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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.
| 1251. |
`La^(3+)` and `Lu^(3+)` areA. Both paramagneticB. Both diamagneticC. `La^(3+)` is diamagnetic and `Lu^(3+)` paramagneticD. None of these |
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Answer» Correct Answer - B `La^(3+) to [Xe] 4f^0 5d^0 6s^0 , Lu^(3+) to [Xe] 4p^14 5d^0 6s^0` No unpaired `e^(-s)` in 4f-orbitals `therefore` diamagnetic |
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| 1252. |
The lanthanoid contraction is related toA. atomic radiiB. atomic as well as ionic radii, `M^(3+)`C. valence electronsD. oxidation states |
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Answer» Correct Answer - B Lanthanoid contraction is related to both atomic and ionic radii, `M^(3+)`. |
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| 1253. |
The lanthanoid contraction is related toA. valence electronsB. densitiesC. ionic radiiD. nucleus of the atom |
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Answer» Correct Answer - C Lanthanoid contraction is related to ionic radii |
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| 1254. |
Why `Zn^(2+)` salts are white while `Ni^(2+)` salts or`Cu^(2+)` salts are blue ? |
| Answer» `Zn^(2+)` has completely filled d-orbitals `( 3d^(10))` while `Ni^(2+)` has incompletely filled d-orbitals`( 3d^(8))` or `Cu^(2+)` has `3d^(9)`.Hence, d-d transiton can take place in `Ni^(2+)` salts or `Cu^(2+)` saltsbut not in `Zn^(2+)` salts. | |
| 1255. |
Among the lanthanides the one obtained by synthetic method isA. LuB. PmC. PrD. No |
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Answer» Correct Answer - B Because it is radioactive element, does not occur in nature. |
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| 1256. |
Among the lanthanides the one obtained by synthetic method isA. LuB. PmC. PrD. Gd |
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Answer» Correct Answer - B Pm (61) - radio active element obtained by synthetic method |
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| 1257. |
The metal which corrodes readily in moist air, isA. goldB. silverC. ironD. nickel |
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Answer» Correct Answer - C Iron is covered by a layer of hydrated iron oxide `(Fe_(2)O_(3).x H_(2)O)`, i.e. rust on exposure to moist air. |
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| 1258. |
Work out the following using chemical equation. In moist air, copper corrodes to produce a green layer on the surface? |
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Answer» In moist air, copper corrodes to produce a green layer on the surface due to the following slow reaction. Basic copper carbonate is formed. `Cu+H_2O+CO_2+O_2toCuCO_3.Cu(OH)_2` |
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| 1259. |
Using IUPAC norms write the systematic names of the `[Pt(NH_(3))_(2)Cl(NH_(2)CH_(3))]Cl` |
| Answer» Diammine chlorido (methyl amine) platinum (II) chloride | |
| 1260. |
Which of the following are amphoteric oxides ? `Mn_(2)O_(7) , CrO_(3) , Cr_(2)O_(3) , CrO, V_(2)O_(5), V_(2) O_(4)`A. `V_(2)O_(5), Cr_(2)o_(3)`B. `Mn_(2)O_(7),CrO_(3)`C. `CrO, V_(2)O_(5)`D. `V_(2)O_(5),V_(2)O_(5)` |
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Answer» Correct Answer - a Oxides in the lower oxidation state are basic and in the higher oxidation state, they are acidic. In the intermediate state, they areamphoteric. |
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| 1261. |
Which of the following reactions are disproportionation reactions? `(A)" " Cu^(+)rarrCu^(2+)+Cu` `(B)" "3 MnO_(4)^(2-)+4H^(+)rarr2MnO_(4)^(-)+MnO_(2)+2H_(2)O` `(C )" "2KMnO_(4)rarrK_(2)MnO_(4)+MnO_(2)+O_(2)` `(D)" "2MnO_(4)^(-)+3Mn^(2+)+2H_(2)O rarr5MnO_(2)+4H^(+)`A. (i),(iii)B. (i),(ii),(iiii)C. (ii),(iii),(iv)D. (i),(iv) |
