94420 + Interview Questions in Chemistry in Class 12

1.	On the basis of the following observations made with aqueous solutions, how many of them have 6 secondary valencies ? {:("Formula","Moles of AgCl precipitated per mole of the compounds with excess" AgNO_(3)),((a) PdCl_(2).4NH_(3),""2),((b) NiCl_(2).6H_(2)O,""2),((c) PtCl_(4).2HCl,""0),((d) CoCl_(3).4NH_(3),""1),((e) PtCl_(2).2NH_(3),""0):}
Answer» SOLUTION :`[Pd(NH_(3))_(4)]Cl_(2) ""[Ni(H_(2)O)_(6)]Cl_(2)`, `H_(2)[PtCl_(6)],"" [Co(NH_(3))_(4)Cl_(2)]CL`, `[PT(NH_(3))_(2)Cl_(2)]`

Discussion

2.	On the basis of the following E^(@) values, the strongest oxidizing agent is [Fe(CN)_(6)]^(4-)to[Fe(CN)_(6)]^(3-)+e^(-1),E^(@)=-0.35V Fe^(2+)toFe^(3+)+e^(-),""E^(@)=-0.77V
Answer» `FE^(3+)` `[Fe(CN)_(6)]^(3-)` `[Fe(CN)_(6)]^(4)` `Fe^(2+)` ANSWER :A

Discussion

3.	On the basis of the following E^(@)values, the strongest oxidizing agent is : [Fe(CN)_(6)]^(4-) to [Fe(CH)_(6)j]^(3)+ +e ^(-),E^(@)=-0.35 V Fe ^(32+)+e ^(-),""E ^(@)=-0.77V
Answer» `,FE(CN)_(6)]^(4)` `Fe ^(2+)` `Fe ^(3+)` `[Fc(CN)_(6)]^(3+)` Solution :From the given data we find `Fe ^(3+)` is strongest oxidising agent. More the POSITIVE value of `E ^(@).` more is the tendency to get oxidized. Thus correct option is (C).

Discussion

4.	On the basis of the following data, explain why Co(III) is not stable in aqueous solution? Co^(3+)+e^(-)toCo^(2+),E^(@)=+1.82V 2H_(2)OtoO_(2)+4H^(+)+4r^(-),E^(@)=1.23V.
Answer» Solution :ADDING the TWO half reaction, EMF comes out to +ve. This means that CO(III) in aqueous solution has the tendency to change to Co(II). Hence, Co (III) is not stable in aqueous solution.

Discussion

5.	On the basis of the electrochemical theory of aqueous corrosion, the reaction occurring at the cathode is:-
Answer» `O_(2(G))+4H_((aq))^(+)+4e^(-)to2H_(2)O_((l))` `Fe_((s))toFe_((aq))^(2+)+2E^(-)` `Fe_((aq))^(2+)toFe_((aq))^(3+)+e^(-)` `H_(2(g))+2OH_((aq))^(-)to2H_(2)O_((l))+2e^(-)` SOLUTION :At cathode: `2H_((aq))^(+)+2e^(-)to2H` `underline(2H+(1)/(2)O_(2)toH_(2)O"")` `2H^(+)+(1)/(2)O_(2)+2e^(-)toH_(2)O`

Discussion

6.	On the basis of the electrochemical theory of aqueous corrosion, the reaction occuring at the cathode is …………. .
Answer» `O_(2(G)) + 4H_((aq))^(+) + 4e^(-) to 2H_2O_((L))` `Fe_((s)) to Fe_((aq))^(2+) + 2E^(-)` `Fe_((aq))^(2+) to Fe_((aq))^(3+) + e^(-)` `H_(2(g)) + 2OH_((aq))^(-) to 2H_2O_((l)) + 2e^(-)` Solution :At cathode, `2H_((aq))^(+) + 2e^(-) to 2H` `2H^(+) + 1//2 O_2 to H_2O` `:. 2H^+ + 1//2 O_2 + 2e^(-) to H_2O` Balancing the above equation `4H^+ + O_2 + 4e^(-) to 2H_2O`

Discussion

7.	On the basis of structure of graphite which of the following is/are true for it ?
Answer» It is a diamagnetic substance It behaves LIKE metallic conductors as well as semiconductor C-C bond length in it is more than the diamond It is thermodynamically more STABLE as well as more dense than diamond Solution :(A),(B) Correct statement (C ) Graphite `sp^3` HYBRIDISATION bond length 1.42 Å, diamond `sp^3` hybridisation bond length 1.54 Å (D)Diamond more dense (3.51 g/ml) than graphite (2.25 gm/ml)

