Saved Bookmarks
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. |
Diethylether and n-propyl methylether are |
|
Answer» Metamers |
|
| 2. |
Diethyl ether is mostly used in solvent extraction due to the following reasons |
|
Answer» Its solvastion CAPACITY is very high (B) Being inert, it does not react with most of the organic compounds (D) Its boiling point is low therefore, it can be easily separated by distillation. |
|
| 3. |
Dienes are of three types. Cumulated, conjugated and isolated diene. A,B and C are three isomeric pentadienes. They differ in their energies which is plotted with their product of hydrogenation. Study the graph and answer the following questions. How many open chain structural isomers are possible for Molecular formula C_(5)H_(8) other than these three pentadienes mentioned above. |
|
Answer» 6 |
|
| 4. |
Dienses are the anem given to compoundswith ……………. |
| Answer» SOLUTION :exactlytwo dublebond | |
| 5. |
Dienes are of three types. Cumulated, conjugated and isolated diene. A,B and C are three isomeric pentadienes. They differ in their energies which is plotted with their product of hydrogenation. Study the graph and answer the following questions. Which one of the following is having two pi bonds on the same carbon. |
|
Answer» A 
|
|
| 6. |
Dienes are of three types. Cumulated, conjugated and isolated diene. A,B and C are three isomeric pentadienes. They differ in their energies which is plotted with their product of hydrogenation. Study the graph and answer the following questions. Which of the following is having maximum sp^(2) carbon atoms in line of after the other in the parent carbon chain. |
|
Answer» A |
|
| 7. |
Diels - Alder reaction will not take place with which of the following reactants? |
|
Answer»
|
|
| 8. |
Diels -Alder reaction is used to synthesise a ring of |
|
Answer» four CARBON atoms |
|
| 9. |
Dichromate ion in aqueous acidic medium reacts with ferrous ion give ferric and chormium ions write th balanced chemical equation corresponding to the reaction |
|
Answer» Solution :Write the skeletal equation of the GIVEN reaction `Cr_(2)O_(7)^(2-)(aq)+Fe^(2+)(aq)rarrCr^(3+)(aq)+Fe^(3+)(aq)` Step 2 Write the O.N of all the elements above their respective symbols  Step 3 Find out hte speciedwhihc hve been oxidised nad reduced and spilt the given skeltal equation in to two HALF reactions Since the O.N of Cr decreases form + 6 `Cr_(2)O_(7)^(2-)` to +3 `Cr^(3+)` and that of Fe increases form + 2 in `Fe^(2+) to +3 in Fe^(3+)` therefreo `Cr_(2)O_(7)^(2-)` gets reduced while `Fe^(2+)` gets oxidised Thus the above sketetal eqation (i) can be DIVIDED in to the following two half reaction equation Oxidation half equation`Fe^(2)+(a) rarr Fe^(3+) (q)` Reduction half equation : `Cr_(2)O_(7)^(2-)(aq) rarr Cr^(3+)(aq)` Step 4 To balance oxidaiton half equation (ii) (a) Balacne all atoms other than O and H Not neeeded because Fe by adding electrons Step 5 To balance the reducation half eqaution (iii) (a)Balance all atoms other thanH and O Since there ar two Cr atoms in `Cr_(2)O_(7)^(2-)` on the L.H.S of Eq (iii) and only on eon R.H.S therefoere multiple `Cr^(3+)` by 2 we have `Cr_(2)O_(7)^(2-)(aq) rarr 2 Cr^(3+)(aq)` (b)Balacne the O.N by adding elctrons Balance charge by adding `H^(+)` ions sice hte reaction occurs in the acidic medium The total charge on L.H.S of Eq (vi) is -8 while on the R.H.S it is +6 Therefore add 14 `H^(+)`to L.H.S