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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.
| 251. |
`H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` behave as acids as well as reducing agents. Which are the correct statements?A. Equivalent weight of `H_(2)C_(2)O_(4)` and `NaHC_(2)O_(4)` are equal to their molecular weights when behaving as reducing agents.B. 100mL of 1N solution of each is neutralised by same volume of 1N `Ca(OH)_(2)`.C. 100mL of 1N solution of each is neutralised by same volume of 1M `Ca(OH)_(2)`.D. 100 mL of 1 N solution of each is oxidised by same volume of 1 N `KMnO_(4)` solution in acidic medium. |
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Answer» Correct Answer - b,d |
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| 252. |
An electrochemical cell is a device in which the oxidation and reduction half rections are taking place separtely and the electrical energy is evloved as a result of the rection.In fact all redox reactions are of spontaneous nature and the free energy change`(triangleG)` is negative. This is vonverted in to the electrical energy. (i)Which electrode acts as negative pole? (ii) which electrode acts as positive pole? (iii) ho does curent flow in the elctrochemical cell? (iv) what is the value associated with the working of electrochemicals cells? |
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Answer» (i) anode acts as negative pole in the electrochemeical cell (ii) Cathode acts as p[ositive pol,e in the electrochemeical cell (iii) Current follows from cathode to anode in the electrochemical cell (iv) All type of batteries , both primary, secondary cells as well as fuel cells are used to lgenerate electric current f.Dry cells and lead storage cells are very commonly used.These are all electrochemical cells in nature. |
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| 253. |
In the following reaction: `Cr(OH)_(3)+OH^(-)+IO_(3)^(-) toCrO_(4)^(2-)+H_(2)O+I^(-)`A. `IO_(3)^(-)` is oxidising agentB. `Cr(OH)_(3)` is oxidisedC. `6e^(-)` are being taken per iodine atomD. none of these |
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Answer» Correct Answer - a,b |
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| 254. |
In the following reaction: `x KMnO_(4)+y NH_(3) rarr KNO_(3)+MnO_(2)+KOH+H_(2)O` `x` and `y` areA. `x=4,y=6`B. `x=3,y=8`C. `x=8,y=6`D. `x=8,y=3` |
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Answer» Correct Answer - D `8KMnO_(4)+3NH_(3)rarr3KNO_(3)+8MnO_(2)+5KOH+2H_(2)O` |
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| 255. |
Equivalent weight of Mohr salt in the titration with `KMnO_(4)` is (M-Molecular weight)A. `M/1`B. `M/4`C. `M/3`D. `M/2` |
| Answer» Correct Answer - A | |
| 256. |
Statement Starch is generally used as absorption indicator in iodimetric or iodometric titrations. Explanation Starch imparts blue colour with iodine.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 The explanation is correct reason of the statement. |
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| 257. |
Statement `2H_(2)O_(2)rarr2H_(2)O+O_(2)` is autoredox change. Explanation One oxygen atom is oxidised and one oxygen atom is reduced.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 The explanation is correct reason: `{:(2O^(1-)rarrO_(2)^(0)+2e),(O^(1-)+erarrO^(2-)):}` |
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| 258. |
