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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Classify the following solids in different based on the nature of intermolecular forces operating in them : Potassium sulphate, tin, benzene, urea, ammonia, water,zine sulphide, graphite, rubidium, argon, silicon carbide. |
| Answer» SOLUTION :Potassium sulphate = Ionic, Tin = Metallic , Benzene = Molecular ( non-polar) , Urea = Molecular ( Polar) , Ammonia =Molecular ( Hydrongen bonded ), water = Molecular ( Hydrongen bonded ) , Zine SULPHIDE = Ionic , Graphite = COVALENT or NETWORK , Rubidium = Metallic, Argon= Molecular( Non- Polar), SILICON carvbide = Covalent or Network. | |
| 2. |
Classify the following salts on the base of solubility. |
Answer» SOLUTION :The SALTS are classify in FOLLOWING THREE SECTION.
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| 3. |
Classify the following reactions in one of the reaction type studied in this unit (i) CH_(3)CH_(2)Br+SH^(-) rarr CH_(3)CH_(2)SH+Br^(-) (ii) (CH_(3))_(2)C=CH_(2)+HCl rarr (CH_(3))_(2)C(Cl)CH_(3) (iii) (CH_(3))_(3)C CH_(2)OH+HBr rarr (CH_(3))_(2)CBrCH_(2)CH_(3) (iv) CH_(3)CH_(2)Br+HO^(-) rarr CH_(2)=CH_(2)+H_(2)O+Br^(-) |
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Answer» Solution :(i) Nucleophilic substitution (II) Electrophilic addition (III) Rearrangement of carbocation INTERMEDIATE FORMED FOLLOWED by nuclephilic substitution. (iv) Elimination. |
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| 4. |
Classify the following reactions in one of the reaction type studied in this unit. (a) CH_(3)CH_(2)Br +HS^(-) rarr CH_(3)CH_(2)SH + Br^(-) (b) (CH_(3))_(2)C = CH_(2) + HCl rarr (CH_(3))_(2)C Cl - CH_(3) (c) CH_(3)CH_(2)Br + HO^(-) rarr CH_(2) + H_(2)O + Br^(-) (d) (CH_(3))_(3)C-CH_(2)OH + HBr rarr (CH_(3))_(2)CBrCH_(2)CH_(3) + H_(2)O |
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Answer» Solution :(a) NUCLEOPHILIC substitution (b) ELECTROPHILIC ADDITION (c) BIMOLECULAR elimination (d) Nucleophilic subtitution with rearrangement. |
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| 5. |
Classify the following salt solution in acid, base and neutral.NaCl, KNO_3 , FeCl_3 , CuSO_4 , CH_3COONa, HCOOK, CH_3COONH_4 , CrCl_3 , K_2SO_4 , Na_3PO_4 , NH_4Cl |
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Answer» Solution :Acidic solution : SALT of strong ACID-weak base. `FeCl_3 , CuSO_4 , CrCl_3 , NH_4Cl` Base solution : Weak acid-strong base `CH_3COONa, HCOOK, CH_3COONH_4, Na_3PO_4` NEUTRAL solution : Salt of strong acid among base `NACL , KNO_3 , K_2SO_4` |
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| 6. |
Classify the following reactions in one of the reaction type studied in this unit. (a) CH_(3)CH_(2)Br + HS^(-) rarr CH_(3)CH_(2)SH + Br^(-) (b) (CH_(3))_(2) C = CH_(2) + HCl rarr (CH_(3))_(2) ClC - CH_(3) (c ) CH_(3)CH_(2)Br + HO^(-) rarr CH_(2) = CH_(2) + H_(2)O + Br^(-) (d) (CH_(3))_(3)C - CH_(2)OH + HBr rarr (CH_(3))_(2) CBrCH_(2)CH_(2)CH_(3) + H_(2)O |
| Answer» Solution :(a) Nucleophilic SUBSTITUTION (b) Electrophilic ADDITION (c ) BIMOLECULAR elimination (d) Re ARRANGEMENT with Nucleophilic substitution | |
| 7. |
Classify the following reactions in one of the reaction type of organic reaction. (a) CH_(3)CH_(2)Br + S^(oplus)H to CH_(3)CH_(2)SH + B^(oplus)r (b) (CH_(3))_(2)C = CH_(2) + HCI to (CH_(3))_(2)CCI - CH_(3) (c) CH_(3)CH_(2)Br + O^(oplus)H to CH_(2) = CH_(2) + H_(2)O + B^(oplus)r |
