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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A metal nitrate reacts with KI solution to give a block precipitate which on addition of excess of KI solution forms an orange coloured solution. The cation of metal nitrate is :A. `Hg^(2+)`B. `Bi^(3+)`C. `Pb^(2+)`D. `Cu^(2+)` |
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Answer» Correct Answer - B It is the correct answer : `Bi(NO_(3))_(3) + 3 Kl rarr underset(("Black"))(Bl_(3) (s)) + 3KNO_(3)` `Bl_(3) + Kl rarr underset("Orange coloured solution")(K [Bl_(4)])` |
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| 2. |
A solution of a metal ion when treated with KI solution gives a red precipitate which dissolves in excess of KI solution to give a colourless solution. Moreover, the solution of the same metal ion on treatment with the solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion isA. `Pb^(2+)`B. `Hg^(2+)`C. `Cu^(2+)`D. `Co^(2+)` |
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Answer» Correct Answer - B It is the correct answer : `{:("Hg"^(2+),+,2l^(-),rarr,Hgl_(2),),(,,(Kl),,("Scarlet red ppt"),),(Hgl_(2),+,2Kl,rarr,underset(("Colourless"))(K_(2)Hgl_(4)),),("Hg"^(2+),+,Co(SCN)_(2),rarr,underset(("Deep blue crystalline ppt."))(Co [Hg (SCN)_(2)]),):}` |
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| 3. |
An queous solution of a substance gives a white precipitate on treating with dilute hydrochloric acid which dissolves on heating. When hydrogen sulphide is passed through the hot solution, a black precipitate is obtained. The substance is :A. `Hg^(2+)` saltB. `Cr^(3+)` saltC. `Ag^(+)` saltD. `Pb^(2+)` salt |
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Answer» Correct Answer - A `Hg^(2+)` salt gives white precipitate with dilute HCl. `HgO + 2HCl rarr underset(("White ppt"))(HgCl_(2)) + H_(2)O` `HgCl_(2) + H_(2)S rarr underset(("Black ppt"))(HgS) + 2HCl` |
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| 4. |
A white crystalline salt [A] reacts with dilute HCl to liberate a suffocatig gas [B] and also forms a yellow precipitate. The gas [B] turns potassium dichromatic acidified with dilute `H_(2)SO_(4)` to a green coloured solution [C]. The compound A, B and C are respectivelyA. `Na_(2)SO_(3), SO_(2), Cr_(2)(SO_(4))_(3)`B. `Na_(2)S_(2)O_(3), SO_(2), Cr_(2)(SO_(4))_(3)`C. `Na_(2)S, SO_(2), Cr_(2)(SO_(4))_(3)`D. `Na_(2)SO_(4), SO_(2), Cr_(2)(SO_(4))_(3)` |
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Answer» Correct Answer - B It is the correct answer . `underset([A])(Na_(2) S_(2) O_(3)) + 2HCl rarr 2NaCl + underset([B])underset(("Yellow"))(SO_(2))+ S + H_(2)O` `K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + 3SO_(2) rarr K_(2)SO_(4) + underset([C])underset(("Green"))(Cr_(2) (SO_(4))_(3)) + H_(2)O` |
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| 5. |
A compound is soluble in water. If ammonia is added to aqueous solution of compound, a reddish brown precipitate appears which is soluble in dill. HCl. The compound is a salt of :A. aluminiumB. zincC. ironD. cadmium |
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Answer» Correct Answer - C It is the correct answer |
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| 6. |
In the second group of qualitative analysis, `H_(2)S` is pass through a solution acidified with HCl in order to :A. limit the concentration of `S^(2-)` ionsB. increase the solubility of `H_(2)S`C. increase the concentration of `S^(2-)` ionsD. provide extra `Cl^(-)` ions |
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Answer» Correct Answer - A HCl suppresses the concentration of `S^(2-)` ions in solution due to common ion effect. `H_(2)S overset((aq))hArr 2H^(+) (aq) + S^(2-) (aq)` `HCl overset((aq))rarr H^(+) (aq) + Cl^(-) (aq)` |
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| 7. |
Dolute nitric acid is generally not used for the preparation of original solution for the basic radicals because it :A. it is a reducing agentB. it is an oxidising agentC. forms insoluble nitratesD. forms soluble nitrates |
