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.
| 1. |
An inorganic salt solution on treatment with HCl gives a white ppt. Which of the following metal ions is possible?A. `Hg_(2)^(2+)`B. `Hg^(2+)`C. `Zn^(2+)`D. `Cd^(2+)` |
| Answer» `Hg_(2)^(2+)` (ous) ions from `Hg_(2)Cl_(2)` (white ppt). | |
| 2. |
An aquous solution of a substance give a white precipitate on tretement with dilute hydrochloric acid, which dissolve on heating. When hydrogen sulphide is passed through the hot acidic solution, a black precipitate is observed.the substance is aA. `Hg_(2)^(2+)` saltB. `Cu^(2+)` saltC. `Ag^(+) ` saltD. `Pb^(2+)` salt |
| Answer» `PbCl_(2)` is soluble only in hot water. `Pb^(2+)` ions give black ppt. of PbS with `H_(2)S`. | |
| 3. |
An aqueous solution `Hg^(2+),Hg_(2)^(2+),Pb^(2+)` and `Cd^(2+)`. The addition of 6N HCl will precipitateA. `Hg_(2)Cl_(2)` onlyB. `PbCl_2` onlyC. `PbCl_2` and `Hg_(2)Cl_(2)`D. `PbCl_(2)` and `HgCl_(2)` |
| Answer» In 6N HCl, `PbCl_(2)` and `Hg_(2)Cl_(2)` are precipitated. | |
| 4. |
Sulphuric acid is not for the preparation of original solution in analysing basic radicals becauseA. it is a reducing agentB. it forms insoluble sulphateC. it forms a soluble complexD. it is viscour in nature |
| Answer» It may form insoluble sulphates with certain basic radicals such as `Pb^(2+),Ba^(2+),Ca^(2+)` etc. | |
| 5. |
Sometimes yellow turbidity appears while passing `H_2 S` gas even in the absence of II group radicals. This is becauseA. Sulphur is present in the mixture as impurityB. IV group radicals are precipitated as sulphidesC. of the oxidation of `H_(2)S` gas by some acid radicalD. III group radicals are precipitated as hydroxide. |
| Answer» `H_2 S + underset("redical")underset("oxidising")underset("From some")([O]) to underset("sulphur")underset("Colloidal")(Sdarr) _H_(2)O` | |
| 6. |
Nitric acid is generally not used for the preparation of original solution in analysis of basic radicals because itA. is an oxidising agentB. is reducing agentC. forms insoluble nitratesD. forms soluble nitrates. |
| Answer» Nitric acid is an oxidising agent it may oxidise basic radicals like `Sn^(2+),Fe^(2+),Hg_(2)^(2+)` etc. | |
| 7. |
`NH_4 CNS` can be used to test one or more out of `Fe^(3+), Co^(2+), Cu^(2+)e`A. `Fe^(3+)` onlyB. `Co^(2+),Cu^(2+)`C. `Fe^(3+),Cu^(2+)`D. All |
|
Answer» `CoCl_(2)+4NH_(4)CNS overset("ether")tounderset("in ethereal layer")underset("Blue colour")((NH_(4))_(2)[Co(CNS)_(4)])+2NH_(4)Cl` `FeCl_(3) +3NH_(4)CNS to underset("Blood red colour")(Fe(CNS)_(3))+3NH_(4)Cl` |
|
| 8. |
When chlorine water is added to an aqueous solution of potassium halide in the presence of chloroform, a violet colour is obtained on adding more of chlorine water, the violet colour disappears, and a colourless solution is obtained. This test confirms the presence of the following in aqueous solutionA. IodideB. BromideC. ChlorideD. Iodide and bromide |
|
Answer» `2I^(-) +Cl_(2) to underset("Violet")(I_(2))+2Cl^(-)` `underset("Excess")(5Cl_(2))+I_(2)+6H_(2)O to underset("Colorless")(2HIO_(3))+10HCl` |
|
| 9. |
Potassium cyanide is used for separatingA. `Co^(2+)` and `Ni^(2+)`B. `Mn^(2+)` and `Zn^(2+)`C. `Ba^(2+)` and `Ca^(2+)`D. `Cu^(2+)` and `Cd^(2+)` |
|
