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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.
| 251. |
Excess of conc. `H_(2)SO_(4)` reacts with `C_(2)H_(2)OH` at 413 K to form.A. `C_(2)H_(5)HSO_(4)`B. `(C_(2)H_(5))_(2)SO_(4)`C. `CH_(2)=CH_(2)`D. `C_(2)H_(6)OC_(2)H_(5)` |
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Answer» Correct Answer - C Alcohol is dehydrated to give alkene. It is a `beta-"elimination"` `CH_(3)-CH_(2)-OH underset(-H_(2)O)overset(H_(2)SO_(4).443K)to CH_(2)=CH_(2)` Note: At 413K, ethyl ether is obtained `C_(2_)H_(5)OH +HCO_(2)H_(5) underset(-H_(2)O) overset(H_(2)SO_(4).443K)to C_(2)H_(5)OC_(2)H_(5)` |
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| 252. |
The reaction of Cu with hot conc. `H_(2)SO_(4)` producA. `SO_(2)`B. `H_(2)S`C. `H_(2)`D. `Cu^(+)`ions |
| Answer» Correct Answer - A | |
| 253. |
On heating `NaCl+K_(2)CrO_(7)+conc.H_(2)SO_(4)`, the gas comes out isA. `O_(2)`B. `Cl_(2)`C. `CrOCl_(2)`D. `CrO_(2)Cl_(2)` |
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Answer» Correct Answer - D `CrO_(2)Cl_(2)` is a strong red gas. |
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| 254. |
Which of the following is magnesium bismuthide ?A. `Bi_(3)Mg`B. `Bi_(3)Mg_(2)`C. `Mg_(3)Bi_(2)`D. `Mg_(3)Bi` |
| Answer» Correct Answer - C | |
| 255. |
Phosphoric acid is prepared byA. hydrolysis of `P_2O_3`B. hydrolysis of `P_4O_10`C. hydrolysis of `PCI_3`D. action of `PCl_(3)` on `H_3PO_3` |
| Answer» Correct Answer - B | |
| 256. |
`H_3PO_4` consist ofA. two P-OH bonds and 2 P=O bondsB. one P-OH bond and 3 P =O bondsC. 2 P-OH bonds and one P-O bondD. 3 P-OH bonds and one P=O bond |
| Answer» Correct Answer - D | |
| 257. |
Which is incorrect statement for `XeF_2?`A. It has linear structureB. It is hydrolysed rapidly in aqueous solution of a baseC. It oxidizes `CI^(-) and I^(-)" to "Cl_2 and I_2` respectivelyD. It cannot act as `F^(-)` donor. |
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Answer» Correct Answer - D It act as a `F^(-)` donor and forms complexes with covalent fluroides like `PF_(6), AsF_(6) and SbF_(5)`. `XeF_(2)+PF_(5) to [XeF]^(+) [PF_(6)]^(-)` (a) The five electron pairs form trigonal bipyramidal with three lone pairs in equatoríal positions. Hence, the structure of `XeF_(2)`, becomes linear in accordance with VSEPR theory. |
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| 258. |
The statement, which prompted Neil Bartlett to prepare the first noble gas compound wasA. Xe-F bond has high bond energyB. `F_2` has exceptionally low bond energyC. `PtF_6` is a strong oxidantD. `O_2` molecule and Xe atom have very similar ionization energies. |
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Answer» Correct Answer - D Both Xe and `O_(2)`, have comparable ionization energies. `Xe to Xe^(-)+e^(-)` lonization energy =1, 170 kJ /mol `O_(2) to O_(2)^(+)+e^(-)`l Ionization energy =1,175 kJ / mol |
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| 259. |
The passivity of iron results due to the formation of a thin protective layer ofA. iron oxideB. ferric hydroxideC. `Fe(NO_(3))_(3)`D. `Fe_(2)O_(4)` |
| Answer» Correct Answer - A | |
| 260. |
Why does `AlCl_(3)` fumes in air? |
| Answer» Correct Answer - A::B::C | |
| 261. |
Which of the following is the correct statement for red lead?A. It is an active form of leadB. Its molecular formula is `Pb_(2) O_(3)`C. It decompose into `PbO` and `CO_(2)`D. It decompose into `PbO` and `O_(2)` |
