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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.
| 1. |
(T) imparts violet colour in the flame test `overset("compound"(U)+conc.H_(2)SO_(4))to(V)_("Red gas")overset(NaOH+AgNO_(3))to(W)_("Red ppt.")overset(NH_(3)" soln")to(X)` `(W)_("Red ppt.")overset(dil.HCl)to(Y)_("White ppt.")` `(U)` Sublimes on heating `underset(Delta)overset(NaOH)to(Z)_(gas)` Identify (T) to (Z).A. `T=KMnO_(4),U=HCl,V=Cl_(2),W=HgI_(2),X=Hg(NH_(2))NO_(3),Y=Hg_(2)Cl_(2),Z=N_(2)`B. `T=K_(2)Cr_(2)O_(7),U=NH_(4)Cl,V=CrO_(2)Cl_(2),W=Ag_(2)CrO_(4),X=[Ag(NH_(3))_(2)]^(+),Y=AgCl,Z=NH_(3)`C. `T=K_(2)CrO_(4),U=KCl,V=CrO_(2)Cl_(2),W=HgI_(2),X=Na_(2)CrO_(4),Y=BaCO_(3),Z=NH_(4)Cl`D. `Y=K_(2)MnO_(4),U=NaCl,V=CrO_(3),W=AgNO_(3),X=(NH_(4))_(2)CrO_(4),Y=CaCO_(3),Z=SO_(2)` |
| Answer» Correct Answer - B | |
| 2. |
Identify (A),(B) and (C). Also explain colour difference between `MCl_(4)` and (B). |
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Answer» Correct Answer - (A) is `TiCl_(4)` as it has no unpaired electron and is liguid at room temperature on account of covalent character because of high polarising power of `Ti^(+4). TiCl_(4)` being covalent gets hydrolysed forming `TiO_(2)(H_(2)O)_(2)` and `HCl` (B) which fumes in air. In `[Ti(H_(2)O)_(6)]Cl_(3)` complex `Ti(III) has one unpaired electron`(3d^(1)) `which gives violet//purple colour due to d-d transition. |
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| 3. |
When `AgNo_(3)`(aq)` reacts with excess of iodine, we get:A. `AgIO_(3)`B. `HIO_(3)`C. `AgO`D. `HI` |
| Answer» Correct Answer - B | |
| 4. |
Which of the following halides react(s) with `AgNO_(3(aq))` to give a precipitate that dissolves in `Na_(2)S_(2)O_(3(aq))`A. `HCl`B. `HF`C. `HBr`D. `Hl` |
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Answer» Correct Answer - ACD `AgNO_(2)+HCl to AgCldownarrow` `AgNO_(3)+HBr to AgBrdownarrow` `AgNO_(3)+HI to AgI downarrow` All these precipitates will get dissolved in hypo forming complex `Na_(3)[Ag(S_(2)O_(3))_(3)]` |
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| 5. |
What happen when silver nitrate solution is added to `Na_(2)S_(2)O_(3)` solution and then content is allowed to keep for a longer period? |
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Answer» Correct Answer - White precipitate of `Ag_(2)S_(2)O_(3)` is obtained which turns yellow , brown and finally black on Keeping. `2Ag_(2)S_(2)O_(3)+Na_(2)S_(2)O_(3) to Ag_(2)S_(2)O_(2)downar row("white")+2NaNO_(3)` `Ag_(2)S_(2)O_(3)+H_(2)O to Ag_(2)Sdownar row(Black)H_(2)SO_(4)` |
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| 6. |
`Na_(2)S_(2)O_(3)` can be prepared by:A. `Na_(2)SO_(3)` solution + `S("powder")overset("Boiled")to`B. `S+NaOHto`C. `Na_(2)S+Na_(2)CO_(3)+SO_(2)to`D. all of the above. |
| Answer» Correct Answer - D | |
| 7. |
What takes place when zinc metal is added to an aqueous solution containing magnesium nitrate and silver nitrate? (P). Zn is oxidized (Q). `Mg^(2+)` is reduced. (R). `Ag^(+)` is reduced. (S). No reaction takes place.A. P and Q onlyB. P and R onlyC. P,Q and R onlyD. S only |
| Answer» Correct Answer - B | |
| 8. |
In the galvanizing process, iron is coated with zinc. The resulting chemical protection is most similar to that provided when:A. a magnesium bar is connected to an iron pipeB. an iron can is plated with tinC. copper pipes are connected using lead solderD. a copper pipe is covered with epoxy paint. |
| Answer» Correct Answer - A | |
| 9. |
Which metal in the first series of transition metals exhibits+1 oxidation state most frequently and why?A. ScB. TiC. CuD. Ag |
