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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. |
Boric acid is used in carom boards for smooth gliding of pawns becauseA. `H_(3)BO_(3)` molecules are loosely chemically bonded and hence softB. its low density makes a fluffyC. it can be powdered to a very small grain sizeD. H-bonding in `H_(3)BO_(3)` gives it a layered structure. |
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Answer» Correct Answer - D H-bonding in `H_(3)BO_(3)` gives it a layered structure which accounts for its slippery nature. |
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| 152. |
White enamel of our teeth isA. `Ca_(3)(PO_(4))_(2)`B. `CaF_(2)`C. `CaCl_(2)`D. `CaBr_(2)` |
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Answer» Correct Answer - B The enamel of our teeth is the hardest substance in the body made up of `CaF_(2)` and dentine below it made of `Ca_(3)(PO_(4))_(2)`. |
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| 153. |
Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides?A. `Al_(2)O_(3) ltMgOltNa_(2)OltK_(2)O`B. `MgOltK_(2)O ltAl_(2)O_(3)ltNa_(2)O`C. `Na_(2)O ltK_(2)OltMgOltAl_(2)O_(3)`D. `K_(2)OltNa_(2)OltAl_(2)O_(3)ltMgO` |
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Answer» Correct Answer - A As metallic character of element attached to oxygen atom increases, the difference between the electronegativity values values of element and oxygen increaes and thus basic character or oxidised increases and vice-versa. Hence the increaseing correct order of basic nature is `Al_(2)O_(3)ltMgOltNa_(2)OltK_(2)O`. |
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| 154. |
When chloride is passed over dry slaked lime at room temperature the main reaction product isA. `Ca(ClO_2)_2`B. `CaCl_2`C. `CaOCl_2`D. `Ca(OC l_2)_2` |
| Answer» Correct Answer - C | |
| 155. |
Among the following, the number of compounds that can react with `PCl_(5)` to give `POCl_(3)` is `O_(2)`,`CO_(2)`,`SO_(2)`,`H_(2)O`,`H_(2)SO_(4)`,`P_(4)O_(10)`. |
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Answer» Correct Answer - 4 |
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| 156. |
Complete and balance the following : (i)`P_(4)O_(10)+PCl_(5)to` (ii)`NH_(3)+NaOCl to ` |
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Answer» Correct Answer - (i)`P_(4)O_(10)+6PCl_(5) to 10POCl_(3)` , (ii)`NH_(3)+NaOCl to NH_(2)Cl + NaOH`(fast) `NH_(3)+NH_(2)Cl to NH_(2)NH_(2)+NH_(4)Cl`(slow) |
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| 157. |
Identify `(A)` to `( E)`. (a) An inorganic iodide `(A)` on heating with a solution of `KOH` gives a gas `(B)` and the solution of a compound `( C)`. (b) The gas `(B)` on ignition air gives a compound `(D)` and water. ( C) Copper sulphate is reduced to the metal on passing `(B)` through the solution. (d) A precipitate of the compound `( E)` is formed on reaction of `( C)` with copper sulphate solution. |
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Answer» Correct Answer - (A)`PH_(4)I`; (B)`PH_(3)`; ( C)`KI` ; (D)`P_(4)O_(10)` ; (E)`Cu_(2)I_(2)` `(A)` is `PH_(4)I`. The given changes are: (i)`underset((A))(PH_(4)l)+KOH to underset((B))(PH_(3))+H_(2)O` (ii)`underset((B))(4PH_(3))+8O_(2) to underset((D))(P_(4)O_(10))+6H_(2)O` (iii)`4Cu^(2+) + PH_(3)+ 4H_(2)O to H_(3)PO_(4) + 4Cu darr + 8H^(+)` (iv)`2CuSO_(4)+4KI to underset((E))(Cu_(2)l_(2))darr+2K_(2)SO_(4) + I_(2)` |
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| 158. |
Point fout in which of the following properties oxygen differs from the rest of the members of its family (Groupi-`VIA`)A. High value of ionisation energiesB. Oxidation states `(2,4,6)`C. PolymorphismD. Formation of hydrides |
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Answer» Correct Answer - B Oxidation states are `2,4,6` |
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| 159. |