Discussion

8.	On the basis of standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. E^(@)"values",Fe^(3+)\|Fe^(2+)=+0.77,I_(2)(s)\|I^(-)=+0.54,Cu^(2+)\|Cu=+0.34,Ag^(+)\|Ag=+0.80V
Answer» `FE^(3+)` `I_(2)(s)` `Cu^(2+)` `AG^(+)` ANSWER :D

Discussion

9.	On the basis of stability of complex extity in the solution, complexes may be of two types, perfect, imperfect complexes, the stabilies depends upon the extent of dissociation which in turn depends on the strength of M-L band. The dissociation of complex may be expressed [ML_(x)P^(+)hArrMy^(+)+xL and eq. constant of this dissociation equilibrium is called instability constant. The stability of complex depends are EAN, charge on central atom, basic nature chelation, nature of metal ion and ligand acc. to HSAB principle Which complex in most stable ?
Answer» `[Cu(CN)_(2)]^(-),K_(d)=1xx10^(-16)` `[Fe(CN)_(6)]^(-4),K_(d)=1xx10^(-37)` `[Fe(CN)_(6)]^(3-),K_(d)=1xx10^(-44)` `[Ag(CN)_(2)]^(-),K_(d)=1xx10^(-20)` Solution :Stability of a complex `prop(1)/(k_(d))`

Discussion

10.	On the basis of stability of complex extity in the solution, complexes may be of two types, perfect, imperfect complexes, the stabilies depends upon the extent of dissociation which in turn depends on the strength of M-L band. The dissociation of complex may be expressed [ML_(x)P^(+)hArrMy^(+)+xL and eq. constant of this dissociation equilibrium is called instability constant. The stability of complex depends are EAN, charge on central atom, basic nature chelation, nature of metal ion and ligand acc. to HSAB principle Which of the following does not follow EAN rule?
Answer» `Fe(CO)_(5)` `V(CO)_(6)` `K_(4)[Fe(CN)_(6)]` `Mn_(2)(CO)_(10)` Solution :`V(CO)_(6)to23-0+6xx2=35`

Discussion

11.	On the basis of stability of complex extity in the solution, complexes may be of two types, perfect, imperfect complexes, the stabilies depends upon the extent of dissociation which in turn depends on the strength of M-L band. The dissociation of complex may be expressed [ML_(x)P^(+)hArrMy^(+)+xL and eq. constant of this dissociation equilibrium is called instability constant. The stability of complex depends are EAN, charge on central atom, basic nature chelation, nature of metal ion and ligand acc. to HSAB principle The EAN of Co in Co(CO)_(4) is 35. Hence it is less stable. It attains stability by
Answer» OXIDATION of CO Reduction of Co Dimerisation Tetramerisation SOLUTION :`2Co(CO)_(4)hArrCo_(2)(CO)_(8)`

Discussion

12.	On the basis of position in the electrochemial series, the metal does not displace hydrogen from water and acids is :
Answer» Hg Al Pb Ba Answer :A

Discussion

13.	On the basis of nature of constituent particles,crystals are classified into four types. Which are they?
Answer» SOLUTION :IONIC crystals, MOLECULAR crystals, Covalent crystals ans METALLIC crystals.

Discussion

14.	On the basis of Lnathanoidcontraction , explain the following : (i)Nature of bonding in La_(2)O_(3) and Lu_(2)O_(3) (ii) Trends in the stability of oxio salts of lanthanoidsfrom La to Lu. (iii) Stability of the complexes of lanthanoids . (iv) Radii of 4d and 5d block elements.(v) Trends in acidiccharacter of lanthanoidoxides.
Answer» SOLUTION :(i) As the size decreases from La to Lu,covalent character increases ( according to Fajan's rule). HENCE, `La_(2)O_(3)` is more ionicand `Lu_(2)O_(3)` is more covalent. (ii) As the size decreasesfrom La to Lu, STABILITY of oxo salts decreases. (iii) As the size of lanthanoid decreases, charge `//`sizeratio increases andhence the stabilityof the complecxes increases. (iv) Due to lanthanoidcontraction, RADII of 4d and 5d BLOCK elements are nearly equal. (v) As explained in (i), covalend character of oxidesincreases from La to Lu, therefore, their basic character decreases or acidiccharacter increases.