The number of electrons required to balanced the following equation- `NO_(3)^(-)+4H^(+)+e^(-)rarr2H_(2)O+NO` is -A. 5B. 4C. 3D. 2 |
| Answer» Correct Answer - C | |
| 259. |
The number of electrons required to balance the following equation are : `NO_(3)^(-)+4H^(+)rarr2H_(2)O+NO`A. 2 on right sideB. 3 on left sideC. 3 on rightD. 5 on left side |
| Answer» Correct Answer - B | |
| 260. |
Statement Oxidation number of `Cu` in `CuH` is -`1` Explanation `Cu` is placed below `H` in electro-chemical series.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 Explanation is correct reason for statement. |
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| 261. |
`H_(2)O_(2)+H_(2)O_(2)rarr2H_(2)O+O_(2)` is an example of dispropotionation because -A. Oxidation number of oxygen only decreasesB. Oxidation number of oxygen only increasesC. Oxidation number of oxygen decreases as well as increasesD. Oxidation number of oxygen neither decreases nor increases |
| Answer» Correct Answer - C | |
| 262. |
During the oxidation of arsenite to arsenate ion in alkaline medium, the number of moles of hydroxide ions involved per mole of arsenite ion areA. `2`B. `3`C. `2//3`D. None of these |
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Answer» Correct Answer - A `overset((3+))(As)overset((3+))(O_(3)) rarr overset((+5))(As)overset((3-))(O_(4))` In the presence of oxidizing agent `(KMnO_(4))` `2` moles of `OH^(-)` is produced in basic medium |
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| 263. |
Which of the following act both as an oxidising as well as reducing agent ?A. `H_(3)PO_(4)`B. `HNO_(3)`C. `HNO_(2)`D. `SO_(3)` |
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Answer» Correct Answer - C O.N. of P in `H_(3)PO_(4)=+5` (Highest) O.N. of N is `HNO_(3)=+5` (Highest) O.N of N in `HNO_(2)=+3` O.N of N in `HNO_(2)=+3` O.N of N is `SO_(3)=+6` (Highest) As O.N of nitrogen in `HNO_(2)` is +3, it means increase or decrease its O.N. Thus `HNO_(2)` acts both as an oxidisingom as well as reducing agent. |
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| 264. |
The value of `x` in the partial redox equation `MnO_(4)^(-)+8H^(+)+xe hArr Mn^(2+)+4H_(2)O` is |
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Answer» Correct Answer - 5 Total charge of `L.H.S ="Total chage on" R.H.S` ` (-1)+8(-n)=+2` or `n=5` |
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| 265. |
The value of `x` in the partial redox equation `MnO_(4)^(-)+8H^(+)+xe hArr Mn^(2+)+4H_(2)O` isA. `5`B. `10`C. `2`D. `3` |
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Answer» Correct Answer - A Balance charges on both side, `-1+8+(-n)=+2 :. n= +5` |
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| 266. |
Reduction never involves:A. gain of electronsB. decrease in oxidation numberC. loss of electronsD. decrease in vlaency of elctropositive component |
| Answer» Correct Answer - C | |
| 267. |
Positive oxidation state of an element indicates that it is -A. Is in elementry formB. OxidisedC. ReducedD. Does not changed. |
| Answer» Correct Answer - B | |
| 268. |
How many mole of electrons are involved in the reduction of one mole of `MnO_(4)^(-)` ion in alkaline medium to `MnO_(3)^(-)`A. 1B. 2C. 5D. 3 |
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Answer» Correct Answer - D In alkaline medium, `KMnO_(4)` acts as an oxidising agent as follows - `MNO_(4)^(-)+2H_()O+3e^(-)rarrMnO_(2)+4OH^(-)` Thus three electrons per molecule of are involved. |
