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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.
| 101. |
Which of the following statements is incrrect ?A. `E _("electrode" )^(Ө)` changes sign whenever we reverse a cell reaction.B. The half-cell reactions are not reversible.C. Changing the stoichoimetric coefficients of a half-cell reaction does not affect the value of `E^(Ө)`D. The more positive the `E^(Ө)` value, the greater the tendency for the substance to be reduced. |
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Answer» Correct Answer - B The half-cell reactions are reversible. Depending upon the conditions, any electrode can act either as an anode or as a cathode. |
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| 102. |
The values of `X, Y` and `Z`s in the reaction are repectively: `XMnO_(4)^(-)+YH_(2)SO_(4)rarrZMn^(2+)+5H_(2)O+9O_(2)+Ze`A. `2,5,6`B. `5,2,9`C. `3,5,5`D. 2,6,6` |
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Answer» Correct Answer - A `2MnO_(4)^(-)+5H_(2)SO_(4)rarr2Mn^(2+)+5H_(2)O+9O_(2)+6e` |
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| 103. |
Assertion: Amongest the halogens, fluorine can oxidise the elements to the highest oxidation- state. Reason: Due to small size of fluoride ion, it is difficult to oxidise fluoride ion to fluorine. Hence reverse reaction takes place more easily.A. If both the assertion and reason are true and reason is the true explanation of the assertion.B. If both the assertion and reason are ture but the reason is not the correct explanation of assertionC. If the assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Due to high electronegativity and high heat of dissociation, fluorine oxidises the elements to their oxidation state. |
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| 104. |
How many mole of electrons are involved balancing the following equations: (a) `H_(2)S+NO_(3)^(-)rarrS+NO` (b) `Mn(OH)_(2)+H_(2)O_(2)rarrMnO_(2)+2H_(2)O` (c ) `Cr_(2)O_(7)^(2-)+Fe^(2+)+C_(2)O_(4)^(2-)rarrCr^(3+)+Fe^(3+)+CO_(2)` (acid medium) (d) `Br_(2)+OH^(-)rarrBrO_(3)^(-)+Br^(-)` (e ) The compound `P_(4)S_(3)` is oxidised by nitrate ions acid medium to give phosphoric acid, sulphate ions and nitric oxide `(NO)`. Write the balanced half reactions and the overall reaction. |
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Answer» `{:((a),,S^(2-), rarr ,S^(0) ,+ 2e,),(,N^(+5), + 3e ,rarr ,N^(2+),, 6N "electron"):}` `{:((b),,Mn^(2+), rarr ,Mn^(4+) ,+ 2e,),(,O_(2)^(1-), + 2e ,rarr ,O_(2)^(2-),, 2N "electron"):}` ( c) `{:((c ),,Cr_(2)^(6-),+3e, rarr, 2Cr^(3+),,3N "electron",,),(,,Fe^(2+), + C_(2)^(3+) ,rarr ,Fe^(3+)+2C^(4+),+3e, ,,):}` `{:((d),,Br_(2)^(0),+2e, rarr, 2Br^(-),,10 N "electron",,),(,,,Br_(2)^(0), rarr ,2Br^(5+) ,+10e,, ,):}` (e) `{:((P^(3//2))_(4)rarr4P^(5+)+14e),((S^(2-))_(3)rarr3S^(6+)+24e),(bar([P_(4)S_(3)rarr4P^(5+)+3S^(6+)+38e]xx3)),(ul([3e+N^(5+)rarrN^(2+)]xx38)):}` `3P_(4)S_(3)+38NO_(3)^(-)rarr12H_(3)PO_(4)+9SO_(4)^(2-)+38NO` `3P_(4)S_(3)+38NO_(3)^(-)+20H^(+)+8H_(2)Orarr12H_(3)PO_(4)+9SO_(4)^(2-)+38NO` |
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| 105. |
For the reaction `KO_(2)+H_(2)O+CO_(2)rarrKHCO_(3)+O_(2)`, the mechanism of reaction suggest.A. Acid-base reactionB. Disproportionation reactionC. HydrolysisD. Redox change |
