InterviewSolution
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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. |
In case of actinidesA. first five shells are completely filledB. first four shells are comletely filledC. last shell is partially filledD. last two shells are partly filled |
| Answer» Correct Answer - B | |
| 252. |
The artificially prepared element from lanthanide series isA. europiumB. thuliumC. promethiumD. gadolinium |
| Answer» Correct Answer - C | |
| 253. |
Which of the following lanthanoide shows stable electronic configuration ?A. Only Eu and GdB. Only Eu, Gd and YbC. Only Eu, Gd, and YbD. Only Gd, Yb and Lu |
| Answer» Correct Answer - C | |
| 254. |
Element of lanthanide series belonging toA. III AB. III BC. IV AD. IV B |
| Answer» Correct Answer - B | |
| 255. |
The element present in after Gd in lanthanide series isA. SmB. TmC. NdD. Pr |
| Answer» Correct Answer - B | |
| 256. |
An element t belonging to lanthanide has at. No. 64, its electronic configuration isA. `4f^(7) 5s^(0) 6s^(2)`B. `4f^(8) 5d^(0) 6s^(2)`C. `4f^(7) 5d^(1) 6s^(2)`D. `4f^(6) 5d^(2) 6s^(2)` |
| Answer» Correct Answer - C | |
| 257. |
Potassium permanganate is prepared from the mineralpyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using it as a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mLof distilled waterand then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooledto `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. Mohr salt, `FeSO_(4). (NH_(4))_(2) SO_(4). 6 H_(2)O` is preferred over `FeSO_(4). 7 H_(2)O` for standardization of `KMnO_(4)` solution because.A. Mohr salt is a double salt while ferrous sulphate is a single salt.B. Mohr salt is not hygroscopic but `FeSO_(4).7 H_(2)O` is hygroscopicC. Mohr slat contains only ferrous ions whereas ferrous sulphate contains some ferric ions.D. Mohr salt solution can be titrated even in the absence of `H_(2)SO_(4)` |
| Answer» `FeSO_(4). 7H_(2)O` contains some ferric ions due to aerial oxidation. | |
| 258. |
Potassium permanganate is prepared from the mineralpyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agentin the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using itas a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mLof distilled water and then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooled to `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. Th evoluem of `KMnO_(4)` solution that must have been added to obtain the faint pink colour at the end point must beA. 100 mLB. 200 mLC. 300 mLD. 400mL |
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Answer» Molesof `( COONa)_(2) = 13.4 // 134 = 0.1` . The balanced equation for the reaction involved is `2MnO_(4)^(-) + 16 H^(+) + 5CrO_(4)^(2-) rarr 2Mn^(2+) + 8 H_(2)O + 10 CO_(2)` `:. ` Moles of `KMnO_(4)` used `= ( 2)/( 5) xx 0.1 = 0.04 ` mol As `KMnO_(4)` solution used is `0.1 M` , i.e., 0.1 mol are present in 1000 mL, therefore , 0.04 mol will be present in 400 mL. |
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| 259. |
Potassium permanganate is prepared from the minera lpyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using itas a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mL of distilled waterand then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooled to `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. If instead of `H_(2)SO_(4)`, HCl or `HNO_(3)` of suitable concentration were used, the volume of`KMnO_(4)` soltuionused would have beenA. less in case of HCl but more in case of `HNO_(3)`B. more in case of HCl but less in case `HNO_(3)`C. more in both casesD. less in both case |
