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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.
| 51. |
A metal ion from the first transition series has a magnitic moment (calculated) or `3.87 B.M.` How man unparied electrons are expected to be present in the ion?A. `1`B. 2`C. `3`D. `4` |
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Answer» Correct Answer - C `3.87=sqrt(n(n+2)),n="numberof unpaired electrons .So" n=3` |
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| 52. |
Which of the following is expected to have highest magnetic moment?A. `Ti^(2+)`B. `Mn^(2+)`C. `Co^(2+)`D. `Cu^(2+)` |
| Answer» Correct Answer - B | |
| 53. |
Which of the following is expected to be coloured?A. `Ag_(2)SO_(4)`B. `CuF_(2)`C. `MgF_(2)`D. `CuCl` |
| Answer» Correct Answer - B | |
| 54. |
Most of transition , metals can display hydrogen from dilute acid. Why? |
| Answer» Correct Answer - Becausee most of the transition metals have negative oxidation potential and lie above hydrogen in electrochemical series. | |
| 55. |
When `CO_(2)` is passed into aqueous:A. `Na_(2)CrO_(4)` solution, its yellow colour change to orangeB. `K_(2)CrO_(4)` solution,it disproportionates to `MnO_(4)` and `MnO_(2)`C. `Na_(2)Cr_(2)O_(7)` solution its orange colour changes to greenD. `KMnO_(4)` solution, its pink colour chanhe to green |
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Answer» Correct Answer - AB (A) `CO_(2)+H_(2)O to H_(2)CO_(3) to 2H^(+) +CO_(3)^(2-)+2H^(+) to Cr_(2)O_(7)^(2-)+H_(2)O` (B) `3K_(2)MnO_(4)+2CO_(2) to 2KMnO_(4)+mnO_(2)+2K_(2)CO_(3)`. (C) No colour change., (D) No colour change. |
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| 56. |
Most transition metals:(I) Form sets of compounds which display different oxidation states of the metal. (II) form coloured ion in solution (III)burn vigorously in oxygen (IV) form complex compound of theseA. I,II,III are correctB. II,III,IV are correctC. I,II are correctD. all the correct |
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Answer» Correct Answer - C (I) show variable oxidaton state as `(n-1)` and ns orbitals have nearly the same energy and ,thus ns as well `(n1-)` orbital electrons can be lost giving variable oxidaton state. (II) Colour of the compound may be attributed to the incomplete `(n-1)` d subshell which may involved in d-d transition of electrons in presence of ligands. (III) They have low volatility because of high enthalpies of atomization on account of strong metallic bonding. |
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| 57. |
Which of the following Statement(s) is//are correct?A. `Transition metals and many of their compounds show paramagnetic behaviour.B. The enthalpies of atomisation of the transition metals are highC. The transition metals generally form coloured compoundsD. Transition metals and their many compounds act as good catalyst. |
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Answer» As metal ions generally contain one or more unpaired electrons in them & hence their complexes are generally paramagnetic ltbrlt (B) Because of having larger number of unpaired electrons in their atoms, they have stronger inter atomic interaction and hence stronger bonding between the atoms. (C ) According to `CFT`, in presence of ligands the colour of the compounds is due to the d-d transition of the electrons. (D) This activity is ascribed to their ability to adopt multiple oxidation state and to form complexes. Therefore (A,B,C,D)` option are correct. |
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| 58. |
When `cO_(2)` is passed into aqueous:A. `Na_(2)CrO_(4)` solution , its yellow colour change to orangeB. `K_(2)MnO_(4)` Solution,it disproportionate to `KMnO_(4)` and `MnO_(2)`C. `Na_(2)Cr_(2)O_(7)` Solution,its orange colour change to greenD. `KmnO_(4)` Solution ,its pink colour change to green. |