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Answer» Correct Answer - A::B::C::D `4KO_(2)+2H_(2)Orarr4KOH+3O_(2)` `["Hydrolysis, disproportionation and redox"]` `4KOH+4CO_(2)rarr4KHCO_(3) ("Acid-base reaction")` |
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| 106. |
Evaluate equivalent weight of reductant or oxidant given on left hand side of each reaction: (a) `As_(2)O_(3)+5H_(2)Orarr2AsO_(4)^(3-)+10H^(+)+4e` (b) `MnO_(4)^(-)+8H^(+)+5erarrMn^(2+)+4H_(2)O` (c ) `Cr_(2)O_(7)^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_(2)O` (d) `C_(2)O_(4)^(2-)rarr2CO_(2)+2e` (e ) `FeC_(2)O_(4)rarrFe^(3+)+2CO_(2)+3e` (f) `2CuSO_(4)+2erarrCu_(2)^(1+)+SO_(4)^(2-)` |
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Answer» `E_("red"//"oxi")={:(ul("Mol.weight of reductant or oxidant")),("Number of electrons gained or lost by"),("one molecule of reductant or oxidant"),("or valence factor"):}` (a) `E_(As_(2)O_(3)) =(M_(As_(2)O_(3)))/(4)` `(As_(2)^(3+) rarr 2As^(5+) + 4e)` (b) `E_(MnO_($)^(-)) = (M_(MnO_(4)^(-)))/(5)` `(Mn^(7+) + 5e rarr Mn^(2+))` ( c) `E_(Cr_(2)O_(7)^(2-)) = (M_(Cr_(2)O_(7)^(2-)))/(6)` `(Cr_(2)^(6+) rarr 2Cr^(3+) + 6e)` (d) `E_(C_(2)O_(4)^(2-)) = (M_(C_(2)O_(4)^(2-)))/(2)` `(C_(2)^(3+) rarr 2C^(4+) + 2e)` (e) `E_(FeC_(2)O_(4)) = (M_(FeC_(2)O_(4)))/(3)` `(Fe^(2+)+C_(2)^(3+) rarr Fe^(3+) + 2C^(4+) + 3e)` (f) `E_(CuSO_(4)) = (M_CuSO_(4))/(1)` `(2Cu^(2+) + 2e rarr Cu_(2)^(+))` |
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| 107. |
Which of the following can be used both as an oxidant and a reductant?A. `HNO_(2)`B. `SO_(2)`C. `O_(2)`D. `CO` |
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Answer» Correct Answer - A::B::C::D The element in a molecule having its oxidation state in the middle (i.e., greater than minimum and less than maximum) can be used as an oxidising agent and a reducing agent both. |
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| 108. |
Which molecule represent by the bold atoms are in their highest oxidation state?A. `H_(2)S_(2)O_(8)`B. `P_(4)O_(10)`C. `F_(2)O`D. `Mn_(2)O_(7)` |
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Answer» Correct Answer - A::B::D The highest oxidation state is by the group number except `F` and `O`. |
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| 109. |
How many mole of `FeSO_(4),H_(2)C_(2)O_(4)` and `FeC_(2)O_(4)` are oxidised separately by one mole of `KMnO_(4)` in acid medium? |
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Answer» `FeSO_(4)` : `Fe^(2+) rarr Fe^(3+) + e` `Mn^(7+) + 5e rarr Mn^(2+)` Thus, `5Fe^(2+) + Mn^(7+) rarr Fe^(3+) + Mn^(2+)` or `5` mole of `FeSO_(4)` are oxidised by `1` mole `KMnO_(4)`. `H_(2)C_(2)O_(4)`: `C_(2)^(3+) rarr 2C^(4+) + 2e` `Mn^(7) + 5e rarr Mn^(2+)` `5C_(2)^(3+) +2Mn^(7+) rarr 10C^(4+) + 2Mn^(2+)` or `5//2` mole of `H_(2)C_(2)O_(4)` are oxidised by `1` mole `KMnO_(4)`. `FeC_(2)O_(4)`: `Fe^(2+) +C_(2)^(3+) rarr Fe^(3+) + 2C^(4+) + 3e` `Mn^(7+) + 5e rarr Mn^(2+)` `5FeC_(2)O_(4) +3Mn^(5+) rarr 5Fe^(5+) rarr 5Fe^(3+) + 10C^(4+) + 3Mn^(2+)` or `5//3` mole of `FeC_(2)O_(4)` are oxidised by `1` mole `KMnO_(4)`. |
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| 110. |
