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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.
| 1001. |
oxidation states available for a particular elements are normally more than the valencies A brown complex has the formula: `[Fe(H_(2))(O)_(5)NO]SO_(4)` the oxidation number of iron is: In which compound,Mn exhibits highest oxidation state?A. `MnO_(2)`B. `Mn_(3)O_(4)`C. `K_(2)MnO_(4)`D. `MnSO_(4)` |
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Answer» Correct Answer - C Mn has miximim (+6) oxidation state in `K_(2)MnO_(4)` |
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| 1002. |
In the following reaction hydrazine is oxidized `N_(2)` `N_(2)H_(4)+OH^(-)toN_(2)+H_(2)O+e` The equivalent weight of `N_(2)H_(4)` (hydrazine) is:A. 8B. 16C. 32D. 64 |
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Answer» Correct Answer - a |
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| 1003. |
What is the difference in oxidation state of nitrogen in between hydroxyl amine `(NH_(2)OH)` and hydrazine `(N_(2)H_(4))` ?A. `+5`B. `+3`C. `-3`D. 1 |
| Answer» Correct Answer - D | |
| 1004. |
The equivalent mass of `MnSO_(4)` is half its molecular mass when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(-)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - b |
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| 1005. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(2)`C. `MnO_(4)^(-)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - B Eq.wt`=M//2` `x-"factor"=2` `:. overset((+2))(MnSO_(4)) rarr overset((+4))(MnO_(2))` |
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| 1006. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_(2)O_(3)`B. `MnO_(4)^(-)`C. `MnO_(2)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - C Equivalent weight `=(molar mass)/("change in ON")=(M)/(2)` Thus, change in oxidation number `=2units` Change in `O.N.` (`a`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+3)underset(uarr)(Mn_(2)O_(3)) 1 unit E=(M)/(1)` (`b`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+4)underset(uarr)(MnO_(2)) 2 unit E=(M)/(2)` (`c`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+7)underset(uarr)(MnO_(4)^(-)) 5 unit E=(M)/(5)` (`d`) `underset(+2)underset(uarr)(MnSO_(4)) rarr underset(+6)underset(uarr)(MnO_(4)^(2-)) 4 unit E=(M)/(4)` |
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| 1007. |
A redox reaction isA. endothermicB. exothermicC. either endothermic or exothermicD. neither endothermic nor exothermic |
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Answer» Correct Answer - B A redox reaction involves complete or partial loss of electrons (an energy consuming process) and complete or partial gain of electrons (an energy releasing process). Since a redox reaction is spontaneous, the amount of energy released must be greater than the amount of energy absorbed. Thus, all redox reactions are exothermic in nature. |
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| 1008. |
Which fo the following metals cannot deplace hydrogen from nonoxidizing acids ? (i) `Au" "` (ii) `Pt` (iii) `Ag" "` (iv) `Cu`A. (i) , (ii), (iii)B. (ii), (iii), (iv)C. (i), (ii)D. (i), (ii), (iii), (iv) |
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Answer» Correct Answer - D Elements such as Sb (a metalloid) , `Cu, Hg, Ag, Pt`, and `Au` (metals ) which lie below H in the activity series cannot displace H from non-oxidizing acids. |
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| 1009. |
Which of the following metals cannot displace hydrogen grom nonoxidizing acids ? `Pb` (ii) `Sn` (iii) `Ni` (iv) `Zn`.A. (i), (ii), (iii)B. (ii), (iii), (iv)C. (i), (ii), (iii) (iv)D. (i), (ii) |