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Answer» Correct Answer - D `2Pb_(3)O_(4) overset (gt 550^(@)C) to 6PbO + O_(2) uarr` |
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| 262. |
Aluminium is a noble metalA. aluminium is a nobel metalB. oxygen forms a protective oxide layer on aluminium surfaceC. iron unergoes reaction easily with waterD. iron forms both mono and divalent ions |
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Answer» Correct Answer - B `2Al+3H_(2)OtoAl_(2)O_(3)+3H_(2)O` (slow reaction) `Al_(2)O_(3)` forms protecting oxide layer on `Al` surface |
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| 263. |
Aluminium does not react with:A. `NaOH`B. conc. `HCI`C. `N_(2)`D. conc.`HNO_(3)` |
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Answer» Correct Answer - D As it becomes passive by the action of conc.`HNO_(3)` forming a protective oxide layer on the surface. |
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| 264. |
Red lead is:A. `PbO`B. `PbO_(2)`C. `Pb_(3)O_(4)`D. `Pb_(2)O_(3)` |
| Answer» Correct Answer - C | |
| 265. |
Pick out the incorrect statementA. `PH_4^(+)` ion is tetrahedral like the `NH_(4)^(+)` ion and is obtained when `PH_3` is bonded to protonB. `PH_4I` is one of the most stable salts containing the phosphonium ion. It is also more stable than ammonium saltsC. `PH_4I` is decomposed by water to form `PH_3`D. `PH_3` converts silver salts in solution to silver phosphide, which subsequently reacts to give free metal. |
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Answer» Correct Answer - B Phosphonium iodide is less thermally stable than ammonium salts, because phophine is less basic than ammonia and decomposition occurs at `60^(@)C`. `PH_(4)I to PH_3+HI` `a. PH_(3)+HI to PH_(3)I` b. `PH_(4)I+H_(2)O to PH_(3)+H_(3)O^(+)+I^(-)` d. `6Ag^(+)+PH_(3)+3H_(2) to 6Ag+H_(3)PO_(3)+6H^(+)` |
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| 266. |
Which of the following gives `PH_3` on treatment with water ?A. `Ca_3P_2`B. `Na_3P`C. AIPD. All of the above |
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Answer» Correct Answer - D All phosphides on treatment with water forms `PH_(3)`. `Ca_3P_(2)+6HOH to 2PH_(3)+3Ca(OH)_(2)` `Na_(3)P+3HOH to PH_(3)+3NaOH` `AIP+3HOH to PH_(3)+Al(OH)_(3)` |
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| 267. |
Ammonium dichromate on heating givesA. 3 and 3B. 4 and 4C. 3 and 4D. 4 and 3 |
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Answer» Correct Answer - D `(NH_(4))_(2)Cr_(2)O_(2) overset(Delta)to N_(2)+ underset(("Green"))(Cr_(2)O_(3))+4H_(2)O` |
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| 268. |
Number of P=O bonds is `P_4O_6`, molecule isA. 1B. 2C. 4D. Nil |
| Answer» Correct Answer - D | |
| 269. |
The acid obtained when `P_4O_6` reacts with water isA. `H_(3)PO_(2)`B. `H_(3)PO_(3)`C. `H_(3)PO_(4)`D. `H_(4)P_(2)O_(7)` |
| Answer» Correct Answer - B | |
| 270. |
Which of the following gives `N_2(g)` on heating ?A. `NH_(4)NO_(2)`B. `NH_(4)NO_(3)`C. `(NH_(4))_(2)Cr_(2)O_(7)`D. Both a and c |
| Answer» Correct Answer - D | |
| 271. |
Phosphine gives black precipitate withA. NaClB. `Cl_(2)`C. `AlCl_(3)`D. `CuSO_(4)` |
| Answer» Correct Answer - D | |
| 272. |
The nature of phosphine isA. acidicB. basicC. neutralD. amphoteric |
| Answer» Correct Answer - B | |
| 273. |
Which of the following acid does not contain +5 oxidation state ?A. Orthophosphoric acidB. Pyrophosphoric acidC. Phosphorus acidD. Metaphosphoric acid |
| Answer» Correct Answer - C | |
| 274. |
Mercuric chloride reacts with phosphine giveA. `Hg_(3)P_(2)`B. `HgP_(2)`C. `Hg_(3)P`D. `HgP` |