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Answer» `Na_(2)CrO_(4)undersetto(H^(+))Na_(2)Cr_(2)O_(7)("oranged colour")` (B)`MnO_(4)^(2-) undersetto(H^(+))MnO_(4)^(-)` (C )False-in acidic medium no colour change take place. (D)`MnO_(4)^(-)+e^(-)undersetto(Oh^(-))MnO_(4)^(2-)`, in strong alkaline medium pink colour of `KMO_(4)` changes to green. Therefore , (A,B) option are correct |
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| 59. |
The correct order of `E_(M^(2+)//M)^(@)` Values with negative sign for the four successive elements `Cr, Mn, Fe` and `Co` is:A. `MngtCrgtFegtCo`B. `CrgtFegtMngtCo`C. `FegtMngtCrgtCo`D. `CrgtMngtFegtCo` |
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Answer» Correct Answer - 1 Generally across the first transition series, the negative value for standard electrode potential decrease (exception `Mn`-due to stable `d^(5)` configuration) Standard electrode potential- `{:(,Mn,Cr,Fe,Co,),(M^(2+)//M,-1.18,-0.90,-0.44,-0.28,eV):}` So, correct order is `MngtCrgtFegtCo.` |
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| 60. |
Why is oxygen superior to fluorine in stabilising high oxidation states of transition metals?A. Because oxygen is less electronegative than fluorineB. Because of larger size of oxygen as compared to fluorineC. because of the ability of oxygen to form multiple bonds to metalsD. both a and c |
| Answer» Correct Answer - C | |
| 61. |
`K_(2)PtCl_(6)` is a well known compound whereas corresponding `Ni` compound is not known. Explain. |
| Answer» Correct Answer - This is because `Pt^(4+)` is more stable than `Ni^(4+)` as the sum of four ionisation energies of `Pt` is less than those of `Ni` | |
| 62. |
Why is the value for `(Mn^(3+))/(Mn^(2+))` couple much more positive than that for `(Cr^(3+))/(Cr^(2+))` or `(Fe^(3+))/(Fe^(2+))`? Explain |
| Answer» Correct Answer - Much larger third ionisation energy of `Mn` (change from 3d^(5) to 3d^(4)`) is responsible for this. This also explains why the `+3` state of `Mn` is of little importance | |
| 63. |
The wrong statement regarding transition metals among the following is:A. `4s` electrons penetrate toward the nucleus more than `3d` electronsB. atomic radii of transition metals increase rapidly with increase in atomic number because of poor shielding of nuclear attraction by `(n-1)` of electronsC. second and third transition series elements have nearly the same sizeD. their densities are higher and densities of the `5d` series element are higher than those of `4d` series elements. |
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Answer» Correct Answer - B (A) The order of penetration of the electrons present in different sub-shells of same energy level is sgtpgtdgtf. (B) The decrease in size is small after mid way. In the beginning , the atomic radius decrease with the increase in nuclear charge (as atomic number increase), Whereas the shielding effect of d-electrons is small. After mid way as the electrons enter the last but one shell. the added d-electron shields the outer most. electrons . Hence with the increase in the d-electrons screening effect increase . this counter balances the increased nuclear charge . As a result , the atomic radii remain pratically same after chromium . (c ) The filling of `4f` before `5d` orbital results in a regular decrease in atomic radii called. Lanthanoids contraction which essentially compensates for the excepted increase in atomic size with increasing atomic number . The net result of the lanthanoid contraction is the seconf and the third d series exhibit similar radii(e.g., `Zr 160 pm, Hf 159 pm` ) (D) In transition elements , the atomic volumes are large because the increase nuclear charge is poorly screened and so attracts all the electron more strongly . In addition , the extra electrons added occupy inner orbitals. Consequently the densities of the transition metal are high . the densities of the second row are high and third row value higher because of lanthanoid contraction. |