| Answer» If HCl were used, the oxygen produced from `KMnO_(4)` will be partly used up for oxidation of HCl. Hence, HCl required will be more. If `HNO_(3)` were used , it itself acts as oxidizing agent and will oxidize the reductant. Hence, `KMnO_(4)` soltuion used will be less. | |
| 260. |
The lanthanide have general electronic configurationA. `[Xe]4f^(1-14), 5s^(2) , 5p^(6) , 6s^(2)`B. `[Xe] (n-1) d ^(1-10) , ns^(1-2)`C. `[Xe] 4f ^(1-14), 5d^(0-1), 6s^(2)`D. `[Xe] 4f^(0-14), 5d^(0-10), 6s^(2)` |
| Answer» Correct Answer - C | |
| 261. |
The most accepted abserved electronic configuration of lanthanide isA. `6s^(2), 4f^(0-14) , 5d^(1)`B. `6s^(2), 4f ^(1-14) , 5d ^(0-1)`C. `6s^(2), 4f^(2-14), 5d ^(0-1)`D. `6s^(2), 4f ^(2-14), 5d^(1)` |
| Answer» Correct Answer - B | |
| 262. |
Potassium permanganate is prepared from the mineral pyrolusite, `MnO_(2)`. Its crystals have deep purple colour. It acts as an oxidizing agent in the neutral, alkaline as well as acidic medium. In acidic medium , it is used in volumetric analysis for estimation of ferrous salts, oxalates etc. The titrations are carried out in presence of `H_(2)SO_(4)`. However ,before using itas a titrant, it is first standized with standard oxalic acid solution or Mohr salt solution. In one of the experiments on titration , 13.4 g of dry pure sodium oxalate ( molar mass `=134 g mol^(-1)` ) was dissolved in 100mLof distilled water and then 100 mL of 2M `H_(2)SO_(4)` were added. The solution, was cooled to `25.30^(@) C`. Now to this solution , 0.1 M `KMnO_(4)` solution was added till a very faint pink colour persisted. The purple colour of `KMnO_(4) ` is due toA. incomplete d-subhsellB. ionic nature of `KMnO_(4)`C. ionic nature of `KMnO_(4)`D. resonance in `MnO_(4)^(-)`ion |
| Answer» Purple colour of `MnO_(4)^(-)` ion is due to charge transfer | |
| 263. |
German silver does not containA. SnB. CuC. ZnD. Ni |
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Answer» Correct Answer - A Garman silver is an alloy of copper (Cu),zinc(Zn) and nickel (Ni) in the ration of 2:1:1. Thus , Sn is not present in it. |
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| 264. |
One of the consituents of German silver isA. `Ag`B. `Cu`C. `Mg`D. `Al` |
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Answer» Correct Answer - B German silver is an alloy which contains copper, zinc and nickel. |
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| 265. |
In the dichromate dianion,A. 4Cr - O bonds are equivalentB. 6Cr - O bonds are equivalentC. all Cr-O bonds are equivalentD. all Cr-O bonds are non-equivalent |
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Answer» Correct Answer - B In the dichromate dianion `Cr_2O_7^(2-)` , six Cr-O bonds are equivalent (bond length 161 pm each ). |
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| 266. |
Passing `H_(2)S` gas into a mixture of `Mn^(2+), Ni^(2+), Cu^(2+)` and `Hg^(2+)` ions in an acidified aqueous solution precipitatesA. CuS and HgSB. MnS and CuSC. MnS and NiSD. NiS and HgS |
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Answer» Correct Answer - A (a) In the acidic medium, only the cations of group (II) is mixture analysis i.e. `Cu^(2+)(aq)" and "Hg^(2+)(aq)` ions are precipitated as their respective sulphides. These are CuS and HgS (both black in colour). |
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| 267. |
Explain why does colour of `KMnO_(4)` disappear when oxalic acid is added to its solution in acidic medium? |