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Answer» Correct Answer - B In the second group of basic radicals, `H_(2)S` gas is passed through the solution. In case, the original solution is prepared in dilute `HNO_(3)` then `H_(2)S` will be oxidised to sulphur which will appear as a yellow precipitate or yellow turbidity. `2HNO_(3) rarr H_(2)O + 2NO_(2) + O` `H_(2)S + O rarr H_(2)O + S` This will interfere with the detection of `Cd^(2+)` and `As^(3+)` ions which also form yellow precipitates of CdS and `As_(2)S_(3)` respectively |
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| 8. |
Silver, mercury (our) and lead are grouped toether in the same group of qualitative analysis because they form:A. nitratesB. carbonates which dissolve in dill. `HNO_(3)`C. insoluble chloridesD. colourless compounds |
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Answer» Correct Answer - C All belong to group I and form insoluble chlorides with dilute HCl |
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| 9. |
Identify the correct order of solubility of `Na_(2)S, CuS` and ZnS in aqueous solution :A. `CuS gt ZnS gt Na_(2)S`B. `ZnS gt Na_(2)S gt CuS`C. `Na_(2)S gt CuS gt ZnS`D. `Na_(2)S gt ZnS gt CuS` |
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Answer» Correct Answer - D Sodium sulphide `(Na_(2)S)` is a strong electrolyte and it readily ionises in aqueous solution. Out of ZnS and CuS, the latter gets precipitated more easily because its `ksp (8.5 xx 10^(-45))` is less than that of `ZnS (2.5 xx 10^(-22))`. Therefore the correct order of solubility is: `Na_(2)S gt ZnS gt CuS` |
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| 10. |
A solution which is `10^(-3) M` each in `Mn^(2+), Fe^(2+), Zn^(2+)` and `Hg^(2+)` is treated with `10^(-16) M` sulphide ion. If `K_(sp)` of MnS, FeS, ZnS and HgS are `10^(-15), 10^(-23), 10^(-20)` and `10^(-54)` respectively, which one will precipitate first ?A. FeSB. MgSC. HgSD. ZnS |
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Answer» Correct Answer - C `HgS` will be precipitated first because [its `K_(sp)`] is minimum. |
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| 11. |
Mark the correct statement :A. Group I basic radicals are precipitated as chloridesB. Group IV basic radicals are precipitated as sulphidesC. Group V basic radicals are precipitated as carbonatesD. All the statement are correct. |
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Answer» Correct Answer - D All the three statement are correct |
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| 12. |
When `H_(2)S` is passed through `Hg_(2)^(2+)`, we get :A. `HgS`B. `HgS + Hg_(2)S`C. `HgS + Hg`D. `Hg_(2)S` |
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Answer» Correct Answer - C `Hg_(2)Cl_(2) + H_(2)S rarr 2HCl + underset((ppt))(HgS) + Hg` |
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| 13. |
How do we differentiate between `Fe^(3+)` and `Cr^(3+)` ions in group III ?A. By adding excess of `NH_(4)OH` solutionB. By increasing `NH_(4)^(+)` ion concentrationC. By decreasing `OH^(-)` ion concentrationD. Both (b) and (c) are correct |
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Answer» Correct Answer - B `k_(sp)` value of `Fe(OH)_(3)` is less that of `Cr(OH)_(3)`. Therefore, the concentration of `OH^(-)` ions in the solution, can be limited by either increasing `NH_(4)^(+)` ion concentration or by decreasing `OH^(-)` ion concentration |
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| 14. |
`Fe(OH)_(3)` can be separated from `Al(OH)_(3)` by the addition of :A. NaOH solutionB. dil. HCl solutionC. NaCl solutionD. NaOH & `NH_(4)Cl` solution |
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Answer» Correct Answer - A `Al(OH)_(3)` dissolves in NaOH while the brown precipitate of `Fe(OH)_(3)` does not dissolve. `Al (OH)_(3) + NaOH rarr underset("Soluble complex")( Na[Al(OH)_(4)])` |
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| 15. |
IN the borax bead test which compound is formed ?A. OrthoborateB. MetaborateC. Double oxideD. Tetraborate |
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Answer» Correct Answer - B A metaborate is formed |
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| 16. |
