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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.
| 151. |
In a reaction, `Cr_(2)O_(7)^(2-)` is reduced to `Cr^(3+)`. What is concentration of `0.1M K_(2)Cr_(2)O_(7)` in equivalent per litre? `Cr_(2)O_(7)^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_(2)O` |
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Answer» `N = M xx " valence factor"` `therefore N = 0.1 xx 6` `(Cr_(2)^(6+) + 6erarr 2Cr_(3+))` or `N_(K_(2)Cr_(2)O_(7)) = 0.6 N` |
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| 152. |
Statement In acidic medium equivalent weight of `K_(2)Cr_(2)O_(7)` is `49`. Explanation `(Cr^(6+))_(2)+6erarr2Cr^(3+)`, Thus `E=(M)/(6)`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 One mole of `K_(2)Cr_(2)O_(7)` shows a change of six `N` electrons. |
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| 153. |
What is the value of n in the following equation : `Cr(OH)_(4)^(-)+OH^(-) rarr CrO_(4)^(2-)+H_(2)O+n e`?A. 3B. 6C. 5D. 2 |
| Answer» Correct Answer - A | |
| 154. |
What mass of `N_(2)H_(4)` can be oxidised to `N_(2)` by `24.0 g K_(2)CrO_(4)`, which is reduced to `Cr(OH)_(4)^(-)`? |
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Answer» `N_(2)^(2-)rarrN_(2)^(0)+4e` `Cr^(6+)=3erarrCr^(3+)` Meq. Of `N_(2)H_(4)=Meq. Of K_(2)CrO_(4)` `(w)/(32//4)xx1000=(24)/(194.2//3)xx1000` `(Etwt.=(M.wt)/("Valence factor"))` `w_(N_(2)H_(4)=2.97g` |
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| 155. |
`12.53 mL` of `0.0509MSeO_(2)` reacted with `25.52 mL 0.1M CrSO_(4)` solution. In the raeaction `Cr^(2+)` was oxidised to `Cr^(3+)`. To what oxidation state selenium was converted in the reaction? Write the redox change for `SeO_(2)`. |
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Answer» `Se^(4+) + (a-4)e rarr Se^(a+)` `n = a-4` ("where `n` is valence factor") or `Cr^(2+) rarr Cr^(3+) + e` `"Meq. of" SeO_(2) = "Meq. of" CrSO_(4)` `0.0509 xx 12.53 = 0.1 xx 1 xx 25.52` `therefore n = 4` `a-4=4` or `a = 0` Thus redox change is : `Se^(4+) + 4e rarr Se^(0)` |
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| 156. |
Which of the following statemetns about tailing of `Hg` is`//`are correct?A. It is due to `Hg_(2)O`.B. It is due to `HgO`C. It is removed by `H_(2)O_(2)`D. It is removed by `O_(3)` |
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Answer» Correct Answer - A::C In tailing of `Hg`, it loses its meniscus and sticks to glass, due to solubility of `Hg_(2)O "in" Hg`, when `Hg` gets oxidised to `Hg_(2)^(2+)` by `O_(3)`. It is removed by `H_(2)O_(2)`. `2Hg+O_(3)rarrHg_(2)O+O_(2)` `Hg_(2)O+H_(2)O_(2)rarr2Hg+H_(2)O+O_(2)`. |
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| 157. |
How many equivalents are there per mole of `H_(2)S` in its oxidation to `SO_(2)`? |
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Answer» `H_(2)S rarrSO_(2)` or `S^(2-) rarrS^(4+) + 6e` `N_(H_(2)S) = M_(H_(2)S) xx 6` `because ("Moles")/("Equivalent") = (N)/(M) = 6` Thus one mole of `H_(2)S` has `6` equivalent in it. |
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| 158. |
Statement The equivalence point refers the condition where equivalents of one species reacts with same number of equivalent of other species. Explanation The end point of titration is exactly equal to equivalence point.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 - A The equivalent point is nearly same but not exactly same to end point. However for all partical purposes the two are taken same. |