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Answer» When oxalic acid is added to acidic solution of `KMnO_(4)`, its colour disappear due to reduction of `MnO_(4)^(-)` ion to `Mn^(2+)`. Chemical reaction occuring during this neutralisation reaction is as follows `5C_(2)O_(4)^(2-)+underset(("Coloured"))(2MnO_(4)^(-))+16H^(+)rarrunderset(("Colourless"))(2Mn^(2+))+8H_(2)O+10CO_(2)` |
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| 268. |
Explain why does coloure of `KMnO_(4)` disppear when oxalic acid is added to its solution in acidic medium. |
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Answer» `KMnO_(4)` oxidizes oxalic acid to `CO_(2)`and half is reduced to `Mn^(2+)` which is colourless. `underset("(Coloured)")(2MnO_(4)^(-)) + 16H^(+) + 5C_(2) O_(4)^(2-) rarr underset("(Colourless)")(2Mn^(2+))+ 8 H_(2)O + 10 CO_(2)` |
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| 269. |
Explain why does colour of `KMnO_(4)` disappear when oxalic acids is added to its solution in acidic medium. |
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Answer» This is a redox reaction in which `MnO_(4)^(-)` ions ( pink colour ) are reduced to `Mn^(2+)` ions (colourless). At the same time oxalate ion in the acidc medium are oxidised to `CO_(2)` `5C_(2)O_(4)^(2-)(aq)+ underset("(Colour)")(2MnO_(4)^(-)(aq))+16H^(+)(aq) to underset("(Colourless)")(2Mn^(2+)(aq))+8H_(2)O(aq)+10CO_(2)(g)` |
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| 270. |
Explain the terms Coordination entity |
| Answer» Coordination entity : A central metal atoms or ion bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example `[CoCl_(3)(NH_(3))_(3)].Ni(CO)_(4)]` , etc. | |
| 271. |
The coordination number in a/an….. Complex may increase to 8.A. cobaltB. osmiumC. nickelD. iron |
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Answer» Correct Answer - B Os`(Z=76):[Xe]4f^(14),5d^(6),6s^(2)` Hence, the coordination number in an osmium complex may increasee to `+8`. |
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| 272. |
Which one of the following is good oxidising agent ?A. `Eu^(2+)`B. `Sm^(2+)`C. `Ce ^(4+)`D. `Yb ^(2+)` |
| Answer» Correct Answer - C | |
| 273. |
Which of the following species is/are paramagnetic? `Fe^(2+),Zn^(0),Hg^(2+),Ti^(4+)`A. Only `Fe^(2+)`B. `Zn^(0) and Ti^(4+)`C. `Fe^(2+) and Hg^(2+)`D. `Zn^(0) and Hg^(2+)` |
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Answer» Correct Answer - A Paramagnetic species contain unpaired electrons. Thus , among the given species only `Fe^(2+)(3d^(6))` is paramagnetic. |
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| 274. |
The density of Sc isA. `2.99 gm //ml`B. `5.96gm//ml`C. `8.99gm//ml`D. `4.90gm//ml` |
| Answer» Correct Answer - D | |
| 275. |
Which metals in the first transition series ( 3d series) exhibits `+1` oxidation state most frequently and why? |
| Answer» Copper in the first transition series exhibits `+1` oxidation statemost frequencly . This is because Cu is `3d^(10) 4s^(1)` . When one electron is lost, the configuration becomes most stable due to fully filled `d^(10)` configuration. | |
| 276. |
In the series Sc`(Z= 21)` to `Zn ( Z= 30)`, the enthalpy of atomisation of zinc is the lowest , i.e., `126 kJ mol^(-1)` . Why ? |
| Answer» In the series , Sc to Zn , all elements have one or more unpaired electrons except zinc which has no unpaired electron as its outer electronic configuration is `3d^(10)4s^(2)`. Hence, atomic intermetallic bonding ( metal - metal bonding) is weakest in zinc. Therefore, enthalpy of atomisation is lowest. | |
| 277. |
The `E^(@) ( M^(@+) //M)` value for copper is positive `( + 0.34V)` . What is possibly the reason for this ? |