The compound formed in the Borax Bead Test of `Cu^(2+)` ions in the oxidising flame is:A. CuB. `Cu(BO_(2))_(2)`C. `CuBO_(2)`D. None of these |
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Answer» Correct Answer - B Copper meta borate, `Cu(BO_(2))_(2)` is formed and has blue colour in the oxidising flame |
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| 17. |
All ammonium salts liberate ammonia when :A. heated aloneB. heated with caustic sodaC. heated with `H_(2)SO_(4)`D. heated with `NaNO_(2)` |
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Answer» Correct Answer - B It is the correct answer. `(NH_(4))_(2) CO_(3) + 2NaOH overset("heat")rarr Na_(2)CO_(3) + 2NH_(3) + 2H_(2)O` (Typical ammoniacal smell) |
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| 18. |
A red solide is insoluble in water. However it becomes soluble if some KI is added to water. On heating the red solid in a test tube, there is liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is :A. `HgO`B. `Pb_(3)O_(4)`C. `(NH_(4))_(2) Cr_(2)O_(7)`D. `Hgl_(2)` |
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Answer» Correct Answer - D The red solide is `Hgl_(2)` `underset(("Red solide"))(Hgl_(2)) + 2 Kl rarr underset(("soluble"))(K_(2) [Hgl_(4)])` `Hgl_(2) overset("heat")rarr underset(("Droplets"))(Hg(l)) + underset(("Violet fumes"))(I_(2)(g))` |
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| 19. |
Ammonia forms the complex `[Cu(NH_(3))_(4)]^(2+)` with copper ion in alkaline solution but not in acidic solution. This is due to the reason that :A. in alkaline solution insoluble `Cu(OH)_(2)` is precipitated which is soluble in excess of any alkaliB. `Cu(OH)_(2)` is amphotericC. in acidic solution hydration protects `Cu^(2+)` ionD. in acidic solution, proton coordinates with `NH_(3)` forming `NH_(4)^(+)` ion and `NH_(3)` is not available |
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Answer» Correct Answer - D It is the correct answer. `overset(..)(N)H_(3) + underset(("Acid"))(H^(+)) rarr NH_(4)^(+)` Lone pair is not available for co-ordination with `Cu^(2+)` ion |
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| 20. |
An aqueous solution of a mixture contains `Br^(-)` and `I^(-)` ions. On passing `Cl_(2)` gas and adding `CHCl_(3)`, then organic layer will acquire :A. violet colourB. reddish brown colourC. colourlessD. blue colour |
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Answer» Correct Answer - A Violet colour will appear in `CS_(2)` layer. It is due to `I^(-)` ions. The violet colour masks the orange colour of `Br^(-)` ions. |
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| 21. |
which among the following pairs of ions cannot be separated by `H_(2)S` in the presence of dilute `HCl` ?A. `Bi^(3+) , Sn^(4+)`B. `Al^(3+), Hg^(2+)`C. `Zn^(2+), Cu^(2+)`D. `Ni^(2+), Cu^(2+)` |
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Answer» Correct Answer - A `Br^(3+)` and `Sn^(4+)` ions belong to the same group (group II). They cannot be separated by passing `H_(2)S` through the mixture solution in the presence of dilute HCl |
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| 22. |
Few drops of salt solution are shaken with chloroform followed by chlorine water. Chloroform layer becomes orange. Solution contains:A. `NO_(2)^(-)` ionsB. `NO_(3)^(-)` ionsC. `Br^(-)` ionsD. `I^(-)` ions |
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Answer» Correct Answer - C The orange colour in chloroform layer is due to `Br_(2)` gas liberated from the solution containing `Br^(-)` ions `2NaBr + Cl_(2) rarr 2 NaCl + underset(("orange colour"))(Br_(2))` |
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| 23. |
A mixture when rubbed with oxalic acid smells like vinegar. It contains salt of :A. sulphateB. nitrateC. nitriteD. acetate |
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Answer» Correct Answer - D Oxalic acid reacts with acetate salt (e.g. sodium acetate) to form acetic acid which has vinegar smell. `{:(2CH_(3)COONa,+,(COOH)_(2),),("Sod. acetate",,"oxalic aicd",),(,,darr,),(2CH_(3)COOH,+,(COONa)_(2),),("Acetic acid.",,"Sod. oxalate",),(("vinegar smell"),,,):}` |
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