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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.
| 801. |
Statement-1. Oxidation number of Ni in `Ni(CO)_(4)`, is zero Statement-2. Oxidation number of CO has been taken as zero.A. Statement -1 is true, statement -2 is also true, statement -2 is the correct explanation of statement-1B. Statement -1 is true, statement -2 is also true, statement-2 is not the corect explanation of statement-1C. Statement -1 is true, statement -2 falseD. Statement -1 is false, statement -2 is true. |
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Answer» Correct Answer - A Statement -2 is the correct explanation for statement -1 |
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| 802. |
In `Ni(CO)_(4)` the oxidation state of Ni is:A. 4B. 0C. 2D. 8 |
| Answer» Correct Answer - B | |
| 803. |
Assertion: Stannous chloride is a powerful oxidising agent which oxidises mercuric chloride to mercury Reason: Stannous chloride gives grey precipitate with mercuric chloride, but stannic chloride does not do so.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - D Here, assertion is false, because stannous chloride is a strong reducing agent not strong oxidising agent. Stannous chlorides gives grey precipitate with mercuric chloride. Hence, reason is true. |
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| 804. |
Assertion : When `SnCl_(2)` solution is added to `HgCl_(2)` solution, a milky white precipitate is obtained and on adding excess `SnCl_(2)`, a black precipitate is formed. Reason : The disproportionation if Hg(II) is easier than its reduction only.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 | |
| 805. |
In the equation, `SnCl_(2)+2HgCl_(2)rarrHgCl_(2)+SnCl_(4)` The equivalent weight of stannous schloride (molecular weight `=190`) will be :A. 190B. 95C. 47.5D. 154.5 |
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Answer» Correct Answer - b |
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| 806. |
Consider the following reaction. `underset(CHO) underset(|) (CHO) + OH^(-) to underset(CH_(2)OH) underset(|)(COO^(-))` Select the incorrect statement.A. It is disproportion reaction.B. It is intramolecular redox reactionC. `OH^(-)` is a reducing as well as oxidising agentD. `underset(CHO) underset(|) (CHO)` is a reducing as well as oxidising agent. |
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Answer» Correct Answer - C Once CHO is oxidised to `COO^(-)` and one CHO is reduced to `CH_(2)OH`. Thus, it is a disproportionation reaction. Thus, (a) and (b) is true and `overset(CHO)overset(|)(CHO)` is reducing as well as oxidising agent. Thus, (d) is also true, Thus , ( c) is incorrect. |
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| 807. |
Which of the following is a disproportion reaction?A. `CH_(4) + 2O_(2) to CO_(2) + 2H_(2)O`B. `CH_(4) + 4Cl_(2) to C Cl_(4) +4HCl`C. `2F_(2) + 2OH^(-) to 2F^(-) + OF_(2) + H_(2)O`D. `2NO_(2) + 2OH^(-) to NO_(2)^(-) + NO_(3)^(-) + H_(2)O` |
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Answer» Correct Answer - D Reactions in which the same substance is oxidised as well as reduced are called disproportionation reactions. Writing the oxidation number of each element above its symbol in the given reactions. (a) `overset(-4)(C)overset(+1)(H_(4)) + 2overset(0)O_(2) to overset(+4)(C)overset(-2)(O_(2)) + 2overset(+1)H_(2)overset(-2)O` (b) `overset(-4)(C)overset(+1)(H_(4)) + 4overset(0)Cl_(2) to overset(+4)Coverset(+1)(Cl_(4)) + 4overset(+1)Hoverset(-1)(Cl)` (c ) `2overset(0)F_(2) + 2overset(-2)Ooverset(+1)H to 2overset(-1)F + overset(+2)Ooverset(-1)F_(2) + overset(+1)H_(2)overset(-2)(O)` (d) `2overset(+4)N overset(-2)(O_(2)) + 2overset(-2)Ooverset(+1)H to overset(+2)Ooverset(-1)F_(2) + overset(+1)(H_(2))overset(-2)O` (d) `2overset(+4)Noverset(-2)O_(2) + 2overset(-2)Ooverset(+1)H to overset(+3)Noverset(-2)(O_(2)^(-)) + overset(+1)H_(2)overset(-2)O` Thus, in reaction (d), N is both oxidised as well as reduced since, the oxidation number increases from +4 in `NO_(2)` to `+5` in `NO_(3)^(-)` and decreases from `+4` in `NO_(2)` to `+3` in `NO_(2)^(-)` |
