InterviewSolution
Saved Bookmarks
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.
| 451. |
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. The missing terms in the reaction : `Cr_(2)O_(7)^(2-) + 14H^(+) + ? rarr 2Cr^(3+) + 7H_(2)O`A. `6e`B. `12e`C. `10 e`D. `3 e` |
|
Answer» Correct Answer - A `Cr_(2)O_(7)^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_(2)O` , Balance charge. |
|
| 452. |
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. Total number of electrons transferred during the change : `3Fe+4H_(2)O rarr Fe_(3)O_(4) + 4H_(2)`A. `2e`B. `4e`C. `6e`D. `8e` |
|
Answer» Correct Answer - D `3Ferarr(Fe^(+8//3))_(3)+8e`, `(H^(1+))_(2)rarr(H^(0))_(2)+2exx4` |
|
| 453. |
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. Select the oxidant in the reaction, `F_(2) + 1//2O_(2) rarr F_(2)O`A. `F_(2)`B. `O_(2)`C. Either of theseD. `+1` |
|
Answer» Correct Answer - A `2e+F_(2)rarr2F^(-)` |
|
| 454. |
Statement In the reactions , `3As_(2)S_(3)rarr28HNO_(3)+4H_(2)Orarr6H_(3)AsO_(4)+9H_(2)SO_(4)+28NO` electrons transferred are `84`. Explanation As is oxidised from `+3` to `+5` and sulphur from `-2` to `+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`. |
|
Answer» Correct Answer - C `{:((As^(3+))_(2)rarr2As^(5+)+4e),((S^(2-))_(3)rarr3S^(6+)+24e),(bar([As_(2)S_(3)rarr2As^(5+)+3S^(6+)+28e]xx3)),([3e+N^(5+)rarrN^(2+)]xx28):}` |
|
| 455. |
Which of the following elements does not show disproportionation tendency?A. CIB. BrC. FeD. I |
| Answer» The element F does not show disproportionation tendency since it can only take up electron (oxidation state=-1) and cannot lose electron . | |
| 456. |
Given `E_(Ag^(+)//Ag)^(@)=+0.80 V, E_(Cu^(2+)//Cu)^(@)=+0.34 V, E_(Fe^(3+)//Fe^(2+))^(@)=+0.76 V, E_(Ce^(4+)//Ce^(3+))^(@)=+1.60 V`Which of the following statements is not correct ?A. `Fe^(3+)` does not oxidise `Ce^(3+)`.B. Cu reduces `Ag^(+) ` to Ag.C. Ag will reduce `Cu^(2+)` to Cu.D. `Fe^(3+)` reduces `Cu^(2+)` to Cu. |
|
Answer» Correct Answer - C Since Ag has higher reduction potential than Cu, Ag will not reduce `Cu^(+)` to Cu. Cu can reduce `Ag^(+) ` to Ag. |
|
| 457. |
निम्नलिखित आयनो को इलेक्ट्रॉन ग्रहण करने की बढ़ती क्षमता के क्रम में लिखो । `H^(+),Mg^(2+),K^(+),Ag^(+),Zn^(2+)` तथा `Cu^(2+)`A. `Ag^(+) gt H^(+) gt Zn^(2+) gt Mg^(2+) gt K^(+)`B. `H^(+) gt Zn^(2+) gt Mg^(2+) gt K^(+) gt Ag^(+)`C. `K^(+)gtMg^(2+)gt Zn^(2+)gt H^(+) gt Ag^(+)`D. `Mg^(2+) gt Zn^(2+) gt K^(+) gt H^(+)` |
|
Answer» Correct Answer - A Based on the values of standard reduction potentials the sequence is followed. |
|
| 458. |
The percentage of copper in copper (II) salt can be determined by using a thisulphate titration. A `0.305g` of copper (II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation: `2Cu^(2+)(aq)+4I^(-)(aq)to2Cul(s)+I_(2)(aq)` The iodine liberated required `24.5cm^(3)` of a 0.1 mole `dm^(-3)`solution of sodium thiosulphate for titration according to reaction: `2S_(2)O_(3)^(2-)+I_(2)(aq)+S_(4)O_(6)^(2-)(aq)` The percentage of copper by mass in the copper (II) salt is (Atomic mass of copper =63.5)A. 64.2B. 51C. 48.4D. 25.5 |
|
Answer» Correct Answer - b |
|
| 459. |