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Answer» Correct Answer - C `Cu^(+)` has `d^(10)` configuration. |
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| 10. |
How would you account for the following: (i) `Cr^(2+)` is reducing in nature while with the same d-orbital configuration `(d^(4)) Mn^(3+)` is an oxidising agent (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation state occurs in the middle of the series. |
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Answer» Correct Answer - (i) It is because `Cr^(3+)` loses electron to become `Cr^(3+)` which is more stable due to half filled `t_(2g)` orbitals. Whereas `Mn^(3+)` will gain electrons to become `Mn^(2+)` which is more stable due to half filled d-orbitals. (ii) It is due to large of unpaired electrons in d-orbitals in middle of the series. |
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| 11. |
Four statement of Cr and Mn are given below. (P). `Cr^(2+)` and `Mn^(3+)` have the same electronic configuration. (Q). `Cr^(2+)` is a reducing agent while `Mn^(3+)` in an oxidizing agent. (S). Both Cr and Mn are oxidizing agent. The correct statement are:A. P,R,SB. P,QC. P,Q,SD. P,S |
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Answer» Correct Answer - B (a). `Cr^(2+)=3d^(4)` `Mn^(3+)=3d^(4)` (b). `Cr^(2+)toCr^(3+)+e^(-)` (b). `Cr^(2+)toCr^(3+)+e^(-)` `Mn^(3+)+e^(-)toMn^(2+)` |
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| 12. |
How would you account for the following: of the `d^(4)` species `Cr^(2+)` is strongly reducing while manganese(III) is strongly oxiding. Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. The `D^(1)` configuration is very unstable in ions. |
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Answer» Correct Answer - (i) `E^(0)` value for `Cr^(3+)//Cr^(2+)` is negative `(-0.41V)` whereas `E^(0)` value for `Mn^(3+)//Mn^(2+)` is positve `(+1.57V)` Hence, `Cr^(2+)` ions can easily undergo oxidation to give `Cr^(3+)` ions and , therefore , act as strong reducing agent whereas `Mn^(3+)` can easily undergo reduction to give `Mn^(2+)` and hence acts as oxidizing agent (ii) `3d^(6)` has higher `CFSE` `(C ) The ions with `D^(1)` configuration have the tendency to lose the only electron present in d-subshell to acquire stable `d^(0)` configuration . Hence , they are unstable and undergo oxidation agent. |
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| 13. |
What is the general electronic configuration of transition elements |
| Answer» Correct Answer - `(n-1)d^(1-10)ns^(1-2)` (palladium is exception); `[Kr]^(36)4d^(10)5s^(0).` | |
| 14. |
Atomic of the transition elements than those of the s-block elements because:A. there is increase in the nuclear charge along the period.B. orbital electrons are added to the penultimate d-subshell rather than to the outer shell of the atom.C. the shielding effect of d-electrons is small.D. All of these. |
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Answer» Correct Answer - D The atomic radii of the transition metals lie in -between those of s- and p-block elements . In the begininh, the atomic radius decrease with the increase in nuclear charge (as atomic number increase), whereas the shielding effect of d-electrons is small and orbital electrons are added to the penultimate d-subshell rather than tot he outer shell of the atom. |
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| 15. |
Comments on the statement that elements of the first transition series posses many properties different from those of heavier transition elements. |