`3` mole of `FeSO_(4)` are oxidised by `a` mole of `KMnO_(4)` in acidic medium whereas `3` moles of `FeC_(2)O_(4)` are oxidised by `b` mole of `KMnO_(4)` in acidic medium, the ratio of `a` and `b` is:A. `1//3`B. `1//2`C. `1//4`D. `1//5` |
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Answer» Correct Answer - A `{:(,Fe^(2+),rarr,Fe^(3+)+,e),(5e+,Mn^(7+),rarr,Mn^(2+),):}` or `5 "mole" Fe^(2+)-=1"mole" KMnO_(4)` `:. 3"mole" Fe^(2+)-=(3)/(5)"mole" KMnO_(4)=a` `{:(Fe^(2+)rarrFe^(3+)+e),((C^(3+))_(2)rarr2C^(4+)+2e),(bar(FeC_(2)O_(4)rarrFe^(3+)+2C^(4+)+3e)),( :. 5 "mole"FeC_(2)O_(4)=3"mole"KMnO_(4)),( :. 3 "mole" FeC_(2)O_(4)=(9)/(5)"mole"KMnO_(4)=b):}` |
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| 111. |
Statement `KMnO_(4)` acts as oxidant as well as self indicator in its titration with Ferrous ammonium sulphate solution in acidic medium. Explanation `KMnO_(4)` reduces itself to `Mn^(2+)` ions and oxidises `Fe^(2+)` to `Fe^(3+)` as well as after redox reaction is complete, the `KMnO_(4)` at the equivalence point imparts pink colour.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 statement. |
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| 112. |
1 mol of `N_(2)H_(4)` loses 14 moles of electrons to from a new compound X. Assuming that the entire nitrogen appear in the new compound, what is the oxidation state of nitrogen in X? |
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Answer» Correct Answer - 5 |
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| 113. |
In ferrous ammonium sulhate oxidation number of `Fe` isA. `+3`B. `+2`C. `+1`D. `-2` |
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Answer» Correct Answer - B In ferrous ammonium sulphate `Fe` shows `+2` oxidation state. |
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| 114. |
The oxidation number of `Cr` in `K_(2)Cr_(2)O_(7)` isA. `-2`B. `-7`C. `+2`D. `+6` |
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Answer» Correct Answer - D `K_(2)overset(*)(C ) r_(2)O_(7)` `2+2x-2xx7=0`, `2x-14+2=0` `2x=12`, `x=(12)/(2)=+6` |
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| 115. |
In compound `HN_(3)` (hydrazoic acid) , oxidation state of N atoms are :A. `0, 0, 3`B. `0, -2, +1`C. `1, 1, -3`D. `-3, -3, -3` |
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Answer» Correct Answer - b |
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| 116. |
The number of moles of `KMnO_(4)` required to oxidise `1 mol` of `Fe(C_(2)O_(4))` in acidic medium isA. `0.6`B. `1.67`C. `0.2`D. `0.4` |
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Answer» Correct Answer - A `underset((n=5))("Equivalent of "MnO_(4)^(ө)) = underset((n=3))("Equivalent of "Fe(C_(2)O_(4)))` `(1)/(5) mol =(1)/(3) mol` `(3)/(5)"mol of" MnO_(4)^(ө) = 1 "mol of" Fe(C_(2)O_(4))` `0.6 "mol of" MnO_(4)^(ө) = 1 mol Fe (C_(2)O_(4))` |
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| 117. |
In the reaction `Ca+H_(2)rarrCaH_(2)` select the incorrect statement.A. Calcium undergoes oxidationB. Hydrogen undergoes reductionC. Calcium acts as oxidising agentD. Hygrogen acts as oxidising agent |
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Answer» Correct Answer - c |
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| 118. |