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| 64. |
Select the correct order of oxidising power in acidic medium.A. `VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)`B. `VO_(2)^(+) lt MnO_(4)^(-) lt Cr_(2)O_(7)^(2-)`C. `MnO_(40^(-) lt Cr_(2)O_(7)^(2-) lt VO_(2)^(+)`D. `Cr_(2)O_(7)^(2-) lt VO_(2)^(+) lt MnO_(4)^(-)` |
| Answer» Correct Answer - A | |
| 65. |
Select the correct statementA. stability of `Cu_((aq))^(+2)`, is greater than `Cu_((aq))^(+1)` due to much more `DeltaH_("hydration")` of `Cu^(+2)`B. `Cu_((aq))^(+2)` is more stable because `IE_(2)` of Cu is less than `IE_(1)`C. Generally salts of `Cu^(+2)` are diamagnetic and colourlessD. SRP `(E^(Ө))` of `Cu^(+2)//Cu` is -ve |
| Answer» Correct Answer - A | |
| 66. |
What is the most common oxidation state of first transition series? |
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Answer» Correct Answer - `+2` Except Scandium (has+3), the most common oxidation state of the first transition series elements is +2 which arises due to loss of `4s-electrons`. |
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| 67. |
Pick out the wrong reaction:A. `2Na_(2)CrO_(4)+2H^(+)toNa_(2)Cr_(2)O_(7)+2Na^(+)+H_(2)O`B. `MnO_(2)+4KOH+O_(2)to4KMnO_(4)+2H_(2)O`C. `MnO_(4)^(-)+8H^(+)+5Fe^(+2)to5Fe^(+3)+Mn^(+2)+4H_(2)O`D. `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16Hto2Mn^(+2)+10CO_(2)+8H_(2)O` |
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Answer» Correct Answer - B (b)`impliesK_(2)MnO_(4)` is formed. |
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| 68. |
Which among the following is the best oxidising agent in acidic medium?A. `Cr_(2)O_(7)^(2-)`B. `CrO_(4)^(2-)`C. `MoO_(3)`D. `WO_(3)` |
| Answer» Correct Answer - A | |
| 69. |
Name the three factors which determine the stability of a particular oxidation state in solution. |
| Answer» Correct Answer - (i) enthalpy of sublimation (ii) ionisation energy (iii) enthalpy of hydrogen. | |
| 70. |
Lowest enthalpy of atomisation among the following is of:A. ScB. CrC. MnD. Zn |
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Answer» Correct Answer - D As number of unpaired `e^(-)uarr,Delta_("atomisation")Huarr`. |
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| 71. |
Which transition element does not exhibit variable oxidation state?A. ScB. CuC. ZnD. Hg |
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Answer» Correct Answer - A Zn is not a transition element. |
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| 72. |
Assign reason for each of the following statement: (i) The largest number of oxidation states are exhibited by the elements in their row transition elements. (ii) The atomic radii decrease in Size with the increasing atomic number in the lanthanoid series |
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Answer» Correct Answer - (i) The maximum number of oxidation state are shown by that element in a transition in a transition series which has maximum number of unpaired electrons. This is so in the middle of the series. (ii) Cause of lanthanoid contraction. |
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| 73. |
The order of stability of +6 oxidation state for group VI follows the order:A. `Cr gt Mo gt W`B. `Cr lt Mo lt W`C. `Mo gt Cr gt W`D. `W lt Cr gt Mo` |
| Answer» Correct Answer - B | |
| 74. |
The higher enthalpy of atomisation can be due to :A. large number of unpaired electronsB. strong interatomic interactionC. strong metallic bondingD. all of the above. |
| Answer» Correct Answer - D | |
| 75. |
Transition elements in lower oxidation states act as Lewis acid because:A. they form complexesB. they are oxidising agentsC. they donate electronsD. they do not show catalytic properties. |