Answer» (A) When heated with KCN solution `Ni^(2+)` forms potassium nickelocyanide `K_(2)[Ni(CN)_(4)]` and `Co^(2+)` forms potassium cobaltocyanide, `K_(2)[co(CN)_(4)]`. On treatement with NaOH and `Br_(2)` water, `K_(2)[co(CN)_(4)]` is oxidised to very stable potassium cobalticyanide `K[Co(CN)_(4)]` and does not yeild `Co^(3+)` ions while potassium nickelcyanide being unstable, decomposes to black nickelic oxide giving rise to Ni ions `2K_(2)[Ni(CN)_(4)]+4NaOH +[O] to Ni_(2)O_(3) +4NaCN +2H_(2)O +KCN ` (D) When treated with KCN, both copper and cadmium forms respectively potassium cuprocyanide `K_(3)[Cu(CN)_(4)]` and potassiun cadmium cyanide, `K_(2)[Cd(CN)_(4)]`. `K_(2)[Cd(CN)_(4)]` is unstable and produces through `Cd^(2+)` ions while `K_(3)[Cu(CN)_(4)]` is stable and does not yield `Cu^(2+)` ions. |
|
| 10. |
Potassium ferrocyanide is used in the detection ofA. `Fe^(2+)` ionB. `Fe^(3+)` ionsC. `Cu^(2+)` ionD. `Cd^(2+)` ion |
|
Answer» `4FeCl_(3) +3K_(4)[Fe(CN)_(6)] to underset("Ferriferrocyanide")underset("Prissaian blue")(Fe_(4)[Fe(CN)_(6)]_(3)+12KCl` `2CuSO_(4)+K_(4)Fe(CN)_(6) to underset("Chocolate")(Cu_(2)[Fe(CN)_(6)]+2K_(2)SO_(4)` |
|
| 11. |
Which of the following is insoluble in dil. `HNO_2` but soluble in aqua regia?A. HgSB. PbSC. `Bi_(2)S_(3)`D. CuS |
|
Answer» HgS is insoluble in dil. `HNO_(3)` but soluble in aqua regia. `3HgS +2HNO_(3) + 6HCl to 3HgCl_(2)+2NO+4H_(2)O+5S` |
|
| 12. |
Potassium ferrocyanide is used in the detection ofA. `Cu^(2+)` ionsB. `Fe^(3+)` ionsC. bothD. none |
|
Answer» Both `Cu^(2+)` and `Fe^(3+)` are detected with potassium ferrocyanide. `2CuSO_(4)+K_(4)[Fe(CN)_(6)] to underset("Chocolate ppt.")(Cu_(2)[Fe(CN)_(6)]+2K_(2)SO_(4)` `4FeCl_(3)+3K_(4)[Fe(CN)_(6)] to underset("(Prussian blue)")underset("ferriferro cyanide")(Fe_(4)[Fe(CN)_(6)])+12KCl` |
|
| 13. |
Silver, mercury (ous) and lead are grouped together in the scheme of qualitative analysis because they formA. soluble nitratesB. carbonates which dissolve in dilute nitric acidC. insoluble chloridesD. All of above |
| Answer» Insoluble chlorides are formed by `Hg_(2)^(2+), Pb^(2+)` and `Ag^(+)` | |
| 14. |
Distinguishing reagent between silver and lead salts isA. `H_(2)S` gasB. Dil. HCl solutionC. `NH_(4)Cl"("Solid") +NH_(4)OH` solutionD. `NH_(4)Cl("solid")+(NH_4)_(2)CO_(3)` solution |
| Answer» Both Pb and Hg form white ppt. of their chlorides, but `PbCl_(2)` is solouble in hot water while `Hg_(2)Cl_(2)` does not dissolve in hot water. | |
| 15. |
Chromyl chloride vapours are dissolved in water and acetic acid and lead acetate solution is added, thenA. The solution will remain colourlessB. the solution will become dark greenC. A yellow solution will be obtainedD. A yellow ppt. will be obtained |
|
Answer» `CrO_(2)Cl_(2)+2H_(2)O to H_(2)CrO_(4)+2HCl` `H_(2)CrO_(4)+(CH_(3)CO O)_(2)Pb` `to underset("Yellow ppt.")(PbCrO_(4) darr) +2CH_(3)CO OH` |
|
| 16. |
Lead has been placed in the group I and II becauseA. it shows the valency of one and twoB. it forms insoluble `PbCl_(2)`C. it forms lead sulphideD. it is partially soluble in water |
| Answer» `PbCl_(2)` is partially soluble in water and dissolves completely only in hot water. So , part of it goes to group II, Where it is precipitated as PbS. | |
| 17. |
A mixture of chlorides of copper, cadmium, chromium, iron and aluminium was dissolved in water acidified with HCl and hydrogen sulphide gas was passed for sufficient time. It was filtered, boiled and a few drops of nitric acid were added while boiling. To this solution ammonium chloride and sodium hydroxide were added and filtered. The filtrate shall give test forA. sodium and ironB. sodium and aluminiumC. aluminium and ironD. sodium, iron , cadmium and aluminium |