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Answer» Correct Answer - C Many metals like Pb, Sn,Ni, Co, and `Cd` which do not react with water are capable of displacing hydrogen from non oxidizing acids : ` Cd(s) + 2HCl(aq.) rarr CdCl_2(aq.) + H_2 (g)` `Sn(s) + 2HCl(aq.) rarr SnCl_2(aq.) + H_2(g)` These metals lie above H in the activity series. |
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| 1010. |
Which fo the following metals can displace hydrogen from cold water, steam, and nonoxidizing acids ?A. `Ni`B. `Li`C. `Mn`D. `Mg` |
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Answer» Correct Answer - B `Li` is the most reactive metal in the activity series. |
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| 1011. |
According to international converntion, standard reduction potentials are now callde standard potential. (i) the oxidizing poweer of the species on the left side fo the reaction dereases (ii) the reducing power fo the species on the right -hand side fo the reaction increased (iii) the oxidizing power of the species on the left side fo the reaction increases (iv) the reducing power of the species on the right side fo the reaction decrease .A. (iii), (iv)B. (i), (ii)C. (i), (iii)D. (ii), (iii) |
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Answer» Correct Answer - B Standard reduction potential measures the tendency to undergo the following reaction : Oxidized form `+n e^(-) hArr` Reduced from. With decrease of ` E^Ө`, the tendency to undergo forward reaction decreases while the tendency to undergo backward reaction increases. |
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| 1012. |
Assertion : A metal having negative reduction potential when dipped in the solution of its own ions has a tendency to pass into solution . Reason : Metals undergo reduction .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. |
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Answer» Correct Answer - C A metal having negative reduction potential has high tendency to get oxidised and hence metal passes into the solution. |
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| 1013. |
How many grams of potassium dichromate are required to oxidise 20.0 g of `Fe^(2+) "in" FeSO_(4) "to" Fe^(3+)` if the reaction is carried out in acidic medium? Molar masses of `K_(2)Cr_(2)O_(7) "and" FeSO_(7)` are 294 and 152 respectivelyA. 6.45 gB. 7.45 gC. 8.45 gD. 9.45 g |
| Answer» Correct Answer - A | |
| 1014. |
Which of the following is not an example of redox reaction?A. `CuO+H_(2)toCu+H_(2)O`B. `Fe_(2)O_(3)+3COto3Fe+3CO_(2)`C. `2K+F_(2)to2KF`D. `BaCl_(2)+H_(2)SO_(4)toBaSO_(4)+2HCl` |
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Answer» Correct Answer - d Following are the examples of redox reaction (a) `CuO+H_(2)toCu+H_(2)O` (b) `Fe_(2)O_(3)+3COto3Fe+3CO_(2)` ( c) `2K+F_(2)to2KF` Option (d) is not an example of redox reaction. |
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| 1015. |
`E^(θ)` values of some redox couples are given below. On the basis of these values choose the correct option. `E^(θ)` values: `Brt_(2)//Br^(-) = +1.90` `Ag^(+)//Ag(s)=+0.80` `Cu^(2+)//Cu(s)=+0.34, I_(2)(s)//I^(-)=+0.54`A. Cu will reduce `Br^(-)`B. Cu will reduce AgC. Cu will reduce `I^(-)`D. Cu will reduce `Br_(2)` |
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Answer» Correct Answer - d Given that `E^(@)` values of `Br//Br^(-)=+1.90V` `Ag//Ag^(+)=-0.80V` `Cu^(2+)//Cu(s)=+0.34V` `I^(-)//I_(2)(s)=0.54V` `Br^(-)//Br_(2)=-1.90V` The `E^(@)` values show that copper will reduce `Br_(2)`, If the `E^(@)` of the following redox reaction is positive. `2Cu+Br_(2)toCuBr_(2)` Now, `CutoCu^(2+)+2e^(-), E^(@)=-0.34V` `(Br_(2)+2e^(-)to2Br^(-), E=+1.09V)/(Cu+2Br_(2)toCuBr_(2),E^(@)=+0.75V)` Since, `E^(@)` of this reaction is positive, therfore, Cu can reduce `Br_(2)`. While other reaction will give negative value. |