Discussion

15.	On the basis of Lc Chatclier's principle, predict which of the following conditions would be unfavourable for the formation of SO_(3) ? Given that 2SO_(2)+O_(2) hArr 2SO_(3), Delta H= -42kcal
Answer» Low pressure and low temperature High pressure and low temperature High temperature and low pressure High concentration of `SO_(2)` Solution :Since REACTION is EXOTHERMIC hence low temperature will favour forward reaction ALSO volume is decreased by APPLYING high pressure.

Discussion

16.	On the basis of intermolecular forces of attraction, polymers, are classified as elastomers, fibres, thermoplastics and thermosetting polymers. Elastomers have the weakest while fibres have the strongest intermolecular forces of attraction. Polymers which can be chemical composition and mechanical strength are called thermoplastics. in contrast, those polymers which can be heated only once when they undergo a permanent change in chemical composition to give a hard, infusible and insoluble mass, are called thermosetting polymers. The linear chains in terylene are held together by
Answer» Covalent BOND Hydrogen BONDS Dipole-dipole INTERACTIONS Van DER WAALS forces Solution :Dipole-dipole interactions

Discussion

17.	On the basis of Lanthanoid contraction, explain the following: (i) Nature of bonding in La_(2)O_(3) and Cu_(2)O_(3) (ii) Trends in the stability of oxo salts of lanthanoids from La to Lu (iii) Stability of the complexes of lanthanoids. (iv) Radii of 4d and 5d-block elements. (v) Trends in acidic character of lanthanoid oxides.
Answer» Solution :(i) From La to Lu, the size of ion decreases.Hence, `La_(2)O_(3)` is ionic while `Lu_(2)O_(3)` is a covalent OXIDE. (ii) Due to decrease in size, the stability of oxo-salts of LANTHANOIDS decreases from La to Lu (iii) As size decreases, the stability of complexes INCREASES. (iv) The 4f-orbitals are filled before the 5d-series of the elements beings. Due to intervention of 4f-orbitals, the lanthanoid contraction takes place. As a result, radii of 4d ADN 5d block elements are nearly of same size. (v) Acidic character of oxides increases from La to Lu.

Discussion

18.	On the basis of intermolecular forces of attraction, polymers, are classified as elastomers, fibres, thermoplastics and thermosetting polymers. Elastomers have the weakest while fibres have the strongest intermolecular forces of attraction. Polymers which can be chemical composition and mechanical strength are called thermoplastics. in contrast, those polymers which can be heated only once when they undergo a permanent change in chemical composition to give a hard, infusible and insoluble mass, are called thermosetting polymers. Which are true for elastomers
Answer» They possess elasticity These possess weak INTERMOLECULAR FORCES of attraction between polymer CHAINS Vulcanized rubber is an EXAMPLE of elastomer All are correct Answer :D

Discussion

19.	On the basis of intermolecular forces of attraction, polymers, are classified as elastomers, fibres, thermoplastics and thermosetting polymers. Elastomers have the weakest while fibres have the strongest intermolecular forces of attraction. Polymers which can be chemical composition and mechanical strength are called thermoplastics. in contrast, those polymers which can be heated only once when they undergo a permanent change in chemical composition to give a hard, infusible and insoluble mass, are called thermosetting polymers. Which of the following can be remelted time and again without producing any change
Answer» PVC Bakelite Melmac Urea-formaldehyde RESIN SOLUTION :PVC is a THERMOPLASTIC

Discussion

20.	On the basis of information given below mark the correct option. Information : On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.
Answer» At specific COMPOSITION methanol - acetone mixture will form MINIMUM boiling azeotrope and will show positive deviation from Raoult's law. At specific composition methanol - acetone mixture FORMS maximum boiling azeotrope and will show positive deviation from Raoult's law. At specific composition methanol - acetone mixture will form minimum boiling azeotrope and will show negative deviation from Raoult's law. At specific composition methanol - acetone mixture will form maximum boiling azeotrope and will show negative deviation from Raoult's law. Solution :Explanation : (A) (A-A) or (B-B) INTERACTIONS are stronger than the (A-B) interactions , where, A is methanol molecule & B represents a molecule of acetone. It means that in this solution molecules of A (or B) will FIND it easier to escape. This will increase the vapour from Rault.s law. Further, (B) Due to this positive deviation the methanol acetone mixture forms minimum boiling azeotrope.