|
Answer» `Al(OH)_(3)` first formed will dissolve in NaOH NaOH +`Al(OH)_(3) to NaAlO_(2) +2H_(2)O` Therefore the filtrate will contain `NaAlO_(2)`. It can be tested for Na and Al. |
|
| 18. |
The compound insoluble in acetic acid isA. Calcium oxideB. calcium dioxideC. calcium oxalateD. calcium hydroxide |
| Answer» Calcium oxalate is insoluble in acetic acid | |
| 19. |
When `H_2 S` is passed through an ammoniacal salt solution X, a white precipitate is obtained. Then X can be aA. cobalt saltB. nickel saltC. maganese saltD. zinc salt |
| Answer» On passing `H_(2)S` through zinc salt solution in the presence of `NH_(4)OH` , a white a dirty white ppt. of ZnS is obtained. | |
| 20. |
Which of the following compound on reaction with NaOH and `Na_2 O_2` gives yellow colour?A. `Cr(OH)_(3)`B. `Zn(OH)_(2)`C. `Al(OH)_(3)`D. none of these |
| Answer» `3Na_(2) O_(2) +2Cr(OH)_(3) overset(Delta) to 2NaOH +2H_(2)O + underset("Yellow")(2Na_(2)CrO_(4))` | |
| 21. |
In a mixture having nitrite and nitrate, nitrite can be destroyed by heating withA. `Na_(2)CO_(3)`B. UreaC. Oxalic acidD. NaCl |
|
Answer» `2NaNO_(2)+H_(2)SO_(4) to Na_(2)SO_(4)+2HNO_(2)` `2HNO_(2) +underset("Urea")(NH_(2)CO NH_(2)) to 3H_(2)O +CO_(2) uarr+2N_(2)uarr` |
|
| 22. |
The ion that cannot be precipitated by both HCl and `H_2 S` isA. `Pb^(2+)`B. `Cu^(+)`C. `Ag^(+)`D. `Sn^(2+)` |
|
Answer» `Pb^(2+)` and `Ag^(+)` can be precipitated both by HCl and `H_(2)S` `Cu^(+)` dissolves in conc. HCl but is precipitated on dilution. It is also precipitated by `H_(2)S`. `Sn^(2+)` is not precipitated by HCl. It is precipitated by `H_(2)S` only. |
|
| 23. |
Which of the following pairs is not distinguished by passing `H_2 S`?A. Hg,PbB. Cd,PbC. As,CdD. Zn,Mn |
| Answer» Both `Hg^(2+)` and `Pb^(2+)` give black ppt. with `H_(2)S` in group II A | |
| 24. |
`H_2 S` will precipitate the sulphide of all the metals from the solution of chlorides of Cu, Zn and Cd ifA. The solution is aqueousB. the solution is acidicC. the solution is dilute acidicD. none of the above solution is present |
| Answer» Sulphides of Cu, Zn and Cd will be precipitated as if the solution is aqueous i.e., it does not contain any acid. | |
| 25. |
A metal chloride solution on mixing with `K_2 CrO_4` solution gives a yellow precipitate, insoluble in acetic acid. The metal may beA. MercuryB. ZincC. silverD. Lead |
| Answer» `PbCl_(2)+K_(2)CrO_(4) to underset("Yellow ppt.")(PbCrO_(4) darr)+2KCl` | |
| 26. |
Sulphide ions react with `Na_2[Fe(NO)(CN)_5]` to form purple coloured compound `Na_4[Fe(CN)_5 (NOS)]` in the reaction, the oxidation state of iron changes fromA. `+2` to +3B. `+3` to +2C. `+2` to +4D. does not change. |
| Answer» Correct Answer - No change | |
| 27. |
A black sulphide is formed by the action of `H_2 S` onA. Cupric chlorideB. Cadmium chlorideC. Zinc chlorideD. Sodium chloride. |
| Answer» `CuCl_(2) +H_(2)S to underset("Black ppt.")(CuS darr)+2HCl` | |
| 28. |
For the test of halides, the soda extract is acidfied withA. Dil `H_(2)SO_(4)`B. Dil. `HNO_(3)`C. Dil. HClD. none of the three |
| Answer» Sodium carbonate extract is treated with dil. `HNO_(3)` to destroy `CO_(3)^(2-)` ion and `CN^(-)` ions (if present ) since both interface with the test. | |