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| 808. |
`HNO_3` acts as .A. acidB. oxidizing agentC. reducing agentD. both (1) and (2) |
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Answer» Correct Answer - D In ` HNO_3`, H atom is attached to the second most electronegative atom. Thus, It can be released as `H^+` : `HNO_3 rarr H^(+) + NO_3^(-)` This imparts acidic character to `HNO_3`, In `HNO_3`, the oxidation state of N is +5 (the highest oxidation state of N) . Thus, it can act as an oxidizing agent only. |
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| 809. |
For decolourisation of `1 "mol of" KMnO_(4)`, the moles of `H_(2)O_(2)` required isA. `1//2`B. `3//2`C. `5//2`D. `7//2` |
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Answer» Correct Answer - C `2KMnO_(4)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]H_(2)O_(2)+OrarrH_(2)O+O_(2)xx5` Thus 1 mole of acidified `KMnO_(4)` requires 5/2 moles of `H_(2)O_(2)` for decolourisation. |
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| 810. |
Among the properties `(A)` reducing`(B)` oxidising `(C )` complexing the set of properties shown by `CN^(Θ)`ion towards metal species is .A. a,bB. a,b,cC. c,aD. b,c |
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Answer» Correct Answer - C `CN^(-)` ion is excellent complexing agent. It can also act as a reducing agent. |
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| 811. |
The largest oxidation number exhibited by an element depends on its outer eletronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number ?A. `3d^(2)4s^(2)`B. `3d^(3)4s^(2)`C. `3d^(5)4s^(1)`D. `3d^(5)4s^(2)` |
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Answer» Correct Answer - d Highest oxidation number of any transition element =(n-1)d electrons + ns electrons. Therefore, large the number of electrons in the 3d-orbitals, higher is the maximum oxidation number (a) `3d^(1)4s^(2)=3``" "`(b)`3d^(3)4s^(2)=3+2=5` ( c)`3d^(5)4s^(1)=5+1=6`and`" "`(d) `3d^(5)4s^(2)=5+2-7` Thus, option (d) is correct |
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| 812. |
In the equation `N_(2)^(ө)+H_(2)OrarrNO_(3)^(ө)+2H^(o+)+ne^(-)` `n` stands forA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B `NO_(2)^(ө)+H_(2)OrarrNO_(3)^(ө)+2H^(o+)+2e^(-)` `Balance charge on both sides. |
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| 813. |
The reaction `5H_(2)O_(2)+XClO_(2)+2OH^(-) rarr XCl^(-)+YO_(2)+6H_(2)O` is balanced ifA. `x=5`, `y=2`B. `x=2`, `y=5`C. `x=4`, `y=10`D. `x=5`, `y=5` |
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Answer» Correct Answer - B `5H_(2)O_(2)+2ClO_(2)+2OH^(-) rarr 2Cl^(-)+5O_(2)+6H_(2)O` |
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| 814. |
`MnO_(4)^(-)` is a good oxidising agent in different medium changing to `MnO_(4)^(-) to Mn^(2+) to MnO_(4)^(2-) to MnO_(2) to Mn_(2)O_(3)` Changes in oxidation number respectively,areA. 1,3,4,5B. 5,4,3,2C. 5,1,3,4D. 2,6,4,3 |
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Answer» Correct Answer - C `overset(+7)(Mn)O_(4)^(-) to Mn^(2+)` change in oxidation number 5 `rArr MnO_(4)^(2-)` change in oxidation number 1 `rArr overset(+4)(MnO_(2))` (change in oxidation number 3) `rArr overset(+3)(Mn_(2)O_(3))` change in oxidation number 4 |
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| 815. |
`2Mn_(2)O_(7)rarr4MnO_(2)+3O_(2)` ( if `M` is m ol.wt. of `Mn_(2)O_(7)`). Find the equivalent weight of ` Mn_(2)O_(7)` in above change. |
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Answer» Correct Answer - 6 `Mn_(2)^(7+)+6e rarr2Mn^(4+)` `2xx"mole of" Mn_(2)O_(7)=4xx"mole of" MnO_(2)` `=4xx3xx"eq.of"MnO_(2)` `12 "eq.of" MnO_(2)` `:.` Mole of `Mn_(2)O_(7)=6 "eq.of" MnO_(2)` `because "Eq.of" Mn_(2)O_(7)="Eq.of" MnO_(2)` `:. "Eq.wt.of" Mn_(2)O_(7)=M//6` |