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| 269. |
An element which never has a positive oxidation number in any of its compoundsA. BoronB. OxygenC. ChlorineD. Fluorine |
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Answer» Correct Answer - D Fluorine always shows `-1` oxidation state. |
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| 270. |
If `HNO_(3)` changes into `N_(2)O`, the oxidation number is changed byA. `+2`B. `-1`C. `0`D. `+4` |
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Answer» Correct Answer - D `underset(x=+5)underset(1+x-6=0)(HNO_(3))rarr underset(x=(2)/(2)=+1)underset(2x-2=0)underset(2x-2=0)(N_(2)O)` |
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| 271. |
What is the oxidation number of chlorine in `ClO_(3)^(-)` ?A. `+5`B. `+3`C. `+4`D. `+2` |
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Answer» Correct Answer - A Let O.N of Cl in `ClO_(3)^(-)` be x `:. X+3 (-2) =-1` `= x =-1 +6 =5`. |
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| 272. |
In the reaction, `KMnO_(4) + 16 HCl rarr 5C1_(2) + 2MnCl_(2) + 2KCl + 8H_(2) O` the reduction product isA. `Cl_(2)`B. `MnCl_(2)`C. `H_(2)O`D. KCl |
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Answer» Correct Answer - B In this reaction O.N of Mn changes from +7 in `KMnO_(4)` to +2 in `MnCl_(2)`. Thus, `MnCl_(2)` is the reduction product. |
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| 273. |
In acidic medium, dichromate ion oxidizes ferrous ion to ferric ion. If the gram molecular weight of potassium dichromate is ` 294 g`, is gram equivalent weight is `"______" g`.A. ` 294`B. ` 127`C. `49`D. ` 24.5` |
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Answer» Correct Answer - C Equivalent mass of oxidizing agent `=("Formula mass")/(("Total change in oxidation number of element"),("oxidized or reduced per mole of compound"))` `overset(+6)(C)r_(2)O_(7)^(2-) rarr 2overset(+3)(C)r^(3+)` The total change in oxidation number (For two Cr atoms ) is of ` 6 ` units ( +12 to 6 ). Thus, Equvalent mass `= (296)/6 = 49` Here students often make the mistake of dividing formula mass by 3 as they calculate the change in oxidation number for just one Cr atom. |
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| 274. |
In acidic medium, dichromate ion oxidizes ferrous ion to ferric ion. If the gram molecular weight of potassium dichromate is ` 294 g`, is gram equivalent weight is `"______" g`.A. 294B. 127C. 49D. 24.5 |
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Answer» Correct Answer - C In acidic medium `K_(2)Cr_(2)O_(7)` acts as a strong O.A and itself reduces to `Cr^(3+)` `K_(2)overset(+6)(Cr_(2))O_(7)+6e^(-)rarroverset(+3)(2Cr^(3+))` `:.` Eq. mass of `K_(2)Cr_(2)O_(7)=("Mol. mass")/(6)` `=(296)/(6) ~~ 49`. |
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| 275. |
Which of the following arrangements represent increaseing oxidation number of the central atom?A. `CrO_(2)^(-),CIO_(3)^(-),CrO_(4)^(-),MnO_(4)^(-)`B. `CIO_(3)^(-),CrO_(4)^(2-),MnO_(4)^(-),CrO_(2)^(-)`C. `CrO_(4)^(-),CIO_(3)^(-),MnO_(4)^(-),CrO_(4)^(2-)`D. `CrO_(4)^(2-),MnO_(4)^(-),CrO_(2)^(-),CIO_(3)^(-)` |
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Answer» Correct Answer - A The O.N of central atom in the different ions is: `CrO_(2)^(-):x-4=-1or x=+3` `CIO_(3):x-6=-1 or x=+5` `CrO_(4)^(2-): x-8=-2orx=+6` `MnO_(4)^(-):x-8=-1or x=+7` therefore is the correct increasing order. |