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| 159. |
Which is the best description of the behaviour of bromine in the reaction given below `H_(2)O+Br_(2) rarr HOBr+HBr`A. Oxidised onlyB. Reduced onlyC. Proton acceptor onlyD. Both oxidised and reduced |
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Answer» Correct Answer - D `H_(2)O+underset(0)(Br_(2)) rarr underset(+1)(HOBr)+underset(-1)(HBr)` In the above reaction the oxidation number of `Br_(2)` increases from zero (in `Br_(2)`) to `+1` (in `HOBr`) and decreases from zero `(Br_(3))` to `-1` (in `HBr`). Thus `Br_(2)` is oxidised as well as as reduced and hence it is a redox reaction. |
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| 160. |
Which is the best description of the behaviour of bromine in the reaction given below `H_(2)O+Br_(2) rarr HOBr+HBr`A. Both oxidized and reducedB. OxidizedC. Reduced onlyD. Proton acceptor only |
| Answer» Correct Answer - A | |
| 161. |
Which is the best description of the behaviour of bromine in the reaction given below `H_(2)O+Br_(2) rarr HOBr+HBr`A. Reduced onlyB. Proton acceptor onlyC. Both oxidised and reducedD. Oxidised only |
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Answer» Correct Answer - C `H_(2)O+overset(0)Br_(2)rarroverset(+1)(HOBr)+overset(-1)(HBr)` In this reaction O.N of bromine increases from 0 (in `Br_(2)`) to +1 (in HOBr). Thus, it is both oxidised and reduced. |
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| 162. |
The oxidation number of `P` in `Mg_(2)P_(2)O_(7)` isA. `+3`B. `+2`C. `+5`D. `-3` |
| Answer» Correct Answer - C | |
| 163. |
Assertion : Displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution. Reason : Fluorine being highly reactive attacks water and displaces the oxygen of water.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 | |
| 164. |
What is the coefficients of `I_(2)` (s) when the reaction below is balanced with smallest whole number coefficients? ____`Cr_2O_(7)^(2-)(aq)+"____" .^(-)(aq)+"____"H^(+)(aq)to"_____"I_(2)(s)+"___"Cr^(3+)(aq)+"___"H_(2)O(l)`A. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - b |
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| 165. |
What is ther coefficient for `H^(+)` when the half equation is balanced with the smallest whole number coefficients?A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - b |
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| 166. |
When the equation below is balanced correctly using the simplest whole number coefficients, what is the coefficient for `CO_(2)(g)` ? __`Cr_(2)O_(7)^(2-)(aq)+` __`H_(2)C_(2)O_(4)(aq)+`__`H_(2)O(l)rarr`__`Cr^(3+)(aq)+`__`CO_(2)(g)+`__`H_(2)O(l)`A. 4B. 6C. 8D. 12 |
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Answer» Correct Answer - b |
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| 167. |
The O.N of N in `(NH_(4)_(2) SO_(4))` is :A. `+3`B. `+2`C. `+5`D. `-3` |
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Answer» Correct Answer - D In `NH_(4)^(+)` ion : `overset(x+1)[NH_(4)]^(+)` x+4(1)=+1,x=-3 O.N of N=-3 |
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| 168. |
What is the coefficient for `O_(2)` when the following reaction __`As_(2)S_(3)+` __`O_(2)rarrAs_(2)O_(3)+`__`SO_(2)` is correctly balanced with the smallest integer coefficientA. 5B. 6C. 8D. 9 |
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Answer» Correct Answer - d |