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| 816. |
Consider the following reactions, (I) `2Mn_(2)O_(7) rarr 4MnO_(2) + 3O_(2)` (II) `SnCl_(2) + 2FeCl_(3) rarr SnCl_(4) + 2FeCl_(2)`A. intermolecular redox reaction and intramolecular redox reactions respectivelyB. Both reaction I and II are intermolecular redox reactionC. Both reaction I and II are intramolecular redox reactionsD. intramolecular redox reactions and intermolecular redox reaction respectively |
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Answer» Correct Answer - D Same compound e.g. `Mn_(2)O_(7)` is dissociated into `O_(2)` and `MnO_(2)`, where oxidation and reduction take place, i.e. `2Mn_(2)O_(7) rarr 4 MNO_(2) + 3O_(2)` It is an example of intramolecular redox reaction, while reaction, such as `SnCl_(2) + 2FeCl_(3) rarr SnCl_(4) + 2FeCl_(2)` is an example of intermolecular redox reaction. |
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| 817. |
Justify that the reaction `2Cu_(2)O_(s)+Cu_(2)S(s)rarr6Cu(s)+SO_(2)(g)` a redox reaction. Identify the species oxidised`//`reduced. Which acts as an oxidanat and which acts as a reductant?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 - D The oxidation states are as : `2overset(+1)(Cu_(2))overset(-2)O_((s))+ overset(+1)(Cu_(2))overset(-2)S_((2))rarr6overset(0)(Cu)_((s))+overset(+4)(S)overset(-2)O_(2(g))` Copper is reduced from +1 state to zero. Sulphur is oxidised from -2 state to +4. `Cu_(2)O` helps sulphur in `Cu_(2)S` to increase its oxidation number, hence Cu(I) is an oxidant and sulphur of `Cu_(2)S` helps copper both in `Cu_(2)S` and `Cu_(2)O` to decrease its oxidation number, hence S of `Cu_(2)S` is a reductant. And the reaction is redox reaction. |
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| 818. |
Oxidation number of P in `PO_(4)^(3-)`, of S in `SO_(4)^(2-)` and that of `Cr_(2)O_(7)^(2-)` are respectivelyA. `+3,+6and+5`B. `+5,+3and+6`C. `+3,+6and+6`D. `+5,+6and+6` |
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Answer» `PO_(4)^(3-),X+4(-2)=-3` `X=+5` `SO_(4)^(2-),X+4(-2)=-2` X=+6 `Cr_(2)O_(7)^(2-),2X+7(-2)=-2` X=-2 |
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| 819. |
Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will beA. 2.5 molB. 0.2 molC. 0.6 molD. 0.4 mol |
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Answer» The oxidising reaction is: `2MnO_(4)^(-)+16H^(+)+5C_(2)O_(4)^(2-)rarr2Mn^(2+)+10CO_(2)` `(2 mol)(5 mol)` No of moles of `MnO_(4) "required oxidise 1 mole of ferrous oxalate" =5//2 mol=2.5 mol` |
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| 820. |
Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will beA. 7.5 molB. 0.2 molC. 0.6 molD. 0.4 mol |
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Answer» Correct Answer - C `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O` `FeC_(2)O_(4)rarrFe^(2+)+C_(2)O_(4)^(2-)` `Fe^(2+)rarrFe^(3+)+e^(-)` `C_(2)O_(4)^(2-)rarr2CO_(2)+2e^(-)` Since one mole of `FeC_(2)O_(4)` loses 3 moles of electrons whole one mole of `MnO_(4)^(-)` accepts five moles of electrons Hence `KMnO_(4)` required to oxidize one mole of `FeC_(2)O_(4)=3//5 = 0.6` mol. |
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| 821. |
Which compound amongst the following gas the highest oxidation number of Mn? `KMnO_(4), K_(2)MnO_(2), MnO_(2) and Mn_(2)O_(3)`A. `KMnO_(4)`B. `K_(2)MnO_(4)`C. `MnO_(2)`D. `Mn_(2)O_(3)` |
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Answer» Correct Answer - A `KMnO_(4) rarr +1+x-8=0 rArr x= +7` `K_(2)MnO_(4)rarr +2+x-8=0 rArr x=+6` `MnO_(2) rarr x-4=0 rArr x=+4` `Mn_(2)O_(3) rarr 2x-6 = 0 rArr x= +3` |
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| 822. |
For the reaction `M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)` if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction is |
