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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. |
When `SO_2` is passed through cupric chloride solutionA. A white precipitate is obtainedB. The solution becomes colourlessC. The solution becomes colourless and a white precipitate of `Cu_2Cl_2` is obtainedD. No visible change takes place |
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Answer» Correct Answer - C The solution becomes colourless and a white precipitate of `Cu_2Cl_2` is obtained . |
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| 2. |
When `SO_2` is passed through acidified solution of `H_2S`A. `H_2SO_4` is formedB. `H_2SO_3` is precipitatedC. Sulphur is precipitatedD. `H_2S` is reduced. |
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Answer» Correct Answer - C `SO_2 + 2H_2S to 2H_2O + 3S` |
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| 3. |
Which of the following is not a true peroxide?A. `PbO_2 `B. `CO_2`C. `MnO_2`D. `BaO_2` |
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Answer» Correct Answer - D `BaO_2` contains `-O^(-) - O^(-) -` group. |
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| 4. |
Which of the following oxides is a peroxide ?A. `Na_2O_2`B. `MnO_2`C. BaOD. `SO_2` |
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Answer» Correct Answer - A `Na_2O_2` is a peroxide (sod. peroxide). |
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| 5. |
Which of the following oxides reacts with HCl and NaOH ?A. CaOB. ZnOC. `N_2O_5`D. `CO_2` |
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Answer» Correct Answer - B CaO is basic, `N_2O_5` and `CO_2` are acidic. However ZnO is amphoteric. Thus, ZnO reacts both with HCl and NaOH `ZnO + 2HCl to ZnCl_2 + H_2O` `ZnO + 2NaOH to Na_2ZnO_2 + H_2O` |
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| 6. |
Sulphur does not exist as `S_2` molecule becauseA. it is less electronegativeB. it is not able to constitute `ppi-ppi` bondC. it has ability to exhibit catenationD. of tendency to show variable oxidation states. |
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Answer» Correct Answer - B S does not form `ppi-ppi` bond due to its larger size and hence does not exist as `S_2` molecules. |
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| 7. |
Oxygen is prepared by the fractional distillation ofA. WaterB. Liquid airC. Hydrogen peroxideD. Heavy water. |
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Answer» Correct Answer - B Liquid air is a mixture of mostly liquid nitrogen and liquid oxygen and the difference in their boiling points is about `12.8^(@)C`. So these two are easily separated by fractional evaporation when nitrogen being more volatile escapes first. |
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| 8. |
`H_2S` reacts with `O_2` to formA. `H_2O +S`B. `H_2O+SO_2`C. `H_2O+ SO_2`D. `H_2SO_4 + S` |
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Answer» Correct Answer - A `2H_2S + O_2 to 2H_2O + 2S` |
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| 9. |
Percentage of `O_2` by volume in the atmosphere isA. 18B. 19C. 24.15D. 20.9 |
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Answer» Correct Answer - D Oxygen is about 21% by volume in the atmosphere. |
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| 10. |
All of the following decompose easily on heating to give `O_2` exceptA. lead nitrateB. potassium chlorateC. mercuric oxideD. manganese dioxide. |
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Answer» Correct Answer - D Manganese dioxide does not decompose on heating to give `O_2` |
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| 11. |
`K_2CS_3` can be called potassiumA. SulphocyanideB. ThiocarbideC. ThiocarbonateD. Thiocyanate. |
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Answer» Correct Answer - C `K_2CS_3` is potassium thiocarbonate (thio-sulphur and `K_2CO_3` is potassium carbonate). |
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| 12. |
Sometimes yellow turbidity appears while passing `H_(2)S` gas even in the absence of II group radicals. This is because ofA. Sulphur is present in the mixture as impurityB. IV group radicals are precipitated as sulphidesC. Of the oxidation of `H_2S` gas by some acid radicalsD. III group radicals are precipitated as hydroxides |
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Answer» Correct Answer - C Such yellow turbidity in group II is due to the oxidation of `H_2S` to S by some oxidising anion (e.g., `NO_3`) present in the mixture. |
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| 13. |