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| 1016. |
If a spoon of copper metal is placed in a solution of `FeSO_(4)`, what will be the correct observation ? A. Copper is dissolved in `FeSO_(4)` to give brown deposit.B. No reaction takes place.C. Iron is deposited on copper spoon.D. Both copper and iron are precipitated. |
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Answer» Correct Answer - B Since reduction potential of copper is higher than iron it does not get oxidised and no reaction takes place. |
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| 1017. |
Which of the following is not a correct statement about electrochemical series of reduction potentials ?A. The standard electrode potential of hydrogen is 0.00 volts.B. Active non-metals have positive reduction potentials.C. Active metals have negative reduction potentials.D. Metals which have positive reduction potentials are good reducing agent. |
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Answer» Correct Answer - D Metals with negative reduction potentials undergo oxidation and act as good reducing agents. |
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| 1018. |
The more positive the value of `E^(θ)`, the greater is the trendency of the species to get reduced. Using the standard electrode potential of redox coples given below find out which of the following is the strongest oxidising agent. `E^(θ)` values: `Fe^(3+)//Fe^(2+) = +0.77` `I_(2)(s)//I^(-) = +0.54`, `Cu^(2+)//Cu = +0.34, Ag^(+)//A = 0.80V`A. `Fe^(3+)`B. `I_(2)(s)`C. `Cu^(2+)`D. `Ag^(+)` |
| Answer» `Ag^(+)` ion has the maximum tendency to get reduced since `E^(@) Ag^(+)//Ag` is the maximum.Therefore ,`Ag^(+)`ion is the strongest oxidising agent | |
| 1019. |
Which of the following is a redox reaction?A. `H_(2)SO_(4)` with NaOHB. In atmosphere `O_(3) "form" O_(2)` by lightiningC. Nitrogen oxide formed from nitrogen and oxygen by lightiningD. Evaporation of water. |
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Answer» It is a redox reactioon as there `overset(0)N_(2)+overset(0)O_(2)rarr2overset(+2)Noverset(-2)O` is loss as well as gain of electrons in the reation |
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| 1020. |
The more positive the value of `E^(θ)`, the greater is the trendency of the species to get reduced. Using the standard electrode potential of redox coples given below find out which of the following is the strongest oxidising agent. `E^(θ)` values: `Fe^(3+)//Fe^(2+) = +0.77` `I_(2)(s)//I^(-) = +0.54`, `Cu^(2+)//Cu = +0.34, Ag^(+)//A = 0.80V`A. `Fe^(3+)`B. `I_(2(s))`C. `Cu^(2+)`D. Ag |
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Answer» Correct Answer - D Strongest oxidising agent means it has greater tendency to get reduced. Thus, `Ag^(+)` having more positive `E^(@)` value, is the strongest oxidising agent. |
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| 1021. |
In the reaction `2Ag+2H_(2)SO_(4)rarrAg_(2)SO_(4)+2H_(2)O+SO_(2),H_(2)SO_(40`acts as `a//an`A. Oxidising agentB. Reducing agentC. CatalystD. Acid as well as oxidant |
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Answer» Correct Answer - D In `2Ag+H_(2)SO_(4)rarrAg_(2)SO_(4)+2H_(2)O+SO_(2)` `H_(2)SO_(4)` act as acid (S retain it O.N. = +6 in `Ag_(2)SO_(4)`) as well as an oxidising agent `(overset(+6)(H_(2)SO_(4))rarroverset(+4)(SO_(2)))`. |