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| 276. |
The following redox reaction occurs in basic medium `NO_(3)^(-)Zn(s)toZn^(2+)+NH_(4)^(+)` when the above reaction is balanced such that the stoichiometric coefficients are in smallest whole number ratio, then the difference of stoichiometric coefficient of Zn (s) and `OH^(-)` ion will be:A. 4B. 10C. 6D. None of these |
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Answer» Correct Answer - c |
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| 277. |
Balance the following by ion electron method is basic medium. `NO_(3)^(ө)+Znrarr Zn^(2+)+NH_(4)^(oplus)`. |
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Answer» Two half reactions are a. `NO_(3)^(ө)rarrNH_(4)^(o+)` b.`ZnrarrZn^(2+)` First balance `O` atoms in basic medium. a. `NO_(3)^(ө)+3H_(2)OrarrNH_(4)^(o+)+6overset(ө)OH` Balancing `H` atoms in basic medium `NO_(3)^(ө)+3H_(2)+4H_(2)OrarrNH_(4)^(o+)+6overset(ө)OH+4overset(ө)OH` Balancing the charge: `NO_(3)^(ө)+8e^(-)+7H_(2)OrarrNH_(4)^(o+)+10overset(ө)OH` b. `ZnrarrZn^(2+)+2e^(-)` Now reaction is `4Zn+NO_(3)^(ө)+7H_(2)Orarr4Zn^(2+)+NH_(4)^(o+)+10overset(ө)OH` |
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| 278. |
The following redox reaction occurs in basic medium `NO_(3)^(-)Zn(s)toZn^(2+)+NH_(4)^(+)` when the above reaction is balanced such that the stoichiometric coefficients are in smallest whole number ratio, then the difference of stoichiometric coefficient of Zn (s) and `OH^(-)` ion will be:A. 4,1,7B. 7,4,1C. 4,1,10D. 1,4,10 |
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Answer» Correct Answer - C `4Zn+NO_(3)^(-)+7H_(2)Orarr4Zn^(2+)+NH_(4)^(+)+10OH^(-)` |
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| 279. |
Balance the equation in basic medium `Al + NO_(3)^(-) to Al(OH)_(4)^(-) + NH_(3)` |
| Answer» `8Al + 3NO_(3)^(-) + 18H_(2)O + 5OHto 8Al(OH)_(4)^(-)+18H_(2)O + 5OH^(-) to 8Al(OH)_(4)^(-)+3NH_(3)` | |
| 280. |
Balance the equation by lon-electron method (vi) `Cl_(2) +OH^(-) to S^(2-) + S_(2)O_(3)^(2-)` (In basic medium) |
| Answer» `Cl_(2) + 2OH^(-) to Cl^(-) + ClO^(-) + H_(2)O` | |
| 281. |
Balance the equation by lon-electron method (viii) `Al + NO_(3)^(-) to Al(OH)_(4)^(-) + NH_(3)` (In basic medium) |
| Answer» `8Al + 3NO_(3)^(-) +18H_(2)O + 5OH^(-) to 8Al(OH)_(4)^(-) + 3NH_(3)` | |
| 282. |
Balance the equation by lon-electron method (x) `Fe_(3)O_(4) + MnO_(4)^(-) to Fe_(2)O_(3) + MnO_(2)` (In basic medium) |
| Answer» `6Fe_(3)O_(4) + 2MnO_(4)^(-) + H_(2)O to 9Fe_(2)O_(3) + 2MnO_(2) + 2OH^(-)` | |
| 283. |
Balance the equation by lon-electron method (ix) `Cr_(2)O_(7)^(2-) + Fe^(2+) to Fe^(3+) +Cr^(3+)`(In acidic medium) |
| Answer» `6Fe^(2+) + Cr_(2)O_(7)^(2-) + 14H^(+) to 6Fe^(3+) + 2Cr^(3+) + 7H_(2)O` | |
| 284. |
Balance the equation by lon-electron method (iv) `N_(2)O_(4) + BrO_(3)^(-) to NO_(3)^(-) + Br^(-)` (In acidic medium) |
| Answer» `3N_(2)O_(4) + BrO_(3)^(-) + 3H_(2)O to 6NO_(3)^(-) + Br^(-) + 6H^(+)` | |
| 285. |
Balance the equation by lon-electron method (v) `Br^(-) + BrO_(3)^(-) to Br_(2)` (In acidic medium) |
| Answer» `5Br^(-) + BrO_(3)^(-) + 6H^(+) to 3Br_(2) + 3H_(2)O` | |
| 286. |
Balance the equation by lon-electron method (iii) `Cu^(2+) + SO_(2) to SO_(4)^(2-) + Cu^(+) ` (In acidic medium) |
| Answer» `2Cu^(2+) + SO_(2) + 2H_(2)O to 2Cu^(+)+ 4H^(+) +SO_(4)^(2-)` | |
| 287. |
Balance the equation by lon-electron method (ii) `MnO_(4)^(-) + H_(2)C_(2)O_(4) to Mn^(2+) + CO_(2)` (In acidic medium) |