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| 169. |
Which of the following act both as oxidant & reductant :-A. `H_(2)S`B. `SO_(3)`C. `H_(2)O_(2)`D. `F_(2)` |
| Answer» Correct Answer - C | |
| 170. |
How does `Cu_(2)O` act as both oxidant and reductant ? |
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Answer» `Cu^(+)` undergoes disproportionation to form `Cu^(2+)` and Cu. `2Cu^(+) (aq) to Cu^(2+) (aq) + Cu(s)` Thus, `Cu^(+)` or `Cu_(2)O` acts both as an oxidant and a reductant |
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| 171. |
STATEMENT-1: Equivalent mass of `KMnO_(4)` in different mediums are different STATEMENT-2 `KMnO_(4)` can act as a oxidising agent . |
| Answer» Correct Answer - 2 | |
| 172. |
What is the coefficient for `Zn` when the equation below is balanced with the smallest whole number coefficient? `Zn+H^(+)(aq)+NO_(3)^(-)(aq)rarrZn^(2+)(aq)+N_(2)O(g)+H_(2)O(l)`A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - b |
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| 173. |
STATEMENT-1: In the reaction `Zn(s) + Cu^(2+)(aq) to Zn^(2+)(aq) + Cu(s) . Cu^(2+)` ions act as oxidising agent and Zn atoms act as a reducing agent. STATEMENT-2: Every redox reaction cannot be splitted into two reactions one being oxidation and the other being reductioin. STATEMENT-3: The oxidation numbers are artifical and are useful as a book keeping device of electrons in reactions.A. T T TB. F F TC. F F FD. T F T |
| Answer» Correct Answer - D | |
| 174. |
Assertion : `Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s)` can be split into following two half reactions `underset(("oxidation half reaction"))(Zn(s)rarrZn^(2+)+2e^(-))` Reason : Every redox reaction can be split into two reactions, one representing loss of electrons and the other representing gain of electrons. `Cu^(2+)(aq)+2e^(-)rarrunderset(("reduction half reaction"))(Cu(s))`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 | |
| 175. |
Identify the oxidant and the reductant in the following reactions: a. `Zn(s)+(1)/(2)O_(2)(g)rarrZn(s)` b. `Zn(s)+2H^(o+)(aq)rarrZn^(2-)(aq)+H_(2)(g)` |
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Answer» Correct Answer - A::B::C::D In reaction (a), `Zn` donates electrons to `O` to give zinc and oxide ions. Therefore, `Zn` acts as a reducing agent while oxygen acts as an oxidising agent. In reaction (b), `Zn` transfers its electrons to `H^(o+)`, and therefore, zinc acts as a reducing agent and `H^(o+)` acts as an oxidising agent. |
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| 176. |
Consider the following reaction : `xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+) to xMn^(2+)+2yCO_(2)+(z)/(2)H_(2)O` The value of x, y and z in the reaction are, respectively.A. 5, 2 and 16B. 2, 5 and 8C. 2, 5 and 16D. 5, 2 and 8 |
| Answer» Correct Answer - C | |
| 177. |
Which of the following is not a rule for calculating oxidation number ?A. For ions, oxidation number is equal to the charge on the ion.B. The oxidation number of oxygen is -2 in all of its compounds.C. The oxidation number of fluorine is -1 in all of its compounds.D. Oxidation number of hydrogen is +1 except in binary hydrides of alkali metals and alkaline earth metals where it is -1. |
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Answer» Correct Answer - B Oxidation number of oxygen is -2 in most of its compounds. The exceptions are -1 in peroxides and -1/2 in superoxides, +2 in `OF_(2)` and +1 in `O_(2)F_(2)`. |