Oxidation state of S in `SO_(4)^(2-)`A. `+8`B. `+6`C. `+4`D. 0 |
|
Answer» Correct Answer - B In `SO_(4)^(2-)` let O.N. of S = x `x + 4(-2)=-2 :. x= +6` |
|
| 460. |
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. an oxidizing agentB. a reducing agentC. a catalystD. an acid as well as an oxidant |
|
Answer» Correct Answer - d |
|
| 461. |
The standard electrode potentials of elements A,B and C are+0.68, -2.50 and -0.50 V respectively.The correct order of their reducing powers is:A. `AgtBgtC`B. `AgtBgtCgtB`C. `CgtBgtA`D. `NBgtCgtA` |
| Answer» is the correct order of reducing power | |
| 462. |
Which of the following is not an example of redox reaction?A. `CuO+H_(2)rarrCu+H_(2)O`B. `Fe_(2)O_(3)+3CO rarr 2Fe + 3 CO_(2)`C. `2K+F_(2) rarr 2KF`D. `BaCl_(2)+H_(2)SO_(4) rarr BaSO_(4)+2HCl` |
|
Answer» Correct Answer - D `BaCl_(2)+H_(2)SO_(4)rarr BaSO_(4)+2HCl` is not a redox reaction as it does not involve any change in oxidation number. It is an example of double decomposition reaction. |
|
| 463. |
The solution in a beaker turns blue ifA. Cu electrode is placed in `ZnSO_(4)` solutionB. Cu electrode is placed in `AgNO_(3)` solutionC. Cu electrode is placed in `Al_(2)(SO_(4))_(3)` solutionD. Cu electrode is placed in `FeSO_(4)` solution |
|
Answer» Correct Answer - B Cu being more reactive than Ag due to lower electrode potential, displaces Ag from `AgNO_(3)` solution. It is dissolved in the solution in the form of `Cu^(2+)` ions. |
|
| 464. |
Write redox couples involved in the reactions (i) to (iv) given in question 31 . |
|
Answer» `(i) Zn(s)|Zn^(2+)(aq),Cu^2+)(aq)|Cu(s)` `(ii) Mg(s)|Mg^(2+)(aq),Fe^(2+)(aq)|Fe(s)` `(iii)CI_(2)(g)|2CI^(-)(aq),2Br^(-)(aq)|Br_(2)(g)` `(iv) Fe(s)|Fe^(2+)(aq),Cd^(2+)(aq)|Cu(s)` |
|
| 465. |
The oxidation number of `Cr` in `K_(2)Cr_(2)O_(7)` isA. `-6`B. `+6`C. `+2`D. `-2` |
|
Answer» `overset(+1)K_(2)overset(x)Cr_(2)overset(-2)O_(7)` `2(+1)+2(x)+7(-2)=0` `2x=+12 or x=+6` |
|
| 466. |
Write redox couples involved in the reactions (a) to (d) given in quesiton 34. |
|
Answer» Given that, `Cu+Zn^(2+)toCu^(2+)+Zn` `Mg+Fe^(2+)toMg^(2+)+Fe` `Br_(2)+2Cl^(-)toCl_(2)+2Br` `Fe+Cd^(2+)toCd+Fe^(2+)` (a) `Cu^(2+)//Cu` and `Zn^(*2+)//Zn``" "`(b)`Mg^(2+)//Mg` and `Fe^(2+)//Fe` ( c) `Br_(2)//Br^(-)` and `Cl_(2)//Cl^(-)``" "`(d) `Fe^(2+)//Fe` and `Cd^(2+)//Cd` |
|
| 467. |
Which of the following is not an example of redox reaction?A. `CuO+H_(2)rarrCu+H_(2)O`B. `Fe_(2)O_(3)+3COrarr2Fe+3CO_(2)`C. `2K+Fe_(2)rarr2KF`D. `BaCI_(2)+H_(2)SO_(4)rarrBaSO_(4)+2HCI` |
| Answer» It is not a redox reaction | |
| 468. |
Assertion : In the species, `S_(4)O_(6)^(2-)` each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero. Reason : The average of four oxidation numbers of sulphurs of the `S_(4)O_(6)^(2-)` is 2.5.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 - B The structure of `S_(4)O_(6)^(2-)` is `.^(-)O-underset(underset(O)("||"))overset(overset(O)(" ||+5"))("S")-overset(0)S-overset(0)S-underset(underset(O)("||"))overset(overset(O)(" ||+5"))("S")-O^(-)` Structural parameters reveal that the element for whom fractional oxidation state is realised, is present in different oxidation states and different oxidation states are due to different bonding situations. |
