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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. |
Aryl bromides can be prepared by reacting silver salt of aromatic acids with Br_(2) in C CI_(4). This reaction is called "Hunsdiecker reaction" // "Balz-Schiemann" reaction. |
| Answer» SOLUTION :Hunsdiecker REACTION | |
| 3. |
Arsenic sulphide sol is negatively charged. Which of the following electrolytes would be most effective in its coagulation? |
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Answer» `BaCl_(2)` |
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| 4. |
Arsenic sulphide is a negative sol. The reagent with least precipitating power is |
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Answer» `AlCl_(3)` |
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| 5. |
Arsenic sulphide (As_(2)S_(3)) reacts with sulphuric acid (H_(2)SO_(4)) to form H_(3)AsO_(4) (Arsenic acid) and sulphur - dioxide (SO_(2)). What will be the coefficient of H_(2)SO_(4),H_(3)AsO_(4)andSO_(2) respectively in the balanced reaction ? |
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Answer» 11, 2 and 14 (x) Oxidation Part : (i) `As_(2)S_(3)toH_(3)AsO_(4)+SO_(2)` (ii) `As_(2)S_(3)to2II_(3)AsO_(4)+2SO_(2)` (Balance As, S) (iii) `underset(2(+3))underset(DARR)(As_(2))underset(+(-2)3)underset(darr)(S_(3))to2H_(3)Asunderset(2(+5))underset(darr)(O_(4))+underset((+4)3)underset(darr)(3SO_(2))` = + 6 - 6= 10 + 12 = Zero `to` = 22 `therefore` Oxidation number is up by (22) (y) Reduction Part : `underset((+6))underset(darr)(H_(2)SO_(4))tounderset((+4))underset(uarr)(SO_(2))` ... (Reduction) Oxidation no. is down by (2) in Reduction part (z) Balance of change of oxidation number oxidation part `xx` 1 and reduction part `xx` 11 `As_(2)S_(3)to2H_(3)AsO_(4)+2SO_(2)""...(i)` `11H_(2)SO_(4)to11SO_(2)` ... (Reduction) ...(ii) `z(i)+z(ii)" "As_(2)S_(3)+11H_(2)SO_(4)to2H_(3)AsO_(4)+14SO_(2)` Balancing H and O by adding `8H_(2)O` we get BALANCED redox EQUATION... `As_(2)S_(3)+11H_(2)SO_(4)to2H_(3)AsO_(4)+14SO_(2)+8H_(2)O` `therefore` Coefficient of `H_(2)SO_(4)` = 11 `therefore` Coefficient of `H_(3)AsO_(4)=2` `therefore` Coefficient of `SO_(2)` = 14 |
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| 6. |
Arsenic forms two oxides, one of which contains 65.2% and the other, 75.7% of the element. Determine the equivalent weights of arsenic in both cases. |
| Answer» SOLUTION :15, 24.9 | |
| 7. |
Arsenic containing medicine used for the treatment of syphilis is : |
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Answer» Erythromycin |
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| 8. |
Arrive at the expressions of K_(P) and K_(C) for the dissociation of PCl_(5). |
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Answer» Solution :Consider that 'a' moles of `PCl_(5)` is taken in a container of volume V. Let 'x' moles of `PCl_(5)` be dissociated into x moles of `PCl_(3)` and x moles of `Cl_(2)`. `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`   Applying law of mass action, `K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)])=(((x)/(V))((x)/(V)))/(((a-x)/(V)))=(x^(2))/((a-x)V)` The EQUILIBRIUM CONSTANT `K_(P)` can also be calculated as follows : We KNOW the relationship between the `K_(C)` and `K_(P)` `K_(P)=K_(c)(RT)^((Deltan_(g)))` Here the `""Deltan_(g)=n_(P)-n_(r)=2-1=1` Hence `K_(P)=K_(C)(RT)` We know that PV=nRT `RT=(PV)/(n)` Where n is the total number of moles at equilibrium `n=(a-x)+x+x=(a+x)` `K_(P)=(x^(2))/((a-x)V)(PV)/(n)` `K_(P)=(x^(2))/((a-x)V)(PV)/((a+x))=(x^(2)P)/((a-x)(a+x))` |
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| 9. |
Arrhenius theory could not explain the acidic nature of |
| Answer» SOLUTION :` CO_2 ` cannot GIVE ` H^(+) ` | |
| 11. |
Arrange X-rays, cosmic rays and radiowaves according to frequency. |
| Answer» SOLUTION :COSMIC RAYS `GT` X-rays `gt` RADIOWAVES | |
| 12. |