Oxygen differs from the other elements of the group. Compounds of oxygen with metals are more ionic in nature and hydrogen bonding is more important for oxygen compounds. Oxygen is never more than divalent beacasue when it hs formed two covalent bonds, there are no low energy orbitals which can be used for form. further bonds. However, the elements S,Se,Te and Po have empty d- orbitals which may be used for bonding, and they can form four or six bonds by unpairing electrons. The higher oxidation states become less stable on decending the group. The bond between S and O, or Se and O, are much shorter than might be expected for a single bond owing to `rho pi - dpi` interaction between the p- orbital of oxygen and d- orbital of S or Se. Which one of the following orders represents the correct order for the properties indicated against them?A. `H_(2)O lt H_(2)S lt H_(2)Se lt H_(2) Te -` "acidic character"B. `H_(2)O lt H_(2)S lt H_(2)Se lt H_(2) Te -` "thermal stability"C. `H_(2)S gt H_(2)Se lt H_(2) Te lt H_(2)O -`"reducing character"D. `H_(2)S gt H_(2)Se lt H_(2)O lt H_(2)Te -`"boiling point " |
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Answer» (A) As bond (H-E) dissociation enthaply decreases down the group, the acidic character increases from `H_(2)O` to `H_(2) Te`. (B) Order of thermal stability is `H_(2)O gt H_(2)S gt H_(2)Se gt H_(2)Te`. (C ) `H_(2)O` does not have reducing property and this character increases from `H_(2)S` to `H_(2)Te`. (D) Water has highest boiling point because of H-bonding and thus the correct order is `H_(2)S lt H_(2)Se lt H_(2)Te lt H_(2)O`. |
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| 160. |
When a compound X reacts with ozone in aqueous medium, a compound Y is prouduced. Ozone also react with Y and produces Z. Z acts as an oxidising agent, then X,Y and Z will be:A. `X=HI,Y=I_(2) " and " Z=HIO_(3)`B. `X=KI,Y=I_(2) " and " Z=HIO_(3)`C. `X=KI,Y=I_(2) " and " Z=HIO_(4)`D. `X=HI,Y=I_(2) " and " Z=HIO_(4)` |
| Answer» Correct Answer - A::B | |
| 161. |
Assertion: `Ti^(3+)` acts as an oxidising agent. Reason: `TI^(+)` is more stable than `TI^(3+)` due to inert pair effect.A. If both assertion and reason are true and the reason is the correct explanation of teh assertion.B. If both assertion and reason are true but reason in not the correct explation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `T1^(3+)` acts as an oxidising agent because it has tendency in reduce to `TI^(+2)` as `+1` oxidation state of `TI` is more stable on account of inert pair effects. |
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| 162. |
`Tl^(3+)` acts an `"………….."` agent. |
| Answer» Correct Answer - oxidising agent | |
| 163. |
Boron forms only `"………….."` compounds while aluminium forms both `"………….."` and `"………….."`compounds. |
| Answer» Correct Answer - covalent, covalent and ionic | |
| 164. |
Aluminium is `"………….."` in nature and dissolves in both dilute hydrochloric acid and sodium hydroxide evolving `"………….."` gas. |
| Answer» Correct Answer - amphoteric, `H_(2)` | |
| 165. |
Hydrolysis of `XeF_(4)` and CaNCN gives respectively :A. `XeO_(3)` and `CaCO_(3)`B. `XeO_(2)` and `CaCN_(2)`C. `XeOF_(3)` and `CaCN_(2)`D. `XeOF_(2)` and `CaCO_(3)` |
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Answer» Correct Answer - A |
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| 166. |
Write down the hydrolysis of `XeF_(6)` in strongly alkaline medium. |
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Answer» Correct Answer - `2XeF_(6)+16OH^(-)toXeO_(6)^(4-) + Xe + O_(2) +12F^(-) + 8H_(2)O` |
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| 167. |
Select the correct statement (s) regarding compounds of noble gases:A. Only compound of Kr, `KrF_(2)` has been studied in detailB. `RnF_(2)` has been identified by radiotracer techniqueC. No true compounds of Ar, Ne of He are yet knownD. Xe forms three binary fluorides `XeF_(2),XeF_(4)` and `XeF_(6)` |
| Answer» Correct Answer - A::B::C::D | |
| 168. |
Which of the following noble gases does not form clathrate compounds?A. `Ne`B. `Kr`C. `Ar`D. `Xe` |
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Answer» Correct Answer - A Ne does not form clathrates as the size of noble gas particle is not in compatison to pore size. |
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| 169. |
Xenon best reacts withA. most elctropositive metalsB. most electronegative metalsC. neutral atomsD. none of these |
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Answer» Correct Answer - B As `Xe` has fully filled stable electronic configuration. Will react only with more electronegative elements like `O` and `F`. |