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| 1022. |
Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overal chemical change is called redox reaction. Redox reactions are of three types : (i) Intermolecular erdox reactions, (ii) Auto-redox or disproportionation reaction, and (iii) Intramolecular redox reactions. What is the percentage of `Fe^(3+)` in `Fe_(0.93)O_(100)` ?A. `15.05%`B. `84.95%`C. `20%`D. `80%` |
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Answer» Correct Answer - A Let `Fe^(3+)` be `a`. Also oxidation no. of `Fe` in `Fe_(0.93)O_(100)=(200)/(93)=2.15` `3xxa+(100-a)xx2=2.15xx100` `:. a=15.05%` |
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| 1023. |
Equivalent weight of `Br_(2)` in the following reaction is `Br_(2)+HgO+H_(2)OtoHgBr_(2).HgO+HBrO` (given `Br=80`)A. `(160)/(3)`B. `80`C. `160`D. `160xx3` |
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Answer» Correct Answer - c |
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| 1024. |
Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overal chemical change is called redox reaction. Redox reactions are of three types : (i) Intermolecular erdox reactions, (ii) Auto-redox or disproportionation reaction, and (iii) Intramolecular redox reactions. Oxidation number of `Y` in `YBa_(2)Cu_(3)O_(7)` is `+3`, then oxidation number of `Cu` is :A. `+7//3`B. `+5//3`C. `+2`D. `+1` |
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Answer» Correct Answer - A `3+2xx2+3xxa+7xx(-2)=0` `:. A= +7//3` |
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| 1025. |
Using the following Latimer diagram for bromine, `pH=0, BrO_(4)^(-) overset(1.82 V)rarr BrO_(3)^(-)overset(1.50 V)rarr HBrO overset(1.595 V) rarr Br_(2) overset(1.06552 V) rarr Br^(-)` the species undergoing disproportionation isA. `BrO_(4)^(-)`B. `BrO_(3)^(-)`C. HBrOD. `Br_(2)` |
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Answer» Correct Answer - C It the potential to the left of a given chemical species is less than that to the right, the species will undergo disproportionation. |
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| 1026. |
In the rection: `2Ag+2H_(2)SO_(4)rarrAg_(2)SO_(4)+2H_(2)O+SO_(2)` Sulphuric acid acts as:A. Oxidising agentB. Reducing agentC. CatalystD. Acid as well as oxidant |
| Answer» `H_(2)SO_(4)` acts both as acid and oxidant | |
| 1027. |
Indentify disproportionation rectionA. `CH_(4)+2O_(2)rarrCO_(2)+2H_(2)O`B. `CH_(4)+4CI_(2)rarrC CI_(4)+4HCI`C. `2F_(2)+2Oh^(-)rarr2F^(-)+OF_(2)+H_(2)O`D. `2NO_(2)+2OH^(-)rarrNO_(2)^(-)+NO_(3)^(-)+H_(2)O` |
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Answer» It is disproportionation reaction `2NO_(2)+2Oh^(-)rarrNO_(3)^(-)+H_(2)O` `NO_(2)`: Oxidation state of `N=+4` `NO_(2)^(-)`: oxidation state of N=+3 `NO_(3)^(-)`: Oxidation state of N=+5 It is disproportion rection since there is decrease as well as increase in oxidation state of N. |
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| 1028. |
Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overal chemical change is called redox reaction. Redox reactions are of three types : (i) Intermolecular erdox reactions, (ii) Auto-redox or disproportionation reaction, and (iii) Intramolecular redox reactions. In which of the following `H_(2)O_(2)` acts as reductant ? (I) `H_(2)O_(2) + O_(3) rarr H_(2)O+2O_(2)` (II) `PbO_(2) + H_(2)O_(2) rarr PbO+H_(2)O + O_(2)` (III) `HCHO + H_(2)O_(2) rarr HCOOH + H_(2)O` (IV) `Cl_(2) + H_(2)O_(2) rarr 2HCl + O_(2)`A. I, II, IVB. I,II,IIIC. I,IVD. II,III |
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Answer» Correct Answer - A Oxidation number of `O` in `H_(2)O_(2)(-1)` increases to zero. |