|
| 469. |
Which of the following cannot displace hydrogen from steam ?A. `Cd`B. `Fe`C. `Cr`D. `Zn` |
|
Answer» Correct Answer - A Less reactive metals, such as `Ma. Al. Mn, Zn, Cr,` and ` Fe`, react with steam to give hydrogen gas : ` 2Fe(s) + 3H_2 O(g) rarr Fe_2O_3(s) + 3H_2(g)` `Cd + H_2O(g) rarr` No reaction. |
|
| 470. |
Which of the following metals cannot displace hydrogen from cold water ?A. `K`B. `Mg`C. `Ca`D. `Na` |
|
Answer» Correct Answer - B All alkali metals and some alkaline earth metals (Ca,Sr, and Ba) which are the most reactive of the metallic elements will displace hydrogen from cold water : `Ca(s) + 2 H_2O(1) rarr Ca (OH)_2(s) + H_2(g)` `Mg(s) + 2H_2O(1) rarr` No reaction Mg reacts with ` H_2O` only at higher temperature. |
|
| 471. |
Assertion: Copper liberates hydrogen from a solution of dilute hydrochloric acid. Reason: Hydrogen is above copper in the electro- chemical series.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. |
|
Answer» Correct Answer - D `Cu` cannot liberate `H_(2)` from a solution of dilute `HCl` as hydrogen is above copper in the electrochemical series. `E_(H^(+//_(1//2)H_(2)))^(@)=+-0.00` , `E_(Cu^(2+//_(Cu))=+0.34`. |
|
| 472. |
Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}` Among the following, identify the correct statement:A. Chloride ion is oxidised by `O_(2)`B. `Fe^(2+)` ios oxidised by iodideC. lodide ion is oxidised by chlorineD. `Mn^(2+)` is odidised by chlorine |
|
Answer» Correct Answer - C On calculating the emf of the cell we find that only the cell involving the oxidation of `1^(-) "ion by " CI_(2)` is positive `(2I^(-)rarrI_(2)+2e^(-),E^(@)=0.54 V)` `{:(CI_(2)+2e^(-)rarr2CI^(-),E^(@)=1.36V),(CI_(2)+2I^(-)rarr2CI^(-)+l_(2),E^(2)_(cell)=1.36-0.54):}`` "In all other cases the" `E^(@)` is -ve |
|
| 473. |
What is the equivalent mass of `S_(2)O_(3)^(2-)` ion as par the following disproportionation reaction. `S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+S_(2)^(2-)` Where the charge above species represents charge on the ion and not oxidation state?A. 132B. 22C. 130.6D. 113.15 |
|
Answer» Correct Answer - c |
|
| 474. |
In which of the following processess is nitrogen oxidised ?A. `NH_(4)^(+) rarr N_(2)`B. `NO_(3)^(-) rarr NO`C. `NO_(2) rarr NO_(2)^(-)`D. `NO_(3) rarr NH_(4)^(+)` |
|
Answer» Correct Answer - A `2N^(3-) rarr N_(2)+6e` |
|
| 475. |
Given below are a set of half-cell reactions (acidic medium) along with their `E_(@)` with respect to normal hydrogen electrode values. Using the data obtain the correct explanation to question given below. `{:(I_(2)+2e^(-)rarr2I^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-),E^(@)=1.36),(Mn^(2+)+e^(-)rarrMn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):}` While `Fe^(2+)` is stable, `Mn^(3+)` is not stable in acid solution because:A. `O_(2) "oxidises lt" n^(2+) "to" Mn^(3+)`B. `O_(2) "oxidises both" Mn^(2+) "to" Mn^(3+) "and" Fe^(2+) to Fe^(3+)`C. `Fe^(3+) "oxidises" H_(2)O to O_(2)`D. `Mn^(3+) "oxidises" H_(2)O "to" O_(2)` |
|