| Answer» Correct Answer - A | |
| 76. |
Transition elements are usually characterised by variable oxidation states but Zn does not show this property because of :A. completion of np-orbitalsB. completion of `(n-1)d` orbitals.C. completion of ns-orbitalsD. inert pair effect. |
| Answer» Correct Answer - B | |
| 77. |
The higher oxidation states of transition elements are found to be in the combination with A and B, which are:A. `F,O`B. `O,N`C. `O,Cl`D. `F,Cl` |
| Answer» Correct Answer - A | |
| 78. |
The variability of oxidation states of transition elements arises out of:A. incomplete filling of d-orbitalsB. oxidation states differing by a unit twoC. greater stability of lower oxidation stateD. greater stability of higher oxidation state |
| Answer» Correct Answer - A | |
| 79. |
How is the variability in oxidation states fo transition metals different from that of the non transition metaals? Illustrate with examples. |
| Answer» Correct Answer - In Transition elements the oxidation states vary from `01` to any highest oxidation state by one. For example, for manganese it may vary as `+2,3,+4,+5,+6,+7,`. In the non-transition element the variation is selective, always differing by `2, e.g. +2,+4 or +3,+5 or +4,+6 etc` | |
| 80. |
How would you account for the following : (i) Among lanthanoids,Ln(III) compounds are predominant. However,occasionally in solution or in solid compound,`+2` and `+4` ions are also obtained. (ii) The `E_(M^(2)//M)^(@)` for copper is positive `(0.34 V)` copper is the only metal in the first series of transition elements showing this behaviour. (iii) The metallic radii of the third `(5d)` series of transition metals are nearly the same as those of the corresponding member of the second series. |
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Answer» Correct Answer - (i) Lanthanoids show +3 oxidation state mostly as `2` electrons from outer `6s` orbital and one electron from `5d` orbital take part in bond formation. Some show `+2` and `4` oxidation state due to stability of half filled and completely filled `4f` orbitals. (ii) It is because of high `Delta_(a)H^(@)` (enthalpy of atomisation) and and low hydration enthalpy `Delta_(hyd)H^(@)`. (iii) it is due to lanthanoids contraction, effective nuclear charge remains almost same therefore , metallic radii are nearly same . |
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| 81. |
Which series of reactions correctly represents chemical rections related to iron and its compounds ?A. `Feoverset(Cl_(2)*heat)toFeCl_(3)overset(heat,air)toFeCl_(2)overset(Zn)toFe`B. `Feoverset(O_(2),heat)toFe_(3)O_(4)overset(CO,600^(@)C)toFeOoverset(CO,700^(@)C)toFe`C. `Feoverset(dil.H_(2)SO_(4))toFeSO_(4)overset(H_(2)SO_(4),O_(2))toFe_(2)(SO_(4))_(3)overset(Delta)toFe`D. `Feoverset(O_(2),heat)toFeOoverset(dil.H_(2)SO_(4))toFeSO_(4)overset(Delta)toFe` |
| Answer» Correct Answer - B | |
| 82. |
Which of the following statement are correct when a mixture of `Nacl` and `K_(2)Cr_(2)O_(7)` is gently warmed with conc.`H_(2)SO_(4)`?A. Deep red vapours are liberatedB. Deep red vapours dissolved in `NaOH(aq)` forming a yellow solution.C. Greenish yellow gas is liberatedD. Deep red vapour dissolve in water forming yellow solution. |
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Answer» Correct Answer - ABD `4Cl^(-)+Cr_(2)O_(7)^(2-)+6H^(+) to 2CrO_(2)Cl_(2)uparrow("deep red")+3H_(2)O` `CrO_(2)Cl_(2)+4OH^(-) to CrO_(4)^(2-)("yellow")+2H_(2)O+2Cl^(-)` `CrO_(2)Cl_(2)+H_(2)O to H_(2)CrO_(4)+HCl` |
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| 83. |
Which of the following reaction(s) is/are incorrect for silver nitrate?A. `6AgNO_(3)+3l_(2)("excees")+3H_(2)OtoAglO_(3)+5Agl+6HNO_(3)`B. `AgNO_(3)("excess")+2KCN toK[Ag(CN)_(2)]+KNO_(3)`C. `2AgNO_(3)+4Na_(2)S_(2)O_(3)("excess")to2Na_(3)[Ag(S_(2)O_(3))_(2)]+2NaNO_(3)`D. `PH_(3)+6AgNO_(3)+3H_(2)OtoAg+6HNO_(3)+H_(3)PO_(3)` |
