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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.
| 151. |
Assertion :When `H_(2)S` is passed through a solution of `CuSO_(4)` no precipitate of `CuS` is abtain until the solution is acidified with `HCI` Reasion: The solution products constant of `CuS` is not so high as to require a high concen tration of `S^(2-)` for the precipitate of `CuS`A. If both (A) and (B) are correct and (R ) is the correct explqanation of (A)B. If both (A) and (B) are correct but (R ) is not the correct explqanation of (A)C. If (A) is correct ,but (R ) is incorrectD. If (A) is incorrect ,but (R ) is correct |
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Answer» Correct Answer - d It is true that precipitate of `CuS` does not requir a high concentration of `S^(2-)` ions But this does not mean that `CuS` will not precipitate with a higher concentration of `S^(2-)` In fact ,the precipitate will be faciliated by a high concentration of `S^(2-)` ,i.e. when the solution is not acidified so , (A) is false ,but (R ) is true |
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| 152. |
Aq solution contains `Zn(CH_(3)COO)_(2) Cd(CH_(3)COO)_(2)` and `Cu(CH_(2)COO)_(2)` on passing `H_(2)S` gas there is precipitate of …….. As sulphide.A. `Zn^(2+),Cd^(2+)`B. `Cu^(2+),Cd^(2+)`C. `Zn^(2+),Cu^(2+)`D. `Zn^(2+),Cu^(2+),Cd^(2+)` |
| Answer» Correct Answer - d | |
| 153. |
`K_(4)[Fe(CN)_(6)]` can be used to detect one or more out of `Fe^(2+),Fe^(3+),Zn^(2+),Cu^(2+),Cd^(2+)`A. `Fe^(2+),Fe^(3+)`B. `Fe^(3+),Zn^(+),Cu^(2+)`C. all but `Fe^(2+)`D. all but `Fe^(2+)` |
| Answer» Correct Answer - d | |
| 154. |
A colourless inorganic salt A decomposes completely as about `250^(@)C` to give only produce B and C liquid at room temprature and neutral to moist paper .white the gas B is a netrual oxide .White phospydras burns in excess of B to produce a strong white dehydrating agent write the balanced equation for the reaction involved in the above process |
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Answer» A colourless inorganic salt A decomposes completely at about `250^(@)C` to give only two products B and C having to residue .The oxide C is a liquid at room temperature and neatral to moist litmous paper is water .Gas B is a neatral oxide , it must be nitrous oxide so, A must be ammonium nitrate .white phosphorus burns in excess of nirrous oxide to produce a strong white debydrating agent phosphorous penoxide `underset((A))(NH_(4)NO_(3)) overset(250^(@)C) rarr underset((B))(N_(2)Ouarr) + underset((C))(2H_(2)uarr)` No residue left. `P_(4) + 10 N_(2)O rarr P_(4)O_(10) +10N_(2)` |
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| 155. |
During the quation analysis of a mixture containing `Cu^(2+) and Zn^(2+) ions H_(2)S` gas is passed therough an acidified solution containing these ions in order to test `Cu^(2+)` alone explane |
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Answer» `H_(2)S hArr 2H^(o+) +S^(2-)` `HCI rarr H +CI^(Theta)` `K_(sp) `(solubility product) of `CuS` is less than `K_(sp)` of `ZnS `.Due to common ion effect ,the ionisation of `H_(2)S` is sappressed and the `S^(2-)` concentration decreases Only those metal sulphides get precipitate which have low `K_(sp)` so ZnS is not precipitate. |
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| 156. |
An aqueous solution containing 1 mol of `HgI_(2)` and 2 mol of Nal is orange in colour .On addition od excess NaI , the solution becomes colourless .The orange colour reappears on subsequent addition of `NaOCl`. |
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Answer» `HgI_(2)` is having orange / scarlet colour it dissolve in sodium iodide due to the formation of a complex . `HgI_(2) + 2Nal rarr Na_(2)HgI_(4)` with `NaOCI` agent a precipitate of `HgI_(2)` is formed `3Na_(2)HgI_(4) + 2NaOCI + 2H_(2)O rarr 3HgI_(2) + 2NaI_(3) + 4NaOH + 2NaCI` |