Answer» Correct Answer - D In this case also, only the e.m.f of the cell involving the oxidation of `H_(2)O to O_(2)` by `Mn^(3+)` ion is positive `{:(Mn^(3+)+e^(-)rarrMn^(2+)xx4,E^(@)=+1.50 V),(2H_(2)Orarr4H^(+)+O_(2)+4e^(-),E^(@)=-1.23V),(Mn^(3+)+2h_(2)Orarr4Mn^(2+)+O_(2)+4H^(+),E^(@)=+0.27V):}` |
|
| 476. |
Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportinates into two oxidation states ( lower and higher) is called disproportionation reaction. Which of the following statements is wrong?A. The algebraic sum of the oxidation numbers of all atoms in an iron is zero.B. The oxidation number is an arbitrary number. It can have positive, negative, zero, or fractional values.C. When a negative ion changes to neutral species, the process is oxidation.D. The oxidation number of phosphorous can very from `-3 t o +5`. |
|
Answer» Correct Answer - A a. Wrong: The algebraic sum of the oxidation numbers of all the atoms in an ion equals the charge present on the ion. b. True statement c. True `(2Cl^(ө)rarrCl_(2)+2e^(-))`(oxidation) True |
|
| 477. |
Assertion (A): A reaction between `Fe` and `I_(2)` occurs, but a reaction between `Fe^(2+)` and `I^(ө)` does not occur. Reason (R ): `Fe` is a better reducing agent than `I^(ө)`.A. If both `(A)` and `(R )` are correct and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct but `(R )` is incorrect.D. If `(A)` and `(R )` are incorrect. |
|
Answer» Correct Answer - A Both `(A)` and `(R )` are correct and `(R )` is the correct explantion for `(A)` `I_(2)+FerarrFe^(2+)+2I^(ө)` `I^(ө)+Fe^(2+)rarr "No reaction"` The oxidation potential of `Fe//Fe^(2+)` is greater than the oxidation potential of `2I^(ө)//I_(2)`. |
|
| 478. |
Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportinates into two oxidation states ( lower and higher) is called disproportionation reaction. Which of the following statements is wrong?A. An acidified `K_(2)Cr_(2)O_(7)` paper on being exposed to `SO_(2)` turns green.B. Mercuric chloride and stannous chloride cannot exist as such.C. Iron turning on addition to `CuSO_(4)` solution decolourises the blue colour.D. `[CuI_(4)]^(2-)` is formed but `[CuCl_(4)]^(2-)` is not. |
|
Answer» Correct Answer - D a. `Cr_(2)O_(7)^(2-)` (orange red) oxidises `SO_(2)` to `SO_(4)^(2-)` and is itself reduced to `Cr^(+3)` (green). `underset(("Orange"))(Cr_(2)O_(7)^(2-)+3H^(o+)+3SO_(2))rarrunderset(("Green"))(2Cr^(3+)+3SO_(4)^(2-)+4H_(2)O` b. Both reacts together `overset(+2)(Hg)Cl_(2)+overset(+2)(SnCl_(2))rarroverset(+1)underset(("White ppt"))(HgCl_(2))darr+overset(+4)(SnCl_(4))` `overset(+1)(Hg_(2))Cl_(2)+overset(+2)(SnCl_(2))rarrunderset((Grey))(2Hg)darr+overset(+4)(SnCl_(4))` c. `overset(0)Fe+underset((Blue))overset(+22)(CuSO_(4))rarrFe^(2+)+overset(0)(Cu)` d. `[CuCl_(4)]^(2-)` is formed but `[CuI_(4)]^(2-)` is not. `I^(ө)`ion reduces `Cu^(2+)` to `overset(+1)(Cul)` and itself undergoes oxidation to form `I_(2)`. However, `CI^(ө)` does not reduce `Cu^(2+)`. |
|
| 479. |
Match the items in column I with relevant items in column II |
|