Arrangement of sulphide ions in zinc blends is ….. |
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Answer» HCP |
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| 13. |
Arrange water vapour, liquid water and ice in the order of increasing entropy: |
| Answer» SOLUTION :ICE lt LIQUID water lt water VAPOUR | |
| 14. |
Arrange - underset(|)overset(|)(C)-underset(|)overset(|)(C)-,underset(|)overset(|)(C)- C equiv C -in decreasing order of bond length and give reason. |
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Answer» Solution :` - C EQUIV C lt underset(|)OVERSET(|)(C) - underset(|)overset(|)(C) - lt - underset(|)overset(|)(C) =underset(|)overset(|)(C) - ` Because, If bond order is LESS, than bond length is more. |
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| 15. |
Arrange the van der Waals constant for the gases . |
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Answer» `I-A, II -D , III -C, IV -B` |
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| 16. |
Arrange the three isotopes of hydrogen in the descending order of their reactivity with chlorine. |
| Answer» Solution :ORDER of reactivity : `H_(2)gtD_(2)gtT_(2)` . As the MASS number of isotopes increases, BOND energy increases and their reactivity DECREASES.reactivity decreases. | |
| 17. |
Arrange the relative stability of following carbocation, (Increasing order) (a)overset(+)CH_3 ""(b )CH_3-overset(+)CH_2 ""(c ) (CH_2)_3overset(+)C |
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Answer» SOLUTION :Increasingorderof therelativestabilityof CARBOCATION. ` OVERSET (+ )CH_3lt CH_3-C overset(+)H_2lt(CH_3)_3 C^(+)` |
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| 18. |
Arrange the molecules H_(2), O_(2), F_(2) and N_(2) in order of increasing bond lengths. |
| Answer» SOLUTION :`{:(H-H lt ""N-=N lt"" O=O lt ""F-F),("(74 PM)(110 pm)(120 PM)(144 PM)"):}` | |
| 19. |
Arrange the halogens F_(2),Cl_(2),Br_(2),I_(2), in order of their increasing reactivity with alkanes. |
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Answer» `I_(2) lt Br_(2) lt Cl_(2) lt F_(2)`<BR>`Br_(2) lt Cl_(2) lt F_(2) lt I_(2)` `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. Reactivity decreases down the group as the ELECTRONEGATIVITY of the HALOGEN decreases down the group. |
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| 20. |
Arrange the ions F^(-),O^(2-) and N^(3-) in the increasing order of their ionic radii. Give reason |
Answer» Solution :`F^(-),O^(3-)` and `N^(3-)` are isoelectronic species.  The anion with the greater negative charge will have a LARGER radius because of the lesser ATTRACTION of the ELECTRONS to the NUCLEUS. Hence the increasing order of ionic radii is, `r_(N^(3-))gtr_(O^(2-))gtr_(F^(-))` |
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| 21. |
Arrange the halogens F_2, Cl_2 , Br_2 , I_2 in order of their increasing reactivity withalkanes. |
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Answer» `I_2 lt Br_2 lt Cl_2 lt F_2` |
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| 22. |
Arrange the halogens F_(2), Cl_(2), Br_(2), I_(2) in order of their increasing reactivity with alkanes. |
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Answer» `I_(2) LT Br_(2) lt Cl_(2) lt F_(2)` |
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| 23. |
Arrange the given species in order of their increasing basic character : H_(2)O, OH^(_), CH_(3)OH, CH_(3)O^(-) |
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Answer» SOLUTION :Applying the concept that a strong ACID has a weak conjugate base, order of basic character is : `H_(2)O lt CH_(3)OH lt OH^(-) lt CH_(3)O^(-)` |
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| 24. |
Arrange the given compounds in the decreasing order of basicity on the basis of Bronsted-Lowry concept : BaO, CO_(2), SO_(3), B_(2)O_(3), Cl_(2)O_(7). |