of Eq (vi) we have (d) Balance O atoms by adding `H_(2)O` molecules since there are seven O atoms on the L.H.S of Eq (vii) but no O atom on the R.H.S therefore and 7 `H_(2)O` to the R.H.S of Eq (vii) we have the H atoms get automatically balanced Thus Eq (viii) represent the balaced reduction half equation Step 6 To balance the electrons lost in Eq (vi) and gained in Eq (viii) multiply Eq (vi) by 6 and add to Eq (viii) We have `6FE^(2+)(aq)rarr6Fe^(3+)+6 e^(-)` `Cr_(2)O_(7)^(2-)(aq)+6 Fe^(2+)(aq)+14^(+)(aq) rarr 2 Cr^(3+)(aq)+6Fe^(3+)(aq)+7 H_(2)O(l)` This gives the final balanced redox equation Step 7 Verification Total charge on L.H.S of Eq (ix) =-2+6(+2)+14(+1)=+24 Total charge on R.H.S of Eq (ix) =2(+3)+6(+3)=+24 Since the magnituede if charge on both sides of eq (ix) equal therefore eq (ix) represent the correct balaced redox equation |
|
| 10. |
Dichlorination of n-butane gives _________number of structural isomers |
|
Answer» |
|
| 11. |
Dichlorination of cyclohexane _______ number of isomers including stereoisomers |
|
Answer» |
|
| 12. |
Diborane react with ammonia under different conditions to give a variety of products. Which one among the following is not formed in these reactions |
|
Answer» `B_(2)H_(6).2NH_3` `3B_(2)H_(6)+6NH_(3)overset(200^@C)rarr2B_(3)N_(3)H_(6)+12H_(2)` `B_(2)H_(6)+NH_(3)overset("very high temp")rarr(BN)_n` `:. B_(12)H_(12)` is not formed |
|
| 13. |
Diborane reacts with carbon monoxide to form |
|
Answer» `BH_3CO` |
|
| 14. |
Diborane on hydrolysis gives |
|
Answer» `BC_3` |
|
| 15. |
Diborane molecule has six hydrogen atoms, but all atoms cannot be substituted in methylation. Why? |
|
Answer» SOLUTION :DIBORANE has two planar `-BH_(2)` groups. Thus it has only four terminal hydrogen atoms. The other two hydrogen atoms are involved in the hydrogen bridge bonds. HENCE they cannot be replaced in methylation Methylation of diborane is limited to tetramethyl DERIVATIVE. Diborane does not form pentamethyl or hexamethyi derivatives. |
|
| 16. |
The number of atoms involved in bridged bonds in one diborane molecule is |
|
Answer» 2 |
|
| 17. |
Diborane is simplest and most familiar hydride of boton. Its chemical formula is B_(2)H_(6). Comparing with ethane, diborane is regarded as electron deficient molecule. In the excited state boron of diborane undergoes sp^(3) hybridisation. Bonding in doborane is described as tricentric two electron bonding.Number of empty sp^(3) hybrid orbitals of each ''B'' atom in B_(2)H_(6) Each B contains one sp^(3) vacant hybrid orbital |
|
Answer» 1  Each B CONTAINS ONE `sp^3`VACANT HYBRID or bital |
|
| 18. |
Diborane is simplest and most familiar hydride of boton. Its chemical formula is B_(2)H_(6). Comparing with ethane, diborane is regarded as electron deficient molecule. In the excited state boron of diborane undergoes sp^(3) hybridisation. Bonding in doborane is described as tricentric two electron bonding.Structure of -BH_(2) group is |
|
Answer» LINEAR |
|
| 19. |
Diborane is a potential rocket fuel which undergoes combustion accordingto the reaction: B_(2)H_(6)(g) + 3O_(2)(g) rarr B_(2)O_(3) (s) + 3H_(2)O(s) From thefollowing data , calculate the enthalpy change for the combustion of diborane. 2B(s) +(3)/(2) O_(2) (g) rarr B_(2)O_(3)(s) , DeltaH= - 1273kJ mol^(-1)......(i) H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l) , DeltaH = - 286kJ mol^(-1)....(ii)H_(2)O(l) rarr H_(2)O(g), Delta H= 44 kJ mol^(-1)....(iii)2B (s) + 3H_(2)(g) rarr B_(2)H_(6)(g), DeltaH = - 36 kJ mol^(-1).....(iv) |