Answer» Correct Answer - `A.to(5)" "B.to(4)" "C.to(3)" "D.to(2)" "E.to(6)` `A.to(5)" "B.to(4)" "C.to(3)" "D.to(2)" "E.to(6)` A. Ions having positive charge - Cation B. The sum of oxidation number of all aotms in a neutral molecule - Zero C. Oxidation number of hydrogen ion`(H^(+))`-+1 D. Oxidation number of fluroint in `NaF-`-1 E. Ions having negative charge - Anion |
|
| 480. |
Fluorine is the best oxidising agent because it hasA. it is most electronegative .B. it has highest reduction potential.C. it has highest oxidation potential.D. it has smallest size. |
|
Answer» Correct Answer - B Higher the reduction potential, stronger is the oxidising agent. |
|
| 481. |
Identify the correct statement(s) in relation to the following reaction. `Zn+2HCl to ZnCl_(2)+H_(2)`A. Zinic is acting as an oxidantB. Chlorine is acting as a reductantC. Hydrogen ion is acting as an oxidantD. Zinc is acting as a reductant |
| Answer» Correct Answer - are both correct | |
| 482. |
`Cl_(2)+H_(2)S to 2HCl+S`, In that above reaction oxidation state of chlorine changes fromA. zero to -1B. 1to zeroC. zero to 1D. remains unchanged. |
| Answer» Correct Answer - A | |
| 483. |
Identify the correct statement(s) in relation to the following reaction. `Zn+2HCl to ZnCl_(2)+H_(2)`A. Zinc is acting as an oxidantB. Chlorine is acting as a reductantC. Hydrogen ion is acting as an oxidantD. Zinc is acting as a reductant |
|
Answer» Correct Answer - c,d Writing the oxidation number of each element above its symbol, so that `overset(0)(Zn)+2overset(+1)(H)overset(-1)(Cl)to overset(+2)(Zn)overset(-1)(Cl_(2))+overset(0)(H_(2))` (a) The oxidation number of Zn increases from 0 in Zn to +2 in `ZnCl_(2)`, therefore, Zn acts as a reductant. Thus, option (a) is incorrect (b) The oxidation number of chlorine does not change, therefore, it neither acts as a reductant nor an oxidant. Therefore, option (b) is incorrect. ( c) The oxidant number of hydrogen decreases from +1 in `H^(+)` to 0 in `H_(2)`. therefore, `H^(+)` acts acts as an oxidant. Thus, option ( c) is correct (d) As explained in option (a), Zn acts as reductant, therefore, it cannot act as an oxidant. Thus, option (d) is correct. |
|
| 484. |
Identify the disproportionation reaction.A. `CH_(4)+2O_(2)toCO_(2)+2H_(2)O`B. `CH_(4)+4Cl_(2)toC Cl_(4)+4HCl`C. `2F_(2)+2OH^(-)to2F^(-)+OF_(2)+H_(2)O`D. `2NO_(2)+2OH^(-)toNO_(2)^(-)+NO_(3)^(-)+H_(2)O` |
|
Answer» Correct Answer - d Reactions in which the same substance is oxidised as well as reduce are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions. (a) `overset(-4)(C)overset(+1)(H_(4))+2overset(@)(O_(2))tooverset(+4)(C)overset(-2)(O_(2))+2overset(+1)(H_(2))overset(-2)(O)` (b) `overset(-4)(C)overset(+1)(H_(4))+4Coverset(0)(l_(2))to overset(+4)(C)overset(-1)(C)l_(2)+4overset(+1)(H_(2))overset(-1)(Cl)` ( c) `2overset(0)(F_(2))+2overset(-2)(O)overset(+1)(H)to2overset(-1)(F^(-))+overset(+2)(O)overset(-1)(F_(2)^(-))+overset(+1)(H_(2))overset(-2)(O)` (d) `2overset(+4)(N)overset(-2)(O_(2))+2overset(-2)(O)overset(+1)(H^(-))tooverset(+3)(N)overset(-2)(O_(2)^(-))+overset(+5)(N)overset(-2)(O_(3)^(-))+overset(+1)(H_(2))overset(-2)(O)` Thus, in reaction (d), N is both oxidised as well as reduced since the O.N. of N increases from +4 in `NO_(2)` to +5 in `NO_(3)^(-)` and decreases from +4 in `NO_(2)` to +3 in `NO_(2)^(-)` |