Discussion

21.	On the basis of information given below mark the correct option. Information :(i) In bromoethane and chloroethane mixture intermolecular interactions of A - A and B - B type are nearly same as A - B type interactions.(ii) In theanol and acetone mixture A - A or B - B type intermolecular interactions are stronger than A - B type interactions.(iii) In chloroform and acetone mixture A - A or B - B type intermolecular interactions are weaker than A - B type interactions.
Answer» Solution (ii) and (iii) will follow Raoult's law. Solution (i) will follow Raoult's law. Solution (ii) will show negative DEVIATION from Raoult's law. Solution (iii) will show POSITIVE deviation from Raoult's klaw. Solution :Explanation : For an ideal solution A - A and B - B intermolecular INTERACTION should be nearly same as A - B type INTERACTIONS.

Discussion

22.	On the basis of information given below mark the correct option. Information : On adding acetone tomethanol some of the hydrogen bonds between methanol molecules breaks.
Answer» At apecific COMPOSITION methanol-ACETONE mixture will FORM minimum boiling azetrope and will show positive deci9ation from Raoul's law. At specific compositon methannol-acetone mixture forms maximuym boiling axetrope and will show positive deviation from Raoult's law. At specific composition methanol-acetone mixture will form minimum biling axetrope and will show negative deviation from Raolt's law. At spectific composition methannol- acetone mixture will from maximum boiling axeotrope and will show negetive from Raoult's law. Solution :Since the hydrogen bonds BREAK, vapour pressure will be HIGHER than the expected value. The solution will show positve deviation for Raoul's law and will form minimum boiling azeotrope.

Discussion

23.	On the basis of information given below mark the correct option. Information : (A) In bromoethane and chloroethane mixture, intermolecular interactions of A-A and B-B type are nearly same as A-B type interactions (B) In ethanol and acetone mixture, A-A or B-B type intermolecular interactions are stronger than A-B type interactions. (C) In chloroform and acetone mixture, A-A or B-B type intermolecular interactions are weaker than A-B type interactions
Answer» Solution (B) and (C) will follow Raoult's law Solution (A) will follow Raoult's law Solution (B) will SHOW NEGATIVE DEVIATION from Raoult's law Solution (C) will show positive deviation from Raoult's law Answer :B

Discussion

24.	On the basis of info0rmation given below mark the correct option.
Answer» In bromoethane and chloroethane mixture intermolecular interactions of A-A and C-C type are nearly same as A-B type ineracitons. In ethanol and acetone mixture A-A of B-B type intermolecular interactions are stronger than A-B type interinteractions. In chloroform and acetone mixtue A-A or B-B type intermolecular interactions are WEAKER than A-B type interactions. Solution (B) and © will follow RAOULT's law. Solution (A) will follow Raoult's law. Solution (B) will show NAGATIVE deviation from Raoult's law. Solution © will show will show positive deviation form Raoult's law. Solution :is the correct answer. Both bromoethane (A)and chloroethane (B) form nearly an ideal solution : A-A and B-B interactions are nearly the same as A-B interaactions

Discussion

25.	On the basis of Hardy-Schulze rule explain why the coagulating power of phosphate is higher than chloride.
Answer» Solution :Greater is the valency of the flocculating (COAGULATING) ion, greater is the coagulating power. Thus, `PO_(4)^(-3)` is more EFFECTIVE in COAGULATION of colloidal particles.

Discussion

26.	On the basis of following thermochemical equations H_(2)O(g)+C(s)toCO(g)+H_(2)(g),DeltaH=131kJ CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g),DeltaH-282kJ H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH=-242kJ C(s)+O_(2)(g)toCO_(2)(g),DeltaH=X kJ:- The value of X will be
Answer» `-393kJ` `-655kJ` `393kJ` `655kJ` SOLUTION :`H_(2)O(g)+C(S)toCO(g)+H_(2)(g)` . . . (1) `CO(g)+(1)/(2)O_(2)(g) to CO_(2)(g)`. . . (2) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O`. . . (3) `C(S)+O_(2)(g) to CO_(2)(g)` . .(4) Eq. (1)+ eq. (2) +eq. (3)=eq. (4) .