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Answer» Correct Answer - AB (A) `6AgNO_(3)+3l_(2)("excess")+3H_(2)O to HlO_(3)+5Agl+5HNO_(3)` (B) `AgNO_(3)("excess")+KCN to AgCN("white")+KNO_(3)` |
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| 84. |
Which of the following reaction(s) is /are used for the preparation of anhydrous `FeCl_(3)`?A. `FeCl_(3).6H_(2)O+6SOCl_(2)toFeCl_(3)+12HCl+6SO_(2)`B. `Fe(OH)_(3)downarrow+3HCl to FeCl_(3)+3H_(2)O`C. `2Fe+4HCl(aq)+Cl_(2)to2FeCl_(3)+2H_(2)`D. `2Fe+3Cl_(2)(dry) to 2FeCl_(3)` |
| Answer» Correct Answer - AD | |
| 85. |
An aqueous solution of inorganic compounds (X) gives following reactions. (i) With an aqueous solution of barium chloride a precipitate insoluble in dilute `HCl` is obtained. (ii) Addition of excess of `Kl` gives a brown precipitate which turns white on additon of excess of hyposolution. (iii) With an aqueous solution of `K_(4)[Fe(CN)_(6)`, a brown coloured precipitate is obtained. Identity (X) and gives equation for the reaction for (i),(ii) and (iii) observations. |
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Answer» Correct Answer - `X=CuSO_(4)` (Step (i) suggested that compound (X) contain `SO_(4)^(2-)` redical. (ii) step (ii) suggested that the compound (X) contain `Cu^(2+)` radical. Reaction (i) `CuSO_(4)(X)+BaCl_(2) to BaSO_(4)downarrow("white")+CuCl_(2),BaSO_(4)` is insolule in `HCl` (ii) `2CuSO_(4)(X)+4KIto2CuI_(2)+2K_(2)SO_(4)` `2CuI_(2)("unstable") to 2Cu_(2)I_(2)downarrow("white")+I_(2)uparrow` `I_(2)+2Na_(2)S_(2)O_(3) toNa_(2)S_(4)O_(6)("colourless")+2NaI("colourless")` `2CuSO_(4)(X)+K_(4)[CN)_(6)] to Cu_(2)[Fe(CN)_(6)]downarrow("Brown")+2K_(2)SO_(4)` |
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| 86. |
`CrO(O_(2))_(2)` is stable in pyridine. Explain? |
| Answer» Correct Answer - It forms adducts with pyridine `CrO(O_(2))_(2)`. | |
| 87. |
Which of the following statement is false?A. `Cr_(2)O_(7)^(2-)` has a `Cr-O-Cr` bondB. `CrO_(4)^(2-)` is tetrahedral in shape.C. `Na_(2)Cr_(2)O_(7)` is a primary standard in volumetryD. `Na_(2)Cr_(2)O_(7)` is less soluble than `K_(2)Cr_(2)O_(7)` |
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Answer» Correct Answer - C The weight of `Na_(2)Cr_(2)O_(7)` cannot be accurately measured. |
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| 88. |
Which of the following statements concern with d-block metals?A. Compounds containing ions of transition elements are usually colouredB. Zinc has lowest melting point among 3d-series elements.C. They show variable oxidation states, which differ by two units onlyD. They easily form complexes. |
| Answer» Correct Answer - A::B::D | |
| 89. |
The metal(s) which does/do not form amalgam is/are:A. `Fe`B. `Pt`C. `Zn`D. `Ag` |
| Answer» Correct Answer - A::B | |
| 90. |
Which of the following compound affects mercury?A. `D_(2)O`B. `H_(2)O`C. `O_(3)`D. dil. HCl |
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Answer» Correct Answer - C `O_(3)` causes tailing of mercury. `Hg+O_(3)toHg_(2)O+O_(2)`. |
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| 91. |
Mercury is a liquid at `0^(@)` C because ofA. very high ionization energyB. weak metallic bonds.C. high heat of hydrationD. high heat of sublimation. |
| Answer» Correct Answer - A::B | |
| 92. |
The ionisation energies of tranition elements are:A. less than p-block elementsB. more than s-block elementsC. less than s-block elementsD. more than p-block elements. |
| Answer» Correct Answer - A::B | |
| 93. |
Write the formula of the compound formed by `K_(2)Cr_(2)O_(7)` in alkaline solution with `30% H_(2)O_(2)`: |
| Answer» Correct Answer - `K_(3)CrO_(8)` | |
| 94. |
The magnetic moment of `._25Mn` in ionic state is `sqrt(15)B.M`, then Mn is in:A. `+2` stateB. `+3` stateC. `+4` stateD. `+5` state |