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| 178. |
Identify the redox reaction taking palce in a beaker. A. `Zn_((s))+Cu_((aq))^(2+)rarrAn_((aq))^(2+)+Cu_((s))`B. `Cu_((s))+2Ag_((aq))rarrCu_((aq))^(2+)+2Ag_((s))`C. `Cu_((s))+Zn_((aq))^(2+)rarr Zn_((s))+Cu_((aq))^(2+)`D. `2Ag_((s))+Cu_((aq))^(2+)rarr2Ag_((aq))^(+)+Cu_((s))` |
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Answer» Correct Answer - B Since copper is more reactive than silver, it displaces `Ag^(+)` ions from its salt solution which get deposited on the copper rod. |
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| 179. |
Assertion :- In the reaction `Zn(s)+cu^(+2)(aq)rarrZn^(+2)(aq)+Cu(s)` `Cu^(+2)` ions act as oxidising agent and Zn atoms act as a reducing agent. Reason :- A substance when readily gain electrons from other, substance is an oxidising agent while reducing agent is a substance which can lose electrons to other susbtances.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 | |
| 180. |
Assertion : In the reaction, `(1)/(2)O_(2)+F_(2)rarrOF_(2)` Fluorine is oxidant. Reason :- Fluorine cannot show positive oxidation state.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 | |
| 181. |
Assertion :- `H_(2)S+Cl_(2)rarr2HCl+S` In the above reaction, Cl has been oxidised to `Cl^(-)` while `S^(-2)` has been reduced to S Reason :- In a reaction the element whose oxidation number decreases is reduced and the element whose oxidation number increases is oxidisedA. 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 - D | |
| 182. |
Identify the species undergoing oxidation and reduction. a. `H_(2)S(g)+Cl_(2)(g)rarr2HCl(g)+S(s)` b. `3Fe_(3)O_(4)(s)+8Al(s)rarr9Fe(s)+4Al_(2)O_(3)(s)` c. `2Na(s)+H_(2)(g)+H_(2)(g)rarr2NaH(s)` |
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Answer» Correct Answer - A a. `H_(2)S` is oxidised because a more electronegative element `Cl_(2)` is added to `H`. Alternatively, a more electropositive element `H` has been removed from S. `Cl_(2)` is reduced due to addition of `H` to it. b. `Al` is oxidised because oxygen is added to it. `Fe_(3)O_(4)` (ferrous ferric oxide) is reduced beacuse oxygen has been removed from it. c. `2overset(0)Na(s)+overset(0)H_(2)(g)rarr2overset(0)Naoverset(-1)H(s)` `2Naimplies2oveset(+1)Na+2e^(-)`(oxidation) `H_(2)+2e^(-)rarr2H^(ө)`(reduction) Therefore, `Na` is oxidised to the concept of electronegativity, the more electronegative element `H` is added to `Na`, so `Na` is oxidised and `H_(2)` is reduced. |
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| 183. |
Identify the species undergoing oxidation and reduction. a. `H_(2)S(g)+Cl_(2)(g)rarr2HCl(g)+S(s)` b. `3Fe_(3)O_(4)(s)+8Al(s)rarr9Fe(s)+4Al_(2)O_(3)(s)` c. `2Na(s)+H_(2)(g)rarr2NaH(s)` |
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Answer» (i) `H_(2)S` is oxidised because a more electronegative element , chlorine is added to hydrogen (or a more electropositive element , hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it . (ii) Aluminium is oxidised because oxygen is added to it . ferrous ferric oxide `(Fe_(3)O_(4))` is reduced because oxygen has beem removed from it . (iii) With the careful application of the concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced . Reaction (iii) chosen here prompts us to think in terms of another way to define redox reactions . |
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| 184. |