Calculate the moles of `KMnO_(4)` required to react completely with 2 m and 1500 mL of `K_(2)C_(2)O_(4)H_(2)C_(2)O_(4)` in acidic medium.A. 0.8B. 0.6C. 1.6D. 2.4 |
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Answer» Correct Answer - d |
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| 119. |
50mL aqueous solution of `Fe_(2)SO_(4)` was neutralized with 100mL of 0.2 M `KMnO_(4)` and 200ml of 1 M `H_(2)SO_(4)` solution. Find the molarity of `FeSO_(4)` solution. |
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Answer» Correct Answer - 2 |
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| 120. |
How many milli grams of `Fe_(0.9)O` reacts completely with 10 mL 0.1 M `KMnO_(4)` solution in acidic conditions ? (Fe =56)A. 47B. 402C. 534D. 570 |
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Answer» Correct Answer - b |
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| 121. |
Permanganate (VII) ion, `MnO_(4)^(-)` oxidises `I^(-)` ion to `I_(2)` and gives manganese (IV) oxide `MnO_(2)` in basic medium. The skeletal ionic equation is given as `pMnO_(4(aq))^(-)+qI_((aq))^(-)+x H_(2)O_((l)) rarr rMnO_(2(s))+sI_(2(s))+yOH_((aq))^(-)` The values of p, q, r and s areA. `{:(,p,q,r,s),(,1,2,8,4):}`B. `{:(,p,q,r,s),(,2,6,2,3):}`C. `{:(,p,q,r,s),(,2,4,2,8):}`D. `{:(,p,q,r,s),(,1,4,8,2):}` |
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Answer» Correct Answer - B `2MnO_(4(aq))^(-)+6I_((aq))^(-)+4H_(2)O_((l))rarr 2 MnO_(2(s))+3I_(2(s))+8OH_((aq))^(-)` |
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| 122. |
write balanced chemical equation for the following reaction (i) Permanganate ion `(MnO_(4)^(-))` reacts with suphur dioxide gas in acidic medium to produce `Mn^2(+)` and hydrogen suphate ion. (Balance by oxidation number method) (ii) Reaction of liquid hydrazine `(N_(2)H_(4))` with chlorate ion `CIO_(3)^(-)` in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method) (iii) Dichlorine heptaoxide`(CI_(2)O_(7))` in gaseous state combines with an aquious sloution of hydrogen peroxide in acidic medium to give chlorite ion `(CIO_(2)^(-))` and oxygen gas. (Balance by ion electron method) |
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Answer» `(i) For answer , consultQ.18 (N.C.E.R.T Exercise)` `(ii) For answer , consultQ.19 (N.C.E.R.T Exercise)` `(iii) For answer , consultQ.20 (N.C.E.R.T Exercise)` |
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| 123. |
Write a balanced ionic equation to describe the oxidation of iodide `(I^-)` in by permanganate `(MnO_(4)^(-))` ion in basic solution to yield molecular iodine `(l_2)` and manganese (IV) oxide `(MnO_2)`. Strategy : We are given the formulas for two reactants and two prodcts. We use these to write the skeletal ionic equatin. We construct and balance the appropriate half-reactions using the rules just described. Then we add the half -reactions and eliminate common terms. |