Discussion

27.	On the basis of Hardy-Schulze rule, explain why the coagulating power of phosphate is higher than chloride.
Answer» Solution :MINIMUM quantity of an electrolyte REQUIRED to cause precipitation of a sol is called its coagulating value. Greater the CHARGE and smaller the AMOUNT of the electrolyte required for precipitation, HIGHER is the coagulating power of the electrolyte.

Discussion

28.	On the basis of Hardy-Schulze rule explain why the coagulating power of phosphate is higher than chloride,
Answer» Solution :COAGULATING power of an electrolyte depends upon the CHARGE on the ion having charge opposite to that of the colloidal particles. Grater the charge on the oppositely CHARGED ion, small is its smount required for coagulation and hance greater is its coagulating power. Hence, for a +vely charged sol, `PO_(4)^(3-)` ion with three units of -ve charge has greater coagulating power than `CI^(-)` ion with one unit -ve charge.

Discussion

29.	On the basis of following equations the heat of dimerisation of NO_(2) will be(i)N_(2)+2O_(2)rarr2NO_(2) ""DeltaH=67.9 kJ(ii)N_(2)+2O_(2)rarrN_(2)O_(4) ""DeltaH=09.3 kJ
Answer» `+77.2 kJ` `-77.2 kJ` `-58.6 kJ` `+58.6 kJ` Solution :Our desirable EQUATION is `2NO_(2)rarrN_(2)O_(4)"...(iii)"` Given `2NO_(2)rarrN_(2)+2O_(2)DeltaH=-67.9 kJ "...(i)"` `N_(2)O_(4)rarrN_(2)+2O_(2)DeltaH=-09.3 kJ "...(ii)"` On substracting equation (ii) from (i) we will GET equation (iii) `2NO_(2)-N_(2)O_(4)rarr0` `2NO_(2)rarrN_(2)O_(4)"....(iii)"` So Heat of dimerisation of `NO_(2)=-67.9-(-09.3)kJ=-67.9+9.3 =-58.6 kJ`

Discussion

30.	On the basis of given part of periodic table, incorrect statement is :
Answer» A is an alkaline earth metal Atomic number of B is 103 which belongs to III B GROUP Atomic number, group no. and period number of D are 72 IVB and `6^(TH)` respectively C is a transuranic element Answer :B

Discussion

31.	On the basis of following E^(@) values , the strongest oxidising agent is [Fe(CN)_(6)]^(4-)to[Fe(CN)_(6)]^(3-)+e^(-)E^(@)=-0.35VFe^(2+)toFe^(3+)+e^(-) "" E^(@)=-0.77V
Answer» `[Fe(CN)_(6)]^(4-)` `Fe^(2+)` `Fe^(3+)` `[Fe(CN)_(6)]^(3-)` Solution :HIGHER the REDUCTION POTENTIAL larger is its TENDENCY to get REDUCED and hence stronger is reducing agent.

Discussion

32.	On the basis of enthalpy of formation, graphite is more stable than diamond, yet diamond does not change into graphite for years. Explain why ?
Answer» Solution :The activation ENERGY for the reaction C (DIAMOND)`to` C (GRAPHITE) is very high which is not AVAILABLE at room TEMPERATURE.

Discussion

33.	On the basis of ellingham diagram which of the following is not correct ?
Answer» Entropy change for all OXIDES is roughly same. Below the BOILING point, `'T DeltaS'` factor DECOMPOSES into metal & oxygen. Above `DeltaG=0` line, oxide decomposes into metal & oxygen. If randomness increases the slope increases. Solution :`(1) (DeltaG)/(T)=-DeltaS,(DeltaG)/(T)` is slope in Ellingham DIAGRAM. Which is same below the boiling point. `(2)` Below the boiling point slope is same as factor `T DeltaS` is same. `(3)` Above `DeltaG=0` line free energy BECOMES positive so oxide decomposes. `(4)` Random increases `i.e. DeltaS` increases, so slope also increases.