| Answer» `2MnO_(2) + 5H_(2)C_(2)O_(4)+6H^(+) to 2Mn^(2+) + 10CO_(2)+ 8H_(2)O` | |
| 288. |
Write correctly balanced half reactions and overall equations for the following skeletal equations: a. `NO_(3)^(ө)+Bi(s)rarrBi^(3+)+NO_(2)` ( in acid solution) b. `Fe(OH)_(3)(s)+H_(2)O_(2)rarrFe(OH)(s)+H_(2)O+O_(2)` (in basic medium) c. `Cr_(2)O_(7)^(2-)+C_(2)H_(4)OrarrC_(2)H_(4)O_(2)+Cr^(3+)` (in acid solutions) d. `MnO_(4)^(ө)+H_(2)C_(2)O_(4)rarrMn^(2+)+CO_(2)uarr` (in acid solutions) e. `Al(s)+NO_(3)^(ө)rarr[Al(OH)_(4)]^(ө)+NH_(3)` (in basic solution) f. `Cr_(2)O_(7)^(2-) + Fe^(2+) rarr Fe^(2+) + Cr^(3+)` (in acid solution) g. `MnO_(4)^(Ө) + Br^(Ө) rarr Mn^(2+) + Br_(2)` (in acid solution) h. `PbO_(2) + Cl^(Ө) rarr ClO^(Ө) +[Pb(OH)_(3)]^(Ө) ` (in basic solution) |
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Answer» Correct Answer - A a. `Bi+6H^(o+)+3NO_(3)^(ө)rarr3NO_(2)+3H_(2)O+Bi^(3+)` b.`2Fe(OH)_(3)+H_(2)O_(2)rarr2Fe+3CO` `2Fe(OH)_(3)+H_(2)O_(2)rarr2Fe(OH)_(2)+2H_(2)O+O_(2)` c. `Cr_(2)O_(7)^(2-)+8H^(o+)+3C_(2)H_(4)Orarr2Cr^(3+)+4H_(2)O+3C_(2)H_(4)O_(2)` d. `2MnO_(4)^(ө)+6H^(o+)+5H_(2)C_(2)O_(4)rarr2Mn^(2+)+8H_(2)O+10CO_(2)` e. `8Al+5overset(ө)OH+3NO_(3)^(ө)+18H_(2)Orarr8Al(OH)_(4)^(ө)+3NH_(3)` f. `Cr_(2)O_(7)^(2-)+14H^(o+)+6Fe^(2+)rarr2Cr^(3+)+7H_(2)O+6Fe^(3+)` g. `2MnO_(4)^(ө)+16H^(o+)+10Br^(ө)rarr2Mn^(2+)+8H_(2)O+5Br_(2)` h. `PbO_(2)+CI^(ө)+overset(ө)OH+H_(2)Orarr[Pb(OH)_(3)]^(ө)+CIO^(ө)` |
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| 289. |
Balance the following half reactions in basis medium: (a) `CrO_(4)^(2-)(aq)rarrCr(OH)_(4)^(Ө)(aq)` (b) `CIO^(Ө)(aq) rarr Cl^(Ө)(aq)` (c ) `Bi^(3+)(aq) rarr BiO_(3)^(Ө)(aq)` |
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Answer» Correct Answer - A Basic medium a. `4H_(2)O+3e^(-)+CrO_(4)^(2-)rarr[Cr(OH)_(4)]^(ө)+4overset(ө)OH` `x-8=-2, x-4= -1` `x=6= ,x=3` b. `2H_(2)O+2e^(-)+ClO^(ө)rarrCl^(ө)+H_(2)O+2overset(ө)OH` `x-2= -1, x= -1` c. `6overset(ө)OH+cancel(3H_(2)O)+Bi^(3+)rarrBiO_(3)^(ө)+2e^(-)+overset(3)cancel(6H_(2)O)` `x=3, x-6= -1` `x=5`. |
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| 290. |
One litre of mixture of `O_2` and `O_3` at STP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 " mL of " `(M)/(10)` sodium thiosulphate solution for titration. What is the mass per cent of ozone in the mixture? Ultraviolet radiation of wavelength 300 nm can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for complete decomposition of ozone in the original mixture? |
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Answer» `O_(3)+2KI+H_(2)Orarr2KOH+I_(2)+O_(2)` `I_(2)+2Na_(2)S_(2)O_(3)rarrNa_(2)S_(4)O_(6)+2Nal` `therefore "Milli-mole of" O_(3)= "milli-mole of"I_(2)` `=(1)/(2)xx"milli-mole of"Na_(2)S_(2)O_(3)` `(milli-mol e=MxxV_("in"mL))` `=(1)/(2)xx40xx(1)/(10)=2 "milli-mole"=0.002"mole"` Total milli-mole of `O_(2)` and `O_(3)` in mixture are calculated from, `PV=nRT` `1xx1=nxx0.0821xx273` `therefore n=0.44"mole"` `therefore "Mole of" O_(2)=0.044-0.002=0.042` Now `"wt.of"O_(2)=0.042xx32g=1.344g` `"wt.of"O_(3)=0.002xx48g=0.096g` `therefore % "of" CO_(3)=(0.096)/(1.44)xx100=6.7%` No.of photon or molecules of ozone `=(0.096xx6.023xx10^(23))/(48)=1.2xx10^(21)` |
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| 291. |
Balance given following half reaction for the unbalanced whole reaction : `CrO_(4)^(2-)rarrCrO_(2)^(-)_OH^(-)` is :A. `CrO_(4)^(-2)+2H_(2)O+3e^(-)rarrCrO_(2)^(-)+4H_(2)O^(-)`B. `2CrO_(4)^(2-)+8H_(2)OrarrCrO_(2)^(-)+4H_(2)O+8OH^(-)`C. `CrO_(4)^(-2)+H_(2)OrarrCrO_(2)^(-)+H_(2)O+OH^(-)`D. `3CrO_(4)^(-2)+4H_(2)O+6e^(-)rarr2CrO_(2)^(-1)+8OH^(-)` |