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| 1029. |
Oxidation is de-electronation whereas reduction is electronation. Oxidants are the substances which oxidise others and reduced themselves. On the other hand reductants are the substances which reduce others and oxidised themselves. The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation-reduction occur simultaneously and the overal chemical change is called redox reaction. Redox reactions are of three types : (i) Intermolecular erdox reactions, (ii) Auto-redox or disproportionation reaction, and (iii) Intramolecular redox reactions. Which statement is wrong about `CrO_(5)` ?A. It has butterfly structureB. Oxidation number of `Cr` is `+ 10`C. Oxidation number of `Cr` is `+6`D. It reacts with `H_(2)SO_(4)` to give `Cr_(2)(SO_(4))_(3)` and `O_(2)` |
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Answer» Correct Answer - B Oxidation no. of `Cr` is `+6` due to four `O`-atoms involved in two peroxide bonds. |
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| 1030. |
Identify the oxidised and reduced species in the following reactions: (a) `CH_(4(g))+4Cl_(2(g))rarrC Cl_(4(g))+4HCl_((g))` (b) `MnO_(2(s))+C_(2)H_(2)O_(4(aq.))overset(2H^(+))rarrMn_((aq.))^(2+)+2CO_(2(g))+2H_(2)O_((l))` (c ) `I_(2(aq.))+2S_(2)O_(3(aq.))^(2-)rarr2I_((aq.))^(-)+S_(4)O_(6)^(2-)` (d) `Cl_(2(g))+2Br_((aq.))^(-)rarr2Cl_((aq.))^(-)+Br_(2(aq.))` |
| Answer» `{:(,"Reduced",,"Oxidised",),((a),Cl_(2),,CH_(4),),((b),MnO_(2),,C_(2)H_(2)O_(4(aq.)),),((c ),I_(2),,S_(2)O_(3)^(2-),),((d),Cl_(2),,Br^(-),):}` | |
| 1031. |
In which of the folowing rection nitrogen is not reduced ?A. `NO_(2)rarrNO^(2^(+))`B. `BNO_(3)^(-)rarrNO`C. `NO_(3)^(-)rarrNH_(4)^(+)`D. `NH_(4)^(+)rarrN_(2)` |
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Answer» `overset(+4)NO_(2)rarroverset(+3)NO_(2)^(+),overset(+5)NO_(3)^(-)rarroverset(+2)NO` `overset(+5)O_(3)^(-)rarroverset(-3)H_(4),overset(-3)NH_(4)^(+)rarroverset(0)N_(2)` In .there is increase in the O.N of N also.Therefore, N atom is oxidi9sed and not reduced N atom is oxidised and not reuced. |
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| 1032. |
`28 NO_(3)^(-)+3As_(2)S_(3)+4H_(2)O rarr 6AsO_(4)^(3-)+28NO+9SO_(4)^(2-)+H^(+)` What will be the equivalent mass of As_(2)S_(3)` in the above reaction?A. `(M.wt.)/(2)`B. `(M.wt.)/(4)`C. `(M.wt.)/(24)`D. `(M.wt.)/(28)` |
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Answer» Correct Answer - D `Eq. mass=(mol ecular mass)/(n-"factor")` `As_(2)^(+3) rarr 2AS^(+5) n-"factor" 4` `S_(2)^(-2) rarr 3 overset(+6)(S)` `24` total `n-"factor"=28` Eq. mass `=(m.wt.)/(28)` |
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| 1033. |
In which of the following reaction, oxidation number of Cr has been affected?A. `2CrO_(4)^(2-) + 2H^(+) to Cr_(2)O_(7)^(2-) + H_(2)O`B. `Cr_(2)O_(7)^(2-) + 2OH^(-) to 2CrO_(4)^(2-) + H_(2)O`C. `(NH_(4))_(2)Cr_(2)O_(7) to N_(2) + Cr_(2)O_(3) + 4H_(2)O`D. `CrO_(2)Cl_(2) + 2OH^(-) to CrO_(4)^(2-) + 2HCl` |
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Answer» Correct Answer - C `(NH_(4))_(2) + underset(+6)underset(uarr)(Cr_(2))O_(7) to underset(+3)underset(uarr)(Cr_(2))O_(3)` |
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| 1034. |
which of the following statements is not correct about the reaction given below? `K_(4)[Fe(CN)_(6)]overset("Oxidation")rarrFe^(3+)+CO_(2)+NO_(3)^(ө)`A. `Fe` is oxidised from `Fe^(2+)` to `Fe^(3+)`B. Carbon is oxidised from `C^(2+)` to `C^(4+)`C. `N` is oxidised from `N^(3-)` to `N^(5+)`D. carbenes |
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Answer» Correct Answer - D In `CN^(ө)`, oxidation of `C` is `+2` , and it changes to `+4` oxidation state in `CO_(2)`. So `C` is also oxidised. |