| 29. |
Action of caustic soda on `Al(OH)_(3)` gives a compound having formulaA. `Na_(3)AlO_(3)`B. `NaAlO_(2)`C. `Na_(2)Al(OH)_(4)`D. `Al_(2)(OH)_(4)` |
| Answer» `Al(OH)_(3) +NaOH to NaAlO_(2) +2H_(2)O` | |
| 30. |
A substance on treatment with dil. `H_2 SO_4` liberates a colourless gas which produces (i) tubidity will baryta water and (ii) turns acidified dichromate solution green. The reaction indicates the presence ofA. `C_(2)O_(3)^(2-)`B. `S^(2-)`C. `SO_(3)^(2-)`D. `NO_(2)^(-)` |
| Answer» `SO_(3)^(2-)` reacts with dil. `H_(2)SO_(4)` to evolve `SO_(2)` which gives turbidity with baryta water , an aqueous solution of `Ba(OH)_(2)` and turns acidified dichromate solution green on its reducrtion to `Cr^(3+)`. | |
| 31. |
When a substance a reacts with water, it produces a combustible gas B and a solution of substance C in water. When another substances D reacts with this solution of C . It also produce gas B on reaction with dilute sulphuric acid at room temperature . A imparts a deep golden yellow colour to the smokless flame of bunsen flame A,B , C and D are respectivelyA. `Na,H_(2),NaOH,Zn`B. `K,H_2,KOH,Al`C. `Ca,H_(2),Ca(OH)_(2),Sn`D. `CaC_(2),C_(2)H_(2),Ca(OH)_(2),Fe` |
|
Answer» Substance A is Na. `underset("A")(2Na) +2H_(2)O to underset("D")(2NaOH) +underset("B")(H_(2))` Substances D is Zn. `underset("D")(Zn)+2NaOH overset(Delta)to Na_(2)ZnO_(2) underset("B")(H_(2))` `underset("D")(Zn)+underset("Dil.")(H_(2)SO_(4)) to ZnSO_(4) +underset("B")(H_(2))` |
|
| 32. |
A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating rod solid in a test tube produces violet coloured fumes and droplets of metal appear on the cooler parts of test tube. The red solid isA. `(NH_(4))_(2)Cr_(2)O_(7)`B. `HgI_(2)`C. HgOD. `Pb_(3)O_(4)` |
| Answer» `HgI_(2)` is scarlet red compound insoluble in water `HgI_(2)+2KI to underset("(soluble)")(K_(2)(HgI_(4))) HgI_(2) overset("Heat")to underset("Droplet")(Hg)+underset("(Violet vapour)")(I_(2))` | |
| 33. |
Ferric ion forms a prussian blue coloured ppt. due toA. `K_(4)[Fe(CN)_(6)]`B. `Fe_(4)[Fe(CN)_(6)]_(3)`C. `KMnO_(4)`D. `Fe(OH)_(3)` |
| Answer» `Fe_(4)[Fe(CN)_(6)]_(3)` is prussian blue. | |
| 34. |
When `AgNO_3` is strongly heated, the products formed areA. NO and `NO_(2)`B. `NO_(2)` and `O_(2)`C. `NO_2` and `N_(2)O`D. `NO_(2)` and `O_(2)` |
| Answer» `2AgNO_(3) overset(Delta)2Ag+2NO_(2)+O_(2)` | |
| 35. |
Consider the following observation: ltbRgt `M^(n+) +HCl to ` white precipitate `overset(Delta)to` water soluble. The metal ion `M^(n+)` will beA. `Hg^(2+)`B. `Ag^(+)`C. `Pb^(+)`D. `Sn^(2+)` |
| Answer» Ppt. of `PbCl_(2)` are soluble in hot water. | |
| 36. |
In the borax test of `Co^(2+)` , the blue colour of bead is due to the formation ofA. `B_(2)O_(3)`B. `Co_(3)B_(2)`C. `Co(BO_(2))_(2)`D. `CoO` |
|
Answer» Correct Answer - B::C::D `Co O +B_(2)O_(3) to underset("Blue bead")(Co(BO_(2))_(2))` |
|
| 37. |
When `I_(2)` is passed through KCl,KF and KBrA. `Cl_(2)` and `Br_(2)` are evolvedB. `Cl_2` is evolvedC. `Cl_(2),F_(2)` and `Br_(2)` are evolvedD. None of these |
| Answer» `I_(2)` cannot oxidised `Cl^(-),Br^(-)` and `F^(-)` to `Cl_(2),Br_(2)` and `F_(2)` respectively. | |
| 38. |
Mercurous ion is represent asA. `Hg_(2)^(2+)`B. `Hg^(2+)`C. `Hg+ Hg^(2+)`D. `3Hg^(2+)` |
|
Answer» Correct Answer - B `[Hg-Hg]^(2+)` i.e. `Hg_(2)^(+)` |
|
| 39. |
A brick red colour is imparted by aA. Ca saltB. Sr saltC. Na saltD. Co salt |