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Answer» Correct Answer - B (b) (i) `[overset(+1)(Cu)]^(+) to [overset(+2)(Cu)]^(2+)+ [overset(0)(Cu)]` (ii) `overset(+6)(3MnO_(4)^(2-))+4H^(+) to overset(+7)(2MnO_(4)^(-))+ overset(+4)(MnO_(2))+2H_(2)O` (iii) `overset(+7)(2KMnO_(4)) to overset(+6)(K_(2)MnO_(4))+overset(+4)(MnO_(2))+O_(2)` |
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| 1262. |
Manganese achieves its hightest oxidation state in its compoundA. `MnO_2`B. `Mn_2O_4`C. `KMnO_4`D. `K_2MnO_4` |
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Answer» Correct Answer - C `{:(KMnO_4,K,Mn,O_4),(,+1,+7,-2):}` Oxidation state of Mn is +7 (highest) |
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| 1263. |
If the colours of salts of transition elements are due to the presence of unpaired electrons in the transition metal ions, which of the following ions will be colourless in aqueous solutionA. `Ti^(3+)`B. `Ti^(4+)`C. `Fr^(2+)`D. `Fr^(3+)` |
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Answer» Correct Answer - B `Ti^(+4)to 3d^(0) 4s^(0)therefore ` no unpaired `e ^(-).` |
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| 1264. |
Assertion: In any transition series the magnetic moment of `M^(2+)` ions first decreases Reason: In a transition series, the number of unpaired electrons first increases and then decreases.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) is incorrect but (R) is correct. |
| Answer» Correct Answer - a | |
| 1265. |
Which of the following pair of transition metal ions, have the same calculated values of magnetic moment?A. `Ti^(2+) and V^(2+)`B. `Fe^(2) and Cu^(2+)`C. `Cr^(2+) and Fe^(2+)`D. `Co^(2+) and Ti^(2+)` |
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Answer» Correct Answer - C `Cr^(2+)(Z=24):[Ar]3d^(4)4s^(0)`, four unapired electrons `Fe^(2+)(Z=26):[Ar]3d^(6)4s^(0)`, foue unpaired electrons. `Cr^(2+)` and `Fe^(2+)` have same number of unpaired electrons. Hence, they have the same value of magnetic moment. |
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| 1266. |
Which one of the following transition metal ions is diamagnetic?A. `Co^(2+)`B. `Ni^(2+)`C. `Cu^(2+)`D. `Zn^(2+)` |
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Answer» Correct Answer - D `Zn^(2+)` ions have all paired electrons. So, it is diamagnetic. |
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| 1267. |
In first inner-transition series, the highest oxidation state oxhibited byA. PmB. EuC. SmD. Nd |
| Answer» Correct Answer - D | |
| 1268. |
Which of the following is not a characteristic of transition elements?A. Variable oxidations stateB. Formation if coloured compoundsC. Formation of intersititial compoundsD. Natural radioactivity |
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Answer» Correct Answer - D Natural radioactivity is not a characteristic of transition elements. |
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| 1269. |
The characteristic colour to the cation of inner-transition element is due toA. absorbed lightB. transmitted lightC. monochromatic lightD. pair of electron |
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Answer» Correct Answer - B Colour imparted to ion is always due to transmitted light |
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| 1270. |
Which of the following is not a characteristic property of transition metal-A. Hight enthalpy of atomisationB. Formation of interstitial compoundsC. DiamagnetismD. Variable oxidation state |
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Answer» Correct Answer - C Transition metals are coloured due to the presence of unpaired `e^(-)`. |