|
Answer» |
|
| 20. |
Diborane is a potential rocket fuel which undergoes combustion according to the reaction: B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g) From the following data, calculate the enthalpy change for the combustion of diborane. 2B(s) + 3/2O_(2)(g) to B_(2)O_(3)(s) DeltaH = -1273 kJ"mol"^(-1) H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l)DeltaH =-286 kJ"mol"^(-1) H_(2)O(l) to H_(2)O(g)DeltaH = 44 kJ"mol"^(-1) 2B(s) + 3H_(2)(g) to B_(2)H_(6)(g)DeltaH = 36 kJ"mol"^(-1). |
|
Answer» SOLUTION :The EQUATION is : `B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g) DeltaH = ?` (i)`3B(s) + 3/2 O_(2)(g)to B_(2)O_(3)(s)DeltaH = -1273 kJ"mol"^(-1)` (ii)` H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l)DeltaH = -286 kJ"mol"^(-1)` (iii) `H_(2)O(l) to H_(2)O(g)DeltaH = 44 kJ"mol"^(-1)` (iv)`2B(s) + 3H_(2)(g) to B_(2)H_(6)(g) DeltaH = 36 kJ"mol"^(-1)` Subtract eq.(iv) from eq.(i) (v) `B_(2)H_(6) (g) + 3/2O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)(g)DeltaH = -1309 kJ"mol"^(-1)` MULTIPLY eq.(ii) by 3 (vi) `3H_(2)(g) + 3/2O_(2)(g) to 3H_(2)O(l)DeltaH = -858 kJ"mol"^(-1)` Add`B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(l)DeltaH = -2167 kJ"mol"^(-1)` Multiply eq.(iii) by 3 (vii) `3H_(2)O(l) to 3H_(2)O(g)DeltaH = 132 kJ"mol"^(-1)` Add `B_(2)H_(6)(g) + 3O_(2)(g) to B_(2)O_(3)(s) + 3H_(2)O(g) Delta = -2035 kJ"mol"^(-1)`. |
|
| 22. |
Diborane forms ionic compound by the addition of which of the following substance ? |
|
Answer» `CO` |
|
| 24. |
Diborane (B_(2)H_(6)) reacts independently with O_(2) and H_(2)O to produce, respectively: |
|
Answer» `H_3 BO_3 and B_2 O_3` |
|
| 25. |
Diaspore is |
|
Answer» `Al_(2)O_(3)` |
|
| 26. |
Diamonds are used in ornaments because of it's high |
|
Answer» density |
|
| 27. |
Diamond is used in glass cutting due to its |
|
Answer» HARD nature |
|
| 28. |
Diamond is insulator, but graphite is a conductor. Why? |
| Answer» Solution :Diamond has no FREE electrons. Hence it does not EXHIBITS electrical CONDUCTIVITY GRAPHITE has free delocalised p-electrons. Hence it is a conductor. | |
| 29. |
Diamondis covalent, yetit has high melting point. Why ? |
| Answer» Solution :Diamondhas threedimensional network structureinvolving strong C-CbondsThese BONDS are DIFFICULT to break and hencethe MELTING point ofdiamondis very HIGH. | |
| 30. |
Diamond is covalent, yet it has high melting point. Why ? |
| Answer» Solution :Diamond has a three-dimensional NETWORK involving strong C - C bonds, which are very DIFFICULT to BREAK and, in turn has high melting POINT. | |
| 31. |
Diamond is covalent yet it has high melting point. Why? |
| Answer» Solution :DUE to strong COVALENT bonds holding all the carbon atoms in diamond TOGETHER. | |
| 32. |
Diamond and solid rhomibcsuplhur both are covalnet solids but the latter has very low melting point than the former. Explain why . |
| Answer» SOLUTION :Diamond is three dimesional net work covalent solid with strong interatomic forces WHEREAS sulphur is a molecular solid consisting of PUCKERED eight membered rings `(S_(8))` held together by weak van DER waals forces. | |
| 33. |
Diamond and solid rhombic sulphur both are covalent solids but the latter has very low melting point than the former. Explain why . |