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Answer» Solution :(a) NUCLEOPHILIC substitution reaction. (b) ELECTROPHILIC addition reaction. (c) ELIMINATION reaction. |
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| 8. |
Classify the following reactions as addition, elimination, substitution, condensation, rearrangement, geometric isomerisation, or oxidation/reduction. . |
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Answer» Solution :(a) and (B) Geometric isomerisation (c) Oxidative cleavage (d) ADDITION ( e) `SN^2` (f) Aldol condensation (g) REARRANGEMENT. |
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| 9. |
Classify the following reactions according to the reaction type : (a) CH_3 CH_2 Br + HS^(-) to CH_3 CH_2 SH+ Br^(-) (b) (CH_3)_2 C= CH_2 + HCI to (CH_5)_2 C (CI)-CH_3 (c) CH_3 CH_2 Br + HO^(-) to CH_2 = CH_2 (d) H_2 C= CH CH_2 CH_3 to H_3 C CH = CH CH_3 (e) CH_5 - underset(underset(C_5 H_5)|)C = NOH to CH_3 - underset(underset(NHC_6 H_5)|)C = O (f) (CH_3 )_3 CCI+ OH^(-) to (CH_3) _2 C = CH_2 + H_2 O + CI^(-) (g) (CH_3 )_3 C - CH_2 OH + HBr to (CH_3)_2 BrCH_2 CH_3 + H_2 O |
| Answer» Solution :`{:((a) CH_3 CH_2 BR + HO^(-), to , CH_3 CH_2 SH+ Br^(-),"Substitution reaction"),((b) (CH_3)_2 C = CH_2 + HS^(-),to,(CH_3)_2 C (CI)- CH_3,"Addition reaction"),((c) CH_3 CH_2 Br + OH^(-),to,CH_2 = CH_2,"ELIMINATION reaction"),((d) H_2 C = CH CH_2 CH_3,to,H_3 C CH = CH CH_3,"Isomerisation reaction"),((e) CH_3 underset(underset(C_6 H_5)|)C=NOH,to,CH_3 - underset(underset(NHC_6 H_5)|)C = O,"Rearrangement reaction"),((f) (CH_3)_3 CCI + OH^(-),to,(CH_2)_2 = CH_2 = H_2 O + CI^(-),"Elimination reaction."),((g) (CH_3)_3 C - CH_2 OH + HBr , to ,(CH_3)_2 CBr CH_2 CH_3 + H_2 O , "Substitution with rearrangement reaction" ):}` | |
| 10. |
Classify the following processes as reversible or irreversible : Mixing of two gases by diffusion. |
| Answer» SOLUTION :IRREVERSIBLE | |
| 11. |
Classify the following processes as reversible or irreversible : Melting of ice without rise in temperature. |
| Answer» SOLUTION :REVERSIBLE | |
| 12. |
Classify the following processes as reversible or irreversible : Evaporation of water at 373 K and 1 atm pressure. |
| Answer» SOLUTION :IRREVERSIBLE | |
| 13. |
Classify the following processes as reversible or irreversible : Dissolution of sodium chloride. |
| Answer» SOLUTION :IRREVERSIBLE | |
| 14. |
Classify the following pH containing solution in Acid, Base and Neutral.(a) 7.0 (b) 7.9 (c) 9.0 (d) 2.0 (e) 6.9 |
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Answer» Solution :(a)7.0 Neutral , (b)7.9 and (C) 9.0 basic `because` pH `gt` 7.0 (d) 2.0 and (E) 6.9 ACIDIC `because` pH `LT` 7.0 |
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| 15. |
Classify the following into metals and non-metals : (i) Helium (ii) Sodium (iii) Hydrogen (iv) Mercury (v) Graphite (vi) Carbon (vii) Lead (viii) Magnesium (ix) Chlorine (x) Phosphorus. |
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| 16. |
Classify the following into ionic, molecular , covalent and metallic crystals. Bronze, Dry ice, Nitre and Diamond |
| Answer» SOLUTION :Bronze=Metallic, DRY ice=Covalent, Nitre=Ionic, Diamond=covalent | |
