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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. |
vii. The stability of dihalides of ` Si, Ge , Sn` and Pb increases steadily in the sequence :A. `GeX_(2) lt SiX_(2) lt SnX_(2) lt PbX_(2)`B. `SiX_(2) lt GeX_(2) lt PbX_(2) lt SnX_(2)`C. `SiX_(2) lt GeX_(2) lt SnX_(2) lt PbX_(2)`D. `PbX_(2) lt SnX_(2) lt GeX_(2) lt SiX_(2)` |
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Answer» Correct Answer - C Due to the inert pair effect (the reluctance of `ns^(2)` electrons of outermost shell to participate in bonding), the stability of `M^(2+)` ions (of group `IV` elements) increases as we go down the group. |
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| 2. |
When `H_2S` is passed through acidified `KMNO_4`, we getA. `K_(2)SO_(4)`B. `MnO_(2)`C. `KHSO_(3)`D. Sulphur |
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Answer» Correct Answer - D `2KMnO_(4)+2H_(2)SO_(4) +5H_(2)S to K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5S` |
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| 3. |
The`3ClO^(-)(aq.)to ClO_(3)^(-)(aq.)+2Cl^(-)(aq.)` is an example ofA. oxidation reactionB. disproportionationC. reduction reactionD. decomposition |
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Answer» Correct Answer - C Oxidation state of Cl in `CIO^(-)` is+1. While oxidation state of Cl in `CIO_(3)^(-)`, is +5 and `Cl^(-)` is -1. Thus, oxidation state of chlorine decreases as well as increases. Hence, chlorine undergoes reduction and this oxidation. Hence, reaction is disproportionation. |
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| 4. |
The reaction: `ClO_(3)^(-) +I_(2) to IO_(3)^(-)+Cl_(2)`A. is possibleB. depends upon the state of productsC. is not possibleD. depends upon the temperature |
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Answer» Correct Answer - A In this reaction, I replaces a from `ClO_(3)^(-)`, because Cl in `ClO_(3)^(-)` is in +5 oxidation state and `l_2` is more electropositive than `Cl_(2)`. Note: The reaction : `I_(2) + 2KCl` is not possible because Cl cannot be replaced by more electropositive `I_(2)`. |
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| 5. |
In the reaction: `3Br_(2)+6OH^(Theta)rarr 5Br^(Theta)+BrO_(3)^(Theta)+3H_(2)O,Br_(2)` isA. is reducedB. is oxidizedC. disproportionatesD. disintegrate |
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Answer» Correct Answer - C `3Br_(2)+6OH^(-) to overset((-1))(5Br^(-))+overset((+5))(BrO_(3)^(-))+3H_(2)O` In this reaction, O.S of bromine changes from (i) 0 to -1, and (ii) 0 to +5. Hence, `Br_(2)` is reduced to `Br^(-)` and also oxidized to `BrO_(3)^(-)`. Thus, `BrO_(3)^(-)` is oxidized as well as reduced. In other words, undergoes disproportionation. |
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| 6. |
`NaClO_(3)` is obtained by reacting `Cl_(2)` andA. cold dil. NaOHB. hot conc. NaOHC. cold `Na_(2)O_(2)`D. hot `Na_(2)O_(2)` |
| Answer» Correct Answer - B | |
| 7. |
Which of the following is acidic oxides ?A. `HClSO_(4)`B. `MgO`C. `Al_(2)O_(3)`D. `CaO` |
| Answer» Correct Answer - A | |
| 8. |
Which of the following is more acidic oxyacid ?A. `HClO_(4)`B. `HClO`C. `HClO_(3)`D. `HClO_(2)` |
| Answer» Correct Answer - A | |
| 9. |
Oxygen orm acidic oxide withA. CaB. NaC. SD. K |
| Answer» Correct Answer - C | |
| 10. |
Of the following, the most acidic isA. `As_(2)O_(3)`B. `P_(2)O_(3)`C. `Sb_(2)O_(3)`D. `Bl_(2)O_(3)` |
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Answer» Correct Answer - B `underset("Addiccharacterdecreasesdownthegroup ")(underset("Acidic")ubrace(N_(2)O_(3)" "P_(2)O_(3)" "As_(2)O_(3))" "underset("Amphoteric")ubrace(Sb_(2)O_(3)) underset("Basic")ubrace(Bi_(2)O_(3)))` |
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| 11. |