| Answer» Correct Answer - A | |
| 292. |
Balance the following by ion electron method (basic medium): `Cr(OH)_(3)+IO_(3)^( ө)rarrI^(ө)+CrO_(4)^(2-)` |
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Answer» Two half reactions are: a. `Cr(OH)_(3)rarrCeO_(4)^(2-)` b. `IO_(3)^(ө)rarrI^(ө)` Balancing `O` atoms: a. `Cr(OH)_(3)+2overset(ө)OHrarrCrO_(4)^(2-)+H_(2)O` b. `IO_(3)^(ө)+3H_(2)OrarrI^(ө)+6overset(ө)OH` Balancing `H` atoms: a. `Cr(OH)_(3)+2overset(ө)OH+3overset(ө)OHrarrCrO_(4)^(2-)+H_(2)O+3H_(2)O` b. `IO_(3)^(ө)+3H_(2)OrarrI^(ө)+6overset(ө)OH` Balancing the charge: a. `Cr(OH)_(3)5overset(ө)OHrarrCrO_(4)^(2-)+4H_(2)O^(+)3e^(-)` b. `IO_(3)^(ө)+3H_(2)O+6e^(-)rarrI^(ө)+6overset(ө)OH` Adding (a) and (b), we get `2Cr(OH)_(3)+IO_(3)^(ө)+4overset(ө)OHrarr2CrO_(4)^(2-)+I^(ө)+5H_(2)O` |
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| 293. |
Balance the following reaction by ion electrons method ( acidic medium). `As_(2)S_(3)-NO_(3)^(ө)rarrS+NO_(2)+AsO_(4)^(3-)` |
| Answer» `3As_(2)S_(3) + 10NO_(3)^(-) + 4H_(2)O to 6AsO_(4)^(-3) + 9S+ 10NO + 8H^(+)` | |
| 294. |
`Cr(OH)_(3)+ClO^(-)+3OH^(-)rarr?+Cl^(-)3H_(2)O`. The missing ion isA. `Cr_(2)O_(7)^(2-)`B. `Cr^(3+)`C. `CrO_(4)^(2-)`D. `Cr_(2)O_(3)` |
| Answer» Correct Answer - C | |
| 295. |
The oxidation number of sulphur in `S_(8), S_(2)F_(2) and H_(2)S` respectively are:A. `0,+1 and -2`B. `0,+2 and -2`C. `+2, +1 and -1`D. `-2, +1 and -2` |
| Answer» Correct Answer - A | |
| 296. |
The oxidation number of sulphur in `S_(8), S_(2)F_(2) and H_(2)S` respectively are:A. 0, +1 and -2B. `+2,+1 and -2`C. 0,+1 and +2D. `-2, +1 and -2` |
| Answer» Correct Answer - A | |
| 297. |
The oxidation states of S atom in `S_(4)O_(6)O^(2-)` from left to right respectively are `O-underset(O)underset(||)overset(O)overset(||)S-S-S-underset(O)underset(||)overset(O)overset(||)S-O`A. `+6,0,0,+6`B. `+3,+1,+1,+3`C. `+5,0,0,+4`D. `+4,+1,+1,+4` |
| Answer» Correct Answer - C | |
| 298. |
Statement Oxidation number of metals in metal carbonyls is zero. Explanation The oxidation number of `CO` has been taken to be zero.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 The explanation is correct reason for statement. |
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| 299. |
Equivalent weight of `H_(3)PO_(2)` in a reaction is found to be half of its molecular weight. It can be due to itsA. oxidation to `H_(3)PO_(3)`B. reaction of two `H^(+)` ionsC. oxidation to `H_(3)PO_(4)`D. reduction to `PH_(3)` |
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Answer» Correct Answer - A (`a`) It is monobasic acid, hence only one `H^(+)` can be neutralised `E=(M)/(1)` (`b`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(+3)underset(uarr)(H_(3)PO_(3))` `E=(M)/(2)` (`c`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(+5)underset(uarr)(H_(3)PO_(4))` `E=(M)/(4)` (`d`) `underset(+1)underset(uarr)(H_(3)PO_(2)) rarr underset(-3)underset(uarr)(PH_(3))` `E=(M)/(4)` |
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| 300. |
Statement The oxidation number of an element in its free or uncombined from is zero. Explanation The oxidation number of a monoatomic cation or anion is equal to its charge.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 - D These are facts. |
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