| Answer» Correct Answer - A | |
| 275. |
Which of the followinf gives dence fumes of `P_(4)O_(10)`A. `PCl_(5)`B. `PCl_(3)`C. White phosphorousD. Red phosphorous |
| Answer» Correct Answer - C | |
| 276. |
Which of the following is more acidic?A. `H_(4)P_(2)O_(5)`B. `H_(4)P_(2)O_(6)`C. `H_(4)P_(2)O_7`D. `H_(3)P_(3)O_(9)` |
| Answer» Correct Answer - C | |
| 277. |
Two -OH bonds are present inA. `H_(4)P_(2)O_(5)`B. `H_(4)P_(2)O_(6)`C. `H_(3)PO_(4)`D. `H_(4)P_(2)O_(7)` |
| Answer» Correct Answer - A | |
| 278. |
Which of the following salts is used in the bead test for basic radicals ?A. `Na(NH_(4))HPO_(4).4H_(2)O`B. `Na_(2)HPO_(4)`C. `(NH_(4))_(2)SO_(4).FeSO_(4).6H_(2)O`D. `(NH_(3))_(2)HPO_4.4H_2O` |
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Answer» Correct Answer - A Microcosmic salt on heating produces a transparent bead of sodium metaphosphate `(NaPO_3).` `Na(NH_(4))HPO_(4).4H_(2) overset(Delta)to NH_(3)+5H_(2)O+NaPO_(3)` `NaPO_(3)` combines with certain metallic oxides to form coloured beads of the corresponding orthophosphates, e.g., `NaPO_(3)+CuO to underset(("Blue"))(CuNaPO_(4))` Hence , microcosmic salt is sometimes used in place of borax for performing borax bead test. |
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| 279. |
Borazole `(B_(3)N_(3)H_(6))` is related to benzene asA. isostructuralB. isoelectronicC. both (1) and (2)D. none |
| Answer» Correct Answer - C | |
| 280. |
The hybridization states of boron and nitrogen in borazole respectivelyA. sp, spB. sp, `sp^(2)`C. `sp^(2),sp^(2)`D. `sp^(2),sp^(3)` |
| Answer» Correct Answer - C | |
| 281. |
Which of the allotropes of sulphur has no sharp melting point and is insoluble in `CS_(2)` ?A. Rhombic sulphurB. Monoclinic sulphurC. Plastic sulphurD. None of the above |
| Answer» Correct Answer - B | |
| 282. |
Sulphur is readily soluble inA. waterB. alcoholC. etherD. `CS_(2)` |
| Answer» Correct Answer - D | |
| 283. |
The `NH_(4)^(+)` ion isA. square planarB. tetrahedralC. a Lewis baseD. planar in solution |
| Answer» Correct Answer - B | |
| 284. |
The formula chloroplatic acidA. `PtCl_(4)`B. `H_(2)PtCl_(6)`C. `HPtCl_(5)`D. `HClO_(2)` |
| Answer» Correct Answer - B | |
| 285. |
In nitrogen cycle, the main products areA. `NO and NO_(2)`B. `NO_(2) and N_(2)O_(3)`C. `NO and N_(2)O_(3)`D. `N_(2)O_(3) and N_(2)O_(5)` |
| Answer» Correct Answer - A | |
| 286. |
Thermodynamically the most stable form of carbon isA. DiamondB. GraphiteC. fullerenesD. Coal |
| Answer» Correct Answer - B | |
| 287. |
Hydrazoic acid isA. `H_(2)NOH`B. `N_(3)H`C. `NH_(2)NH_(2)`D. `C_(6)H_(5)-N=N-OH` |
| Answer» Correct Answer - B | |
| 288. |
Phosphorus has the oxidation state of + 3 inA. orthophosphoric acidB. hypophosphoric acidC. metaphosphoric acidD. orthophosphorus acid |
| Answer» Correct Answer - D | |
| 289. |
Sulphir combines with.A. MgB. AuC. PtD. Te |
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Answer» Correct Answer - A Mg is most reactive of these metals. |
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| 290. |
Which of the following is the strongest oxidant?A. `F_(2)`B. `Br_(2)`C. `Cl_(2)`D. `I_(2)` |
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Answer» Correct Answer - A `F_2` is strongest oxidant, because i) `F_2` has exceptionally low dissociation energy of F-F bond. ii) `F2` has high energy of hydration arising from the smaller size of the F ion |
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| 291. |