| 17. |
Classify the following into ionic, molecular,cvalent and metallic crystals. Bronze, Dryice,Nitre and Diamond |
| Answer» Solution :Bronze = Metallice , DRY ICE = Coalent , NITRE = Ionic , DIAMOND = covalent | |
| 18. |
Classify the following into elements, compounds and mixtures (i) Marble (ii) Honey (iii) Touch paste (iv) Sugar (v) Gold (vi) Brass (vii) Nitre (viii) Slakes lime (ix) Nitrogen (x) Fruit juice. |
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| 19. |
Classify the following in carbocation, free radical and caranion. overset(.)(C )H_(3) overset(+)(C )H_(3): bar(C )H_(3) (CH_(3))_(3) overset(.)(C ), (CH_(3))_(3) overset(+)(C ), (CH_(3))_(2) overset(+)( C)H, (CH_(3))_(3) bar(C ): , (CH_(3))_(3) overset(..)(C )H, C_(6)H_(5) overset(.)(C )H_(2) CH_(3) overset(..)(C )H_(2), C_(6)H_(5) C overset(+)(H)_(2), CH_(2)= overset(.)(C )H_(2), CH_(2)= CH_(2)- overset(+)(C )H_(2) |
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Answer» Solution :CARBOCATION: `overset(+)(C )H_(3), (CH_(3))_(3) overset(+)(C ), (CH_(3))_(2) C overset(+)(H), C_(6)H_(5)C overset(+)(H)_(2), CH_(2)= CH - overset(+)(C )H_(2)` Free radical: `overset(.)(C )H_(3), (CH_(3))_(3) overset(.)(C ), C_(6)H_(5) overset(.)(C )H_(2), CH_(2)= CH- overset(.)(C )H_(2)` Caranion: `bar(C )H_(3), (CH_(3))_(3) bar(C ):. CH_(3) baroverset(..)(C )H_(2)` |
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| 20. |
Classify the following groups as ortho, para or meta directing when present on benzene nucleus. (i) -CH_(3),(ii)-NH_(2), (iii) -NO_(2), (iv)-COOH, (v) -OH, (vi) -SO_(3)H, (vii) -X("halogen"), (viii) OCH_(3), (ix) -CHO, (x) -CN (B) Classify the following goups as activating or deactivating with respect to further electrophilicsubstitution of the aromatic ring. (i) -NH_(2) (ii) -NO_(2) (iii) -SO_(3)H, (iv) -CH_(3) (v) -CI, (vi)-CN. ( C) For each of the following substituents indicate whether in donate electron or withdraw electron. |
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Answer» (i), (ii), (v), (vii) and (viii) meta-directing groups : (iii), (IV), (VI), (ix) and (x) Activating grups : (i) and (iv) Deacting groups : (ii), (iii), (v) and (vi) EDG : (ii), (v), and (viii) EWG : (i), (iii), (iv) and (vi) |
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| 21. |
Classify the following compounds on the basis of the oxidation number of xenon. XeF_(4), XeOF_(2), XeO_(2)F_(2), XeF_(6) |
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| 22. |
Classify the following compounds in the form of alkyl, allylie, vinyl, benzylic halides, (a) CH_(2)=CH-CH_(2)Cl (b) C_(6)H_(5)CH_(2)I (c) CH_(3)-underset(Br)underset(|)(CH)-CH_(3) (d) CH_(2)=CH-Cl |
| Answer» Solution :`CH_(3)-CH=CH-Cl` ALLYLIC halide (b) `C_(6)H_(5)CH_(2)` Benzylic halide (c) `CH_(3)-underset(Br)underset(|)(CH)-CH_(3)` Alkyl halide (d) `CH_(2)=CHCL` Vinyl halide | |
| 23. |
Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides CH_(3)-CH=CH-Cl |
| Answer» SOLUTION :`CH_(3)-CH=CH-Cl` | |
| 24. |
Classify the following compounds in the form of alkyl. allylic. vinyl. benzylic halides: (a) CH_3-CH = CH -Cl(b) C_6H_5CH_2I (c ) CH_3 - underset(Br)underset(|)CH - CH_3(d) CH_2 = CH - Cl |
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Answer» Solution :(a) `CH_3-CH = CH -Cl`Allylic halide (b) `C_6H_5CH_2I`Benzylic halide (c ) `CH_3 - UNDERSET(Br)underset(|)CH - CH_3`Alkyl halide (d) `CH_2 = CH - Cl`VINYL halide |