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Answer» Correct Answer - C `sqrt(15)=sqrt(n(n+2)),n=3,"and three unpaired electrons are found when Mn is in" Mn^(4+) i.e,3d^(3)4s^(0)` "configuration as its metal electrons configuration is "[Ar]^(18)3d^(5)4s^(2)` |
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| 95. |
ActinidesA. Are all synthetic elementsB. includes element`104`C. Have only short lived isotopesD. have variable valency |
| Answer» Correct Answer - D | |
| 96. |
The `+3` ion of which one of the following has half filled `4f` subshell?A. `La`B. `Lu`C. `Gd`D. `Ac` |
| Answer» Correct Answer - C | |
| 97. |
which of the following statement is correct?A. The lesser number of oxidation state in `3d`-series in the beginning of the series is due to the presence of too few electrons to loose or shareB. The lesser number of oxidation states in `3d-series` towards the end of the series is due to the presence of too many electrons and thus fewer empty orbital to share electrons with the ligandsC. (A) and (B) bothD. None is correct |
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Answer» Correct Answer - C The lesser number of oxidation states in the begining of series can be due to the presence of smaller number of electrons to lose or share `(Sc,Ti)`. On the other hand, at the extreme right hand side end `(Cu,Zn)` lesser number of oxidation state is due to large number of d electrons so that only a fewer orbitals are available in which the electron can share with other for higher valence. |
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| 98. |
Which of the following statement is/are correct?A. transition metals and their many compounds act as good catalyst.B. The enthalpies of atomisation of the transition metals are highC. The transition metals generally form interstitial compounds with small atoms like `C,B,H` etc.D. All transition metal compounds are not paramagnetic. |
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Answer» Correct Answer - ABCD (A) This activity is ascribed to their ability to adopt multiple oxidation state and to form complexes. (B) Because of having larger number of unpaired electrons in their atoms, they have stronger interatomic interaction and hence stronger bonding between the atoms. (C) Transition metals like `Fe,Co,Ni,Cu etc`. Form interstitial compounds with elements such as hydrogen, boron, carbon and nitrogen . The small atoms of these non-metallic elementss `(H,B,C,N,etc)` get trapped in vacant spaces of the lattices of the transition metal atoms. |
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| 99. |
Which of the following is true for the speical having `3d^(4)` configuration ?A. `Cr^(2+)` is reducing in nature.B. `Mn^(3+)` is oxidising in nature.C. Both (A) and (B)D. None of those |
| Answer» `Cr^(2+)` is reducing as its configuration changes from `d^(4)` to `3d^(4)`, the latter having a halg-filled t^(3)`_(2g)` energy level of `3d` orbitals in octahedral crystal field soliting . On the other hand , the change from `Mn^(3+)`result in the half-filled `(d^(5))` configuration which has extra stability. Therefore(C) is correct. | |
| 100. |
Choose the correct statement(s).A. Successive enthalpies of transition elements do not increase as steeply as in the main group elements.B. The magnitude of the increase in the second and third ionsiation enthalpies for the successive elements of transition series, in general, is much higher.C. The lowest common oxidation state of transition metals is `+2`.D. The second ionisation enthalpy of Cr and Cu are usually high. |
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Answer» Correct Answer - A::B::C::D (d). `d^(5) and d^(10)` configurations of `M^(+)`, ions are disrupted, with considerable loss of exchange energy. |
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