The two oxygen-oxygen bond lengths in ozone areA. 110 pm , 148 pmB. 110 pm , 128 pmC. 128 pm , 128 pmD. 128 pm , 148 pm. |
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Answer» Correct Answer - C 128 pm. The two C-O bond lengths are same due to resonance. |
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| 14. |
The percentage of ozone in ozonised oxygen is aboutA. 10B. 40C. 80D. 100 |
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Answer» Correct Answer - A The percentage of ozone in ozonised oxygen is 10-15%. |
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| 15. |
Which of the following hydrides shows the highest boiling point ?A. `H_2O`B. `H_2S`C. `H_2Se`D. `H_2Te.` |
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Answer» Correct Answer - A In hydrides of group 16 , `H_2O` has highest boiling point . It is due to extensive H--bonding which exist even in liquid state. |
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| 16. |
Which of the following hydrides of the oxygen family shows the lowest boiling point?A. `H_2O`B. `H_2S`C. `H_2Se`D. `H_2Te` |
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Answer» Correct Answer - B Among group 16 hydrides, `H_2O` has the highest boiling point. It is due to H-bonding which exist even in liquid state. Due to absence of H-bonding and lower molecular mass `H_2S` has least boiling point. |
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| 17. |
Normal melting point and boiling point of rhombic sulphur are 387.5 K and 717.6 K respectively. When sulphur is heated in a test tube to 455 K and the test tube in verted, the content which pour out isA. plastic sulphurB. molten sulphurC. monoclinic sulphurD. None of these |
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Answer» Correct Answer - D Nothing will come out of the tube |
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| 18. |
Assertion (A) : Among the hydrides of group 16, water has the lowest melting point and boiling point. Reason (R) : It is due to least molecular mass.A. Both A and R true and R is the correct explanation of AB. Both A and R true and R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false |
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Answer» Correct Answer - D Assertion. Among the hydrides of group 16. Water has the highest m.p. and b.p. Reason. It is due to extensive H-bonding in water |
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| 19. |
Which of the following has the highest boiling point?A. `H_2O`B. `H_2S`C. `H_2Se`D. `H_2Te` |
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Answer» Correct Answer - A Among all the hydrides of Group 16, `H_2O` has the highest b.p. This is due to extensive H-bonding in water which persist even in liquid state. |
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| 20. |
Oleum isA. Castor oilB. Oil of vitriolC. Fuming of `H_2SO_4`D. None of them |
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Answer» Correct Answer - C Oleum is obtained by dissolving `SO_3` in `H_2SO_4` `SO_3 +H_2SO_4 (conc.) to underset("Oleum")(H_2S_2O_7)` Oleum is also called fuming sulphuric acid because it fumes in moist air due to `SO_3`. |
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| 21. |
The oxidation state of O in `Na_2O_2` isA. `+2`B. -2C. -1D. `+1` |
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Answer» Correct Answer - C `overset(1)(Na_2)overset(x)(O_2)` 2 x (+1) +2x =0 or 2x +2 =0 2x = -2 or `x = - 2/2 = -1` |
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| 22. |
What is not applicable to `TeCl_4` ?A. Te has one lone pairB. It is tetrahedral in shapeC. It reacts with HCl to form `H_2[TeCl_6]`D. The hybrid state of Te is `sp^3d`. |
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Answer» Correct Answer - B `TeCl_4` has see saw shape like `SF_4`. |
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| 23. |
Which of the following molecular species has unpaired electrons(s) ? .A. `N_2`B. `F_2`C. `O_2`D. `O_2^2` |
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Answer» Correct Answer - C `O_2` has two unpaired electrons and `O^(2-)` has one unpaired electron. |
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| 24. |
When thiosulphate ion is oxidised by iodine. which one of the following ion is produced ?A. `SO_3^(2-)`B. `SO_4^(2-)`C. `S_4O_6^(2-)`D. `S_2O_6^(2-)` |
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Answer» Correct Answer - C `2S_2O_3^(2-)toI_2 to underset("Tetrathionate ion")(S_4O_6^(2-) +2I^(-))` |