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| 157. |
`NH_(4)CI` can be replaced by `(NH_(4))_(2)SO_(4)` in group III. |
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Answer» Correct Answer - T `NHCI` is used along with `NH_(4)OH` in group II to decreases ionicsation of `HN_(4)OH` by common the effect (of `NH_(4)^(oplus)),(NH_(4))_(2)SO_(4)` is also a source of `NH_(4)^(Theta)` ion but it common be used since `SO_(4)^(2-)` ion will precipitate `Zn^(2+),Ca^(2+)Ba^(2+),Sr^(2+)` as their suilphate hence false |
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| 158. |
Which of the following ppt (s) of sulphite ion have white colour ?A. `Ag_(2)SO_(3)`B. `PbSO_(3)`C. `CaSO_(4)`D. `BaSO_(3)` |
| Answer» Correct Answer - a,b,c,d | |
| 159. |
Reddish brown colouration when neutral `FeCl_(3)` is added to the `CH_(3)CO O^(Θ)` aq solution is due to the formation of (a) ____. |
| Answer» Correct Answer - a.`[Fe_(3)(OH)_(2)(CH_(2)COO)_(4)]^(Theta)` | |
| 160. |
`Pb` has been placed in groups I and II becauseA. It shows the valency of one and twoB. It is partly soluble in `H_(2)O`C. It forms insoluble `PnCI_(2)`D. It from lead sulphide |
| Answer» Correct Answer - b | |
| 161. |
`Cu^(2+) `gives white ppt. of (a)____ with (b) ____ and deep blue colour of _____(c ) ______with (d) ____. |
| Answer» Correct Answer - a.`Cu_(2)I_(2)` b. `KJ`, c. `[Cu(NH_(3))_94)]^(2+)` | |
| 162. |
With `Fe^(3+) `ions `[Fe(CN)_(6)]^(4-)` gives prussian blue colouration due to the formation of ferri- ferro cyanide `Fe[Fe(CN)_(6)]_(2)` white with `NH_(4)SCN,Fe^(3+)` ion gives……. ColourationA. Deep redB. BlueC. BrownD. Green |
| Answer» Correct Answer - a | |
| 163. |
When `KNO_(2)` and `CH_(2)COOH` is added as `CoCl_(2)` solution, yellow ppt of `K_(4)[Cu(NO_(2))_(6)]` is formed. |
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Answer» Correct Answer - F Yellow ppt of `K_(3)[Co^(III)(NO_(2))_(6)]` and not of `K_(4)[Co(NO_(2))_(6)]` are obtained `Co^(2+)+ 7NO_(2)^(Theta) + 2H^(oplus) + 3K^(oplus) rarr K_(3)[Co(No_(2))_(6)] darr + NO uarr + H_(2)O` |
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| 164. |
With `Cu^(2+)` ions `[F(CN)_(6)]^(4-)` gives a ……..ppt of `Cu_(2)[Fe(CN)_(6)]` (Cupric ferro cyanide)A. BlueB. GreenC. chocolateD. White |
| Answer» Correct Answer - c | |
| 165. |
If orange turbidity appears on dilution with `H_(2)O` of the solution in dil HCl , it is due to (a) ___and (b) _____ ion is assumed confirmed. |
| Answer» Correct Answer - `SbOCI` b.` Sb^(3+)` | |
| 166. |
`BaBr_(2)` in aq solution give yellow ppt ,with (a)____ as well as with (b)___. |
| Answer» Correct Answer - a.`K_(2)CrO_(4)` b. `AgNO_(3)` | |
| 167. |
`Fe(OH)_(3)` and `Cr(OH)_(3)` ppt are sepurated byA. Aq `NH_(3)`B. `HCI`C. `NaOH//H_(2)O_(2)`D. `H_(2)SO_(4)` |
| Answer» Correct Answer - c | |
| 168. |
A compound give violet flame rest and gives a white ppt with `AgNO_(3)` .The compound isA. `NaCI`B. `KCI`C. `BaCI_(2)`D. `CaCI_(2)` |
| Answer» Correct Answer - b | |
| 169. |
White ppt of `PbCl_(2)` is soluble in aq `NH_(3)`. |
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Answer» Correct Answer - F white ppt of `PbCI_(2)` is soluble in hot water hence false |
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| 170. |
`Fe(OH)_(3)` and `Al(OH)_(3)` ppt. can be separated by (a) ___when (b)___ becomes soluble due to the formation of (c ) ___ and (d) ____ remain insoluble. |
| Answer» Correct Answer - `NaOH`,b. `AI(OH)_(2)` c. `NaAIO_(2)` d.`Fe(OH)_(3)` | |
| 171. |
What ppt (B) `+ NH_(3) rarr` Black ppt . (H). Hence, (H) is due to the formula ofA. `Hg(NH_(2))CI`B. `Hg`C. `Hg(NH_(2))CI + Hg`D. `Hg(NH_(2))` |
| Answer» Correct Answer - c | |
| 172. |
`AgNO_(3)`given white ppt hypo changing to black ofter some time .Black ppt is ofA. `Ag_(2)S_(2)O_(3)`B. `Ag_(2)SO_(4)`C. `Ag_(2)S_(4)O_(6)`D. `Ag_(2)S` |
| Answer» Correct Answer - d | |
| 173. |