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Answer» Step 1: First we write the skeletal ionic equation, which is `MnO_(4)^(-) (aq) + I^(-) (aq) to MnO_(2)(s) + I_(2)(s)` Step 2: The two half-reactions are: Oxidation half : `overset(-1)(I^(-))(aq) to overset(0)(I_(2))(s)` Reduction half : `overset(+7)(MnO_(4)^(-)) (aq) to MnO_(2)(s)` Step 3 : To balance the I atoms in the oxidation half reaction , we rewrite it as : `2I^(-) (aq) to I_(2)(s)` Step 4 : To balance the O atoms in the reduction half reaction , we add two water molecules on the right : `MnO_(4)^(-) (aq) + 4 H^(+) (aq) to MnO_(2)(s) + 2H_(2)O(1)` To balance the H atoms , we add four `H^(+)` ions on the left : `MnO_(4)^(-) (aq) + 4 H^(+) (aq) to MnO_(2)(s) + 2H_(2)O (1)` As the reaction takes place in a basic solution , therefore , for four `H^(+)` ions , we add four `OH^(-)` ions to both sides of the equation : `MnO_(4)^(-) (aq) + 4 H^(+) (aq) + 4 OH^(-) (aq) to MnO_(2) (s) + 2H_(2)O (1) + 4 OH^(-) (aq)` Replacing the `H^(+)` and `OH^(-)` ions with the water , the resultant equation is : `MnO_(4)^(-)(aq) + 2H_(2)O(1) to MnO_(2)(s) + 4 OH^(-)(aq)` Step 5 : In this step we balance the charges of the two half - reactions in the manner depicted as : `2I^(-) (aq) to I_(2) (s) + 2e^(-)` `MnO_(4)^(-) + 2H_(2)O (1) + 3 e^(-) to MnO_(2)(s) + 4 OH^(-) (aq)` Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. `6I^(-) (aq) to 3 I_(2) (s) + 6 e^(-)` `2 MnO_(4)^(-) (aq) + 4 H_(2)O(1) + 6 e^(-) to 2 MnO_(2) (s) + 8 OH^(-) (aq)` Step 6 : Add two half-reactions to obtain the net reactions after cancelling electrons on both sides. `6I^(-)(aq) + 2MnO_(4)^(-) (aq) + 4H_(2)O(1) to 3I_(2)(s) + 2 MnO_(2) (s) + 8 OH^(-) (aq)` Step 7 : : A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides. |
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| 124. |
The oxidation number of hyrogen is (i) `0` (ii) ` +1` (iii) ` -1` (iv) `+ 1` only.A. (i), (ii), (iii)B. (i), (ii), (iii), (iv )C. (i), (ii)D. (i),(iii) |
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Answer» Correct Answer - A Because hydrogen is always univalent , its oxidation number cannot be `+2`. In `H_2` (free, uncombined element), the oxidation number of H is zero. The oxidation number of H in its compounds is usually `+1`, except when it is bonded to highly reactive metals in binary cmpounds (that is, compouns containing two elements ). Here, its oxidation number is `-1`. |
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| 125. |
Following reaction describes the rusting of iron `4Fe+3O_(2) rarr 4Fe^(3+)+6O_(2-)` Which one of the following statements is incorrect?A. This is an example of a redox reactionB. Metallic iron is reduced to `Fe^(3+)`C. `Fe^(3+)` is an oxidising agentD. Metallic iron is a reducing agent |
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Answer» Correct Answer - B Metallic iron is oxidised to `Fe^(+3)`. |
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| 126. |
In the context of the reaction, `4Fe+3O_(2)rarr4Fe^(3+)+6O^(2-)` , which of the following statements is`//`are correct?A. It is a redox reactionB. Metallic iron is a reducing agentC. `Fe^(3)` is an oxidising agentD. Metallic iron is reduced to `Fe^(3+)` |
| Answer» Correct Answer - D | |
| 127. |
In the context of the reaction, `4Fe+3O_(2)rarr4Fe^(3+)+6O^(2-)` , which of the following statements is`//`are correct?A. It is redox reactionB. `Fe_((s))` is a reducing agentC. `Fe_((aq.))^(3+)` is an oxidising agentD. `Fe_((s))` is reduced to `Fe_((aq.))^(3+)` |
| Answer» Correct Answer - A::B::C | |
| 128. |