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| 25. |
Classify the following compounds in saturated, unsaturated and aromatic hydrocarbon. Ethane, Ethene, Ethyne, Propene, Propane, Benzene, Toluene, Anthracene, Cyclopropane. |
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Answer» Solution :(a) SATURATED hydrocarbon : ETHANE, Propane, CYCLOPROPANE (b) UNSATURATED hydrocarbon : Ethene, Ethyne, Propene. (c) Aromatc hydrocarbon : Benzene, Toluene, Anthracene |
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| 26. |
Classify the following compounds in Acyclic, Alicyclic, Benzenoid and non-benzenoid: Tropolone, Propane, Benzene, Butene, ethane, acetic acid, tetrahydrofuran, cyclohexene, Nepthalene, cyclobutane |
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Answer» Solution :Acyclic: Acitic acid, Ethane, Butene, Propane ALICYCLIC : Tetrahydrofouran, CYCLOHEXENE, CYCLOBUTANE BENZENOID: Nepyhelone, Benzene Nonbenzenoid: Tropolone |
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| 27. |
Classify the following as prue substance, homogeneous mixture, heterogeneous mixture, element and compound: (i) Milk, (ii) Air, (iii) Petrol (iv) distilled water (v) Common salt (vi) graphite (vii) Tap-water (viii) Smoke (ix) Dry ice (x) Cold drinks (xi) gun powder. |
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Answer» Solution :(i) Mixture (HOMOGENEOUS), (II) Mixuture (homogenous), (iii) Mixture (homogenous), (iv) COMPOUND, (V) mixture (heterogeneous), (vi) ELEMENT, (vii) mixture (homogeneous), (viii) mixture (heterogeneous), (ix) compound, (x) mixture (homogeneous), (xi) mixture (heterogeneous). |
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| 28. |
Classify the following as pure substances or mixtures. Separate the pure substances into elements, compounds and divide the mixtures into homogeneous and heterogeneous : (i) Air (ii) Milk (iii) Graphite (iv) Gasoline (v) Diamond (vi) Tap water (vii) Distilled water (viii) Oxygen (ix) Brass (x) 22 Carat gold (xi) Steel (xii) Iron (xiii) Sodium chloride (xiv) Iodised table salt. |
| Answer» SOLUTION :`{:("(i)","AIR",:,"Mixture (Homogeneous)"),("(ii)","Milk",:,"Mixture (Homogeneous)"),("(iii)","GRAPHITE",:,"Pure substance (Element)"),("(IV)","Gasoline",:,"Mixture (Homogeneous)"),("(v)","Diamond",:,"Pure substance (Element)"),("(vi)","Tap water",:,"Mixture (Homogeneous)"),("(vii)","Distilled water",:,"Pure substance (Compound)"),("(viii)","Oxygen",:,"Pure substance (Element)"),("(ix)","Brass",:,"Mixture (Homogeneous)"),("(x)","22 Carat gold",:,"Mixture (Homogeneous)"),("(xi)","Steel",:,"Mixture (Homogeneous)"),("(xii)","Iron",:,"Pure substance (Elemenet)"),("(xiii)","SODIUM chloride",:,"Pure substance (Compound)"),("(xiv)","Iodized table salt",:,"Mixture (Heterogeneous)"):}` | |
| 29. |
Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper. |
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Answer» Solution :Amorphous solids. POLYURETHANE, teflon. CELLOPHANE, polyvinyl CHLORIDE, fibre glass. CRYSTALLINE solids. Benzoic acid, potassium nitrate, COPPER |
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| 30. |