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Answer» Solution :`BaO+H_(2)O hArr Ba(OH)_(2) "(Basic)" " " CO_(2)+H_(2)O hArr H_(2)CO_(3)` (Weakly ACIDIC) `SO_(3)+H_(2)OhArr H_(2)SO_(4) "(Strongly acidic)" " " B_(2)O_(3)+3H_(2)O hArr 2 H_(2)BO_(3)` (Very weakly acidic) `Al_(2)O_(7)+H_(2)O hArr 2 HClO_(4)` (Very strongly acidic) Hence, in the decreasing ORDER of basicity, we have `BaO gt B_(2)O_(3) gt CO_(2) gt SO_(3) gt Cl_(2)O_(7)`. |
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| 25. |
Arrange the given dicarbon species in order of their bond lengths giving reasons L C_(2), C_(2)^(-), C_(2)^(2-) . |
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Answer» Solution :Total number of electrons in `C_(2) = 12` Its electronic configuration is `sigma _(1s)^(2) sigma _(1s)^(**2) sigma _(2s)^(2) sigma _(2s)^(**2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)`"" BOND order = `(1)/(2) (8-4) = 2` E.C. of `C_(2)^(-)=sigma _(1s)^(2) sigma _(1s)^(**2) sigma _(2s)^(2) sigma _(2s)^(**2) pi_(2p_(x))^(2)pi_(2p_(y))^(2) sigma _(2p_(z))^(1)`"" Bond order `= (1)/(2) (9-4) = 2.5` E.C. of `C_(2)^(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)`"" Bond order ` = (1)/(2) (10 -4)=3.0` As GREATER the bond order ,SHORTER is the bond LENGTH , therefore , order of bond length will be `C_(2)^(2-) lt C_(2)^(-) lt C_(2)` . |
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| 26. |
Arrange the followingCaH_2, BeH_2 and TiH_2 in order of increasing electrical conductance |
| Answer» SOLUTION :`BeH_2(COVALENT)ltCaH_2ltTiH_2` | |
| 27. |
Arrange the followingH-H, D_D, and F-F in order of increasing bond dissociation enthalpy |
| Answer» SOLUTION :F-FltH-HltD-D | |
| 28. |
Arrange the followingg hydrogen halides in order of their decreasin reactivity with propene. |
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Answer» HClgtHBrgtHI |
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| 29. |
Arrange the following types of radiations in increasing order of frequency: (a) radiation from mincrowave oven (b) amber light from traffic signal (c) radiation fromFM radio (d) cosmic rays from outer space and (e) X-rays |
| Answer» Solution :Cosmic RAYS `lt X`-rays `lt` AMBER colour `lt` microwaves `lt` RADIATION from FM radio | |
| 30. |
Arrange the followingtypeof radiationinincreasingorderof frequency : ( aradiationfrommicrowaveoven ( b)amberlightfromtrafficsignal ( c)radiationfrom FM radio (d )cosmicraysfrom outerspaceand(e ) X- rays . |
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Answer» Solution :Tyesof RADIATION inincreasingorder of FREQUENCY areas under : FM radiation `LT`microwave radiation `lt ` amberligth`lt X` - rays `lt `cosmic rays : Becauseitsfrequencyis as under : 
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| 31. |
Arrange the following type of radiations in increasing order of frequency: a. Radiation from microwave oven b. Amber light from traffic signal c. Radiation from FM ragio d. Consmic rays from outer space and e. X-rays |
| Answer» SOLUTION :Cosmic rays `lt "X-rays" lt "amber colour" lt "MICROWAVE" lt "FM"` | |
| 32. |
Arrange the following sulphates of alkaline earth metals in order of decreasing thermal stability: BeSO_(4),MgSO_(4),CaSO_(4),SrSO_(4). |
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Answer» Solution :Sulphate decompose on HEATING to give the corresponding oxide. `MgSO_(4)OVERSET(Delta)(rarr)MgO+SO_(2)+(1)/(2)O_(2)` The stability increases as the electropositive character of the metal or basic nature of the metal increases down the group `(darr)`. This is clear form the INCREASING decomposition temperature down the group. `1647 K 1422 K 1168 K773 Krarr` Temperature of decomposition `SrSO_(4) gt CaSO_(4) gt MgSO_(4) gt BeSO_(4)` |
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| 33. |
Arrange the following single bonds in order of bond energy giving reasons :C - C , N - N, O-O, F-F |
| Answer» SOLUTION :` C - C GT N- N gt O - O gt F - F` | |
| 34. |