A,B and C are three elements forming a compound in which their oxidation state are +2,+5, and -2 respectively. Which could not be the formula of compound?A. `A_(2)(BC)`B. `A_(2)(BC_(4))_(3)`C. `A_(3)(BC_(4))_(2)`D. `ABC_(2)` |
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Answer» Correct Answer - a,b,d |
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| 129. |
A,B and C are three elements forming a compound in which their oxidation state are +2,+5, and -2 respectively. Which could not be the formula of compound?A. `A_(2)(BC)_(2)`B. `A_(2)(BC_(4))_(3)`C. `A_(3)(BC_(4))_(2)`D. `ABC` |
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Answer» Correct Answer - C `A_(3)(BC_(4))_(2)`. The sum of oxidation no. of A, B & C in the compound should be zero as the molecule is neutral. |
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| 130. |
Oxidation state of in `Fe_(3)O_(4)` isA. `+2`B. `+3`C. `8//3`D. `2//3` |
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Answer» Correct Answer - C In `Fe_(3)O_(4)` let O.N. of Fe = x `3x + 4(-2) =0 or x =8//3`. |
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| 131. |
When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes fromA. `+2` to `+7`B. `+4` to `+7`C. `+7` to `+2`D. `+6` to `+2` |
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Answer» Correct Answer - C `2overset(+7)MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr 2 overset(+2)Mn^(2+)+10CO_(2)+8H_(2)O` |
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| 132. |
When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes fromA. From 7 to 2B. From 6 to 2C. From 5 to 2D. From 7 to 4 |
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Answer» Correct Answer - A Mn changes from oxidation no. of +7 in `KMnO_(4)` to +2 in `Mn^(+2)` |
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| 133. |
Among es the following, the pair having both the metals in their highest oxidation state is :A. `[Fe(CN)_(6)]^(3-)` and `[Co(CN)_(6)]^(3-)`B. `CrO_(2)Cl_(2)` and `MnO_(4)^(-)`C. `TiO_(2)` and `MnO_(2)`D. `[MnCl_(4)]^(2-)` and `[NiF_(6)]^(2-)` |
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Answer» Correct Answer - b |
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| 134. |
The pair of the compounds in which both the metals are in the highest possible oxidation state isA. `[Fe(CN)_(6)]^(3-),[Co(CN)_(6)]^(3-)`B. `CrO_(2)Cl_(2),MnO_(4)^(-)`C. `TiO_(3),MnO_(2)`D. `[Co(CN)_(6)]^(3-),MnO_(2)` |
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Answer» Correct Answer - B::C (A) In `[Fe(CN)_(6)]^(3-)`,O.S. of Fe = +3 In `[Co(CN)_(6)]^(3-)`,O.S. of Co = +3 (B) In `CrO_(2)Cl_(2)` , O.S of Cr = +6 In `MnO_(4)^(-)` , O.S. of Mn = +7 (C) In `TiO_(3)` , O.S. of Ti = +6 In `MnO_(2)` , O.S of Mn = +4 (D) In `[Co(CN)_(6)]^(3-)` , O.S of Co = +3 In `MnO_(2)`, O.S. of Mn = +4. |
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| 135. |
Which of the following reaction depicts the oxidsing behaviour of `H_(2)SO_(4)`?A. `2HI+H_(2)SO_(4)rarrI_(2)+SO_(2)+2H_(2)O`B. `Ca(OH)_(2)+H_(2)SO_(4)rarrCaSO_(4)+2H_(2)O`C. `NaCl+H_(2)SO_(4)rarrNaHSO_(4)+HCl`D. `2PCl_(5)+H_(2)SO_(4)rarr2POCl_(3)+2HCl+SO_(2)Cl_(2)` |
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Answer» Correct Answer - a |
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| 136. |