|
| 485. |
Mark the correct statement from the following :A. Copper metal can be oxidised by `Zn^(2+)` ions.B. Oxidation number of phosphorus in `P_(4)` is 4C. An element in the highest oxidation state acts only as a reducing agent.D. The element which shows highest oxidation number of +8 is Os in `OsO_(4)` |
|
Answer» Correct Answer - D Cu cannot be oxidised by `Zn^(2+)` ion, oxidation number of phosphorus is zero. An element in its highest oxidation state acts as an oxidising agent. |
|
| 486. |
The oxidation state of nitrogen in `N_(3)H` isA. `+1//2`B. `+3`C. `-1`D. `-1//3` |
|
Answer» `overset(x)N_(3)overset(+1)H` O.N of `N=-1//3` |
|
| 487. |
Suggest a list of the substances where carbon can exhibit oxidation states from `-4` to `+4` and nitrogen from `-3` to `+5`. |
|
Answer» variable oxidation states of carbon: `CH_(4) CH_(3)CI CH=CH HCHO CO K_(2)C_(2)O_(4) CO_(2)` `(-4) (-3) (-2) (-1) (0) (+2) (+3) (+4)` variable oxidation states of nitrogen: `NH_(3) NH_(2)NH_(2) NH_(2)OH N_(2) N_(2)O NO N_(2)N_(2)O_(3) NO_(2) N_(2)O_(5)` `(-3) (-02) (-1) (0) (+1) (+2) (+3) (=4) (+5)` |
|
| 488. |
The oxidation number of nitrogen in `(N_(2)H_(5))^(+)` isA. `-2`B. `+2`C. `+3`D. `-3` |
|
Answer» Correct Answer - A `(N_(2)H_(5))^(+)rArr 2x+5=+1 rArr 2x=-4 rArr ` or, `x=-2` |
|
| 489. |
The oxidation state of nitrogen in `N_(3)H` isA. `+(1)/(2)`B. `+3`C. `-1`D. `-(1)/(3)` |
|
Answer» Correct Answer - D The oxidation state of hydrogen is always `+1` except in ionic hydrides (where ON of hydrogen is `-1`). `N_(3)H` (hydrogen acid) is not an ionic compound, hence `3x + (+1) = 0` `3x = -1, x = -1//3` |
|
| 490. |
Which of the following is redox reaction ?A. `H_(2)SO_(4)` with `NaOH`B. In atmosphere, `O_(3)` from `O_(2)` by lightningC. Evaporation of `H_(2)O`D. Nitrogen oxides from nitrogen and oxygen by lightning |
|
Answer» Correct Answer - D `overset(0)(2N_(2))+overset(0)(O_(2)) rarr overset(+2-2)(2NO)` Here `O.N`. Of `N` increases from `O` in `N_(2)` to `+2` in `NO`, `2-` and that of decreased from `O` in `O_(2)` to `-2` in `O`, therefore, it is a redox reaction. |
|
| 491. |
Assertion: Oxygen atom in both `O_(2)` and `O_(3)` has oxidation number zero. Reason: In `F_(2)O`, oxidation number of `O` is `+2`.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 - D In elements, in the free state each atom bears an oxidation number of zero. Oxidation number of oxygen in most of its compounds is -2 with two exceptions : in peroxides, oxidation number of oxygen is `-(1)/(2)`. |
|
| 492. |
Oxidation number of oxygen in `F_(2)O` isA. `-1`B. `+1`C. `+2`D. `-2` |
|
Answer» Correct Answer - C Fluorine is the most electronegative element and oxygen is the second one. Hence, in `F_(2)O`, oxygen is the positive element `x + 2 (-1) = 0` `rArr x = +2` Oxidation state of oxygen in `F_(2)O` (or `OF_(2)`) is `+2` |
|
| 493. |
When sodium metal is dissolved in liquid ammonia, blue colour solution is formed. The blue colour is due toA. Solvated `Na^(+) ions`B. Solvated electronsC. Solvated `NH_(2)^(-)` ionsD. Solvated protons |