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Answer» Correct Answer - 2 `Mn^(7+) + 5e rarr Mn^(2+)` `:. 1` mole of `MnO_(4)^(-)` accepts `5` moles of electrons. `:. 5` mole of electron are lost by `1.67` mole of `M^(x+)` `:. 1` mole of `M^(x+)` will lose electron `= (5)/(165) ~~ 3` moles Since `M^(x+)` change to `MO_(3)^(-)` (oxidation no. of `M = +5`) by accepting `3` electron `:. x = +5 - 3=+2` |
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| 823. |
For the reaction `M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)` if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction isA. `5`B. `3`C. `2`D. `1` |
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Answer» Correct Answer - C `5e^(-)+MnO_(4)^(ө)rarrMn^(2+)` `M^(x+)rarrMO_(3)^(ө)+(5-x)` `x-6 = -1` `x=5` `"Equivalent of" MnO_(4)^(ө) = "Equivalent of" M^(x+)` `1 mol xx 5= 1.67 mol xx (5-x)` `:. X=2` |
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| 824. |
In the reaction `3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr5Br^(ө)+BrO_(3)^(ө)+6HCO_(3)^(ө)`A. Bromine is oxidized and the carbonate radical is reduced.B. Bromine is reduced and the carbonate radical is oxidized.C. Bromine is neither reduced nor oxidized .D. Bromine is both reduced and oxidzed. |
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Answer» Correct Answer - d |
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| 825. |
Amount of oxalic acid present in a solution can be determined by its titration with `KMnO_(4)` solution in the presence of `H_(2)SO_(4)`. The titration gives unsatisfactory result when carried out in the presence of `HCl`, because `HCl`:A. gets oxidised by oxalic acid to chlorineB. furnishes `H^(+)` ions in addition to those from oxalic acidC. reduces permangante to `Mn^(2+)`D. oxidises oxalic acid to carbon dioxide and water |
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Answer» Correct Answer - C `HCl` being stronger reducing agent it reduces `MnO_(4)^(-)` to `Mn^(2+)` and result of titration become unsatisfactory. |
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| 826. |
`Cu_(2)S+MnO_(4)^(ө)rarrCu^(2+)+Mn^(2+)+SO_(2)` The equivalent weight of `Cu_(2)` isA. `(M)/(2)`B. `(M)/(6)`C. `(M)/(8)`D. `(M)/(4)` |
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Answer» Correct Answer - C Both `Cu^(1+)` and `S^(2-)` in `Cu_(2)S` undergo oxidation. `Cu_(2)^(+1xx2)rarr2Cu^(2+)+2e^(-)(x=2)`(oxidation) `2x=2, 2x=4` `S^(2-)rarrSO_(2)+6e^(-)(x=6)`(oxidation) `x= -2 , x-4=0` `x=4` Total `e^(-)= 8` `Ew=(M)/(8)` |
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| 827. |
Balance the following reaction by the oxidation number method - `MnO_(4)^(-)+Fe^(+2)rarrMn^(+2)+Fe^(+3)` |
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Answer» Write the oxidation number of all the atoms. `overset(+7-2)(MnO_(4)^(-))+" "Fe^(+2)rarrMn^(+2)+Fe^(+3)` Change in oxidation number has occurred in Mn and Fe. `{:(overset(+7)(MnO_(4)^(-))rarroverset(+2)(Mn^(+2))," ..................(1) (Decrement in oxidation number by 5)"),(Fe^(+2)rarrFe^(+3)," ..................(2) (Increment in oxidation number by 1)"):}` To make increase and decrease equal, eq. (2) is multiplied by 5. `MnO_(4)^(-)+5Fe^(+2)rarrMn^(+2)+5Fe^(+3)` To balance oxygen, `4H_(2)O` are added to R.H.S. and to balance hydrogen, `8H^(+)` are added to L.H.S. `MnO_(4)^(-)+5Fe^(+2)+8H^(+)rarrMn^(+2)+5Fe^(+3)+4H_(2)O` This is the balanced equation. |
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| 828. |
A studend gets fingerprints on a cuvette before using it to determine the concentration of a coloured species using its known extinction coefficient. What is the effect on the absorbance and reported concentration ?A. `{:("Absorbance",,"Reported concentration"),("Increased",,"too law"):}`B. `{:("Absorbance",,"Reported concentration"),("Increased",,"too high"):}`C. `{:("Absorbance",,"Reported concentration"),("decreased",,"too low"):}`D. `{:("Absorbance",,"Reported concentration"),("decreased",,"too high"):}` |