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| 1035. |
which of the following statements is not correct about the reaction given below? `K_(4)[Fe(CN)_(6)]overset("Oxidation")rarrFe^(3+)+CO_(2)+NO_(3)^(ө)`A. Fe is oxidised from `Fe^(2+)` to `Fe^(3+)`.B. Carbon is oxidised from `C^(2+)` to `C^(4+)`.C. N is oxidised from `N^(3-)` to `N^(5+)`.D. Carbon is not oxidised. |
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Answer» Correct Answer - D In `CN^(-)` , oxidation of C is +2, and it changes to +4 oxidation state in `CO_(2)`. So C is also oxidised. |
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| 1036. |
`C_(2)H_(6)(g)+nO_(2) rarr CO_(2)(g)+H_(2)O(l)` In this equation, the ratio of the coefficients of `CO_(2)` and `H_(2)O` isA. `1:1`B. `2:3`C. `3:2`D. `1:3` |
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Answer» Correct Answer - B The balanced equation is `2C_(2)H_(6) + 5O_(2) to 4CO_(2) + 6H_(2)O` Ratioof the coefficient of `CO_(2)` and `H_(2)O` is `4:6` and `2:3` |
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| 1037. |
`[Co(H_(2)N-CHCH_(2)_NH_(2))_(3)]_(2)S_(3)overset("oxidation")(rarr)Co^(+4)+CO_(3)^(-2)+NO_(3)^(-)+SO_(4)^(-2)` What is the equivalent weight of the reactant in the avobe reaction?A. `(3M)/(182)`B. `(M)/(182)`C. `(11M)/(182)`D. `(7M)/(182)` |
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Answer» Correct Answer - b |
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| 1038. |
Consider the following reactions, `C_(2) H_(6)(g) + nO_(2) rarr CO_(2) (g) + H_(2) O(l)` In this equation, ratio of the coefficient of `CO_(2)` and `H_(2)O` isA. `1 : 1`B. `2 : 3`C. `3 : 2`D. `1 : 3` |
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Answer» Correct Answer - B The balancd equation is `2C_(2)H_(6) + 7O_(2) rarr 4CO_(2) + 6H_(2)O` |
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| 1039. |
Assertion :- Oxidation state of Hydrogen is `+1` in `H_(2)O` while `-1` in `CaH_(2)`. Reason :- `CaH_(2)` is a metal hydride and for metal hydrides, hydrogen is assigned the oxidation number of `-1`.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 | |
| 1040. |
Assertion: Oxidation number of carbon in `CH_(2)O` is zero. Reason: `CH_(2)O` formaldehyde, is a covalent compound.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 and reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If the assertion false and 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. Oxidation number can be calculated using some rules. `H` is assigned `+1` oxidation state and `O` has oxidation number `-2` `:. ` O.No. of `C` in `CH_(2)O`: O.no. of `C+2(+1)+(-2)=0` `:. ` O.NO. of `C=0` |
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| 1041. |
Assertion :- Oxidation number of carbon in `CH_(2)O` is zero. Reason :- `CH_(2)O` (formaldehyde) is a covalent compound.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 - B | |
| 1042. |
When potassium permanganate, `KMnO_(4)` , is added to an acidified solution of oxalic acid, `H_(2)C_(2)O_(4)` , the products are `CO_(2)` gas and `Mn^(2+)` ions. What is the reducing agent in this reaction?A. `KMnO_(4)`B. `H_(2)C_(2)O_(4)`C. `H_(3)O^(+)`D. `CO_(2)` |
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Answer» Correct Answer - b |
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| 1043. |
Balance the following reaction by oxidation number and ion electron method: `KMnO_(4)+H_(2)SO_(4)+K_(2)C_(2)O_(4)rarrMnSO_(4)+CO_(2)+K_(2)SO_(4)` |