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Answer» Correct Answer - The given statement is true as explained below: (i) Atomic radii of the heavier transition elements `(4d and 5d series`) are larger than those of the corresponding elements of the first transition series through of `(4d and 5d series)` are very close to each other. (ii) For first transition series,+2 anxd +3` oxidation state are more common whereas for heavier transition elements, higher oxidation state are more common. `(iii) Melting and boiling points of heavier elements are greater than those of the first transition series due to strong intermettalic bonding (M-M bonding). (iv) Enthalpies of atomisation of `4d` and `5d` series are higher than the corresponding element of the first series. (V) lonisation enthalpies of `5d` series are higher than the corresponding elements of `3d` and `4d` series. `(Vi)` The elements of the first transition series from low spin or high spin complex depending upon the strength of the ligand field . However, the heavier element from low spin complex irrespective of the strength of the ligand field. |
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| 16. |
Write the balanced chemical equation for developing photographic films. |
| Answer» Correct Answer - `C_(6)H_(4)(OH)_(2)+2AgBr to 2Ag+C_(6)H_(4)O_(2)+2HBr+2Na_(2)S_(2)O_(3) to Na_(3)[Ag(S_(2)O_(3))_(2)]+NaBr`. | |
| 17. |
Assertion: Tungsten has very high melting point. Reason: Tungsten is a covalent compound.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - C | |
| 18. |
Assertion: Tungsten has very high melting point. Reason: Tungsten is a covalent compound.A. Statement-1 is true , Statement-2 is true and Statement-2 is correct explanation for Statement-1B. Statement-1 is true, Statement-2 is True and Statement-2 is Not correct explanation for statement-1C. Statement-1 is true, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - C Statement-1: The maxima at about the middle of each indicate that one unpaired electron per d orbital is particularly favourable of strong interatomic intersection. In general , greater he number of valence electrons, Stronger is the resultant bonding. Cr,Mo and W have maximum number of unpaired electrons and therefore , these are very hard metals and have maximum enthalpies of atomization. So tungsten has very melting point. Statement-2: It is not a covalent compound but is a metal i.e an element. |
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| 19. |
Which set of reactants produces a gaseous product? (P). `6MHNO_(3)(aq)+Cu(s)` (Q). `6MHCl(aq)+CaCO_(3)(s)`A. P onlyB. Q onlyC. Both P and QD. neither P nor Q |
| Answer» Correct Answer - C | |
| 20. |
For a stoichiometric mixture of reactants, which statement best describes the changes that occur when this reaction goes to completion? `Zn+4HNO_(3)toZn(NO_(3))_(2)+2NO_(2)+2H_(2)O`A. All of the zinc is oxidized and some f the nitrogen is reduced.B. All of the zinc is oxidised and all of the nitrogen is reduced.C. Some of the zinc is oxidised and all of the nitrogen is reduced.D. Some of the zinc is oxidized and some of the nitrogen is reduced. |
| Answer» Correct Answer - A | |
| 21. |
Which two sets of reactants best represent the amphoteric character of `Zn(OH)_(2)`? Set `1: Zn(OH)_(2) & OH^(-)(aq)` Set `2: Zn(OH)_(2)(s)& H_(2)O(l)` Set `3: Zn(OH)_(2)(s)&H^(+)(aq)` Set `4: Zn(OH)_(2)(s)& NH_(3)(aq)`A. Sets 1 and 2B. Sets 1 and 3C. Sets 2 and 4D. Sets 3 and 4 |
| Answer» Correct Answer - B | |
| 22. |