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| 170. |
`XeF_(6)` on reaciton with `KF` yieldsA. `[XeF_(5)]+[KF_(2)]^(-)`B. `K^(+)[XeF_(7)]^(-)`C. `[XeF_(4)]^(+2)[KF_(3)]^(-2)`D. none of these. |
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Answer» Correct Answer - B `KF+XeF_(6)rarrK^(+)[XeF_(7)]^(-)` |
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| 171. |
The compound that attacks pyrex glass isA. `XeF_(2)`B. `XeF_(4)`C. `XeF_(6)`D. Both `A` and `B` |
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Answer» Correct Answer - C `XeF_(6)` can be sublimed at room temperature it is more susceptible to hydrolysis `XeF_(6)` melt to give yellow liquid, since it attacks pyrex glass because silica react with `XeF_(6)`. `2XeF_(6)+3SiO_(2)rarr3SiF_(4)+2XeO_(3)` |
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| 172. |
Which of the following noble gases is used in the treatment of cancer ?A. HeliumB. ArgonC. KryptonD. Radon |
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Answer» Correct Answer - D Rn is used in cancer treatment |
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| 173. |
The compound that attacks pyrex glass isA. `XeF_2`B. `XeF_4`C. `XeF_6`D. All |
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Answer» Correct Answer - C `XeF_6` is high reactive and attack glass materials `2XeF_6 + SIO_2 to 2XeOF + SIF_4` |
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| 174. |
Which of the following is not the correct uses of clathrates ?A. In the separation of noble gasesB. In transporting of isotopes of noble gasesC. Kr - 85 clathrate provide a useful source of `beta`- radiationsD. Clathrates compounds are used for producing compounds of noble gases |
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Answer» Correct Answer - D All except (d ) are uses of clathrates |
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| 175. |
Molecules of a noble gas do not posses vibrational energy because a noble gasA. MonoatomicB. ChemicallyinertC. Complete filled shellsD. Diatomic |
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Answer» Correct Answer - A Monoatomic atoms do not have vibratory motions. |
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| 176. |
Which of the following posses the highest bond energy ?A. `F_(2)`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
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Answer» Correct Answer - B |
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| 177. |
Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation. |
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Answer» 1) Electronic configurations : Oxygen (O) - [He] `2s^(2)2p^(2)` Sulphur (S) - [Ne] `3s^(2)3p^(2)` Selenium (Se) - [Ar] `3d^(10)4s^(2)4p^(4)` Tellurium (Te) - [Kr] `4d^(10)5s^(2)sp^(4)` Polonium (Po) - [Xe] `4f^(14)5d^(10)6s^(2)6p^(4)` `to` All the above elements has general outer electronic configuration `ns^(2)np^(4)`. 2) Oxidation states : All the gives elements (chalcogens) exhibits common oxidation state of - 2 `to O^(-2),S^(-2,)Se^(-2)` etc. 3) Hydride formation : All these elements (chalcogens) forms hydrides of type `EH_(2)` (E = chalcogen) Eg : `H_(2)O,H_(2)S,H_(2)Se,H_(2)Te, H_(2)Po`. The above mentioned concepts evident that the elements O, S, Se, Te and Po are present in the same group of periodic table. |
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| 178. |
What happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt ? |
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Answer» When `SO_(2)` is passed through an aqueous solution of Fe (III), i.e., ferric salt, it is reduced to Fe (II) i.e., ferrous salt. Here, `SO_(2)` acts as a reducing agent. `underset(("Oxide agent"))underset("Ferric ion")(2"Fe"^(3+))+underset("agent")underset("Reducing")(SO_(2))+2H_(2)O tounderset("ion")underset("Ferrous")(2"Fe"^(2+))+SO_(4)^(2-)+4H^(+)` |
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| 179. |
What happens when (i)Hydrogen sulphide is bubbled through an aqueous solution of sulphur dioxide. (ii)Hydrogen sulphide is passed through acidified ferric chloride. |
| Answer» Correct Answer - (i)`SO_(2) ("moist") + 2H_(2)S to 2H_(2)O + 3S darr` , (ii)`2FeCl_(3) + H_(2)S to 2FeCl_(2) + 2HCl + S` | |