In laboratory silicon can be prepared by the reactionA. Silica with magnesiumB. By heating carbon in electric furnaceC. By heating potassium fluosilicate with potassiumD. None of these |
| Answer» Correct Answer - C | |
| 12. |
Which has the lowest boiling point?A. `NH_(3)`B. `PH_(3)`C. `SbH_(3)`D. `AsH_(3)` |
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Answer» Correct Answer - B `{:(,"Hybride",NH_(3)" "PH_(3)" "AsH_(3)" "SbH_(3)" "BiH_(3)),(,"Boiling point",238.5" "185.5" "210.6" "254.6" "290):}` |
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| 13. |
Which of the following hydeides has the lowest boiling point?A. `SbH_(3)`B. `AsH_(3)`C. `PH_(3)`D. `NH_(3)` |
| Answer» Correct Answer - C | |
| 14. |
`O_2` molecule isA. ferrimagneticB. ferromagneticC. paramagneticD. diamagnetic |
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Answer» Correct Answer - C `O_(2), sigma 1s^(2),sigma^(**),1s^(2),sigma 2s^(2) sigma^(**) 2s^(2), sigma 2p_(z)^(2), pi2p_(xx)^(2), pi2p_(y)^(2), pi^(**)2p_(x)^(1), pi^(**) 2p_(y)^(-1)` Since `O_2`, molecule has two unpaired electrons, therefore, it is paramagnetic. |
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| 15. |
Which of the following reactions is employed to produce ozone in the laboratory ?A. Exposure of air to U.V. light leaB. Reaction of `F_2` with `H_O` at low temperatureC. Reaction of `SO_2" with "H_2O_2`D. Passage of silent electric discharge through `O_2` |
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Answer» Correct Answer - D Ozone is formed, when energy is supplied to dioxygen in the form of UV radiation or silent electric discharge. `3O_(2_((g))) Leftrightarrow 2O_(3_((g))),DeltaH=142.7" kJ mol"^(-1)` Since `O_3` is an endothermic molecule, so it is necessary to use a silent electric discharge, otherwise sparking would generate heat and decompose it. This is the laboratory method. a) Exposure of `O_2` of air UV light also form `O_3`. This occurs in atmosphere, when photochemical smog is formed over cities like Delhi, Los Angeles or Tokyo. (b). `2H_(2)O+2F_(2) to 4HF+O_(2)` `3H_(2)O+3F_(2) to 6HF+O_(3)` (c). `SO_(2)+H_(2)O_(2) to H_(2)SO_(4)` |
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| 16. |
Whichg of the following hydrides has the lowest boiling point?A. `H_(2)O`B. `H_(2)S`C. `H_(2)Se`D. `H_(2)Te` |
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Answer» Correct Answer - B `{:(,H_(2)O,H_(2)S,H_(2)Se,H_(2)Te),(,373K,213K,269K, 232K):}` `H_(2)S` has lowest boiling point and `H_(2)O` has highest boiling point because due to highest boiling point of `H_(2)O` is high. |
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| 17. |
Arrange the hydrides of group `15` in the correct order of reducing natureA. `NH_3 lt PH_3 lt ASH_3 lt SbH_3 lt BiH_3`B. `NH_3 gt PH_3 gt ASH_3 gt SbH_3 gt BiH_3`C. `PH_3 lt ASH_3 lt SbH_3 lt BiH_3 lt NH_3`D. `PH_3 gt ASH_3 gt SbH_3 gt BIH_3 gt NH_3` |
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Answer» Correct Answer - A The least stable hydride `(BiH_3)` acts as the strongest reducing agent, where as the most stable hydride `(NH_3)` as a weakest reducing agent. Consequently, reducing nature of hydrides increases on moving down the group. |
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| 18. |
The boiling points of hydrides of group 16 are in the orderA. `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te`B. `H_(2)O gt H_(2)Se gt H_(2)Te gt H_(2)O`C. `H_(2)O gt H_(2)Te gt H_(2)Se gt H_(2)S`D. `H_(2)Te gt H_(2)Se gt H_(2)O gt H_(2)S` |
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Answer» Correct Answer - C Boiling point increases in moving from `H_2S` to `H_2Te,` owing to increase in vander Waals forces, due to the increase in molecular mass. `H_2O` has UPI exceptionally high b.p. because it form strong ba intermolecular H-bonds. |