Which of the following chemical reactions depicts the oxidizing behaviour of `H_(2)SO_(4)` ?A. `NaCl+H_(2)SO_(4) rarr NaHSO_(4) + HCl`B. `2PCl_(5) + H_(2)SO_(4) rarr 2POCl_(3) + 2HCl + SO_(2)Cl_(2)`C. `2HI + H_(2)SO_(4) rarr I_(2)+SO_(2)+2H_(2)O`D. `Ca(OH)_(2)+H_(2)SO_(4) rarr CaSO_(4)+2H_(2)O` |
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Answer» Correct Answer - C Only in reaction (C), the O.N of I increases from -1 is HI to zero in `I_(2)`. |
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| 137. |
Among the following elements, what is the total number of elements having the lowest oxidation state of zero? `Ta b. `Te` c. `Tc` d. `Ti` e. `TI` |
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Answer» Correct Answer - D The maximum oxidation state is the group number. The maximum oxidation state of metals is zero, for non-metal it is equal to the group number minus `8`. The maximum and the minimum oxidation states are a. `+5`, b. `+6, -2` , c. `+7 , 0` , d. `+4,0` , e. `+3, 0` |
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| 138. |
What is O.N of Na in `Na_(2)O_(2)`? |
| Answer» `overset(x)Naoverset(-2)O_(2):2x+2(-2)=0 or x = +2` | |
| 139. |
The values of x, y and z in the reaction are respectively: `4KO_(2)+H_(2)O+xH_(2)O+yCO_(2)rarr4KHCO_(3)+zO_(2)`A. 3, 6, 6B. 2, 4, 3C. 3, 2, 5D. 4, 3, 6 |
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Answer» Correct Answer - b |
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| 140. |
Which of the following equation is correctly balanced ?A. `5BiO_(3)^(-)+22H^(+)+Mn^(2+)rarr5BiO^(3+)+7H_(2)O+MnO_(4)^(-)`B. `5BiO_(3)^(-)+14H^(+)+2Mn^(2+)rarr5BiO^(3+)+7H_(2)O+2MnO_(4)^(-)`C. `5BiO_(3)^(-)+4H^(+)+Mn^(2+)rarr2BiO^(3+)+2H_(2)O+MnO_(4)^(-)`D. `5BiO_(3)^(-)+12H^(+)+3Mn^(2+)rarr6BiO^(3+)+6H_(2)O+3MnO_(4)^(-)` |
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Answer» Correct Answer - b |
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| 141. |
Which of the following colour changes shown during redox titrations is not correct ?A. `Cr_(2)O_(7)^(2-)` oxidises the indicator diphenylamine to produce blue colour showing end point.B. Iodine formed by oxidation of `I^(-)` ions gives blue colour with starch showing end point.C. `KMnO_(4)` in the form of `MnO_(4)^(-)` ions gives pink colour showing end point.D. Thiosulphate ions `(S_(2)O_(3)^(2-))` give blue colour showing end point. |
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Answer» Correct Answer - D Thiosulphate ions do not act as indicator and do not indicate end point by change in colour. |
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| 142. |
Which solution can serve as both reactant and indicator when it is used in redox titrations ?A. `FeNH_(4)(SO_(4))_(2)`B. `KMnO_(4)`C. `H_(2)C_(2)O_(4)`D. `Na_(2)S_(2)O_(3)` |
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Answer» Correct Answer - b |
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| 143. |
Assertion : In titrations involving potassium permanganate no indicator is used. Reason : `MnO_(4)^(-)` acts as the self-indicator.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 | |
| 144. |