Classify the following as amorphous or crystalline solids:Polyurethane, naphthalene, benzoic acid , teflon , potassium niturate, cellophone, polyvinyl chloride , fibre glass, copper. |
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Answer» Solution :Amorphous solids, polyurethane, TEFLON, cellophone, POLYVINYL chloride, fibre GLASS, Crystalline solids:BENZOIC acid, potassium nitrate , copper. |
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| 31. |
Classify the followiing reactions in one of the reaction type studied in this unit . (CH_(3))_(3) C - CH_(2) OH + HB to (CH_(3))_(2) CBr - CH_(2) - CH_(3) + H_(2) O |
| Answer» SOLUTION :Nucleophilic SUBSTITUTION reaction with earrangement - ` overset (-)(O)H ` US SUBSTITUTED by Br . | |
| 32. |
Classify the followiing reactions in one of the reaction type studied in this unit . (CH_(3))_(2) C = CH_(2) + HCl to (CH_(3))_(2) C C l - CH_(3) |
| Answer» SOLUTION :ELECTROPHILIC addition reaction - HCL is ADDED to the C = C bond . | |
| 33. |
Classify the followiing reactions in one of the reaction type studied in this unit . CH_(3) CH_(2) Br + H overset (-) S to CH_(3) CH_(2) SH + Br |
| Answer» SOLUTION :Nucleophilic SUBSTITUTION REACTION - BR is SUBSTITUTED by ` H overset (-) S ` . | |
| 34. |
Classify the followiing reactions in one of the reaction type studied in this unit . CH_(3) - CH_(2) Br + H overset (-)O to CH_(2) = CH_(2) + H_(2) O + Br |
| Answer» SOLUTION :` beta `-ELIMINATION reaction -H and Br are ELIMINATED from successive carbon atoms . | |
| 35. |
Classify the elementshaving atomicnumberas givenbelowinto threeseparate pairs onthe basisof similarchemicalproperties: 9, 12, 16 , 34 , 53 , 56 Givetheirbriefelectronicexplanation. |
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Answer» Solution :The secondperiodends atatomic number10while the thirdperiodends atatomicnumber 18. Therefore9, 12and 16are the firstelementsin theirrespective GROUPS. Theatomicnumber ofthe otherelementsofthe same groupscan bededuced byadding magic numbers of 8 ,18 ,18and 32to the atomicnumbersoftheelementsof 2ndperiodor byaddingmagic NUMBER of 18, 18 and 32to theelementsof 3rdperiod.Thus `9 + 8 + 18+ 18 = 53` `12 + 8+ 18+ 18 =56` ` 16 + 18=34` elements withatomicnumbers9 ( F )and 53 ( I ) belongto halogenfamily (Group 17) , elements withatomic numbers 12(Mg )and 56 (Ba )belongto alkaline earthmetals(Group 2 ) while elementswithatomicnumbers16(S)and 34 (Se )belong to oxygenfamily(Group 16 ) Forelectronicconfigurationarefer to thetext. |
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| 36. |
Classify the element bromine, in different methods. |
| Answer» SOLUTION :BROMINE is the element present in GROUP 17 and period 4. It is a p-block element. It is liquid element at room temperature. It is a non-metal. | |
| 37. |
Classify matter according to physical state. |
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Answer» (i) SOLID(ii) LIQUID(iii) Gas The constituent particles of matter in these three states can be represented as shown in the figure.  Figure : ARRANGEMENT of particles in solid, liquid and gaseous state In solids, these particles are held very close to each other in an orderly fashion and there is not much freedom of movement. In liquids, the particles are close to each other but they can move around. However, in gases, the particles are far apart as compared to those present in solid or liquid states and their movement is EASY and fast. |