| Answer» Calcium salts impart brick red colour to the flame | |
| 40. |
When borax is heated in a platinum loop, the transparent bead formed containsA. `Na_(2)B_(4)O_(7)`B. Sodium metaborate +`B_(2)O_(3)`C. WaterD. CaO |
| Answer» `underset("Borax")(Na_(2)B_(4)O_(7)).10H_(2)O overset(Delta)to underset("sodium metaborate")(2NaBO_(2))+underset("Boric anhydride")(B_(2)O_(3))` | |
| 41. |
Colour of `KMnO_(4)` is decolourised without evolution of any gas. The radical present may beA. `SO_(4)^(2-)`B. `SO_(3)^(2-)`C. `Sn^(2+)`D. Both (b) and (C) |
| Answer» Both `SO_(3)^(2-)` and `Sn^(2+)` reduce `KMnO_(4)` without evolution of any gas. | |
| 42. |
$ Addition of `NH_(4)OH` to an aqueous solution of `BaCl_(2)` in the presence of `NH_(4)Cl` (excess) precipitates `Ba(OH)_(2)`. ! `Ba(OH)_(2)` is insoluble in water.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 of AC. A is true but R is falseD. A is false but R is true. |
| Answer» Addition of `NH_(4)OH` to an aqueous solution of `BaCl_2` in the presence `NH_(4)Cl` (excess) does not give a precipitate of `Ba(OH)_(2)` as `Ba(OH)_2` is soluble in water. | |
| 43. |
Which one of the following metals will give blue ash when its salt is heated with `Na_2 O_3` solid and `Co(NO_3)_2` on a charcoal piece?A. CuB. MgC. AlD. Zn |
| Answer» Due to the formation of `Al_(2)O_(3)Co O` (blue in colour) `Al^(3+)` will form blue mass of this test. | |
| 44. |
In `KMnO_4` titrations the indicator used isA. phenolphthaleinB. methyl orangeC. Both (A) and (B)D. none of these |
| Answer» In `KMnO_(4)` titrations , `KMnO_(4)` itself acts as an indicator. | |
| 45. |
The most suitable indicator for the titration of HBr against KOH isA. phenolphthaleinB. methyl orangeC. bromothymol blueD. all are correct |
| Answer» HBr is a strong acid and KOH is a strong base, therefore, both all the three can be used as an indicator in the titration. | |
| 46. |
The most suitable indicator for the titration of `Ba(OH)_2` against HI isA. phenolphthaleinB. methyl orangeC. Both (A) and (B)D. none of these |
| Answer» `Ba(OH)_(2)` is a strong base and HI is a strong acid, therefore both phenolphthalein and methyl orange can be used as an indicator in their titration. | |
| 47. |
The most suitable indicator for the titration of `NH_4OH` against `HNO_3` isA. phenolphthaleinB. methyl redC. thymol phthaleinD. none of these |
| Answer» `NH_(4)OH` is weak base and `HNO_(3)` is a strong base, therefore, most suitable indicator is methyl orange. | |
| 48. |
If `pK_("In")` of an indicator is 10.5, the pH transition range for which it is most suitable isA. `8.5-10.5`B. `10.5-12.5`C. `10.0-11.0`D. `9.5-11.5` |
|
Answer» The pH transition range for an indicator `=pK_("in") pm1` `=10.5 pm 1` Therefore, suitable range is , 9.5-11.5 |
|
| 49. |
The most suitable indicator for the titration of HCOOH against NaOH isA. methyl orangeB. phenolphthaleinC. thymol phthaleinD. Both (b) and (C) |
| Answer» HCOOH is a weak acid and NaOH is a strong base, therefore, both, phenolphathalein and thymol phthalein can be used as indicator. | |
| 50. |
The colour of the dye obtained by coupling benzene diazonium chloride with aniline isA. redB. orangeC. yellowD. none of these |
| Answer» When benzendiazonium chloride is coupled with aniline under acidic conditions (pH: 4-5) a yellow dye-aniline yellow is obtained. | |