The table below shows the data for three titrations to determine the concentration of a NaOH solution With standard 0.200 M HCl solution using phenolphthalein as the indicator Which explanation best accounts for the lower value of the NaOH M in Trial 3 ?A. Some of the neutralized solution from Trial 2 was left in the flask for Trial 3.B. The numner of drops of phenolphtalein was doubled in Trial 3.C. The HCl concentration was used as 0.250 M in the NaOH molarity calculation .D. A few drops of NaOH solution were spilled on the desktop in Trial 3. |
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Answer» Correct Answer - d |
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| 145. |
`KI` and ` CuSO_4` solution when mixed give .A. `CuI_(2)+K_(2)SO_(4)`B. `Cu_(2)I_(2)+K_(2)SO_(4)`C. `K_(2)SO_(4)+Cu_(2)I_(2)+I_(2)`D. `K_(2)SO_(4)+CuI_(2)+I_(2)` |
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Answer» Correct Answer - C `2CuSO_(4)+4KIrarr underset(("White ppt."))(Cu_(2)I_(2))+I_(2)+K_(2)SO_(4)` |
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| 146. |
In the iodometric estimation in the laboratory which process is involved?A. `Cr_(2)O_(7)^(2-)+H^(+)+I^(-)rarr2Cr^(3+)+I_(2),I_(2)+S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+I^(-)`B. `MnO_(4)^(2-)+H^(+)+I^(-)rarrMn^(2+)+I_(2),I_(2)+S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+I^(-)`C. `MnO_(4)^(-)+OH^(-)+I^(-)rarrMnO_(2)+I_(2),I_(2)+S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+I^(-)`D. `Cr_(2)O_(7)^(2-)+OH^(-)+I^(-)rarr2Cr^(3+)+I_(2),I_(2)+S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+I^(-)` |
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Answer» Correct Answer - A Iodometry involves oxidation of iodide ion (usually KI) by strong oxidising agent in neutral or acidic solution to `I_(2)`. The `I_(2)` evolved is then titrated with standard solution of a reducing agent (generally `Na_(2)S_(2)O_(3)`). In choices (C), and (D), medium is alkaline and in choice (B), `MnO_(4)^(2-)` (mangnate) is not a strong oxidising agent. |
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| 147. |
In the following reaction, `4P+3KOH+3H_(2)O rarr 3KH_(2)PO_(2)+PH_(3)`A. `P` is oxidised as well as reducedB. `P` is reduced onlyC. `P` is oxidised onlyD. None of these |
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Answer» Correct Answer - A `P` is oxidised as well as reduced (as in option a). |
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| 148. |
For reaction : `Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O`, How many `e^(-)s` are requiredA. 3B. 4C. 5D. 6 |
| Answer» Correct Answer - D | |
| 149. |
In the reaction: `Cr_(2)O_(7)^(2-) + 14H^(o+) + 6I^(Ө) rarr2Cr^(3+) + 3H_(2)O + 3I_(2)` Which element is reduced?A. `Cr`B. `H`C. `O`D. `I` |
| Answer» Correct Answer - A | |
| 150. |
The following equations are balanced atomwise and chargewise. (p) `Cr_(2)O_(7)^(2-)+8H^(+)3H_(2)O_(2)rarr2Cr^(3+)+7H_(2)O+3O_(2)` (q) `Cr_(2)O_(7)^(2-)+8H^(+)5H_(2)O_(2)rarr2Cr^(3+)+9H_(2)O+4O_(2)` (r) `Cr_(2)O_(7)^(2-)+8H^(+)7H_(2)O_(2)rarr2Cr^(3+)+11H_(2)O+5O_(2)` The precies equation/equations representing the oxidation of `H_(2)O_(2)` is/are:A. (P) onlyB. (Q) onlyC. (R) onlyD. all the three |
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Answer» Correct Answer - a |
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