Discussion

34.	On the basis of Ellingham's diagram explain the principle of extraction of iron from its oxide ore.
Answer» Solution :1. Thermodynamcs helps us to understand how coke REDUCES the oxides of IRON. The Ellingham diagram for FE. ` to FeOFe to Fe_(2)O_(3) to CO` and `CO to CO_(2)` are given above. 2. From the above Ellingham diagram, it is known that `Delta G^@` vs T plot for the reaction `Fe to FeO` goes up and `C to CO` line joins the `Fe to FeO` line. 3. Above 1073K. `AG^@` of `C to CO` becomes less than `AG^@` of `Fe toFeO` and hence carbonitselfisoxidizedtocarbonmonoxide. Therefore, carbonbecomesverygoodreducing agent which reduces FeO to Fe at the temperature range 900-1500K. `FeO(s) + C(s) to Fe(s//I) + CO(g)` 4.At temperature below 800K, standard free energy change for `CO to CO_(2)` line is less than `Fe to Fe_(2)O` line. Therefore, carbon monoxide is the best REDUCING agent and is oxidised to carbon dioxide by reducing `Fe_(2)O_(3)` and `Fe_93)O_(4)` to Fe. `3 Fe_(2)O_(3) + CO to 2Fe_(3)O_(4) + 4CO_(9)`. `Fe_(3)O_(4) + 4CO to 3Fe + 4CO_(2)`. `Fe_(2)O_(3) + CO to 2FeO + CO_(2)` 5.At temperature below 1073K, CO reduces FeO to Fe `FeO to Fe +CO_(2)`.

Discussion

35.	On the basis of data given below predict which of the following gases shows least adsorption on a difinite amount of charcoal? {:(Gas , CO_(2), SO_(2), CH_(4), H_(2)),("Critical temp/K", 304, 630,190,33):}
Answer» `CO_(2)` `SO_(2)` `CH_(4)` `H_(2)` Solution :Higher the critical TEMPERATURE, greater is the adsorption. `H_(2)` has lowest critical temperature and hence is least adsorbed.

Discussion

36.	On the basis of data given below, E_(Sc^(3+)//Sc^(2+))^(@)=-0.37V,E_(Mn^(3+)//Mn^(2+))^(@)=+1.57V E_(Cr^(2+)//Cr)^(@)=-0.90V,E_(Cu^(2+)//Cu)^(@)=0.34V which of the following statements is incorrect?
Answer» `Sc^(3+)` has good stability due to `[Ar]3D^(0)4s^(0)` configuration `Mn^(3+)` is more stable than `Mn^(2+)` `CR^(2+)` is reducing in nature. Copper does not give `H_(2)` on REACTION wwith dil `H_(2)SO_(4)` Solution :`Mn^(2+)(d^(5))`1 is more stable than `Mn^(3+)(d^(4))`,, thus `E_(Mn^(3+)//Mn^(2+))^(@)=+ve`

Discussion

37.	On the basis of data below predict which of the following gases shows least adsorption on a definite amount of charcoal ?
Answer» `CO_2` `SO_2` `CH_4` `H_2` SOLUTION :Higher the CRITICAL TEMPERATURE, stronger are the attractive forces and thus stronger is the ADSORPTION. `H_2` due to its lowest critical temperature is LEAST adsorbed.

Discussion

38.	On the basis of data given below predict which of the following gases shows least adsorption on a definite amount of charcoal ?
Answer» `CO_2` `SO_2` `CH_4` `H_2` SOLUTION :Higher the CRITICAL TEMPERATURE, stronger are the ATTRACTIVE forces and thus stronger is the adsorption. `H_2` DUE to its lowest critical temperature is least adsorbed.

Discussion

39.	On the basis of crystal field theory, write the electronic configuration for d^(4) ion if Delta_(0)gtP.
Answer» Solution :In this CASE, the 4TH ELECTRON will enter into `t_(2g)`, i.e., pairing in one of the `t_(2g)` orbitals will take place. Hence, electronic configuration will be `t_(2g)^(4)`.

Discussion

40.	On the basis of crystal field theory, write the electronic configuration for d^(4) ion if Delta_(0) < P.
Answer» Solution :If `Delta_(0)ltP`, the configration is :`t_(2g)^(3)e_(g)^(1)`

Discussion

41.	On the basis of crystal field theory, write the electronic configuration for d^(4) ion, if Delta_(0) > P.
Answer» SOLUTION :If `Delta_(0)> P`, the configuration for `d^(4)` ION is `t_(2G)^(4)e_(g)^(0)`

Discussion

42.	On the basis of crystal field theory, write the electronic configuration for d^(4) ion if Delta_(0) > P.
Answer» SOLUTION :If `Delta_(0)> P`, the configuration of `d^(4)` will be `t_(2G)^(4)e_(G)^(0)`

Discussion

43.	On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
Answer» Solution :With WEAK field LIGANDS, `Delta_(0)ltP`. Electronic configuration of Co(III) will be `t_(2g)^(4)e_(G)^(2)`. Thus, it has 4 unpaired electrons and is PARAMAGNETIC. With strong field ligands, `Delta_(0)gtP`. Electronic configuration will be `t_(2g)^(6)e_(g)^(0)` has no unpaired electron and is diamagnetic.