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E^+. (a)Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene. (b)Toluene, p-H_3C-C_6H_4-NO_2, p-O_2N-C_6H_4-NO_2 |
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Answer» Solution :The typical reactions of benzene are electrophilic substitution reactions. Higher the electron-density in the benzene ring , more reactive is the COMPOUND towards these reactions. Since, `NO_2` is a more POWERFUL electron-withdrawing group than Cl, therefore, more the number of NITRO groups, less reactive is the compound . THUS, the overall reactivity decreases in the order : Chlorobenzene gt p-nitrochlorobenzene gt 2,4-dinitrochlorobenzene. (b)Here, `CH_3` group is electron DONATING but `NO_2` group is electron-withdrawing. Therefore, the maximum electron-density will be in the toluene, followed by p-nitrotoluene followed by p-dinitrobenzene . Thus, overall reactivity decreases in the order : Toluene gt `p-H_3C - C_6H_4-NO_2 gt p-O_2N - C_6H_4-NO_2` |
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| 35. |
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+ (a)Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H_3C–C_6H_4–NO_(2), p-O_2N–C_6H_4–NO_2. |
| Answer» Solution : Since-`NO_2`is a more powerful electron-withdrawing group than - CI, it deactivates the ring to a GREATER extent. HIGHER the number of `-NO_2`groups, greater is the deactivating effect. Hence, the decreasing order of reactivity of the given compounds FOLLOWS the following order : Chlorobenzene > p-nitrochlorobenzene > 2, 4-dinitrochlorobenzene | |
| 36. |
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E^(+) (a) Chlorobenzene, 2, 4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H_(3)C-C_(6)H_(4)-NO_(2) p-O_(2)N-C_(6)H_(4)-NO_(2) |
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Answer» Solution :Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to bond to nucleophiles. The higher the electron density on a benzene ring the more reactive is the compound towards an electrophile, `E^(+)` (Electrophilic reaction). (a) The presence of an electron withdrawing group (i.e., `NO_(2)^(-)` and `Cl^(-)`) deactivates the aromatic ring by decreasing the electron density. Since `NO_(2)^(-)` group is more electorn withdrawing (due to resonance EFFECT) than the `Cl^(-)` group (due toinductive effect), the decreasing order of reactivity is as follows : Chlorobenzene `gt` p-Nitroclorobenzene `gt` 2, 4-Dinitrochlorobenzene  (b) While `CH_(3)^(-)` is an electron donating group `NO_(2)^(-)` group is electron with drawing. HENCE, toluene will have the maximum electron density and is most easily attacked by `E^(+)`. `NO_(2)^(-)` is an electron withdrawing group. Hence, when the number of `NO_(2)^(-)` substitution is GREATER, the order is as follows : Toluene `lt p-H_(3)C-C_(6)H_(4)-NO_(2), p-O_(2)N-C_(6)H_(4)-NO_(2)`  (Because the no. of inactive `-NO_(2)` groups decreases.) |
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| 37. |
Arrange the following resonating structures of formic acid in order of decreasing stability: underset((I))(H-overset(O)overset(||)(C)-OH)harrunderset((II))(H-overset(O^(-))overset(||)(C)-overset(+)(O)H)harr underset((III))(Hoverset(O^(-))overset(|)underset(+)(C)-OH)harrunderset((IV))(H-overset(O^(-))overset(|)underset(-)(C)-O) |
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Answer» `II GT I gt III gt IV` |
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| 38. |
Arrange the following pkb in increasing order |
| Answer» Solution :`pK_(2) LT pK_(1)lt pK_(4) ltpK_(3)` | |
| 39. |
Arrange the following oxides in order of increasing acidic character: SO_3, Cl_2O_7, CaO andPbO_2. |
| Answer» Solution :More the positive oxidation state of the central atom, more the acidic NATURE. So, `CaO lt PbO_(1) lt SO_(3) lt Cl_(2)O_(7)` | |
| 40. |
Arrange the following NaH, MgH_2 and H_2O in order of increasing reducing property |