| 180. |
Distinguishing reagent between silver and lead salt isA. `H_(2)S` gasB. dil. HCl solutionC. `[NH_(2)Cl(solid)+NH_(4)OH]` solutionD. `[NH_(2)Cl (solid)+(NH_(4))CO_(3)]` solution |
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Answer» Correct Answer - B On adding dil. HCL to silver and lead salts , while precipitate of chlorides of Ag and Pb are obtained which on heating with water are separated from each other as `PbCl_(2)` is soluble in hot water but AgCl is not . |
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| 181. |
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?A. `HNO_(3),NO,NH_(4),Cl,N_(2)`B. `HNO_(3),NO,N_(2),NH_(4),Cl,C. `HNO_(3),NH_(4),Cl,NO,N_(2)`D. `NO,HNO_(3),NH_(4),Cl,N_(2)` |
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Answer» Correct Answer - B `{:("Compound","oxidation state of Nitrogen"),(HNO_(3),=+5),(NO,=+2),(NH_(4)Cl,=-3),(N_(2),=0):}` So, correct order will `HNO_(3),NO,N_(2),NH_(4)Cl` |
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| 182. |
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?A. `HNO_(3),NO,NH_(4)CI, N_(2)`B. `HNO_(3),NO,N_(2),NH_(4)CI`C. `HNO_(3),NH_(4)CI,NO,N_(2)`D. `NO,HNO_(3),NH_(4)CI,N_(2)` |
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Answer» Correct Answer - B |
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| 183. |
Which ordering of compounds is accrding ot decreasing order of the oxidation state of nitrogen?A. `HNO_(3),NO,NH_(4)Cl,N_(2)`B. `HNO_(3),NO,N_(2),NH_(4)Cl`C. `HNO_(3),NH_(4)Cl,NO,N_(2),`D. `NO,HNO_(3),NH_(4)Cl,N_(2)` |
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Answer» Correct Answer - B `Hoverset(+5)(NO_(3)),overset(+2)(NO),overset(0)N_(2),overset(-3)NH_(4)Cl` |
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| 184. |
The product formed in the reaction of `SOCl_2` (thionyl chloride) with white phosphorous is.A. `PCI_(3)`B. `SO_(2)CI_(2)`C. `SCI_(2)`D. `POCI_(3)` |
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Answer» Correct Answer - A |
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| 185. |
Concentrated nitric acid upon long standing turns yellowish-brown due to the formation of :A. `NO `B. `NO_(2)`C. `N_(2)O`D. `N_(2)O_(4)` |
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Answer» Correct Answer - B `NO_2` is a brown coloured gas and imparts this colour to concentrated `HNO_3` upon long standing . `4HNO_(3) to 2H_2O +4NO_(2)+O_(2)` |
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| 186. |
Name the oxide of chlorine which has odd number of electrons and paramagnetic in nature. |
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Answer» Correct Answer - Chlorine dioxide, `CIO_(2)` |
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| 187. |
Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation ofA. NOB. `NO_(2)`C. `N_(2)O`D. `N_(2)O_(4)` |
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Answer» Correct Answer - B `4HNO_(5)to4NO_(2)+O_(2)+2H_(2)O` |
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| 188. |
Out of `O_(2)` and `O_(3)`, which is paramagnetic ? |
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Answer» `to O_(2)` is paramagnetic due to presence of unpaired electrons `to O_(3)` (gaseous) is diamagnetic due to absence of unpaired electrons. |
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| 189. |
An aqueous solution of sodium carbonate absorbs `NO` and `NO_(2)` to yieldA. `CO_(2) + NaNO^(3)`B. `CO_(2) + NaNO_(2)`C. `NaNO_(2) + CO`D. `NaNO_(3) + CO` |
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Answer» Correct Answer - C `2Na_(2)CO_(3) +NO + 3NHO_(2) rarr 4NaNO_(2) + CO` |
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| 190. |
Which among the following is paramagnetic ?A. `Cl_(2)O`B. `ClO_(2)`C. `Cl_(2)O_(7)`D. `Cl_(2)O_(6)` |
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Answer» Correct Answer - B `ClO_(2)` contains total valence electrons 19. `7` valence electrons of `Cl` 6 valence electrons of one oxygen atom. So there must be unpaired electron, thus it is paramagnetic in nature. |
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| 191. |
