions is not a redox reaction ?A. ` CH_4(g) +2O_2 (g) rarr CO_2(g) +2 H_2O(l)`B. ` 3 Mg (s) + N_2 (g) rarr Mg_3N_2(s)`C. ` S(s) + O_2(g) rarr SO_2(g)`D. `CaO(s) + CO_2 (g) rarr CaCO_3(s)` |
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Answer» Correct Answer - D A combination reaction may be represented by `A+ B rarr C` If either A or B or both A and B are in the elemental form, then the reaction is redox in nature. Thus, all combination reactions which involve at least one element as a reactant are redox reactions. Reaction 4 is not a redox reaction as no element is involved as a reactant, it is a combination of two compounds to form a single new compound. |
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| 380. |
The are `"_______"` general tytes of redox reactions .A. threeB. fiveC. fourD. two |
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Answer» Correct Answer - C They are (i) combination reactions, (ii) decomposition reactions, (iii) displacement reactions, and (iv) disproportionation reactions. |
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| 381. |
Which of the noble gases exhibits the maximum number fo different oxidation numbers ?A. `Kr`B. `Xe`C. `Ar`D. `Ne` |
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Answer» Correct Answer - B `Xe` forms maximum number of compounds. Its oxidation numbers are `+2 (XeF_2), + 4 (XeF_4)`, and `+6 (XeF_6)`. |
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| 382. |
Identify the disproportionation reaction.A. `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2)O`B. `CH_(4) + 4Cl_(2) rarr C Cl_(4) + 4HCl`C. `2F_(2) + 2OH^(-) rarr 2F^(-) + OF_(2) + H_(2)O`D. `2NO_(2) + 2OH^(-) rarr NO_(2)^(-) + NO_(3)^(-) + H_(2)O` |
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Answer» Correct Answer - D Disproportionation reaction, `2overset(+4)(N)O_(2) + 2OH^(-) rarr overset(+3)(N)O_(2)^(-) + overset(+5)(N)O_(3)^(-) + H_(2)O` |
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| 383. |
How many gram of `I_(2)` are present in a solution which requires `40 mL` of `0.11N Na_(2)S_(2)O_(3)` to react with it, `S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I`A. `12.7 g`B. `0.558 g`C. `25.4 g`D. `11.4 g` |
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Answer» Correct Answer - B Meq.of `I_(2)="Meq.of" Na_(2)S_(2)O_(3)` `=40xx0.11` `:. (w)/(254//2)xx1000=40xx0.11` `w_(I_(2))=0.558g` |
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| 384. |
If 10.0 g `V_2O_5` is dissolbed in acid and reduced to `V^(2+)` by treatment with tin (Sn) metal how many moles of `I_2` could be reduced by the resulting `V^(2+)` solution as it is oxidised to `V^(4+)`? (Atomic weight of V is 51) |
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Answer» `V_(2)^(5+)+6erarr2V^(2+)` `ZnrarrZn^(2+)+2e` `V^(2+)rarrV^(4+)+2e` `I_(2)+2erarr2I^(-)` and Meq.of `V^(2+)(v.f=3)=` Meq.of ` V_(2)O_(5)(v.f=6)` `=(10)/(182//6)xx1000` Meq.of `V^(2+)(v.f.=2)=(10)/(182//6)xx1000xx(2)/(3)` Meq.of `V^(2+)(v.f=2)=` Meq.of `I_(2)` `(10xx6)/(182)xx1000xx(2)/(3)=` Meq.of `I_(2)` or Meq.of `I_(2)=219.78` `m` Mole of `I_(2)=(219.78)/(2)=109.89` Mole of `I_(2)=(109.89)/(1000)` Mole of `I_(2)=0.1098` |
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| 385. |
In which of the following compounds iron has lowest oxidation state?A. `FeSO_(4).(NH_(4))_(2)SO_(4).6H_(2)O`B. `K_(4)Fe(CN)_(6)`C. `Fe(CO)_(5)`D. `Fe_(2)O` |