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Answer» Correct Answer - b |
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| 829. |
Amount of oxalic acid present in a solution can be determined by its titration with `KMnO_(4)` solution in the presence of `H_(2)SO_(4)`. The titration gives unsatisfactory result when carried out in the presence of HCl because HCl:A. furnishes `H^(+)` ions in addition to those from oxalic acid.B. reduced permanganate to `Mn^(2+)`C. oxidised oxalic acid to carbon dioxide and water.D. gets oxidised by oxalic acid to chlorine. |
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Answer» Correct Answer - b |
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| 830. |
`Cu` reacts with `HNO_(3)` according to the equation `Cu+HNO_(3)rarrCu(NO_(3))_(2)+NO_(2)+H_(2)O` If `NO` and `NO_(2)` are formed in a 2:3 ratio, what is the coefficient for Cu when the equation is balanced with the simplest whole numbers ?A. 2B. 3C. 6D. 9 |
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Answer» Correct Answer - d |
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| 831. |
Select the nature or type of redox change in the following reactions: (a) `2Cu^(+)rarrCu^(2+)+Cu^(0)` (b) `Cl_(2)rarrClO^(-)+Cl^(-)` (c ) `2KClO_(3)overset(Delta)rarr2KCl+3O_(2)` (d) `(NH_(4))_(2)Cr_(2)O_(7)rarrN_(2)+Cr_(2)O_(3)+4H_(2)O` (e ) `10FeSO_(4)+2KMnO_(4)+8H_(2)SO_(4)rarrK_(2)SO_(4)rarr2MnSO_(4)+5Fe_(2)(SO_(4))_(3)+K_(2)SO_(4)+8H_(2)O` (f) `5H_(2)C_(2)O_(4)+2KMnO_(4)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O` |
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Answer» (a) and (b) represents auto-redox or disproportionation in which same substance is oxidised and reduced as well. (ii) (c ) and (d) represents intramolecular redox change in which one element of a compound is oxdised and the other elements is reduced. (iii) (e ) and (f) represents intermolecular redox in which one of the two reactant is oxidised and other is reduced. |
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| 832. |
Standard reduction potentails of the half reactions are given below: `F_(2)(g)+2e^(-) rarr 2F^(-)(aq.),, E^(ɵ)= +2.87` `Cl_(2)(g)+2e^(-) rarr 2Cl^(-)(aq.),, E^(ɵ)= +1.36 V` `Br_(2)(g)+2e^(-) rarr 2Br^(-)(aq.),, E^(ɵ)= +1.09 V` `I_(2)(s)+2e^(-) rarr 2l^(-)(aq.),, E^(ɵ)= +0.54 V` The strongest oxidizing and reducing agents respectively are:A. `F_(2)` and `I^(-)`B. `Br_(2)` and `Cl^(-)`C. `Cl_(2)` and `Br^(-)`D. `Cl_(2)` and `I_(2)` |
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Answer» Correct Answer - A The higher the value of reduction potential, the higher will be the oxidising power whereas the lower the value of reduction potential the higher will be the reducing power. |
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| 833. |
For which substance is the oxiadation number of vandium the same as that in the `VO_(3)^(-)` ion ?A. VNB. `VCl_(3)`C. `VOSO_(4)`D. `VF_(5)` |
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Answer» Correct Answer - d |
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| 834. |
When the equation `Sn^(2+)` (aq) `+IO_(3)^(-)(aq)+H^(+)(aq)rarrSn^(4+)(aq)+I_(2)(aq)+H_(2)O(l)` is balanced, what is the `Sn^(2+)(aq)//IO_(3)^(-)(aq)` mole ratio?A. `(1)/(1)`B. `(2)/(1)`C. `(1)/(2)`D. `(5)/(2)` |
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Answer» Correct Answer - d |
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| 835. |
In the balanced chemical reaction `IO_(3)^(ө)+al^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)` `a, b,c`, and `d`, respectively, correspond toA. 5,6,3,3B. 5,3,6,3C. 3,5,3,6D. 5,6,5,5 |