The `Mn^(3+)` ion is unstable in solution and undergoes disproportionation reaction to give `Mn^(+2), MnO_(2)`, and `H^(o+)` ion. Write a balanced ionic equation for the reaction. |
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Answer» The give reaction can be represented as : `Mn_((aq)^(3+) to Mn_((aq))^(2+) + MnO_(2 (s)) + H_((aq))^(+)` The oxidation half equation is : `overset(+3)(M)n_((aq))^(3+) to overset(+4)(M)n O_(2 (s))` The oxidation number is balanced by adding one electron as : `Mn_((aq))^(3+) to MnO_(2 (s)) + e^(-)` The charge is balanced by adding `4H^(+)` ions as : `Mn_((aq))^(3+) to MnO_(2 (s)) + 4 H_((aq))^(+) + e^(-)` The O atoms and `H^(+)` ions are balanced by adding `2H_(2)O` molecules as : `Mn_((aq))^(3+) + 2H_(2)O_((l)) to MnO_(2 (s)) + 4 H_((aq))^(+) + e^(-) .... (i)` The reduction half equation is : `Mn_((aq))^(3+) to Mn_((aq))^(2+)` The oxidation number is balanced by adding one electron as : `Mn_((aq))^(3+) e^(-) to Mn_((aq))^(2+) ....... (ii) ` The balanced chemical equation can be obtained by adding equation (i) and (ii) as : `2Mn_((aq))^(3+) + 2H_(2)O_((l)) to Mn_((aq))^(2+) + 4 H_((aq))^(+)` |
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| 185. |
The oxidation number of phosphorus in `PH_(4)^(+),PO_(2)^(3-),PO_(4)^(3-) and PO_(3)^(3-)` are respectively :-A. `-3,+1,+3,+5`B. `-3,+3,+5,+1`C. `+3,-3,+5,+1`D. `-3,+1,+5,+3` |
| Answer» Correct Answer - D | |
| 186. |
Reaction (A) `S^(-2)+4H_(2)O_(2)rarrSO_(4)^(2-)+4H_(2)O` (B) `Cl_(2)+H_(2)O_(2)rarr2HCl+O_(2)` The true statement regarding the above reactions is :A. `H_(2)O_(2)` acts as reductant in both the reactions.B. `H_(2)O_(2)` acts as oxidant in reaction (A) and reductant in reaction (B).C. `H_(2)O_(2)` acts as an oxidant in both the reactions.D. `H_(2)O_(2)` acts as reductant in reaction (A) and oxidant in reactin (B) |
| Answer» Correct Answer - B | |
| 187. |
When the equation ______`MnO_(4)^(-)+`___`SO_(3)^(2-)+`_______`H^(+)to`______`Mn^(2+)+`_______`SO_(4)^(2-)+`________`H_(2)O` is balanced correctly with the smallest whole number coefficients, what is the coefficient for `H_(2)O`?A. 3B. 5C. 8D. 10 |
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Answer» Correct Answer - a |
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| 188. |
The oxidation number of `Fe` in `Fe_(0.94)O` isA. `200`B. `200//94`C. `94//200`D. None |
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Answer» Correct Answer - B `Fe_(0.94)O: x xx 0.94-2 =0` `:. X =(2)/(0.94)=(200)/(94)` |
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| 189. |
What are the oxidation states of phosphorus in the following compounds ? `H_(3)PO_(2),H_(3)PO_(4),Mg_(2)P_(2)O_(7),PH_(3),HPO_(3)`A. `+1, +3, +3, +3, +5`B. `+3, +3, +5, +5, +5`C. `+1, +2, +3, +5, +5`D. `+1, +5, +5, -3, +5` |
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Answer» Correct Answer - D `H_(3)PO_(2):+3+X-4=0rArrx=+1` `H_(3)PO_(4):+3+x-8=0 rArr x=+5` `Mg_(2)P_(2)O_(7):+4+2x-14=0 rArr x=+5` `PH_(3):x+3=0 rArr x=-3` `HPO_(3):+1+x-6=0 rArr x=+5` |
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| 190. |
Consider the following reaction `x MnO_(4)^(-) + C_(2)O_(4)^(2-) + zH^(+) rarr x Mn^(2+) + 2y CO_(2) + (z)/(2)H_(2)O` The value of x, y and z in the reaction are respectivelyA. 5,2 and 8B. 5,2 and 6C. 2,5 and 8D. 2,5 and 6 |
| Answer» Correct Answer - D | |
| 191. |
Which of the following act both as an oxidising as well as reducing agent ?A. `HNO_(3)`B. `H_(2)SO_(4)`C. `H_(2)O_(2)`D. `HNO_(2)` |