Which element commonly exhibits both +1 and +3 oxidation states?A. `Al(Z=13)`B. `Sc(Z=21)`C. `Sn(Z=50)`D. `Tl(Z=81)` |
| Answer» Correct Answer - D | |
| 23. |
Iron exhibits `+2 and +3` oxidation states. Which of the following statements about iron is incorrect?A. Ferrous oxide is more basic in nature than the ferric oxide.B. Ferrous compounds are relatively more ionic than the corresponding ferric compoundsC. Ferous compounds are less volatile than the corresponding ferric compoundsD. Ferrous compounds are more easily hydrolysed than the corresponding feric compounds. |
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Answer» Correct Answer - 4 `Fe^(3+)` is easily hydrolysed than `Fe^(2+)` due to more positive charge. |
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| 24. |
The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is becauseA. The actinoids are more reactive than the lanthanoidsB. The `5f` orbitals exhibit farther from the lanthanoids.C. The `5f` orbital are more buried than `4f` orbitalsD. There is a similarity between `4f` and `5f` orbitals in their angular part of the wave function |
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Answer» Correct Answer - 2 The distance between the nucleus and `5f` orbital(actinides) is more than the distance between the nucleus and `4f` orvbitals (lanthanides). Hence the hold of nucleus on valence electron decrease in actinides. For this statement-2 the actinoids exhibit more number of oxidation states in general. |
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| 25. |
Identify the incorrect statement among the following.A. The chemistry of varius lanthanoids is very similar.B. `4f` and `5f` orbitals are equally shieldedC. d-block element show irregular and erratic chemical properties among themselves.D. La and Lu have partially filled d orbitals and no other partially filled orbitals. |
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Answer» Correct Answer - 2 The decrease in the force of attraction exerted by the nucleus on the valency electrons due to presence o electrons in the inner shells is called shielding effect.An `4f` orbital is nearer to the nucleus than `5f` orbitals. In addition , tha `20` electrons of `3d` and `4d` orbitals contributed the shielding to `4f` electron while `44` electrons of `3d,4d,5d` and `4f` contributed the shielding to `5f`. Hence shielding of `5f` is more than `4f`. |
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| 26. |
Iron is rendered passive by treatment with concentrated :A. `HCl`B. `H_(2)SO_(4)`C. `Cu`D. `Ag` |
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Answer» Correct Answer - D `HNO_(3)` being strong oxiding agent oxidises iron to its oxides `(Fe_(3)O_(4))` which forms a thin protective layer over the metal. This makes the iron passive |
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| 27. |
When `H_(2)O_(2)` is added to an acidified solution of `K_(2)Cr_(2)O_(7)`:A. Solution turns green due to formation of `Cr_(2)O_(3)`B. Solution turns yellow due to formation of `K_(2)CrO_(4)`C. Solution gives green ppt of `Cr(OH)_(3)`D. |
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Answer» Correct Answer - C `Cr_(2)O_(7)^(2-)+2H^(+)+4H_(2)O_(2) to 2CrO(O_(2))_(2)("blue coloured")+5H_(2)O`. |
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| 28. |
Statement-1: `K_(2)Cr_(2)O_(7)+2NaCl toNa_(2)Cr_(2)O_(7)+2KCl` Statement-2: `K_(2)Cr_(2)O_(7)` is less soluble than `Na_(2)Cr_(2)O_(7)`.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - D | |
| 29. |
`H_(2)S` gas is passed through an acidic solution of `K_(2)Cr_(2)O_(7)` . The solution turns milky, why?A. `K_(2)SO_(4)`B. `Cr_(2)(SO_(4))_(3)`C. `S`D. `CrSO_(4)` |