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| 38. |
Classify in acids and bases of following CO_2, BCl_3, NH_3, CH_3NH_2 , NO_2^(+) and C_6H_5NH_3^(+) |
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Answer» SOLUTION :ACID: `CO_2, BCl_3 , NO_2^(+ ) , C_6H_5NH_3^(+)` BASE : `NH_3` and `CH_3NH_2` |
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| 39. |
Classify following substances in non electrolyte, strong electrolytes and weak electrolytes. (i)NaCl , (ii) Sugar , (iii) Glucose , (iv)CH_3COOH , (v)CH_3COONa , (vi)HCl , (vii)HNO_3 , (viii)NH_3 , (ix)NaOH |
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Answer» Solution :NON electrolyte : (II), (iii) Strong electrolyte : (i), (V), (vii), (VI), (ix) WEAK electrolyte : (iv), (viii) |
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| 40. |
Classify following oxides as neutral, acidic, basic or amphoteric : CO,B_2O_3 , SiO_2, CO_2 , Al_2O_3 , PbO_2, Tl_2O_3 |
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Answer» Solution :Neutral OXIDE : CO ACIDIC oxide : `B_2O_3 , SiO_2 ,CO_2` AMPHOTERIC oxide : `Al_2O_3 , PbO_2` BASIC oxide: `Tl_2O_3` |
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| 41. |
Classify following oxides as neutral, acidic, basic or amphoteric: CO, B_2 O_3 , SiO_2, CO_2 , Al_2 O_3 , PbO_2 , Tl_2 O_3 |
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Answer» Solution :Neutral OXIDES : CO : ACIDIC oxides : `B_2 O_3 , SiO_2, CO_2` , Amphoteric oxides: `Al_2 O_3 , PbO_2 ` , BASIC oxides : `Tl_2 O_3`. |
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| 42. |
Classify following in strong and weak base. NaOH, KOH, Mg(OH)_2, Cu(OH)_2, Al(OH)_3, CsOH, Ba(OH)_2, Ca(OH)_2 |
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Answer» Solution :STRONG Base :NaOH, KOH, CSOH, `Ba(OH)_2` WEAK Base : `Mg(OH)_2 , Cu(OH)_2 , Al(OH)_3, Ca(OH)_2` |
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| 43. |
Classify following in strong and weak acid. HClO_4, HCN, HI, HBr, CH_3COOH, H_2S , H_3PO_4, HNO_3, HNO_2 |
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Answer» SOLUTION :Strong Acid: `HClO_4, HNO_3,HBr` WEAK Acid: `HCN, HI, CH_3COOH, H_2S, H_3PO_4, HNO_2` |
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| 44. |
Classify following in acid, base and salt. HCl, NaOH, NH_3, Na_2CO_3, CH_3COOH, HCOOH, NaCl , KOH, BaSO_4, NaNO_3, HF, H_2SO_4, HClO_4, Calcium hydroxide , orange juice , nitrous acid , HCN, NaCN , NH_4OH , C_6H_5NH_2, CH_3NH_2, CO(NH_2)_2, Sucrose CH_3COONa , NH_4Cl, CH_3COONH_4, Zn_3(PO_4)_2, H_3PO_4 etc. |
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Answer» Solution :Acid : HCl, `CH_3COOH`,HCOOH, HF, `H_2SO_4 , HClO_4`, Orange juice , Nitrous acid, HCN, `H_3PO_4` Base: `NaOH, NH_3 , KOH` , CALCIUM hydroxide, `NH_4OH, C_6H_5NH_2 , CH_3NH_2` Salt: `Na_2CO_3 , NaCl , BaSO_4 , NaNO_3 , NACN , CH_3COONa, NH_4Cl , CH_3COONH_4, Zn_3(PO_4)_2` Other: `CO(NH_2)_2`, Sucrose |
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| 45. |
Classify following: (i) Give classification in Nucleophilic and Electrophilic overset(+)(N) O_(2), bar(O)H, CH_(3) overset(..)(N)H_(2), overset(..)(N)H_(3), Br^(+), CH_(3) - underset(underset(O)(||))(C )- CH_(3) |