A white ppt obtained in the anylsis of a mixture becomes black on treatment with `NH_(4)OH` it may beA. `Hg_(2)CI_(2)`B. `HgCI_(2)`C. `PbCI_(2)`D. `AgCI` |
| Answer» Correct Answer - a | |
| 174. |
A white ppt obtained in the analysis of a mixture becomes black on treatment with `NH_(3)` or `NH_(4)OH` due to the formation of finely divided Hg and Hg `(NH_(2))CI` i.e. `[Hg+Hg (NH_(2))CI]` The salt may beA. `PbCI_(2)`B. `AgCI`C. `Hg_(2)CI_(2)`D. `Hg_(2)CI_(2)` |
| Answer» Correct Answer - d | |
| 175. |
`HgCI_(2)+ "excess of" KI rarr (A) underset(NH_(3)) to (B)` ,(A) and (B) respectively areA. B. `(Y),(X)`C. both (X)D. both (Y) |
| Answer» Correct Answer - a | |
| 176. |
When excess of `SnCI_(2)` is added to a soin of `HgCI_(2)` a white ppt turning gray is obtained the hrey colour is due to the formation ofA. `Hg_(2)CI_(2)`B. `SnCI_(4)`C. `Sn`D. `Hg` |
| Answer» Correct Answer - d | |
| 177. |
Addition of `SnCl_(2)` to `HgCI_(2)` gives ppt. :A. white turning to greyB. Black turning to whiteC. white turning to redD. None of these |
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Answer» Correct Answer - a `HgCI_(2)` formed is white which later on turns to Hg grey due to father reduction |
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| 178. |
`Na_(2)SO_(3),Na_(2)S_(2)O_(3),Na_(2)CO_(3),Na_(2)CrO_(4)` are separately treated with `AgNO_(3)` solution in how many cases is/are red ppt abtained ? |
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Answer» Correct Answer - 1 `Na_(2)CrO_(6) `with `AgNO_(2)` gives red ppt of `Ag_(2)CrO_(2)` but other gives white ppt of `Ag_(2)SO_(3)` and `Ag_(2)CO_(3)` |
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| 179. |
Yellow coloured solution of `FeCI_(3)` changes in light green whenA. `SnCI_(2)` is addedB. `Zn`is addedC. `H_(2)S` gas is addedD. All true |
| Answer» Correct Answer - d | |
| 180. |
When `H_(2)S` gas is passed through HCI containing aqueous solution of `CuCI_(3),HGCI_(2),BaCI_(2),BiCI_(3) and CoCI_(2)` then which of the following precipitate out ?A. CuSB. HgSC. `Bi_(2)S_(3)`D. CoS |
| Answer» Correct Answer - a,b,c | |
| 181. |
`MCI_(2) + K_(2)CrO_(4) rarr `yellow ppt what can be `MCI_(2)` a. If it is soluble in hot water? b. If is gives green colour in flame ? |
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Answer» (A) `PbCI_(2)` (B) `BaCI_(2)` |
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| 182. |
a. (A) (yellow coloured solution ) changes to light green coloured solution (B) on passing `H_(2)S` gas (A) and (B) both give white ppt. with `BaCI_(2)` solution, insoluble in conc. `HCI`(A) given blue coloured ppt (C ) with `K_(4)[F2(CN)_(6)]` B does not .What are (A),(B) and (C )? b. Identify A,B,C and D in the following reactions Bauxite + chemical `+ CI_(2) overset(Delta) to A + CO` `A + H_(2)O rarr B + HCI, B + H_(2)SO_(4) rarr C + H_(2)O ` `B + NaOH rarr D + H_(2)O` |
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Answer» (A) `FeCI_(3)` (B) `FeCI_(2)` (C ) `KFe^(III)[Fe^(II)(CN)_(6)]` prussian blue b. `underset("Bauxite")(AI_(2)O_(3)) + 3C +3CI_(2) overset(Delta ) to underset((A))(2AICI_(3)) + 3CO` `underset((A))(AICI_(3)) + 3H_(2)O rarr underset((B))(AI(OH)_(3)) + 3HCI` `underset((B))(2AI(OH)_(3)) + 3H_(2)SO_(4) rarr underset((C))(AI_(2)(SO_(4))_(3)) + 6H_(2)O` `underset((B))(AI(OH)_(3)) + NaOH rarr underset((D))(NaAIO_(2)) + 2H_(2)O` |
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| 183. |
Prussion blue is formed when :A. Ferrous sulphate reacts with `FeCI_(3)`B. Forric sulphatee reacts with `K_(4)[Fe(CN)_(6)]`C. Ferrous ammonium sulphatee reacts with `FeCI_(3)`D. Ammonium sulphate reacts with `FeCI_(3)` |
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Answer» Correct Answer - b `2Fe_(2)(SO_(4))_(3) + 3K_(4)[Fe(CN)_(6)] rarr underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))+6K_(2)SO_(4)` |
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| 184. |