The reddish brown coloured gas formed when nitric oxide is oxidised by air isA. `N_(2)O_(5)`B. `N_(2)O_(4)`C. `NO_(2)`D. `N_(2)O_(3)` |
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Answer» Correct Answer - C `2NO+O_(2)to2NO_(2)` |
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| 192. |
Which of the following oxides of chlorine is obtained by passing dry chlorine over silver chlorate at `90^(@)C`.A. `Cl_(2)O`B. `ClO_(3)`C. `ClO_(2)`D. `ClO_(4)` |
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Answer» Correct Answer - C `2AgClO_(3) + Cl_(2)("dry") overset("Heat")rarr 2AgCl + 2ClO_(2) + O_(2)` |
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| 193. |
Give reasons for the following : (i) NO (Nitric oxide) is paramagnetic in the gaseous state but diamagnetic in the liquid and solid states. Why ? (ii) Nitric oxide becomes brown when released in air. |
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Answer» (i) NO has an odd number of electrons (7+8=15 electrons) and hence is paramagetic in the gaseous state. But in liquid and solid states, it exists as a dimer (Table 11.14, page 11//19) and hence is diamagnetic in these states. (ii) Nitric oxide has one unpaired electron and hence is very reactive. As a result, it readily combies with `O_(2)` of the air to form nitrogen dioxide `(NO_(2))` which has brown colour. `underset("Colourles")(2NO)+O_(2)tounderset("Brown")(2NO_(2))` |
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| 194. |
Which of the following product is formed when sulphur dioxide gas is passed through sodium chlorate in strongly acidic solution?A. `NaClO_(4)`B. `ClO_(2)`C. `Na_(2)SO_(3)`D. `SO_(3)` |
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Answer» Correct Answer - B `underset("oxidant")(2ClO_(3)^(-)(aq))+underset("reductant")(SO_(2)(g)) overset("Acid")to2ClO_(2)(g)+SO_(4)^(2-)(aq)` |
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| 195. |
A certain element is a metalloid that forms an acidic oxide with the formula `R_(2)O_(5)`. Identify the elements |
| Answer» Since the elements forms as axide with formula, `R_(2)O_(5)`, it must be an element of group 15. Now group 1 has two metalloids, As and Sb. Since the acidic character of pentoxides of group 15 elements decrease while the basic character increases down the group, therefore, `As_(2)O_(5)` is acidic while `Sb_(2)O_(5)` is amphoteric. Thus, the metalloid of group 15 elements which forms an acidic pentoxide `(As_(2)O_(5))` is As. | |
| 196. |
In elements of group 14. (a) which forms the most acidic oxide (b) Which is normally found in +2 oxidation state (c) which is used as semi conductor? |
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Answer» (a) Carbon (typical non-metal) forms most acidic oxide `(CO_(2))`. (b) Lead is found in +2 oxidation staet because of maximum inert pair effect. (c) Silicon (or germainium) can be used in semi-conductor. |
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| 197. |
which of the following elements forms a strongly acidic oxide?A. `P`B. `As`C. `Sb`D. `B` |
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Answer» Correct Answer - A Acidic character of oxides decrease down the group. |
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| 198. |
An element forms a gaseous oxide which on dissolving in water gives an acidic solution. The element isA. HydrogenB. SodiumC. MagnesiumD. Sulphur |
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Answer» Correct Answer - D Sulphur form gaseous oxide and on dissolving in water gives an acidic solution. `SO_(3)+H_(2)OrarrH_(2)SO_(4)` |
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| 199. |
The element that does not form acidic oxide isA. CarbonB. PhosphorusC. ChlorineD. Barium |
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Answer» Correct Answer - D Metals do not form acidic oxides |
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| 200. |
Interhalogen compounds can be used as I. non-aqueous solvents, II. Flurinating agents. The correct use(s) is/areA. Only IB. Only IIC. Both I and IID. Neither I nor II |
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Answer» Correct Answer - C Interhalogen compounds can be used as non-aqueous solvents and fluorinating agents . |
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