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Answer» `E^(@) ( M^(2+) //M) ` for any metal is related to the sum of the enthalpy changes taking place in the following steps `: ` `M(s) + Delta_(a) H rarr M(g)`, `(Delta_(a) H =` enthalpy of atomisation sublimation ) `M(g) + Delta+(i) H rarr M^(2+) (g) `, `( Delta_(i) H =` ionization enthalpy ) `M^(2+) (g) + aq) rarr M^(2+) ( aq) + Delta_(hyd) H` `( Delta_(hyd) H =` hydration enthalpy ) Copper has high enthalpy of atomisation (i.e., energy absorbed ) and low enthalpy of hydration , ( i.e., energy released ). The high energy required to transform Cu(s) to `Cu^(2+) (aq)` , i.e., sum of enthalpies of sublimation and ionization is not balanced by its hydration enthalpy . Hence, `E^(@) (Cu^(2+) //Cu) ` is positive. |
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| 278. |
Dipositive Pm is isoelectronic withA. `Nd ^(3+)`B. `Ce^(4+)`C. `Eu ^(3+)`D. `Sm^(3+)` |
| Answer» Correct Answer - D | |
| 279. |
High negative value of raduction potential, indicates lanthanides areA. oxidising agentB. reducing agnetC. high complex forming tendencyD. high magnetic property |
| Answer» Correct Answer - B | |
| 280. |
The shield effect of electron decreases in the orderA. `f gt d gt p gt s`B. `s gt p gt d gt f `C. `s gt d gt p gt f `D. `p gt d gt s gt f ` |
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Answer» Correct Answer - B We know that, shielding effect of orbitals depends upon shape of orbitals. F-orbitals has diffused shape, hence f-orbitals has poor shielding effect. It decreases in th order, `s gt p gt d gt 4f gt 5t.` |
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| 281. |
Which of the following ions is colourless in solution?A. `V^(3+)`B. `Cr^(3+)`C. `Co^(2+)`D. `Sc^(3+)` |
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Answer» Correct Answer - D `V(23=[Ar]3d^(3),4s^(2)` `V^(3+)=[Ar]3d^(2),4s^(0)` (Two unpaired electrons) `Cr(24)=[Ar]3d^(5),4s^(1)` `Cr^(3+)=[Ar]3d^(3+),4s^(0)` (Three unpaired electrons) `Co(27)=[Ar]3d^(7),4s^(2)` `Co^(2+)=[Ar]3d^(7),4s^(0)` (Three unpaired electrons) `Sc(21)=[Ar]3d^(1),4s^(2)` `Sc^(3+)=[Ar]3d^(0),4s^(0)` (No unpaired electrons) Thus, in `Sc^(3+)` no unpaired d-electrons is present. Hence, no d-d transition is possible and it is colourless ion . |
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| 282. |
Which of the following species is expected to be paramagnetic ?A. `Zn^(2+)`B. `Cu^(+)`C. `Cu^(2+)`D. `H_(2)` |
| Answer» Correct Answer - C | |
| 283. |
Which of the following ions has `d^(5)` electronic configureation ?A. `Cr ^(2+)`B. `Co^(3+)`C. `Mn^(3+)`D. `Fe^(3+)` |
| Answer» Correct Answer - D | |
| 284. |
All `Ln^(3+)` ions are strong reducing agent in solid and aqueous medium. Which is due toA. high I.P.B. high lattice energyC. high hydration energy and high negative vlue of reduction potentialD. all of these |
| Answer» Correct Answer - D | |
| 285. |
Basic characters of `Ln ^(3+)` ion isA. increases with ionic size increasesB. decreases with ionic size decreasesC. increases with nuclear charge increasesD. increases with atomic number increases |
| Answer» Correct Answer - B | |
| 286. |
`Ln^(3+)` ions areA. reducing agentB. oxidising agentC. eitherD. neither |
| Answer» Correct Answer - A | |
| 287. |
`Ln ^(4+)` ions are used asA. oxidising agentB. reducing agentC. eitherD. neither |
| Answer» Correct Answer - A | |
| 288. |
Lanthanides does not show `+4` O. S. in which of the following configurationA. `f^(0)`B. `f ^(7)`C. `f ^(14)`D. all of these |
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Answer» Correct Answer - C Lanthanides show +4 O.S. in `f ^(0), and f^(7)` configuration, whereas it is possible to attend stable electronic configuration. i.e., `(+3` to `+4`) But they does not show +4 O.S. in `4f^(14)` configuration because, by lose of three electrons they already acquire `4f^(14)` configuration i.e. `(+3` to `+4)` O.S. Hence `+4` O.S. is not possible in `4f^(14)` configuration. |