| Answer» Correct Answer - C::D | |
| 192. |
Which can act as an oxidising as well as a reducing agent ?A. `HClO_(4)`B. `HNO_(3)`C. `H_(2)SO_(4)`D. `H_(2)O_(2)` |
| Answer» Correct Answer - D | |
| 193. |
The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. `1//5`B. `1//2`C. `1//4`D. `1//5` |
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Answer» Correct Answer - D `{:(2KMnO_(4)+,KI+,H_(2)O,rarr,KIO_(3)+,2MnO_(2)+,2KOH),(,Mn^(7+),+3e,rarr,Mn^(4+),,),(,,I^(-),rarr,I^(5+),+6e,):}` `:. 2 "mole of KMnO_(4)` reduced by `1` mole `KI` |
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| 194. |
The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium isA. oneB. twoC. fiveD. one-fifthe |
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Answer» Correct Answer - B In alkaline medium, ` KMnO_4` oxidixes iodide to iodate and gets reduced to manganese dioxide : `overset(+7)(M)nO_(4)^(-) rarr overset(+4)(M)n O_(2)` `overset(-1)(I^(-)) rarr overset(+5)(I)O_(3)^(-)` Thus, 1 mol of `KMnO_4` contains `3` equivalents (because its oxidation number decreases nu `3` unts : ` +7 rarr +4`). On the other hand, 1 mol of `KI` contains `6` equivalents (because oxidation number of iodine increases ny `6` units, ` -1 rarr +5`) According to the law of equivalents, we need `6` equivalents (2 mol) of `KMnO_4` to react with `6` requivalents (1 mol) of ` KI`. Alternatively, we can get the answer by means fo chemical equations : `{:(2KMnO(4)+H_(2)O rarr 2KOH+2MnO_(2)+3[O]),(KI+3[O] rarr KIO_(3)),(bar(2KMnO_(4) +KI +H_(2)O rarr 2KOH +2MnO_(2) +KIO_(3))):}` |
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| 195. |
In the reaction `MnO_(4)^(-)+SO_(3)^(-2)+H^(+)rarrSO_(4)^(-2)+Mn^(2+)+H_(2)O`A. `MnO_(4)^(-) and H^(+)` both are reducedB. `MnO_(4)^(-)` is reduced and `H^(+)` is oxidisedC. `MnO_(4)^(-)` is reduced and `SO_(3)^(2-)` is oxidisedD. `MnO_(4)^(-)` is oxidised and `SO_(3)^(2-)` is reduced |
| Answer» Correct Answer - C | |
| 196. |
The oxidation sates of metal in the compounds `Fe_(0.94)O` and `[Cr(P Ph_(3))_(3)(CO)_(3)]` respectively areA. `(200)/(94),0`B. `0,(94)/(200)`C. `2,1`D. `1, (200)/(94)` |
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Answer» Correct Answer - A `Fe_(0.94)O:x xx0.94-2=0` `:.x=(2)/(0.94)=(200)/(94)` `:.` Oxidation state of Fe`=(200)/(94)` `[Cr(P Ph_(3))_(3)(CO)_(3)]:x+(0xx3)+(0xx3)=0,x=0` `:.` Oxidation state of `Cr=0` |
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| 197. |
Which of the following halogen always show only one oxidation state -A. ClB. FC. BrD. I |
| Answer» Correct Answer - B | |
| 198. |
The oxidation state of `Cr` in `CrO_(5)` is:A. `+ 10`B. `+8`C. `+ 6`D. `+4` |
| Answer» Correct Answer - C | |
| 199. |
The oxidation state of `Cr` in `CrO_(5)` is:A. `+3`B. `+5`C. `+6`D. 0 |
| Answer» Correct Answer - C | |
| 200. |
Assertion: In queous solution, `SO_(2)` reacts with `H_(2)S` liberating sulphur Reason : `SO_(2)` is an effective reducing agent.A. If both assetion and reson are corect and reason is correct explanation for assertionB. If both assetion and reason are correct but reason is not correct explanation for assertionC. If asertion is correct but reason is incorrectD. If both assetion and reason are incorrect |
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Answer» Correct Answer - B Correct explanation. `SO_(2)` is an effective oxidising agent in this case. |
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