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Answer» Oxidation number method: Follow the step given below to balance the reaction given above. Assign oxidation state to the atoms that the oxidised or reduced. Note that oxidation state of `O,K,H` and `S` is same on both sides. `KMoverset(+7)nO_(4)+H_(2)SO_(4)+K_(2)overset(+3)C_(2)O_(4)rarroverset(+2)MnSO_(4)+overset(+4)CO_(2)` Write two half reactions and balance the atoms as follows: a. `Mn^(7+)rarrMn^(2+)`("reduction") b. `C_(2)^(3+)rarr2C^(4+)("oxidation")` Balance the charge by adding `5e^(-)` to left of (a) and `2e^(-)` to the right of (b). a. `Mn^(7+)+5e^(-)rarrMn^(2+)`(reduction) b. `C_(2)^(3+)rarr2C^(4+)+2e^(-)`(oxidation) Add two half reactions after multiplying (a) by `2` and (b) by `5` in order that electrons from both sides cancel each other. `2Mn^(7+)+5C_(2)^(3-)rarr2Mn^(2+)+10C^(4+)` Now compare this balanced equation with the molecular unbalanced equation as follows: `2KMnO_(4)+5K_(2)C_(2)O_(4)rarr2MnSO_(4)+10CO_(2)` As the charges balanced, now balance `K` atoms on both sides by adding `6K_(2)SO_(2)` on the rignt. `2KMnO_(4)+5K_(2)C_(2)O_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)` Now to balance `SO_(4)^(2-)` ions on both sides, and `8H_(2)SO_(4)` on the left. `2KMnO_(4)+5K_(2)C_(2)O_(4)+8H_(2)SO_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)` Finally add `8H_(2)O` on the right to balance. `O` and `H` atoms to get balanced equation. `2KMnO_(4)+5K_(2)C_(2)O_(4)+8H_(2)SO_(4)rarr2MnSO_(4)+10CO_(2)+6K_(2)SO_(4)+8H_(2)O` Note: Make a final check by counting `O` atoms on both sides. |
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| 1044. |
The oxidation nunber of P in `HP_(2)O_(7)^(-)` ion isA. `+5`B. `+6`C. `+7`D. `+3` |
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Answer» Correct Answer - B Let O.N. of P in `HP_(2)O_(7)^(-)` ion = x `:. 1+2x + 7 (-2) = -1` `2x = -1 + 13 or x =6`. |
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| 1045. |
Balance the chemical equation by the oxidation number method. (v) `SnO_(2) + C to Sn + CO` |
| Answer» `SnO_(2) + 2C to Sn + 2CO` | |
| 1046. |
Balance the chemical equation by the oxidation number method. (vi) `FeCl_(3) + H_(2)S to FeCl_(2) + S + HCl ` |
| Answer» `2FeCl_(3) + H_(2)S to 2FeCl_(2) + S+ 2HCl` | |
| 1047. |
When `NaCl` is dissolved in water the sodium ion becomesA. OxidisedB. ReducedC. HydrolysedD. Hydrated |
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Answer» Correct Answer - D `NaCl+H_(2)O rarr NaOH+HCl` Sodium ion hydrated in water. |
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| 1048. |
When `Cl_(2)` gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes fromA. Zero to `-1` and zero to `+3`B. Zero to `+1` and zero to `-3`C. Zero to `+1` and zero to `-5`D. Zero to `-1` and zero to `+5` |
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Answer» Correct Answer - D `underset("O.N.=0)(Cl_(2))+NaOHrarrunderset(-1)(Cl)^(Theta)+underset(+5)(ClO_(3))^(Theta)` |
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| 1049. |
When `CI_(2)` gas reacts wioth hot and concentrated solution of sodium hydroxide the oxidation number of chloirine changes from:A. `Zero to +1 and zsero to -5`B. `Zero to -1 and zsero to +5`C. `Zero to -1 and zsero to +3`D. `Zero to +1 and zsero to -3` |
| Answer» `3overset(CI_(2))+6NaOHrarr5Naoverset(-1)CI+Naoverset(+5)CIO_(3)+3H_(2)O` | |
| 1050. |
In the ions equatio, `BrO_(3)^(-) + 6H^(+) + xe^(-) rarr Br^(3+) + 3H_(2)O`, the value of x isA. 6B. 2C. 4D. 3 |
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Answer» Correct Answer - B Balanced reaction is `BrO_(3)^(-) + 6H^(+) + 2e^(-) rarr Br^(3+) + 3H_(2)O` `:.` Value of x is 2 |
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