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| 19. |
The basic character of hydrides of the 15th-group elements decreases in the orderA. `SbH_(3) gt PH_(3) gt AsH_(3) gt NH_(3)`B. `NH_(3) gt SbH_(3) gt PH_(3) gt AsH_(3)`C. `NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)`D. `SbH_(3) gt AsH_(3) gt PH_(3) gt NH_(3)` |
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Answer» Correct Answer - C `NH_(3) gt PH_(3) gt AsH_(3) gt SbH_(3)` On moving down the group atomic size increases and availability of lone pair decreases. Hence, basic character decreases, |
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| 20. |
When chloride is passed over dry slaked lime at room temperature the main reaction product isA. `Ca(ClO_(2))_(2)`B. `CaOCl_(2)`C. `Ca(OCl)_(2)`D. `CaCl_(2)` |
| Answer» Correct Answer - B | |
| 21. |
`2E+N_(2)to2EN` (very hard substance) `EN+H_(2)Oto` Acid + pungent smelling gas Acid is:A. `HNO_(3)`B. `H_(3)BO_(3)`C. `HNO_(2)`D. can be A & B |
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Answer» Correct Answer - B `2B+N_(2)to2BN` `BN+3H_(2)O to H_(3)BO_(3)+NH_(3)`(pungent smelling gas) |
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| 22. |
Which of the following is the correct graph for `EN` values of carbon family :A. B. C. D. |
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Answer» Correct Answer - B The `EN` values of Carbon family are : `{:("Element",C,Si,Ge,Sn,Pb),("EN",2.5,1.8,1.8,1.8,1.9):}` |
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| 23. |
Consider a prototypical fullerene, `C_(60)` Let,` a` = Number of `5`-membered rings: `b` = Number of `6`-membered rings `c` = Number of `pi` bonds in `C_(60)` Find the value of `(3a-2b+c)` |
| Answer» Correct Answer - B | |
| 24. |
For boron family (B,Al,Ga,In and Tl) x : Number of elements which are solid at `40^(@)` `y` : period number of element which has greater ionization energy than element just above and below it in periodic table. ` Z`: Period number of most abundant element of group `13`. Report your answer `x+2y+3z` |
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Answer» Correct Answer - A::B `x=4` `B,Al, In` & `Tl` are solid at `40^(@)C`.Melting point for Gallium is `30^(@)C`. `y=4` `I.E. : B gt Al gt Ga gt In gt Tl` `z=3` `Al` is third most abundant element after oxygen and silicon.So it has to be most abundant element in the family. `implies x=2y+3z=4+(2xx4)+(3xx3)=21` |
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| 25. |
How many of the following statements are correct regarding allotropes of carbon: (a) Graphite is not a good conductor of electicity in perpendicular direction of layers at ordinary temperatures. (b) Coke is the impure form of carbon. ( c) Anthracite is the purest form of carbon. (d) Buckminister fullerence contains `12` five membered rings and `20` six-membered rings. (e) Diamond is a good conductor of heat. (f) Graphite is diamagnetic in nature. (g) Graphite is thermodynamically more stable than diamond |
| Answer» Correct Answer - N//A | |
| 26. |
Which one of the gas dissolves in `H_(2)SO_(4)` to give oleum?A. `SO_(2)`B. `H_(2)S`C. `S_(2)O`D. `SO_(3)` |
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Answer» Correct Answer - D `H_(2)SO_(4)+SO_(3) to underset("Oleum or funning "H_(2)SO_(4))(H_(2)S_(2)O_(7))` |
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| 27. |
Which of the following resembles water in the liquid state, where it is a solvent for many electrolytes, and even undergoes autoionization as water does ?A. `N_(2)`B. `Cl_(2)`C. `NH_(3)`D. `N_(2)O` |
| Answer» Correct Answer - C | |
| 28. |
Nitric oxide (NO) is paramagnetic in gaseous stateA. gaseous stateB. solid stateC. liquid stateD. polymeric state |