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Answer» Correct Answer - C Iron has zero oxidation state in carbonyl complexes. |
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| 386. |
The coordination number and oxidation number of `Cr` in `K_(3)[Cr(C_(2)O_(4))_(3)]` are, respectively,A. `4 and +2`B. `6 and +3`C. `3` and `-3`D. `3` and `0` |
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Answer» Correct Answer - B Coodination number = Number of ligands `xx` denticity `3xx2` ( bidendate) `=6` `overset(+3)K[overset(+3)(Cr)overset(-2xx3)((C_(2)O_(4))_(3))]^(3-) :. "Oxidation state of " Cr= +3`. |
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| 387. |
In the reaction `H_(2)S+NO_(2) rarr H_(2)O+NO+S.H_(2)S` isA. OxidisedB. ReducedC. PrecipitatedD. None of these |
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Answer» Correct Answer - A In this reaction `H_(2)S` is oxidised because the oxidation state of `S` change from `-2` to `0`. |
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| 388. |
On combustion of `CH_(4)` to `CO_(2)` and `H_(2)`, the oxidation number of carbon changes by: |
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Answer» Correct Answer - 8 `overset(-4)(CH_(4))+ 2O_(2) rarroverset(+4-4)(CO_(2))+2H_(2)O` oxidation no. change `= +4 -(-4) = +8` |
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| 389. |
What weight of `FeSO_(4)` ( mol.wt. `=152`) will be oxidised by `200 mL` of normal `KMnO_(4)` solution in acid solution?A. `30.4g`B. `60.8 g`C. `121.6g`D. `15.8 g` |
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Answer» Correct Answer - A Meq.of `FeSO_(4)="Meq.of" KMnO_(4)=200xx1` `:. (w)/(152//1)xx1000=200` `w=30.4 g` |
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| 390. |
What volume of `O_(2)` measured at standard condition will be formed by the action of `100 mL` of `0.5N KMnO_(4)` on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is, `KMnO_(4)+H_(2)SO_(4)+H_(2)O_(2)rarrKHSO_(4)+MnSO_(4)+H_(2)O+O_(2)`A. `0.12 L`B. `0.28 L`C. `0.56 L`D. `1.12 L` |
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Answer» Correct Answer - B Eq. of `O_(2)=` Eq. of `KMnO_(4)` ` implies (w)/(8)=0.1xx0.5` ` implies w=0.4 g` `:. 32 g` of `O_(2)=22.4 L` at `STP` `:. 0.4 g` of `O_(2)=(22.4)/(32)xx0.4=0.28` |
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| 391. |
When `Sn^(2+)` changes to `Sn^(4+)` in a reactionA. It loses two protonsB. It gains two electronsC. It loses two electronsD. It gains two protons |
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Answer» Correct Answer - C `Sn^(2+) rarr Sn^(4+)+2e^(-)`. In this reaction `Sn^(2+)` changes to `Sn^(4+)`, it is called an oxidation reaction. |
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| 392. |
`25 mL` of `0.50M H_(2)O_(2)` solution is added to `50 mL` of `0.20 M KMnO_(4)` is acid solution. Which of the following statements is true?A. `0.010` mole of oxygen is liberatedB. `0.005` mole of `KMnO_(4)` are leftC. `0.030 g` atom of oxygen gas is evolvedD. `0.0025` mole `H_(2)O_(2)` does not react with `KMnO_(4)` |
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Answer» Correct Answer - B `M` of eq. of `KMnO_(4)=.2xx50xx5=50` `M` eq. of `H_(2)O_(2)=2xx25xx.5=25` `M` eq. of `KMnO_(4)` remaining `=(50-25)=25` Mole of `KMnO_(4)=(25)/(5)xx10^(-3)=5xx10^(-3)=.005` |
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| 393. |