The strongest acid isA. `HNO_(2)`B. `HNO_(3)`C. `H_(2)N_(2)O_(2)`D. `HNOS` |
| Answer» Correct Answer - B | |
| 292. |
When `SO_2` is passed through an aqueous solution of `I_(2)`, it becomes colourless. This is due to ingA. bleaching action of `SO_2`B. formation of HIC. combination of `SO_2 and I_(2)`D. formation of `HNO_3` |
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Answer» Correct Answer - C `SO_(2)+2H_(2)O to H_(2)SO_(4)+2H` `I_(2)+2H to 2HI` =`SO_(2)+2H_(2)O+I_(2) to H_(2)SO_(4)+2HI` |
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| 293. |
`SO_(2)`, reacts with `Cl_(2)`, to formA. `SOCl_(2)`B. `SO_(2)Cl_(2)`C. `SCl_(2)`D. `S_(2)Cl_(2)` |
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Answer» Correct Answer - B `SO_(2)+Cl_(2) to SO_(2)Cl_(2)` |
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| 294. |
Which of the following is the strongest oxidant?A. `F_2`B. `Cl_(2)`C. `Br_(2)`D. `I_(2)` |
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Answer» Correct Answer - A `F_2` is strongest oxidant, because (i) `F_2` has exceptionally low dissociation energy of F-F bond, (ii) `F_2` has high energy of hydration arising from the smaller size of the `F^(-)` ion. |
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| 295. |
In the following reaction oxidation state of fluorine changes to `2F_(2)+H_(2)O to 4HF+O_(2)`A. 0 to -01B. 0 to +1C. `-1 "to "0`D. `+1" to "0` |
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Answer» Correct Answer - A In `F_2`, fluorine has O.S. zero. In HF, fluorine has O.S.-1. |
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| 296. |
Which member of the halogen family (X) does not show oxidation state of +1?A. FluorineB. BromineC. ChlorineD. Iodine |
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Answer» Correct Answer - A Since fluorine is most electronegative element, so it always shows the oxidation state of (-I). With CI, Br and I the oxidation state may be positíve or negative, depending on which atom in the molecule has greater electronegativity. Covalence of one is most common for CI, Br and I, CI, Br and I also exhibit higher valencies, with oxidation states of +1, +3, +5 and +7. |
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| 297. |
Which of the following is a suboxide ?A. `CsO_(2)`B. `MnO_(2)`C. `C_(3)O_(2)`D. `Fe_(3)O_(4)` |
| Answer» Correct Answer - C | |
| 298. |
Sulphur reacts with `HNO_(3)` to formA. `H_(2)SO_(3)`B. `H_(2)SO_(4)`C. `SO_(2)`D. `SO_(3)` |
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Answer» Correct Answer - B `2HI+H_(2)SO_(4) to 6NO_(2)+H_(2)SO_(4)+2H_(2)O` `S+H_(2)O+3O to H_(2)SO_(4)` =`6HNO_(3)+S to 6NO_(2)+H_(2)SO_(4)+2H_(2)O` |
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| 299. |
Which is the best reducing agent ?A. `F^(-)`B. `I_(2)`C. `Br_(2)`D. `I^(-)` |
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Answer» Correct Answer - D lodine is strongest reducing agent than fluorine, chlorine and bormine. Due to an excess of electrons to the iodine anion acts as a stronger reducing agent then neutral atom. |
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| 300. |
Charge distribution in iodine monochloride is best represented asA. `I^(-) Cl^(+)`B. `I^(delta)+C^(delta-)`C. `I^(+) Cl^(-)`D. `I^(delta-) Cl^(delta-)` |
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Answer» Correct Answer - C Cl is more electronegative than I atom, therefore, Cl atom attains slight negatives charge and I atom acquires slight positive charge. |
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