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| 289. |
Which of the following ions is colourless ?A. `V^(3+)(Z=23)`B. `Cu^(2+)(Z=29)`C. `Ti^(4+)(Z=22)`D. `Fe ^(2+)(Z=26)` |
| Answer» Correct Answer - C | |
| 290. |
Which of the following metals exhibits more than one oxidation state ?A. FeB. ZnC. CrD. Both (a) and (x c ) |
| Answer» Correct Answer - D | |
| 291. |
In aqueous solution, cerium shows stable O.S.A. `+2`B. `+4`C. `+3`D. `+5` |
| Answer» Correct Answer - B | |
| 292. |
`Ln^(2+)` ions one used asA. oxidising agentB. reducing agentC. eitherD. neither |
| Answer» Correct Answer - B | |
| 293. |
In the transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. Explain. |
| Answer» In the middle of the series ( `d^(5)` configuration ) there is a participation of two ns electrons and (n-1)d electrons in the bond formation. Therefore, the elements in the middle of the transition sereis exhibit maximum oxidation state. (e.g., Mn present in 3d series . | |
| 294. |
General electronic configuration of lanthanides isA. `4f^(0-14) 5d^(0 or 1) 6s^2`B. `5f^(0-14) 6d^(0 or 1) 7s^2`C. `4f^(1-14) 5d^(0 or 1) 6s^2`D. `5f^(1-14) 6d^(0 or 1) 7s^2` |
| Answer» Correct Answer - C | |
| 295. |
maximum number of variable oxidation state in a transition series are exhibited byA. extreme left elementsB. exterme right elementsC. middle elementsD. all transition elements |
| Answer» Correct Answer - C | |
| 296. |
Why `Sm^(2+), Eu^(2+)` and `Yb^(2+)` ions in solutions are good reducing agents but an aqueous solution of `Ce^(4+)` is a good oxidizing agent ? |
| Answer» The most stable oxidation state of langthanides is `+3` . Hence, ions in `+2` state tend to change to `+3` state by loss of electron and those in `+4` state tend to change to `+3` state by gain of electron. | |
| 297. |
In aqueous solution, Eu shows stable O.S.A. `+3`B. `+4`C. `+3`D. `+5` |
| Answer» Correct Answer - B | |
| 298. |
Lanthanidwes shows variable oxidation state, because they release electron from the following orbitalsA. 4fB. 6sC. 5dD. all of these |
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Answer» Correct Answer - D We know that, lanthanide shows `+2,+3,+4.` O.S. by loss of two, three and four electrons. These electrons may be rmoved from 6s, 4f and 5d orbitals. |
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| 299. |
Iron exhibits `+2 and +3` oxidation states. Which of the following statements about iron is incorrect?A. Ferrous oxide is more basic in nature than the ferric oxideB. Ferrous compounds are relatively more ionic that the corresponding ferric compoundsC. Ferrous compounds are less volatile thanthe corresponding ferric compoundsD. Feerous compounds are more easily hydrolysed than the corresponding ferric compounds. |
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Answer» Correct Answer - D `Fe^(3+)` (ferric) compounds are more easily hydrolysed as compared to `Fe^(2+)` (ferrous) compareds since `Fe^(3+)` ion has greataer magnitude of positive chanrge. |
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| 300. |
One among the langthanides ,Ce(III), can be easily oxidized to Ce(IV)(At. No. of `Ce= 58)` . Explain why ? Or Of the lanthanides, cerium (At. No. 58) forms a tetrapositive ion, `Ce^94+)` in aqueous solution. Why? |
| Answer» Ce(III) having the configuration `4f^(1) 5d^(-) 6s^(0)` can easily lose an electron to acquire the configuration `4f^(0)` and form Ce(IV) . In fact , this is the only `( + IV)` lanthanide which exists in solution. | |