| Answer» SOLUTION :`H_2OltMgH_2ltNaH` | |
| 41. |
Arrange the following molecules in increasing order of their dipole moments: NH_(3),NF_(3),CB r_(4). |
| Answer» SOLUTION :`CB r_(4) LT NF_(3) lt NF_(3)` | |
| 42. |
Arrange the following molecules in order ionic character of their bondsLIF , K_(2)O , N_(2),SO_(2), CIF_(3) |
| Answer» SOLUTION : `N_(2) LT SO_(2) lt CIF_(3) lt K_(2)O lt LiF` . | |
| 43. |
Arrange the following molecules in increasing order of intermolecular forces : C_(2)H_(6),CH_(3)NH_(2),CH_(3)F and CH_(2)F_(2) |
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Answer» Solution :GREATER the intermolecular forces, greater is the boiling point. `C_(2)H_(6)` has only London forces,`CH_(3)NH_(2)` has hydrogen BONDING and London forces. `CH_(3)F` and `CH_(2)F_(2)` have dipole-dipole attractions besides London forces, but dipole-dipole attractions in `CH_(2)F_(2)` are greater than those in `CH_(3)F`. Hence, ORDER of boiling point is `C_(2)H_(6)(-88^(@)C)ltCH_(3)F(-78^(@)C)ltCH_(2)F_(2)(-52^(@)C)ltCH_(3)NH_(2)(-6^(@C)`. |
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| 44. |
Arrange the following molecules in decreasing order of bond length. |
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Answer» `O_(2)GT O_(2)^(-)gt O_(2)^(+)gtO_(2)^(2-)` |
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| 45. |
Arrange the following molecular species in increasing order of stability .N_(2), N_(2)^(+) , N_(2)^(-), N_(2)^(2-) |
| Answer» SOLUTION : `N_(2)ltN_(2)^(+) = N_(2)^(-)gtN_(2)^(2-)` | |
| 46. |
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al,Cu,Fe,Mg and Zn |
| Answer» SOLUTION :`MggtAlgtZngtFegtCu`. | |
| 47. |
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn. |
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Answer» Solution :`E^(@)Al^(+3)//Al=-1.66"volt"` `E^(@)Cu^(+2)//Cu=+0.34"volt"` `E^(@)FE^(+2)//Fe=-0.44"volt"` `E^(@)Mg^(+2)//Mg=-2.36"volt"` `E^(@)Zn^(+2)//Zn=-0.76"volt"` Reduction potential of metal is more positive then it occurs as STRONG reducing agent. Therefore Mg displaces all the other metals. Al displaces all the other atom in aqueous medium EXCEPT Mg. Zn displaces all the other ATOMS in aqueous medium except Mg and Al. Fe displaces only Cu. |
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| 48. |
Arrange the following metals in the order in which they displace each other from the solution of their saltsAI,Cu,Fe,Mg and Zn |
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Answer» SOLUTION :From thethe `E_(AI^(3+)//AI)^(@)=-1.66 V E_(Cu^(2+)//Cu)^(@)=+0.34 V` `E_(Fe^(2+)//Fe)^(@)=-0.44 V E_(Mg^(2+)//Mg)^(@)=-2.36 v` and `E_(Zn^(2+)//Zn)^(@)=-0.76 V` since a metal with more -ve electrode POTENTIAL is a strongr reducing agent than the one haveing less neagtive or +ve electrode potential therefore Mg can displace all the above metals from their aquiesoue solutoin AI can displace all metals except Mg form the aquieoius solution of their salts working on similarlines we can SHOW that zn can displaceall metals except Mg and AI from the aqueous solution of their salts while fe can displace only Cu from the aqueous solution of its salts THUS the order in which they can displace each other from the solution of their salts is Mg AI ,An Fe Cu |
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| 49. |
Arrange the following metal in which they displace each other form the solution of their salts AI,Cu,Fe, Mg,Ag and Zn |
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Answer» Solution :Metals which have lower electrode potentials can DISPLACE other which have higher electrode potentials from the solution of their salts thus the ORDER is Mg (-2.37 v),AI(-1.66 V),Zn(-0.76 V)Fe (-0.44 V),cu(+0.34 V),AG(+0.80 V) |
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| 50. |
Arrange the following LiH, NaH and CsH in order of increasing ionic nature |
| Answer» SOLUTION :LiHltNaHltCsH | |