Discussion

44.	On the basis fo the standard electrode potential values stated for acid solution, predict whther Ti^(4+) species may be used to oxidize Fe^(II) to Fe^(III). {:("Reaction","E/V"),(Ti^(IV)+e^(-)toTi^(3+):,+0.01),(Fe^(3+)+e^(-)toFe^(2+):,+0.77):}
Answer» Solution :We want to CHECK the spontaneity of the reaction: `Ti^(4+)+Fe^(2+)toTi^(3+)+Fe^(3+)` E.M.F. of this reaction `=+0.01+(-0.77)=-0.76V` As EMF is -ve, the reaction is non-spontaneous, i.e., `Ti^(4+)` cannot be used to oxidize `Fe^(II)` to `Fe^(III)`.

Discussion

45.	On the basic of stability of complex ion in the solution, complexes may be of two types, perfect and imperfect complexes. The stability depends upon the extent of dissociation which in turn depends upon charge on central metal atom, basic nature of ligand, chelation, and nature of metal ion and ligand according to HSAB principle. Q> Which complex is most stable?
Answer» `[CU(CN)_(2)]^(-) K_(d)=1XX10^(-16)` `[FE(CN)_(6)]^(4-)K_(d)=1xx10^(-37)` `[Fe(CN)_(6)]^(3-)=1x10^(-44)` `[Ag(CN)_(2)]^(-)K_(d)=1x10^(-20)` Answer :C

Discussion

46.	On the basic of stability of complex ion in the solution, complexes may be of two types, perfect and imperfect complexes. The stability depends upon the extent of dissociation which in turn depends upon charge on central metal atom, basic nature of ligand, chelation, and nature of metal ion and ligand according to HSAB principle. Q> Which one of the following does not follow EAN rule?
Answer» `Fe(CO)_(5)` `V(CO)_(6)` `K_(4)[Fe(CN)_(6)]` `Mn_(2)(CO)_(10)` Answer :B

Discussion

47.	On the basis of information given below mark the correct option. Information : On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.
Answer» At specific composition, methanol - ACETONE mixture will form minimum boiling AZEOTROPE and will show positive deviation from Raoult's law At specific composition, methanol-acetone mixture forms maximum boiling azeotrope and will show positive deviation from Raoult's law At specific composition methanol-acetone mixture will form minimum boiling azeotrope and will show NEGATIVE deviation from Raoult's law At specific composition methanol-acetone mixture will form maximumboiling azeotrope and will show negative deviation from Raoult's law Solution :As hydrogen BONDS break, vapour pressure will be higher, i.e. it wll show `+ve` it will show `+ve` deviation from Raoult's law and form a minimum boiling azeotrope.

Discussion

48.	On the basic of forces between their molecules in a polymer to which class does nylon-6, 6 belong?
Answer» SOLUTION :It BELONGS to be CLASS of FIBRES.

Discussion

49.	On the addition of a solution containing CrO_(4)^(2-) ions to the solution of Ba^(2+), Sr^(2+) and Ca^(2+) ions, the precipitate obtained first will be of
Answer» `CaCrO_(4)` `SrCrO_(4)` `BaCrO_(4)` MIXTURE of (a), (b), (c) Solution :When we added barium ion in chromate ion solution we obtained YELLOW ppt of `BaCrO_(4)`. `BaCl_(2)+ K_(2)CrO_(4) rarr underset("Yellow ppt".)(BaCrO_(4)) darr + 2KCl`

Discussion

50.	On the addition of few drops of a solution containing CrO_(4)^(2-) ions to the solution of Ba^(2+), Sr^(2+) and Ca^(2+) ions, the precipitate obtained first will be of:
Answer» `CaCrO_(4)` `SrCrO_(4)` `BaCrO_(4)` A MIXTURE of all the three Answer :C

Discussion

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