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Answer» Solution :Nucleophilic : `BAR(O)H, CH_(3) overset(..)(N)H_(2), overset(..)(N)H_(3)` Electrophilic: `overset(+)(N)O_(2), Br^(+), CH_(3) - underset(underset(O)(||))(C )- CH_(3)` |
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| 46. |
Classify each of the following substances into an acid or base or both and mention the concept/concepts on the basis of which you can do so. (i) HCl(aq) " " (ii) NH_(3)(g) " " (iii) Na_(2)CO_(3)(aq) " " (iv)CH_(3)CO OH(aq) (v)CO_(2)(g) " " (vi)BF_(3) " " (vii)Ag^(+) " " (viii)CN^(-) (ix)H_(2)O " " (x)H_(2)SO_(4) "(xi)" HCO_(3)^(-) "(xii)"SiF_(4) |
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Answer» Solution :(i) HCL (aq) - Acid (ARRHENIUS conecpt and Bronsted - LOWRY concept) (ii) `NH_(3)(g) `-Base (Bronsted concept and Lewis concept) (iii) `Na_(2)CO_(3)(aq)`- Base (Bronsted concept) (iv) `CH_(3)CO OH(aq)`-Acid (Arrhenius concept and Bronsted concept) (V) `CO_(2)(g)`-Acid (Bronsted concept and Lewis concept) (vi) `BF_(3)`-Acid (Lewis concept) (vii) `Ag^(+)`- Acid (Lewis concept) (viii) `CN^(-)`- Base (Lewis concept) (ix) `H_(2)O`- Both acid and base, i.e., amphoteric (Bronsted concept) (x) `H_(2)SO_(4)`-Both acid and base, i.e., amphoteric (Bronsted concept) (xi) `CHO_(3)^(-)`-Both acid and base, i.e., amphoteric (Bronsted concept) (xii) `SiF_(4) ` - Acid (Lewis concept), as silicon can expand its octet . |
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| 47. |
Classify each of the following solids as lonic, metallic , molecular, network( convalent) or amorphous: (i) Tetraphosophours decoxide( P_(4) O_(10))(ii) Ammonium phosphate,(NH_(4))_(3) PO_(4)(iii) SIC(iv) I_(2) (v)P_(4)(vi) plastics (vii) Graphite(viii) Brass (ix)Rb (x) LiBr(xi) Si |
| Answer» SOLUTION :IONIC = `( NH_(4))_(3) PO_(4)` and LiBr , Metallic = Brass , Rb , Molecular = ` P_(4)O_(10), I_(2), P_(4)` ,Network (covalent) = Graphite ,SIC, Si , Amorphous = Plastics . | |
| 48. |
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous (i) Tetraphosphorus decoxide (P_4O_10) , (ii) Ammonium phosphate, (NH_4)_3PO_4 ,(iii)SiC ,(iv)I_2 , (v) P_4 , (vi) Plastics ,(vii) Graphite ,(viii) Brass ,(ix) Rb ,(x) LiBr ,(xi) Si |
| Answer» Solution :Ionic =`(NH_4)_3PO_4` and LIBR , Metallic =Brass, Rb , MOLECULAR =`P_4O_10, I_2, P_4` , Network (covalent)=Graphite , SIC, Si, AMORPHOUS =Plastics | |
| 49. |
Classify each of the following as being either a p-type or n-type semiconductor: (i) Ge doped with In (ii) B doped with Si. |
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Answer» Solution :(i)Ge is Group 14 element and In is Group 13 element. Hence, an electron deficit HOLE CREATED and therefore, it is p-type. (ii) B is Group 13 element and Si is Group 14 element, there will be a free electron. Hence, it is n-type |
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| 50. |
Classify each of the following as being either a p-type or n-typesemiconductor : (i)Ge doped with in (ii) B doped with Si. |
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Answer» Solution :(i) Ge is Group 14 element and In is Group 13 element. Hence, an electron deficit HOLE is carrated and therefore, it is p-type. (II)B is Group 13 element and Si is Group 14 element , there will be a FREE electron.Hence , it is n -type |
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