| Answer» Correct Answer - A | |
| 29. |
`FeCl_2` reacts with `SO_2`A. to give FeSB. to give FeOC. Fe will be oxidizedD. Fe will be reduced |
| Answer» Correct Answer - C | |
| 30. |
All the rare gases, except helium, have for their outermost shell, the general electronic configurationA. `np^(2), nd^(6)`B. `ns^(2), np^(6), nd^(10)`C. `ns^(2), np^(6), nd^(6)`D. `ns^(2),np^(6)` |
| Answer» Correct Answer - A | |
| 31. |
The outermost electronic configuration of the most electronegative element isA. `ns^(2) np^(3)`B. `ns^(2) np^(5)`C. `ns^(2) np^(4)`D. `ns^(2) np^(6)` |
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Answer» Correct Answer - C The configuration of most electronegative element is (c), which is one electron short of noble gas configuration. |
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| 32. |
Ozone oxidises PbS toA. PbOB. `PbSO_(3)`C. `PbSO_(4)`D. Pb |
| Answer» Correct Answer - C | |
| 33. |
Ozone layer is depleted by aenA. NOB. `NO_(2)`C. `NO_(3)`D. `N_(2)O_(5)` |
| Answer» Correct Answer - A | |
| 34. |
Ozone with `K` solution producesA. `KIO_(3)`B. `I_(2)`C. `KI_(3)`D. HI |
| Answer» Correct Answer - B | |
| 35. |
Ozone belong to which group of the periodic table ?A. 15B. 16C. 17D. none |
| Answer» Correct Answer - D | |
| 36. |
Oxygen is denser than air, so it is collected overA. waterB. spiritC. mercuryD. kerosene |
| Answer» Correct Answer - A | |
| 37. |
Boron cannot from which one of the following anions?A. `BF_(6)^(3-)`B. `BH_(4)^(-)`C. `B(OH)_(4)^(-)`D. `BO_(2)^(-)` |
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Answer» Correct Answer - A Due to non-availability of `d`-orbitals, boron is unable to expand its octet.Therefore, the maximum covalence of boron cannot exceed `4`. |
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| 38. |
`KO_(2)+CO_(2)rarr?("gas")`A. `H_(2)`B. `N_(2)`C. `O_(2)`D. CO |
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Answer» Correct Answer - C `2KO_(2)+CO_2 to K_(2)CO_(3)+3/2O_(2)` |
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| 39. |
A gas that cannot be collected over water is.A. `N_(2)`B. `O_(2)`C. `SO_(2)`D. `PH_(3)` |
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Answer» Correct Answer - C `SO_(2)` is soluble in water `H_(2)O+SO_(2) to underset("Sulphurus acid")(H_(2)SO_(3))` |
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| 40. |
The small size and high charge of `Al^(3+)` ion gives it a high charge density which is responsible for its tendency to show (a) covalency in its compounds in the gaseous state (b) high hydration energy which stabilizes its compounds in solution, and (c) high lattice energy of its compounds in the solid state. Thus aluminium can forms both covalent and ionic bond. Like halides of boron, halides of aluminium do not show back bonding because of increase in size of aluminium. Actually aluminium atoms complete their octets by forming dimers. Thus chloride and bromide of aluminium exist as dimers, both in the vapour state and in polar-solvents like benzene while the corresponding boron halides exists as monomer. In boron trihalides the extent of back bonding decreases with increases with increase in size of halogens and thus lewis acid character increases. All `BX_(3)` are hydrolysed by water but `BF_(3)` shows a different behaviour. Which of the following statements about anhydrous aluminium chloride is correct ?A. It is an ionic compound.B. It is not easily hydrolysed.C. It sublimes at `100 ^(@)C` under vacuum.D. It is a strong lewis base. |
| Answer» Correct Answer - C | |
| 41. |
Catenation tendency in group `14` is:A. `CgtgtSigtGe=Pb` due to bond energies `C-HgtSi-H gt Ge-H gt Sn-H`B. `CgtgtSigtGe=Sn gt Pb` due to bond energies `C-CgtSi-C gt Ge-C gt Sn-C`C. `C=Si=Ge=Sn = Pb` due to bond energies `C-HgtSi-H gt Ge-H gt Sn-H`D. `CgtgtSigtGe=Sn gt Pb` due to bond energies `C-CgtSi-Si gt Ge-Ge gt Sn-Sn` |