The reaction of `H_(2)S+H_(2)O_(2) rarr S+2H_(2)O` manifestsA. Oxidising action of `H_(2)O_(2)`B. Reducing nature of `H_(2)O_(2)`C. Acidic nature of `H_(2)O_(2)`D. Alkaline nature of `H_(2)O_(2)` |
| Answer» Correct Answer - A | |
| 394. |
`25 mL` of `0.50M H_(2)O_(2)` solution is added to `50 mL` of `0.20 M KMnO_(4)` is acid solution. Which of the following statements is true?A. `0.010` mole of oxygen is liberatedB. `0.005` mole of `KMnO_(4)` are leftC. `0.030 g` atom of oxygen is liberatedD. `0.0025` mole of `H_(2)O_(2)` does not react with `KMnO_(4)` |
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Answer» Correct Answer - B Meq.of `H_(2)O_(2)=25xx0.5xx2=25`, Meq.of `KMnO_(4)=50xx0.2xx5=50`, `:. 25 "Meq." or 5` milli-mole of `KMnO_(4)` are left. |
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| 395. |
What volume of `O_(2)` measured at standard condition will be formed by the action of `100 mL` of `0.5N KMnO_(4)` on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is, `KMnO_(4)+H_(2)SO_(4)+H_(2)O_(2)rarrKHSO_(4)+MnSO_(4)+H_(2)O+O_(2)`A. `0.12"litre"`B. `0.28 "litre"`C. `0.56"litre"`D. `1.12"litre"` |
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Answer» Correct Answer - B Meq.of `O_(2)=Meq.of KMnO_(4)=100xx0.5` `(w)/(8)xx1000=50` `:. w_(O_(2))=0.4g` `:. VO_(2)=(22.4xx0.4)/(32)=0.28` litre |
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| 396. |
`N_(2) + 3 H_(2) rarr 2NH_(3)` Molecular weight of `NH_(3)` and `N_(2)` are `x_(1)` and `x_(2)`, respectively. Their equivalent weights are `y_(1)` and `y_(2)`, respectively. Then `(y_(1) - y_(2))`A. `((2X_(1)-X_(2))/(6))`B. `(X_(1)-X_(2))`C. `(3X_(1)-X_(2))`D. `(X_(1)-3X_(2))` |
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Answer» Correct Answer - A `6e+N_(2)^(0)rarr2N^(3-)` `:. E_(N_(2))=(X_(2))/(6)=Y_(2):. E_(NH_(3))=(X_(1))/(3)=Y_(1)` `:. Y_(1)-Y_(2)=(X_(1))/(3)-(X_(2))/(6)=(6X_(1)-3X_(2))/(18)` `=(2X_(1)-X_(2))/(6)` |
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| 397. |
What is the equivalent weight of `NH_(3)` in the given reaction? `3CuO+2NH_(3) rarr 3Cu+N_(2)+3H_(2)O`A. `17`B. `(17)/(4)`C. `(17)/(2)`D. `(17)/(3)` |
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Answer» Correct Answer - D `2N^(3-) rarr (N_(2))^(0)+6e^(-)` `2` mole of `NH_(3)=1` mole `N_(2)` Thus equivalents `2xxn=1xx6` `n=6//2=3` `:.` Eq. wt. `= (M)/(3)=(17)/(3)` |
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| 398. |
Which of the following is the strongest oxidising agent ?A. HOClB. `HClO_(2)`C. `HClO_(3)`D. `HClO_(4)` |
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Answer» Correct Answer - A Out of HOCl, `HClO_(2) HClO_(3) and HClO_(4),HOCl` is the strongest oxidising agent. |
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| 399. |
Which of the following is the strongest reducing agent in aqueous medium?A. `Mg`B. `Na`C. `Li`D. `Ca |
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Answer» Correct Answer - C In an aqueous medium, `Li` is the strongest reducing agent, since the high negative enthalpy of hydration compoensates high `IE_(1)`. |
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| 400. |
Which of the following is the strongest oxidising agent?A. `I_(2)`B. `F_(2)`C. `Cl_(2)`D. `Br_(2)` |
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Answer» Correct Answer - B Reduction potential of `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)` |
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