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| 1271. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. `1//3`B. 3C. `1//4`D. 4 |
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Answer» Correct Answer - A (a) `Cr_(2)O_(7)^(2-)+3Sn^(2+)+14H^(+)rarr2Cr^(3+)+3Sn^(4+)+7H_(2)O` 3 moles of `Sn^(2+)` ions are oxidised by `Cr_(2)O_(7)^(2-)` ions =1 mole. 1 mole of `Sn^(2+)` ions are oxidised by `Cr_(2)O_(7)^(2-)` ions = `1//3` mole. |
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| 1272. |
`Cr_(2)O_(7)^(2-)underset(Y)overset(X)hArr` `2CrO_(4)^(2-)`X and Y are respectively:A. `X=OH^(-),Y=H^(+)`B. `X=H^(+),Y=OH^(-)`C. `X=OH^(-),Y=H_(2)O_(2)`D. `X=H_(2)O_(2),Y=OH^(-)` |
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Answer» Correct Answer - A `Cr_(2)O_(7)""^(-2) underset(H^(+))overset(OH^(-))(hArr) CrO_(4)""^(2-)` |
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| 1273. |
The radius of `La^(3+)(Z=57)` is 106 pm. Which one of the following given values will be closest to the radius of `Lu^(3+)(Z=71)`?A. 160 pmB. 140 pmC. 106 pmD. 85 pm |
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Answer» Correct Answer - d Since ionic radii decreases from `La^(3+)`, to `Lu^(3+)` due to lanthanoid contraction, so `Lu^(3+)` have least ionic radii out of the given radii. Hence answer is 85 pm. |
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| 1274. |
To measure the quantity of `MnCl_(2)` dissolved in an queous solution, it was completely converted to `KMnO_(4)` using the reaction `MnCl_(2)+K_(2)S_(2)O_(8)+H_(2)O to KMnO_(4)+K_(2)SO_(4)+HCl`(equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further , oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. Calculate the quantity of `MnCl_(2)` (in mg) presence in the initial solution. ( Atomic weights in g `mol^(-1)`: Mn=55,Cl=35.5) |
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Answer» The redox equations involved are given : `2MnCl_(2)+5K_(2)S_(2)O_(8)+8H_(2)O ot 2KMnO_(4)+4K_(2)SO_(4)+6H_(2)SO_(4)+4HCl` ……(i) `2KMnO_(4)+5H_(2)C_(2)O_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+10CO_(2)`……(ii) Mass of oxalic acid added =225 mg Millimoles of oxalic acid added `=(225)/(90)=2.5` Millimoles of `KMnO_(4)` use to react with oxalic acid `=(5)/(2)xx(2)/(5)=1` Millimoles of `MnCl_(2)` present in the initial solution =1 Mass of `MnCl_(2)` presen in the initial solution =55+71=126 mg |
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| 1275. |
Complete the given reactions : `(A)2MnO_4^(-) + H_2O + I^(-) to 2 (i) + 2OH^(-) + IO_3^(-)` (B)`MnO_4^(-) + 5Fe^(2+) + 8H^(+) to (ii) + 5(iii) + 4H_2O`A. (i)`MnO_2` , (ii)`Mn^(2+)` , (iii)`Fe^(3+)`B. (i)`Mn^(2+)` , (ii)`Mn^(2+)` , (iii)`Fe^(3+)`C. (i)`MnO_2` ,(ii) `MnO_4^(2-)` ,(iii) `Fe(OH)_3`D. (i)`MnO_4^(2-) ,(ii) Mn^(2+) ,(iii) Fe_2O_3` |
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Answer» Correct Answer - A (A)`2MnO_4^(-) + H_2O + I^(-) to 2MnO_2 + 2OH^(-) + IO_3^(-)` (B)`MnO_4^(-) + 5Fe^(2+) + 8H^(+) to Mn^(2+) + 5Fe^(3+) + 4H_2O` |
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| 1276. |
Assertion. Cu, Ag and Au lie in Group II of the periodic table. Hence, their atomci radii are in the order `Cu lt Ag lt Au` Reason . In any group of transition elements,the atomic radii increaseas we move down the group. |
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Answer» Correct A. The atomic radii are in the order `:` `Cu lt Ag ~= Au`. Correct R. Ag and Au have same atomic radii due to the effect of lanthanoide contraction. |
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| 1277. |
Compare the chemistry of actinoids with that of lanthanoids with special reference to (i) electronic configuration (ii) oxidation state (iii) atomci and ionic sizes (iv) chemical reactivity. |