| Answer» Solution :Diamond is three dimensional net work COVALENT solid with strong INTERATOMIC forces whereas sulphur is a molecular solid consisting of PUCKERED eight membered rings `(S_8)` HELD together by weak van der WAALS forces. | |
| 34. |
Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic ? |
|
Answer» `N_(2)` |
|
| 35. |
Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic ? |
|
Answer» `N_(2)` Electronic configuration of `NO_(2) = sigma 1S^(2), sigma^(**) 1s^(2), sigma 2S^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(Z)^(2)` It has no unpaired electron indicates DIAMAGNETIC species. Electronic configuration of `N_(2)^(2-) " ion " = sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(z)^(2) pi^(**)2p_(x)^(1) = pi^(**) 2p_(x)^(1)` It has two unpaired electrons, so it is PARAMAGNETIC in nature. (C) Electronic configuration of `O_(2) = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma^(**) 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)` The presence of two unpaired electrons shows it is paramagnetic nature. (D) Electronic configuration of `O_(2)^(2-) "ion" = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)` It contains no unpaired electron, therefore, it is diamagnetic in nature. |
|
| 36. |
Dialysis can be used to separate |
|
Answer» glucose and fructose |
|
| 37. |
Diagram shows a graph between pressure and density for an ideal gas at two temperatures T_1 and T_2which is correct |
|
Answer» `T_(1) gt T_(2)` |
|
| 38. |
Diagonal relationship of elements can be explain by which of the following properties? |
|
Answer» IONIC size |
|
| 39. |
Diagonal relationship is shown by elements of |
|
Answer» 1st period |
|
| 40. |
Digonal relationship is shown by the elements |
|
Answer» Be and MG |
|
| 41. |
Diagonal relationship is shown by |
|
Answer» all elements with their DIAGONALLY opposite elements |
|
| 42. |
Diagonal relationship is present between the lighter elements of periods |
|
Answer» SECOND, THIRD |
|
| 43. |
Diacetone alcohol is obtained from the reaction of |
|
Answer» Acetone and ethanol |
|
| 44. |
Di - Isobutyl alluminium hydride (DIBAL-H) can be used to carry out which of the following conversions, |
|
Answer» ESTER to ALDEHYDE |
|
| 45. |
Device a scheme for the synthesis of n-butane using CH_3I as the only carbon source. Can you employ the reactions in your scheme to synthesis propane in fairly pure state ? Explain |
|
Answer» Solution :`underset"Methyl IODIDE"(2CH_3I) underset"Wurtz reaction"overset"Na, dry ether"to underset"ETHANE"(CH_3-CH_3) underset(or SO_2Cl_2 //ROOR)overset((i)Cl_2//hv)to underset"ETHYL chloride "(CH_3CH_2Cl)underset"Wurtz reaction"overset"Na, dry ether"tounderset"BUTANE"(CH_3CH_2-CH_2CH_3)` Propane cannot be synthesized by the above scheme in fairly pure state because Wurtz reaction reaction between `CH_3I` and `C_2H_5I` will give three PRODUCTS, i.e., ethane, propane and butane. |
|
| 46. |
Development of priodic table have made the study of elements and their compounds easiere Give the name of the element with otomic number 112 |
| Answer» SOLUTION :UNUNBIUM (UUB) | |
| 47. |
Deutero methane is obtained by the deuterolysis of |
|
Answer» `Mg_3 N_2 ` `CaC_2+ D_2 OtoC_2D_2 + Ca (OD)_2 ` `Al_4 C_3 + D_2O to CD_4+ Al(OD)_3` `Ca_3 P_3 + D_2O to PD_3 + Ca(OD)_2` |
|
| 49. |
Deuterium is heavy hydrogen, which is used in |
|
Answer» Artificial transmutation of ELEMENTS |
|
| 50. |
Deuter meme is obtained hy the deutomlysis of |
|
Answer» `Mg_(3)N_(2)` |
|