A metal salt solution forms a yellow precipitate with potassium chromate in acetic acid , a white precipitate with dil sulphuric acid but gives no precipitate with sodium chloride or iodate .The white precipitate abtained when sodium carbonate is added to the metal salt solution consiste ofA. Lead carbonateB. basic lead carbonateC. Barium carbonateD. Strontium carbonate |
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Answer» Correct Answer - c `BaCrO_(*4)` is yellow ,`BaSO_(4)` is white |
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| 185. |
Write the balanced equition for the following 'Potassium permanganate is reacted with warm solution of oxalic acid in the oresence of sulphuric acid' |
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Answer» Ionic equation `2MnO_(4)^(Theta) + 5C_(2)O_(4)^(2-) + 16H^(Theta) rarr 2Mn^(2+) + 8H_(2)O + 10CO_(2)` |
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| 186. |
Lead has been placed in qualitative group analysis Ist and 2nd becauseA. It shown the valency one and twoB. it forms insoluble `PbCI_(2)`C. It form lead salphideD. `PbCI_(2)` is parially soluble in water |
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Answer» Correct Answer - a Pb form `PbCI_(2)` in group I and PbS in group II Lead salt are partially soluble in water |
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| 187. |
In qualitative inorganic amalysis phosphan , if present is to be elemenated in the apperopriate grest in order to detect the radical :A. `Pb^(2+)`B. `As^(3+)`C. `Ca^(2+)`D. `Cd^(2+)` |
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Answer» Correct Answer - c Phosphate interferes in the usual inorganic analysis after group II. |
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| 188. |
Cations are classified into varius group on the basis of their behaviour against some reagents .The group reagent used for the classifaction of most common cation are `HCI,H_(2)S,NH_(4)OH,(NH_(4))_(2)CO_(3)`. Classification is based on whether a cation reacts with these reagents by the formation of precipitates or not . An aqucous solution which is sightly acids contains cattions `Fe^(2+),Zn^(2+) and Cu^(2+)`. THe rengent added in excess to this solution would identify the separate `Fe^(2+)` ion in one step isA. `2M HCI`B. `6M NH_(3)`C. `6M NaOH`D. `H_(2)S gas ` |
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Answer» Correct Answer - b `Fe^(3+) + 3NH_(4)OH rarr Fe(OH)_(3) darr ("reddish brown") +3NH_(4)^(o+)` `Cu^(2+) + 4NH_(3) rarr underset(("blue colour soluble complex"))([Cu(NH_(3))_(4)]^(2+))` ` Zn^(2+) + 4NH_(3) rarr underset(("colourless soluble complex"))([Zn(NH_(3))_(4)]^(2+))` |
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| 189. |
Cations are classified into varius group on the basis of their behaviour against some reagents .The group reagent used for the classifaction of most common cation are `HCI,H_(2)S,NH_(4)OH,(NH_(4))_(2)CO_(3)` Classification is based on whether a cation reacts with these reagents by the formation of precipitates or not . An aqueous solution contain `Hg^(2+),Hg_(2)^(2+),Pb^(2+)`.THe addition of `2M HCI` will procipitate.A. `HgCI_(2)`onlyB. `PbCI_(2)"only"^(-)`C. `PbCI_(2)` and `Hg_(3)CI_(2)`D. `PbCI_(2)` and `CdCI_(2)` |
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Answer» Correct Answer - c `2M MCI` is group reagent for reagent for group 1 cation `PbCI_(2)` And `HgCI_(2)` will get precipitated, is their solubility products `(K_(sp))` are less than that of other radical |
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| 190. |
Riddish brown gas is obtain with the following are treated with cone `H_(2)SO_(4)`A. `Br^(Θ)`B. `NO_(2)^(Θ)`C. `NO_(3)^(Θ)`D. `SO_(3)^(2-)` |
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Answer» Correct Answer - a,b,c `Br^(Theta)` ions gives `Br_(2),NO_(2)^(Theta)` and `NO_(3)^(Theta)` give `NO_(2)` gas wich are brown in colour |