| Answer» Correct Answer - D | |
| 30. |
Refer to the graph for trends in M.P. of transition elements of 3d,4d and 5d series. Q. A metal having high boiling point:A. has high enthalpy of atomizationB. tends to be noble in their reactionsC. has strong metallic bondingD. all of the above. |
| Answer» Correct Answer - D | |
| 31. |
Refer to the graph for trends in M.P. of transition elements of 3d,4d and 5d series. Q. Choose the correct statement(s).A. III refers to 3rd transition series of periodic table.B. X corresponds to tungstenC. Y" corresponds to manganese.D. all of the above. |
| Answer» Correct Answer - D | |
| 32. |
Acidified `KMnO_(4)` can be decolourised by:A. `SO_(2)`B. `H_(2)O_(2)`C. `FeSO_(4)`D. `Fe_(2)(SO_(4))_(3)` |
| Answer» Correct Answer - A::B::C | |
| 33. |
Which of the following alloys contain (s) Cu and Zn?A. BronzeB. Brass is much less dense that copper.C. Gun metalD. Type metal. |
| Answer» Correct Answer - B::C | |
| 34. |
Interstitial compounds are formed by:A. `Co`B. `Ni`C. `Fe`D. `Ca` |
| Answer» Correct Answer - A::B::C | |
| 35. |
To an acidified dichromate solution, a pinch of `Na_(2)O_(2)` is added and shaken. What is observed ?A. Blue colourB. Orange colour changing to oxygenC. Copious evolution of oxygen.D. Bluish-green precipitate. |
| Answer» Correct Answer - A::C | |
| 36. |
Which reaction is not possible using acidified `K_(2)Cr_(2)O_(7)`?A. `I^(-)toIO_(3)^-`B. `Sn^(2+) to Sn^(4+)`C. `H_(2)StoS`D. `Fe^(2+) to Fe^(3+)` |
| Answer» Correct Answer - A | |
| 37. |
Which of the following is correct?A. `CrO_(3)` has peroxide bond in the structure.B. Heating `(NH_(4))_(2)Cr_(2)O_(7)` produces `Cr_(2)O_(3)` along with `O_(2)`.C. Acidification of `K_(2)Cr_(2)O_(7)` turns it yellow.D. `Cl_(2)` is liberated in little amount when `K_(2)Cr_(2)O_(7)`, KCl and conc. `H_(2)SO_(4)` are heated together. |
| Answer» Correct Answer - D | |
| 38. |
When acidified solution of `K_(2)Cr_(2)O_(7)` is shaken with aqeous solution of `FeSO_(4)`, Then:A. `Cr_(2)O_(7)^(2-)` ion is reduced to `Cr^(3+)` ionsB. `Cr_(2)O_(7)^(2-)` ion is converted to `CrO_(4)^(2-)` ionsC. `Cr_(2)O_(7)^(2-)` ion is reduced to CrD. `Cr_(2)O_(7)^(2-)` ion is converted to `CrO_(3)` |
| Answer» Correct Answer - A | |
| 39. |
Assertion : Ammoniacal silver nitrate converts glucose to gluconic acid and metallic is precipitated. Reason : Glucose acts as a week reducing is precipitated.A. Statement-1 is True, Statement -2 is True , Statement-2 is a correct explanation for statement-1B. Statement-1 is True, Statement -2 is True , Statement-2 is Not a correct explanation for statement-1.C. Statement-1 is True, Statement -2 is FalseD. Statement-1 is False, Statement -2 is True |
| Answer» `Ag_(2)O+C_(6)H_(12)O_(6)to2Ag+C_(6)H_(12)O_(7)` | |
| 40. |
Which of the following statements (s) is (are) correct with reference to the ferrous and ferric ions? (a). `Fe^(3+)` gives brown colour with potassium ferricyanide. (b). `Fe^(2+)` gives blue precipitate with potassium ferricyanide. (c). `Fe^(3+)` gives red colour with potassium thiocyanate. (d). `Fe^(2+)` gives brown colour with ammonium thiocyanate.A. `Fe^(3+)` gives brown with potassium ferricyanideB. `Fe^(3+)` gives blue precipitate with potassium ferricyanideC. `Fe^(3+)` gives red colour with potassium sulphocyanideD. `Fe^(2+)` gives brown olour with potassium sulphocyanide |
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Answer» `Fe^(3+)` produces red colouration with `KSCN` but `Fe^(2+)` does not give brown colour with `KSCN. Therefore, (D) option is correct |