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Answer» (i) Electronic configuration. The general electronic configuration of langthanoids is `[Xe]^(54) 4f^(1-14) 5d^(0-1) 6s^(2)` whereas that of actinoids is `[Rn]^(86) 5f^(1-14) 6 s^(0-1) 7s^(2)` . Thus, langthanoids belong to 4f- series whereas actinoids belong to 5f - series. (ii) Oxidation staes. Lnathanoids show limited oxidation states`( + 2, +3, +4)` out of which `+3` is most common. This is becasue of large energy gap between 4 f and 5d subshells. On the other hand, actinoids show a large number of oxidation staes becasue of small energy gap between 5f , 6d and 7s subshell. (iii) Atomic and ionic sizes. Both show decrease in size of their atoms or ions in `+3` oxidation state. In langthanoids, the decrease is called langthanoid contraction whereas in actinoids, it is called actinoid contraction . However , the contraction is greater from element to element in actinoids due to poorer shielding by 5f electrons than that by 4f electrons in langthanoids. (iv) Chemical reactivity. Refer to page. |
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| 1278. |
Which element is not naturally occuring ?A. ThB. UC. AcD. Am |
| Answer» Correct Answer - D | |
| 1279. |
Given reasons for the following features of transition meatal chemistry `:` (i) The lowest oxide of a transition metal ( say, chromium , atomic number 24 ) is basic whereas the highest oxide is usualy acidic. (ii) Transition metals sometimes exhibit very low oxidation states such as `+1` and 0. |
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Answer» (i) Lowest oxide of Cr is CrO which is basic. The highest oxide is `CrO_(3)` which is acidic. ( In between `Cr_(2)O_(3)` is amphoteric ). Higher the oxidation state of the metal, more easily it can accept electron and hence greater is the acidic character. (ii) ` + 1` oxidation state is shown by Cu because after loss of one electron, it acqires stable configuration of`3s^(10)`. Zero oxidation state is shown in forming metal carbonyls, e.g., `Ni(CO)_(4)` because`pi-` electrons donated by the ligands are accepted into the empty d orbitals. |
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| 1280. |
For the first row transition metals, the `E^(@)` values are `{:(e^(@),V,Cr,Mn,Fe,Co,Ni,Cu),((M^(2+)//M),-1.18,-0.91,-1.18,-0.44,-0.28,-0.25,+0.34):}` Explain the irregularity in the above values. |
| Answer» This is because `E^(@)` values are the sum of sublimation enthalpy , ionization enthalpy , hydration enthalpy etc. The irregularity is due to the irregular variation of ionization enthalpies `(IE_(1)+IE_(2))` and also the sublimation enthalpies which are relatively much lower for Mn `( 240 kJ mol^(-1))` and `V ( 470 kJ mol^(-1))` | |
| 1281. |
`{:(E_(M^(2+)//M)^(@),Cr,Mn, Fe, Co, Ni, Cu),(,-0.91,-1.18,-0.44,-0.28,-0.25,+0.34):}` From the given data of `E^(@)` values, answer the following questions : (i) Why is `E_((Cu^(2+)//Cu))^(@)` value exceptionally positive ? (ii) Why is `E_((M^(2+)//M))^(@)` vlue highly negative as compared to other elements ? (iii) Which is a stronger reducing agent , `Cr^(2+)` or `Fe^(2+)` ? Give reason. |
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Answer» (i) This is due to the reason that `Cu^(2+)` ion has high value of `Delta_(a)H^(@)` and low `Delta_(hy)H^(@)`. (ii) This is due to the reason that `Mn^(2+)(3d^(5))` configuration is exceptionally stable. (iii) `Cr^(2+)` is a stronger reducing agent because of +3 oxidation state of Cr ( `t_(2g)^(3)` orbital) is more stable. |
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| 1282. |