|
Answer» Correct Answer - B When sodium metal is dissolved in liquid ammonia to form coloured solution. Dilute solutions are bright blue in colour due to the presence of solvated electrons. `Na+(x+y)NH_(3) rarr [Na(NH_(3))_(x)]^(+)+[e(NH_(3))_(y)]^(-)` Blue colour |
|
| 494. |
The brown ring complex compound is formulated as `[Fe(H_(2)O_(5))No]SO_(4)`. The oxidation state of `Fe` isA. `1`B. `2`C. `3`D. `0` |
|
Answer» Correct Answer - A `[overset(+1)(Fe)(H_(2)overset(0)O)_(5)overset(1+)(NO)]^(2+) SO_(4)^(2-)` Oxidation state of `Fe= +1` |
|
| 495. |
Cartain materials such as turpentine oil, unsaturated organic compound, phosophorus, metals such as `Zn`, and `Pb`, etc., can absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to `H_(2)O_(2)`. This is called autoxidation. Intermolecular redox reactions are those in which one molecule is oxidised and the other is reduced. Intramolecular redox reactions are those in which oen atom of a molecule is oxidised and the other atom is reduced. Which of the following reactions is//are intramolecular redox reactions (s) ?A. `2Mn_(2)O_(7)rarr4MnO_(2)+3O_(2)`B. `K_(3)[Fe(CN)_(6)]+30H_(2)OrarrFe^(3+)+6CO_(2)+6NO_(3)^(ө)+60H^(o+)+60e^(-)`C. `2HgOrarr2Hg+O_(2)`D. `PhCHOoverset(NaOH)rarrPhCH_(2)OH+PhCOONa` |
|
Answer» Correct Answer - A::C Hence, it is an intramolecular redox reaction. b. It is an intramolecular redox reaction. c. It is an intramolecular redox reaction. `{:(2e^(-)+,Hg^(2+),rarr,Hg,("Reduction")),(,2O_(2)^(2-),rarr,O_(2)+4e^(-),("Oxidation")):}` d. It is a disproportionation reaction. |
|
| 496. |
The oxidation number of `C` in `CH_(2)O` isA. `-2`B. `+2`C. `0`D. `+4` |
|
Answer» Correct Answer - C `CH_(2)O:x+2-2=0impliesx=0` |
|
| 497. |
Cartain materials such as turpentine oil, unsaturated organic compound, phosophorus, metals such as `Zn`, and `Pb`, etc., can absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to `H_(2)O_(2)`. This is called autoxidation. Intermolecular redox reactions are those in which one molecule is oxidised and the other is reduced. Intramolecular redox reactions are those in which oen atom of a molecule is oxidised and the other atom is reduced. Which of the following reactions has`//`have spectator ions?A. `Zn+CuSO_(4)rarrZnSO_(4)+Cu`B. `KIO_(3)+KI+H_(2)SO_(4)rarrKI_(3)+K_(2)SO_(4)+H_(2)O`C. `2KMnO_(4)+10KCl+8H_(2)SO_(4)rarr5Cl_(2)+2MnSO_(4)+8H_(2)O+6K_(2)SO_(4)`D. `[CrCl_(6)]^(3-)+Zn[ZnCl_(4)]^(2-)+Cr^(2+)` |
|
Answer» Correct Answer - A::B::C::D Spectator ions in `(a), (b)`, and `(c )` are `SO_(4)^(2-)` ions while in `(d)`, it is `Cl^(ө)` ion, since all of them remain unchanged |
|
| 498. |
Cartain materials such as turpentine oil, unsaturated organic compound, phosophorus, metals such as `Zn`, and `Pb`, etc., can absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to `H_(2)O_(2)`. This is called autoxidation. Intermolecular redox reactions are those in which one molecule is oxidised and the other is reduced. Intramolecular redox reactions are those in which oen atom of a molecule is oxidised and the other atom is reduced. Which of the following reactions is`//`are disproportionation reactions `(s)`?A. `Cl_(2)+2overset(ө)OHrarrCl^(ө)+ClO^(ө)+H_(2)O`B. `2HCuCl_(2)rarrCu+Cu^(2+)4Cl^(ө)+2H^(o+)`C. `HCHO+overset(ө)OHrarrCH_(3)OH+HCOO^(ө)`D. `MgCO_(3)rarrMgO+CO_(2)` |