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| 41. |
Ferric sulphate on heating gives:A. `SO_(2)` and `SO_(3)`B. `SO_(2)` onlyC. `SO_(3)` onlyD. `S` |
| Answer» Correct Answer - C | |
| 42. |
Anhydrous ferric chloride is prepared byA. Heating hydrated ferric chloride at a high temperature in a stream of air.B. heating metallic iron in a strem of dry chloride gas.C. reaction of ferric oxide with HCl.D. reaction of metallic iron with HCl. |
| Answer» Correct Answer - B | |
| 43. |
Ferric iodide is very unstable but ferric chloride is not |
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Answer» Correct Answer - `I^(-)` ion is a stronger reducing agent in comparison to `Cl^(-)` ion. `Fe^(3+)` is easily reduced by iodide ion. `2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2)` |
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| 44. |
Which of the following ions in solution undergoes disproportionation?A. `Fe^(+2)`B. `Cr^(+3)`C. `Cu^(+)`D. `Zn^(+2)` |
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Answer» Correct Answer - C `Cu^(+)toCu^(2+)+Cu^(0)` |
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| 45. |
What is meant by disproportionation of an oxidation state ? Give an example |
| Answer» Correct Answer - When a particular oxidation state become less stable relative to other oxidation state , one lower, one higher. It is said to undergo disproportionation,e.g., `3Mn^(VI)O_(4)^(2-)+4H^(+) to 2Mn^(VII)O_(4)^(-)+Mn^(IV)O_(2)+2H_(2)O` `Mn(VI)` is unstable relative to `Mn(VII)` and `Mn(Iv)`. | |
| 46. |
Statement-1: Many copper (I) compound are unstable in aqueous solution and undergo disproportionation. Statement-2: `Cu^(2+)` I mre stable than `Cu^(+)` because of much more negative `Delta_(hyd)H^(@)` of `Cu^(2+)` than `Cu^(+)`, which more than compensates for seconds ionisation enthalpy.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true, statement-2 is true, statement-2 is NOT a correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true. |
| Answer» Correct Answer - A | |
| 47. |
Following reaction (s) is /are involved in the iodometric estimationA. `Cr_(2)O_(7)^(2-) +H^(+)+I^(-) to 2Cr^(3+)+I_(2),I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)`B. `MnO_(4)^(-)+H^(+)+I^(-) toMn^(2+)+I_(2),I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)`C. `MnO_(4)^(-)+OH^(-)+I^(-)to MnO^(2)+I_(2),I_(2)+S_(2)O_(3)^(2-) to S_(4)O_(6)^(2-)+I^(-)`D. `Cr_(2)O_(7)^(2-)+OH^(-)+I^(-) to 2Cr^(3+)+I_(2),I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)` |
| Answer» Correct Answer - AB | |
| 48. |
Amongst the following species, maximum covalent character is exhibited by:A. `FeCl_(2)`B. `ZnCl_(2)`C. `HgCl_(2)`D. `CdCl_(2)` |
| Answer» Correct Answer - C | |
| 49. |
Purple of Cassius isA. pure goldB. colliodal solution of goldC. gold (I) hydroxideD. gold (III) chloride |
| Answer» Correct Answer - B | |
| 50. |
The melting point of `Zn` is lower as compared to those of the other elements of `3d` series because:A. The orbitals are completely filled.B. the d-orbitals are partially filledC. d-electrons do not participate in metallic bondingD. size of `Zn` atoms is smaller |
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Answer» Correct Answer - AC Strong metallic bonds between the atoms of transition elements attribute to their high melting and boiling points . Zinc has all electrons paired `([Ar] 3d^(10) 4s^(2))` and thus do not participate in metallic bonding . SO accordingly its melting point is least. |
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