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| 25. |
Which of the following is acidic?A. `SO_3`B. `N_2O`C. BeOD. HgO |
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Answer» Correct Answer - A HgO is a basic oxide (oxide of a metal). BeO is amphoteric (oxide of a metalloid) `N_2O` is a neutral oxide. `SO_3` is an acidic oxide (oxide of a non-metal). |
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| 26. |
iodine oxidises the `S_2O_3^(2-)` ion toA. `SO_3^(2-)`B. `SO_4^(2-)`C. `S_4O_6^(2-)`D. `S^(2-)` |
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Answer» Correct Answer - C `I_2+S_2O_3^(2-)tounderset("Tetrathionate ion")(S_4O_6^(2-)) +2I^(-)` |
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| 27. |
Peroxy linkage is present inA. `H_2S_2O_6`B. `H_2SO_3`C. `H_2S_2O_7`D. `H_2S_2O_8` |
| Answer» Correct Answer - A | |
| 28. |
In contact process inpurities of arsenic is removed byA. `Al(OH)_3`B. `Fe(OH)_3`C. `Cr(OH)_3`D. `Fe_2O_3` |
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Answer» Correct Answer - B `Fe(OH)_3` removes the impurities of Arsenic. |
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| 29. |
Which of the following oxides is most acidic and most basic respectively? I(CaO), II`(K_2O)`, III`(H_2O)`, IV`(SO_3)`, V`(N_2O_5)`, VI`(SO_2)`.A. IV, IIB. V, IC. V, VI, IIID. V, II |
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Answer» Correct Answer - A `SO_3`(VI) is the most acidic while `K_2O` (II) is the most basic |
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| 30. |
`SCl_2` is the best known dihalice of sulphur, hybrid state of sulphur in `SCl_2` isA. `sp^2`B. `sp^3`C. `sp^3d`D. `sp^3 d^2` |
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Answer» Correct Answer - B In `SCl_2`, there are two lone pairs and two bond pairs around S. It has bent shape and hybrid state of sulphur is `sp^3`. |
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| 31. |
The hybrid state and oxidation state of S in `SF_4` are respectivelyA. `sp^2`, +4B. `sp^3`, +6C. `sp^3d`, +4D. `dsp^2`, +6 |
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Answer» Correct Answer - C In `SF_4`, S atom is `sp^3d`-hybridized and its oxidation state is +4. |
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| 32. |
`Na_2S_2O_3` is prepared byA. Reacting `H_2SO_3` with NaOHB. Reducing `Na_2SO_4` with S in alkaline mediumC. Heating NaOH and SD. Reducing `Na_2SO_4` with S in acidic medium. |
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Answer» Correct Answer - C `6NaOH(aq)+4S(s)overset(Delta)toNa_2S_2O_3(aq)+4Na_2S(aq)+3H_2O(l)` |
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| 33. |
Which one of the following is an oxyacid ?A. `Ba(OH)_2`B. `Mg(OH)_2`C. `H_3PO_3`D. HCl |
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Answer» Correct Answer - C `Ba(OH)_2` and `Mg(OH)_2` are bases. HCl is a non oxyacid. Only `H_3PO_3` is an oxy-acid. |
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| 34. |
The products obtained by passing chlorine through hypo solution areA. `Na_2SO_3 + HCl + S `B. `Na_2SO_3 + SO_3 + HCl`C. `Na_2SO_4 + HCl + S`D. `Na_2SO_4 + HCl + SO_2` |
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Answer» Correct Answer - C `({:(Cl_2+H_2Oto,2HCl+O),(Na_2S_2O_3+Oto,Na_2SO_4+S):})/ul(Na_2S_2O_3+Cl_2+H_2Oto2HCl+S+Na_2SO_4)` |
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| 35. |
Polyanion formation is maximum inA. NitrogenB. OxygenC. SulphurD. Boron |
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Answer» Correct Answer - C Polyanion formation is maximum in sulphur. This is due to the fact that sulphur shows maximum catenation in the group. |
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| 36. |
Copper turnings when heated with concebtracted sulphuric acid will giveA. `SO_2`B. `SO_3`C. `H_2S`D. `O_2` |
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Answer» Correct Answer - C `Cu+2H_2SO_4 to CuSO_4 + SO_2 + 2H_2O` |
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| 37. |
Which element burns to form a gaseous oxide at room temperature ?A. HydrogenB. HeliumC. SodiumD. Sulphur |
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Answer» Correct Answer - D `S+O_2 to SO_2(g)` `H_2O` is liquid `Na_2O` is solid |
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| 38. |
The metal with highest electrical resistance at room temperature isA. PbB. TeC. PoD. Fe. |
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Answer» Correct Answer - B The metal with highest electrical resistance at room temperature is Te. |
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| 39. |