|
Answer» Correct Answer - A::B::C `{:(2e^(-)+Cl_(2) rarr2Cl^(ө)("Reduction")),(" "2ClO^(ө)+2e^(-) ("Oxidation")),(" "[Disproportionation]):}` It is also a disproportional reaction. `[overset(+1)Hoverset(+1)(Cu)overset(-1xx2)(Cl_(2))]` `e^-)+Cu^(1+)rarrCu ("Reduction")` `Cu^(1+)rarrCu^(2+)+e^(-)("oxidation")` c. It is a also a disproportionation reduction. `HCHO` is oxidised to `HCOO^(ө)` and reduced to `CH_(3)OH`. d. It is none since the oxidation state of `Mg` is `+2` in both reactant and product. Similarly, the oxidation states of `C` and `O` are `+4` and `-2` in both reactant and product. |
|
| 499. |
Cartain materials such as turpentine oil, unsaturated organic compound, phosophorus, metals such as `Zn`, and `Pb`, etc., can absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to `H_(2)O_(2)`. This is called autoxidation. Intermolecular redox reactions are those in which one molecule is oxidised and the other is reduced. Intramolecular redox reactions are those in which oen atom of a molecule is oxidised and the other atom is reduced. Which of the following statements about the reaction is`//`are correct? `2AsCl_(4)^(ө)+3Znrarr2Au+3Zn^(2+)+8Cl^(ө)`A. `AuCl_(4)^(ө)` is reduced to `Au`B. `Zn` is oxidised to `Zn^(2+)`C. `Cl^(ө)` is a spectator ion.D. It is an intermolecular redox reaction. |
|
Answer» Correct Answer - A::B::C::D All statements are correct. `Cl^(ө)` ion remains unchanged. It is a spectator ion. |
|
| 500. |
Cartain materials such as turpentine oil, unsaturated organic compound, phosophorus, metals such as `Zn`, and `Pb`, etc., can absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to absorb `O_(2)` from air in the presence of `H_(2)O`, which is converted to `H_(2)O_(2)`. This is called autoxidation. Intermolecular redox reactions are those in which one molecule is oxidised and the other is reduced. Intramolecular redox reactions are those in which oen atom of a molecule is oxidised and the other atom is reduced. Which of the following reactions is`//`are intermolecular redox reaction `(s)`A. `5KI+KIO_(3)+6HClrarr3I_(2)+6KCl+3H_(2)O`B. `Fe+N_(2)H_(4)rarrNH_(3)+Fe(OH)_(2)`C. `NO_(3)^(ө)+H_(2)S+H_(2)O+H^(o+)rarrNH_(4)^(o+)+HSO_(4)^(ө)`D. `CrO_(7)^(2-)+2overset(ө)OHrarr2CrO_(4)^(2-)+H_(2)O`s |
|
Answer» Correct Answer - A::B::C It is also an intermolecular redox reaction. a. It is an intermolecular redox reaction. `{:(,2I^(ө),rarr,I_(2)+2e^(-),("Oxidation")),(10e^(-)+,2IO_(3)^(ө),rarr,I_(2),("Reduction")):}` b. It is also an intermolecular redox reaction. `{:(,Fe,rarr,Fe^(2+)+2e^(-),("Oxidation")),(2e^(-)+,N_(2)H_(4),rarr,2NH_(3),("Reduction")),(,2x = -4,,2x = -6,):}` c. It is also an intermolecular redox reaction. `{:(8e^(-)+,NO_(3)^(ө),rarr,overset(o+)(NH_(4)),("Oxidation")),(,x=5,,x+4=+1,),(,,,x = -3,),(,S^(2-),rarr,HSO_(4)^(ө)+8e^(-),("Oxidation")),(,x = -2,,1+x-8 = -1,),(,,,x=6,):}` d. It is, since the oxidation state of `Cr` in both reactant and product is `+6`. |
|