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Answer» Correct Answer - A `overset(+5)(I)O_(3)^(-)+overset(-1)(I)+H^(+)rarrH_(2)O+I_(2)^(0)` Decrease in O.N of I =5 per atom Increase in O.N of I = 1 per atom `IO_(3)^(-)+5I^(-)+H^(+)rarrH_(2)O+3I_(2)` `IO_(3)^(-)+5I^(-)+6H^(+)rarr3H_(2)O+3I_(2)`. |
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| 836. |
In the balanced chemical reaction `IO_(3)^(ө)+al^(ө)+bH^(ө)rarrcH_(2)O+dI_(2)` `a, b,c`, and `d`, respectively, correspond toA. 5,1,6B. 1,5,6C. 6,1,5D. 5,6,1 |
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Answer» Correct Answer - A Balanced equation is as follows `5I^(-)+IO_(3)^(-)+6H^(+)rarr3I_(2)+3H_(2)O` |
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| 837. |
Oxidarion number of Mn in `K_(2)MnO_(4)` isA. 2B. 4C. 6D. 7 |
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Answer» Correct Answer - C `K_(2)MnO_(4),2(+1)+x+4(-2)=0orx=+6` |
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| 838. |
Oxidarion number of Mn in `K_(2)MnO_(4)` is |
| Answer» Correct Answer - `+6` | |
| 839. |
Determine the oxidation number of the element as indicated (iii) I in `KIO_(3)` |
| Answer» Correct Answer - `+5` | |
| 840. |
The oxidiant state of iodine in `H_(4)IO_(6)^(ө)` isA. `+7`B. `-1`C. `+5`D. `+1` |
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Answer» Correct Answer - A `H_(4)IO_(6)^(ө): 4+x-12= -1` `impliesx=7` Oxidation state of `I= +7` |
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| 841. |
Determine the oxidation number of the element as indicated (i) P in `NaH_(2)PO_(4)` |
| Answer» Correct Answer - `+5` | |
| 842. |
What is the oxidation number of iodine in `Kl_(3)`? |
| Answer» Correct Answer - `-(1)/(3)` | |
| 843. |
The oxidation state of iodine in `ICI_(3)` isA. `+1`B. `+3`C. `+5`D. `+7` |
| Answer» Correct Answer - B | |
| 844. |
STATEMENT-1: Volume required for 0.1 M solution is in order `V_(KMnO_(4)) lt V_(K_(2)Cr_(2)O_(7)) lt V_(H_(2)O_(2))` STATEMENT-2: The number of equivalents required will be in order `H_(2)O_(2) gt KMnO_(4) gt K_(2)Cr_(2)O_(7)` STATEMENT-3: The `n_("factor")` is in order `n_(H_(2)O_(2)) lt n_(KMnO_(4)) lt n_(K_(2)Cr_(2)O_(7)`A. T T FB. T F TC. F F TD. F T F |
| Answer» Correct Answer - C | |
| 845. |
Which one of the following has the highest oxidation number of iodine?A. `K_(3)I`B. `KI`C. `IF_(5)`D. `KIO_(4)` |
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Answer» Correct Answer - D `KIO_(4)` `1+x-2xx4=0, x=8-1=+7` |
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| 846. |
For a given reductant , ratio of volumes of 0.2 M `KMnO_(4)` and `1M K_(2)Cr_(2)O_(7)` in acidic medium will be |
| Answer» Correct Answer - 6 | |
| 847. |
`0.05` moles of `NaHCO_(3)` will react with how many equivalent of `Mg(OH)_(2)`?A. `0.2` equivalentB. `0.05` equivalentC. `0.02` equivalentD. `0.01` equivalent |
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Answer» Correct Answer - B Eq. of `NaHCO_(3)=` Eq. of `Mg(OH)_(2)` `0.05=` equation of `Mg(OH)_(2)` `:.` Equivalent of `Mg(OH)_(2)=0.05` |
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| 848. |
STATEMENT-1: For the reaction `NaOH + H_(2)CO_(3) to NaHCO_(3) + H_(2)O` equivalent weight of `H_(2)CO_(3)` is 62. STATEMENT-2: n factor of `H_(2)CO_(3)` is 1 (in above reaction) and equivalent mass `=("Molecular mass")/("n factor")` |
| Answer» Correct Answer - 1 | |
| 849. |
0.144 g of pure `FeC_(2)O_(4)` was dissolvedin dilute `H_(2)SO_(4)` and the solution was diluted to 100 ml . What volume in ml of 0.1 M `KMnO_(4)` will be needed to oxidise `FeC_(2)O_(4)` solution |
| Answer» Correct Answer - 6 | |
| 850. |
Equivalent weight of `FeC_(2)O_(4)` during its reaction with `KMnO_(4)` is:A. `M//3`B. `M//1`C. `M//2`D. `M//4` |
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Answer» Correct Answer - A `{:(Fe^(2+)rarrFe^(3+)+e),((C^(3+))_(2)rarr2C^(4+)+2e),(bar(FeC_(2)O_(4)rarrFe^(3+)+2CO_(2)+3e)):}` |
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