A chalcogen combines directly with hydrogen with great difficulty to form a hydride. This chalcogen also burns in air to form a solid polymeric dioxide and has got the highest electrical resistance amongst metals. The chalcogen isA. OB. SC. TeD. Se |
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Answer» Correct Answer - C Out of these four chalcogens, Se and Te reacts with hydrogen directly with great difficulty. Since Te has got the highest electrical resistance. Hence the metal is Te. |
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| 40. |
One of the following metals with which sulphur combines, isA. MagnesiumB. GoldC. PlatinumD. Iodine |
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Answer» Correct Answer - A Mg |
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| 41. |
The oxidation states of the most electronegative elements in the products of the reaction between `BaO_(2)` and `H_(2)SO_(4)` areA. O and -1B. -1 and -2C. -2 and 0D. -2 and +1 |
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Answer» Correct Answer - B `BaO_2+H_2SO_4 to BaSO_4 +H_2O_2` The most electronegative element in the products of the reaction is O. Its oxidation state in `BaSO_4` is -2 and in `H_2O_2` it is -1. |
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| 42. |
The most electronegative element in Group 16 isA. OB. SC. SeD. Po |
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Answer» Correct Answer - A Oxygen is the most electronegative element in group 16. |
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| 43. |
Which of the non metals reacts with sulphur ?A. Noble gasesB. ChlorineC. IodineD. Nitrogen. |
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Answer» Correct Answer - B `2S+Cl_2toS_2Cl_2` |
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| 44. |
The second most highly electronegative elements in the periodic table isA. SulphurB. OxygenC. SeleniumD. Polonium. |
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Answer» Correct Answer - B Oxygen (3.5) The first being fluorine (4.0). |
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| 45. |
On heating ozone, its volume.A. decreases to halfB. becomes doubleC. increases to 3/2 timesD. remains unchanged. |
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Answer» Correct Answer - C `2O_3 to 3O_2` 2 vol, 3 vol, 1 vol . 3/2 col. |
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| 46. |
A certain compound when burned gives three oxides. The first turned lime water milky, the second turned anhydrous `CuSO_4` blue and the third formed an aqueous solution of low pH. The elements present in the compound areA. C, O and SB. C, H and CaC. C, H and NaD. C, H and S |
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Answer» Correct Answer - D Compound contains C, H and S and on burning gives `CO_2, H_2O and SO_2`. `CO_2` turns lime water milky `H_2O` turns anhydrous `CuSO_4` blue. `SO_2` forms an acidic solution of low pH. |
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| 47. |
The first ionisation potential in electron volts of nitrogen and oxygen atoms are respectively given byA. 14.6,13.6B. 13.6,14.6C. 13.6,13.6D. 14.6,14.6 |
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Answer» Correct Answer - A First ionisation potential of nitrogen is more than oxygen. It is due to exactly half filled p-subshell in N.Thus ionisation potentials (in eV) of N , O are 14.6, 13.6 respectively. |
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| 48. |
Which of the following gases turns lead acetate paper black?A. `SO_2`B. `SO_3`C. `H_2S`D. `H_2SO_4` |
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Answer» Correct Answer - C `H_2S` gas turns lead acetate paper black. |
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| 49. |
Which of the following has highest thermal stability and maximum acid strength 2A. `H_2S`B. `H_2O`C. `H_2Se`D. `H_2Te` |
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Answer» Correct Answer - A Thermal stability decreases in the order `H_2O gt H_2S gt H_2Se gt H_2Te` However, `H_2O` is neutral and `H_2S` is distinctly acidic. Thus, highest thermal stability and maximum acid strength is in case of `H_2S` |
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| 50. |
The reaction between `NH_2^(Ө)` and `N_2 O` givesA. NOB. `N_3`C. `N_2O_5`D. `NH_2NH_2` |
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Answer» Correct Answer - B `underset("Amide ions")(NH_2^(-))+N_2O to underset("Azide ions")(N_3^(-))+H_2O` |
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