For the first row trasition metals the `E^(ɵ)` value are: `{:(underset(-1.18)(V),,underset(-0.91)(Cr),,underset(-1.18)(Mn),,underset(-0.44)(Fe),,underset(-0.28)(Co),,underset(-0.25)(Ni),underset(+0.34)(Cu)):}` Explain the irregularity in the abpve values. |
| Answer» The `E^(ɵ)` `((M^(2+))/(M))` values are not regular which can be explained from the regular variation of ionisation enthalpies `(triangle_iH_1+triangle_iH_2)` and also the sublimation enthalpies which are relatively much less for manganese aned vanadium. | |
| 1283. |
Describe giving reason which one of the following pairs has the property indicated ? (i) Fe or Cu has higher melting point. (ii) `Co^92+)` and `Ni^(2+)` has lowe magnetic moment. |
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Answer» (i) Fe has higher melting point than Cu. This is because Fe has four unpaired electrons in 3d -subshell while Cu has only one electron in the 4s-subshell. Hence, metallic bonds in Fe are much stronger than those in CU. (ii) `._(27) Co = [Ar] 3d^(7) 4s^(2) ,Co^(2+)= [ Ar] 3d^(7) ( 3 ` unpaired electron) `. _(28) (Ni) =[Ar] 3d^(8) 4s^(2) , Ni^(2+) = [Ar] 3d^(8) ( 2` unpaired electrons) Hence, Ni has lower magnetic moment than Co. |
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| 1284. |
For the first row transition metals, the `E^(@)` values are `{:(,E^(@),V,Cr,Mn,Fe,Co,Ni,Cu),(,(M^(2+)//M),-1.18,0.90,-1.18,-0.44,-0.28,-0.25,+0.34):}` What is the reason for the non regularity in the above values?A. non regular variation of ionization enthalpiesB. different number of electrons present in `M^(2+)` ionsC. non-regular variation of ionic radiiD. the variation in densities of transition metals. |
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Answer» Correct Answer - A A it is the correct reason. For details consult section 8:4 |
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| 1285. |
Describe giving reason which one of the following pairs has the property indicated? (a). `Fe` or `Cu` has higher melting point. (b). `Co^(2+)` or `Ni^(2+)` has lower magnetic moment. |
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Answer» (a). Fe has higher melting point than Cu. This is because Fe has four unpaired electrons in 3d-subshell while Cu has only one electron electron in the 4s-soubshell. Hence, metallic bonds in Fe are much stronger than those in Cu. (b). `Co(Z=27)implies3d^74s^2,Ni^(2+)=3d^8` (2 unpaired electron). Hence, Ni has lower magnetic moment than Co, |
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| 1286. |
Decide giving reason which one of the following pairs exhibits the property indicated `:` (i) `Sc^(3+)` or `Cr^(3+)` exhibits paramagnetism(ii) V or Mn exhibits more number of oxidation states ( Atomic numbers `: Sc = 21, Cr= 24, V = 23 , Mn = 25)` |
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Answer» (i)` . _(21) Sc = [ Ar] 3d^(1) 4s^(2) , Sc^(3+) = [Ar] -` No unparied electron `. _(24)Cr)=[Ar] 3d^(5) 4s^(1) , Cr^(3+)= [Ar] 3d^(3) - `Three unpaired electron Hence, `Cr^(3+)` exhibits paramagnetism. (ii) Mn exhibits more number of oxidation states. Reason. `._(23) V = 3d^(3) 4s^(2)`. Its ox. states can be` + 2, + 3, +4, +5` `. _(25) Mn = 3d^(5) 4s^(2)` . Its ox. states can be ` +2, +3,+4,+5,+6,+7` |
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| 1287. |
Which of the following ions exhibits d-d transitions and paramagnetism as well?A. `CrO_(4)^(2-)`B. `Cr_(2)O_(7)^(2-)`C. `MnO_(4)^(-)`D. `MnO_(4)^(2-)` |
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Answer» Correct Answer - D The no. of unpaired electrons in all the ions are listed. In `CrO_(4)^(2-),Cr^(+6)(n=0)` diamagnetic In `Cr_(2)O_(7)^(2-),Cr^(+6)(n=0)` diamagnetic In `MnO_(4)^(-),Mn^(+7)(n=0)` diamagnetic In `MnO_(4)^(2-),Mn^(+6)(n=1)` diamagnetic Thus `MnO_(4)^(2-)` ion is paramagnetic and also exhibits d-d transition. |