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| 191. |
What happens when chloride, bromide and iodide are separately heated with conc. `H_(2)SO_(4)`? |
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Answer» a. Chloride gives HCl gas which gives thick white fumes with aqueous ammonia. b. Bromide gives reddish-brown `Br_(2)` vapours. c. Iodide gives violet vapours of iodine which turns starch paper blue. |
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| 192. |
A solution of white crystals with a soluble of `Na_(2)CO_(3)` .The action of cone `H_(2)SO_(4)` on the crystals yieds a brown gas .The crystals are ofA. `NaNO_(3)`B. `KCI`C. `Ca(ON_(3))_(2)`D. `NaBr` |
| Answer» Correct Answer - d | |
| 193. |
Given salt is a bromide or iodide. How will you identify it by treating the salt with chlorine water and `CS_(2)`? |
| Answer» If `CS_(2)` layer assumes an orange colour, it is `Br^(ɵ)` if `CS_(2)` layer assumes a violet colour it is `I^(ɵ)` | |
| 194. |
Some pule green crystals are strongly heated .The gases then off are passed into a container surrounded by ice and then through a solution of acidified `KMnO_(4)` The `KMnO_(4)` is decolrised, a waxy white solid iws formed in the ice container this is dissolvesd in water .The solution willA. Give a precipitate with silver nitric solutionB. Give a precipitate with burium chloride solutionC. Turn red litmus blueD. Give blue colour with strach solution |
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Answer» Correct Answer - b Assume green crystal of `FeSO_(4)` `FeSO_(4) overset(Delta) to Fe_(2)O_(3) + SO_(2) + SO_(3)` `SO_(3) + MnO_(4)^(Theta) rarr Mn^(2-) + SO_(4)^(2-)` `SO_(3)` forms waxy white solid `(H_(2)SO_(4))` in ice container white is dissolved in water and gives `SO_(4)^(2-)` ion .When `BaCI_(2)` reacts `SO_(4)^(2+)` ion it forms ppt of `BeSO_(4)` |
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| 195. |
Why does the chromyl chloride test fail with `Br^(ɵ)` and `I^(ɵ)`? |
| Answer» Because both chromyl bromide and chromyl iodide are not volatile in nature. | |
| 196. |
For testing sodium carbonate solution for the present of suplhiate ions as impurities one should add :A. Excess hydrochloric acid and silver nitrate solutionB. Excess sulpharic acid and silver nitrate solutionC. Excess nitric acid and silver nitrate solutionD. Excess hydrochloric acid and barium chliride solution |
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Answer» Correct Answer - d `SO_(4)^(2-) +BaCI_(2) rarr underset(("White ppt"))(BaSO_(4))+2CI^(Theta)` |
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| 197. |
Why sodium corbonate extract is used for testing acid radicals? |
| Answer» The scheme followed for the systematic analysis of acid radicals is based on sodium salts of acid radicals. Other basic if present may interfere with their salts. | |
| 198. |
For testing `SO_(4)^(2-)` with `BaCl_(2)` solution why should sodium carbonate not be acidified with too much of conc. HCl. |
| Answer» In the presence of conc. HCl, `BaCl_(2)` itself will be precipitated the precipitated `BaCl_(2)` is mistaken as `BaSO_(4)` | |
| 199. |
An inorganic salt solution paires on treatment with HCI will not give a white precipitate of which metal ions?A. `Hg_(2)^(2+)`B. `Hg^(2+)`C. `Zn^(2+)`D. `Al^(3+)` |
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Answer» Correct Answer - b,c,d `HgCI_(2)` is insoluble |
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| 200. |
Why is sodium carbonate extract acidified before perfoming the confirmatory tests for anions? |
| Answer» sodium carbonate extract in addition to the sodium salts of anions contain carbonate also. On heating with the test reagent carbonates of certain metals precipitate which interere in te detection of acid redicals. Because of this, `Na_(2)CO_(3)` is decomosed by adding HCl, `HNO_(3),H_(2)SO_(4),` depending upon the nature of test. | |