| Answer» Correct Answer - D | |
| 42. |
Which of the following Statement (s) is/are correct?A. `B_(2)O_(3)` and `SiO_(2)` are acidic in nature and are important constituents of glass.B. Borides and Silicid are hydrolysed by water forming boranes and Silanes respectively.C. Diborane on reaction with chlorine (g) forms `B_(2)H_(5)Cl`D. `SiO_(4)^(4-)` gets hydrolysed by acid or water and form `Si_(2)O_(7)^(6-)` |
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Answer» Correct Answer - A,B and D `(A),(B)` and `(D)` are correct statement but `( C)` is incorrect. `B_(2)H_(6)+6Cl_(2)to2BCl_(3)+6HCl` |
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| 43. |
What do you understand by (a) inert pair effect (b) allotropy and (c) catenation? |
| Answer» Correct Answer - A::B::C::D | |
| 44. |
Why does silicon not form an analogue of graphite? |
| Answer» Correct Answer - A::B::C::D | |
| 45. |
Diamond and Graphite areA. isomersB. isotopesC. allotropesD. none of the above |
| Answer» Correct Answer - C | |
| 46. |
Give reasons: (i) Graphite is used as a lubricant. (ii) Diamond is used as an abrasive. |
| Answer» Correct Answer - A::B::C::D | |
| 47. |
Pick out the incorrect statement regarding hydrogen halides.A. Hydrogen chloride can be prepared by the reaction of NaCl with conc. `H_2SO_4`B. Reactions of respective ionic halides, i.e., NaBr and KI with conc. `H_2SO_4` are employed to produce HBr and HIC. Hdrogen halides (X=Cl, Br, I) are prepared by action of phosphorus trihalides with water.D. A solution of hydrogen chlorides in toluene does affect blue litmus paper. |
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Answer» Correct Answer - B Reactions of NaBr and KI with conc. `H_2SO_4` cannot be used to prepare HBr and EI, since these two hydrides are reducing agents and are readily oxidized by conc. `H_2SO_4` `2HX +H_(2)SO_(4)to SO_(2)+2H_(2)O+X_(2), X=Br,I` (a) (i) `NaCl+H_(2)SO_(4) to NaHSO_(4)+HCl("in the cold")` (ii) `NaHSO_(4)+NaCl to NaHSO_(4)+HCl("on heating")` (c) `PX_(3)+3H_(2)O to H_(2)PO_(3)+3HX` For example `PCl_(3)+3H_(2)O to H_(3)PO_(3)+3HCl` `underset(("Red"))(2P)+3Br_(2) to 2PBr_(3)` `underset(("Red"))(2P)+3I_(2) to 2PI_(3)` `PBr_(3)+3H_(2)O to H_(3)PO_(3)+3HBr_(2)` `and PI_(3)+3H_(2)O to H_(3)PO_(3)+3HI` |
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| 48. |
Pick out the incorrect statement for `XeF_(6)`A. ` XeF_(6)` is hydrolysed partially to form `XeOF_4`B. It reacts with `SiO_(2)` to form `XEOF_(4)`C. On complete hydrolysis, it forms `XeO_3`D. It acts as F acceptor when treated with alkali metal fluoride, but cannot act as F donor to form complexes. |
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Answer» Correct Answer - A It also act as `F^(-)` donor with covalent pentafluorides. `XeF_(6)+SbF_(5) to [XeF_(5)]^(+) [SbF_(6)]^(-)` (a) `XeF_(6)+H_(2)O to XeOF_(4)+2HF` `2XeF_(6)+SiO_(2) to 2XeOF_(4)+SiF_(4)` (c) `XeF_(6)+3H_(2)O to XeO_(3)+6HF` |
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| 49. |
Among the following nitrates, silver nitrates, lead nitrate, silver nitrate and ammonium nitrate, the one that decomposes without leaving any solid residue isA. `Pb(NO_(3))_(2)`B. `NH_(4)NO_(3)`C. `AgNO_(3)`D. `NaNO_(3)` |
| Answer» Correct Answer - B | |
| 50. |
Amongest `H_(2)O,H_(2)S,H_(2)Se` and `H_(2)Te` the one with highest boiling point is :A. `H_(2)O` because of hydrogen bondingB. `H_(2)` Te because of higher molecular weightC. `H_(2)S` because of hydrogen bondingD. `H_(2)Se` because of lower molecular weight |
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Answer» Correct Answer - A `H_2O` containing hydrogen bond. |
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