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| 1288. |
Which one of the following compounds does not decolourise an acidified aqueous solution of `KMnO_(4)`A. Sulphur dioxideB. Ferric chlorideC. Hydrogen peroxideD. Ferrous sulphate |
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Answer» Correct Answer - B Oxidising agent `KMnO_4` and no colour is discharged |
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| 1289. |
A solution of `KMnO_(4)` on reduction yields either a coloureless solution or a brown precipitate or a green solution depending on Ph of the solution. What different stages of the reduction do these represent and how are they carried out ? |
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Answer» `KMnO_(4)` is anoxidizing agent. Its oxidising behaviour depends upon pH of the solution. In acidic medium `(pH lt 7) , MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr underset("(Coloureless)")(Mn^(2+)) + 4H_(2)O` In alkaline medium `( pH gt 7) , MnO_(4)^(-) + e^(-) rarr underset("(Green)")(MnO_()^(2-))("manganese")` In neutral medium `( pH = 7 ) , MnO_(4)^(-) + 2H_(2)O + 3e^(-) rarr underset("(Brown ppt.)")(MnO_(2)) + 4OH^(-)` |
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| 1290. |
A metal `X` on heating in nitrogen gas gives `Y,Y` on treatment with `H_(2)O` gives a colourless gas which when passed through `CuSO_(4)` solution gives a blue colour. `Y` is:A. `Mg(NO_(3))_(2)`B. `Mg_(3)N_(2)`C. `NH_(3)`D. MgO |
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Answer» Correct Answer - B Only `NH_(3)` gas gives deep blue solution with `CuSO_(4).NH_(3)` is obtained by the re5action of nitride of metal with water. `underset(X)(3Mg)+N_(2)rarrunderset(Y)(Mg_(3)N_(2))overset(6H_(2)O)rarrunderset("Colourless")(2NH_(3))uarr+3Mg(OH)_(2)` `CuSO_(4)+4NH_(3)rarrunderset("Deep blue colour")([Cu(NH_(3))_(4)]SO_(4)` |
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| 1291. |
One mole of `FeC_(2)O_(4)` is oxidised by KMnO_(4) in acidic medium. Number of moles of KMnO_(4) used areA. 0.6molB. 1.2molC. 0.4molD. 1mol |
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Answer» Correct Answer - A `Fe^(2) and C_(2)O_(4)^(2-)` both are oxidised by `MnO_(4)^(-)` `5Fe^(2+)+MnO_(4)^(-)rarrMn^(2+)+5Fe^(3+)` `5C_(2)O_(4)^(2-)+2MnO_(4)^(-)rarr10CO_(2)+2Mn^(2+)` `5FeC_(2)O_(4)-=3MnO_(4)^(-)` `1FeC_(2)O(4)-=0.6MnO_(4)^(-)` |
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| 1292. |
`2mNo_(4)^(2-)+cl _(2) rarr Mn^(2+)+5Fe^(3+) ` `MnO_(4)^(2-)` can be converted to `MnO_(4)^(-)`A. by oxidation with `Cl_(2)`B. by electrochemical oxidationat anodeC. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - C `2MnO_(4)^(2-)+Cl_(2)rarr2MnO_(4)^(-)+2Cl^(-)` `mnO_(4)^(2-)+e^(-)rarrMnO_(4)^(-)` |
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| 1293. |
Consider the followinh reaction of permaganate ions. `10^(-)+2MnO_(4)^(-)+16H^(+)rarr2Mn^(2+)8H_(2)O+5l_(2)` (II) `5Fe^(2+)+MnO_(4)^(-)+8H^(+)rarrMn^(2+)+4H_(2)O+5Fe^(3+)``5NO^(2+)+2MnO_(4)^(-)+6H^(+)rarr2Mn^(2+)+5NO_(3)^(-)+3H_(2)O` `2MnO_(4)^(-)+3Mn^(2+)+2H_(2)Orarr5MnO_(2)+4H+``5NO^(2+)+2MnO_(4)^(-)+6H^(+)rarr2Mn^(2+)+5NO_(3)^(-)+3H_(2)O` `2MnO_(4)^(-)+3Mn^(2+)+2H_(2)Orarr5MnO_(2)+4H+``5NO^(2+)+2MnO_(4)^(-)+6H^(+)rarr2Mn^(2+)+5NO_(3)^(-)+3H_(2)O` (III) `2MnO_(4)^(-)+3Mn^(2+)+2H_(2)Orarr5MnO_(2)+4H+` In the above reaction, Which reaction occur in acidic solution?A. I and IIB. II and IIIC. III,II and ID. IV and II |
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Answer» Correct Answer - C Reactions occur in acidic solution are `10l^(-)+2MnO_(4)^(-)+16H^(+)rarr2Mn^(2+)+8H_(2)O+5l_(2)` `5Fe^(2+)+MnO_(4)^(-)+8H^(+)rarrMn^(2+)+4H_(2)O+5Fe^(3+)` `5NO_(2)^(-)+2MnO_(4)^(-)+6H^(+)rarr2Mn^(2+)5NO_(3)^(-)+3H_(2)O` |
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