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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. |
Identify the correct structure of dichromate ion.A. B. C. D. |
| Answer» Correct Answer - A | |
| 52. |
Which of the following compounds is used as the starting material for the preparation of potassium dichromate?A. `K_2SO_4Cr_2(SO_4)_324H_2O` (Chrome alum)B. `Pb CrO_4` (Chromite yellow)C. `FeCr_2O_4` (Chromite)D. `PbCrO_4PbO` (Chrome red) |
| Answer» Correct Answer - C | |
| 53. |
Ammonium dichromate on heating givesA. `CrO_(3)`B. `Cr_(2)O_(3)`C. CrOD. `CrO_(2)` |
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Answer» Correct Answer - B Ammonium dichromate, dichromate, `(NH_(4))_(2)Cr_(2)O_(7)`, is often used in 'volcano' demonstrations. If a red - hot wire or a lit match is touched to a pile of ammonium dichromate exothermic decomposition is intiated, emitting sparks and water vapor in a spectacular way. However, this is not a safe demonstration because a dust containing carcinogenic chromium `(+6)` copounds is usually relesed. The reaction is nonstoichlimetric, producing chromium (III) oxide, water vapor, nitrogen gas, and some ammonia gas. It is commonly represented as `(NH_(4))_(2) Cr_(2)O_(7)(s)` `overset (Delta) rarr` `Cr_(2)O_(3)(s) + N_(2)(g) + 4H_(2)(s)` |
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| 54. |
When intimate mixture of potassium dichromate and potassium chloride is heated with conc.`H_(2)SO_(4)` which of the following is produced in the form of red vapours ?A. `CrO_(3)`B. `Cr_(2)O_(3)`C. `CrO_(2)Cl_(2)`D. `CrCl_(2)` |
| Answer» Correct Answer - C | |
| 55. |
Assertion : Solid potassium dichromate gives greenish yellow vapour with concentrated `H_(2) SO_(4)` and solid ammonium chloride. Reason : The reaction of ammonium chloride with solide `K_(2) Cr_(2) O_(7)` and concentrated `H_(2) XO_(4)` produces chromyl chloride.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
| Answer» Correct Answer - D | |
| 56. |
Assertion : Potassium dichromates gives deep red vapours with concentrated `H_(2)SO_(4)` and sodium chloride. Reason : The reaction of sodium chloride with solid `K_(2) Cr_(2) O_(7)` and concentrated `H_(2) SO_(4)` produces chromyl cholride.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `K_(2) Cr_(2) O_(7) + 2H_(2)SO_(4) rarr KHSO_(4) + 2 CrO_(3) + H_(2) O` `[NaC1 + H_(2)SO_(4) rarr NaHSO_(4) + HC1] xx 4` `[CrO_(3) + 2HC1 rarr CrO_(2) C1_(2) + H_(2) O] xx 2` `bar(K_(2) Cr_(2) O_(7) + 6 H_(2) SO_(4) + 4NaC1 rarr 2KHSO_(4) + 4 NHSOH_(4) +)` `2 CrO_(2) C1_(2) + 3H_(2) O` `CrO_(2) C1_(2)` (Chromy1 chloride) is liberated as deep red vapour |
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| 57. |
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.A. 5.9 B.MB. 2.9 B.MC. 6.9 B.MD. 9.9 B.M |
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Answer» Correct Answer - A `mu=sqrt(n(n+2))` B.M. Electronic configuration of ion (25) = `1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2` No. of unpaired electrons (n) = 5 `mu=sqrt(5(5+2))=sqrt35=5.9 B.M` |
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| 58. |
The radiocative lanthanide isA. ytterbium `(Yb)`B. iron `(Fe)`C. promethium `(Pm)`D. copper `(Cu)` |
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Answer» Correct Answer - C promethium is synthetic and radioactive element. |
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| 59. |
The pair of lanthanides with the highest third ionization energy isA. `Lu` and `Yb`B. `Eu` and `Gd`C. `Eu` and `Yb`D. `Dy` and `Yb` |
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Answer» Correct Answer - C `Eu = [Xe] 4f^(7) 5d^(0) 6s^(2)` and `Yb = [Xe] 4f^(14) 5d^(0) 6s^(2)`. Removal of 3rd electron requires more energy from the elements. |
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| 60. |
Which one of the following is an electronic configuration of thorium?A. `[Rn] 5f^(2) 6 d^(0) 7 s^(2)`B. `[Rn] 5f^(1) 6 d^(1) 7 s^(2)`C. `[Rn] 5f^(0) 6 d^(2) 7 s^(2)`D. `[Rn] 5f^(2) 6 d^(2) 7 s^(2)` |
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Answer» Correct Answer - C Thorium `(Z = 90)` have `[Rn] 5f^(0) 6d^(2) 7 s^(2)` configuration. |
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| 61. |
Consider the following statement in respect of Lantinanoids : (i) The basic strength of hydroxides of lanthanoids increases from `La (OH)_(3)` to `Lu (OH)_(3)`. (ii) The lanthanoid ions `Lu^(3+), Yb^(2+)` and `Ce^(4+)` are diamagnetic. Which of the statements (s) given above is /are correct ?A. 1 onlyB. 2 onlyC. Both 1 and 2D. neither 1 not 2 |
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Answer» Correct Answer - B Basic strength decreases from `La(OH)_3` to `Lu(OH)_3`. Hence, (1) is incorrect. `Ce=[Xe] 4f^1 5d^1 6s^2, Ce^(4+)=[Xe]` `Yb =[Xe]4f^14 6s^2, Yb^(2+)=[Xe] 4f^14` `Lu=[Xe] 4f^14 5d^1 6s^2, Lu^(3+)=[Xe]4f^14` |
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| 62. |
(a). `CuSO_4 .5H_2Ooverset(100^@)to(A)overset(230^@)to(B)overset(800^@)to(C)+(D)` (b). `AgNO_3overset(red hot)to(E)+(F)+O_2` |
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Answer» (i) ` CuSO_(4).5H_(2)O overset(100^(@)C) to underset("[A]")(CuSO_(4) ). H_(2)O overset(280^(@)C) to underset([B])(CuSO_(4)) overset(800^(@)C) to underset([C])(CuO)+ overset([D])(SO_(2))+(1)/(2)O_(2)` (ii) `2AgNO_(3) overset("red hot")to underset([E])(2Ag)+ underset([F])(2NO_(2))+O_(2)` |
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| 63. |
What can be inferred from the magnetic moment values of the following complex species ?`{:("Example","Magnetic",(BM)),(K_(4)[Mn(CN)_(6)),2.2),([Fe(H_(2)O)_(6)]^(2+),5.3),(K_(2)[MnCl_(4)],5.9):}` |
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Answer» Magnetic moment `(mu)` is given as `mu=sqrt(n(n+2))` For value n = 1, `mu=sqrt(1(1+2))=sqrt(3)=1.732` For value n = 2, `mu=sqrt(2(2+2))=sqrt(8)=2.83` For value n = 3, `mu=sqrt(3(3+2))=sqrt(15)=2.87` For value n = 4, `mu=sqrt(4(4+2))=sqrt(24)=4.899` For value n = 5, `mu=sqrt(5(5+2))=sqrt(35)=5.92` (i) `K_(4)[Mn(CN)_(6)]` For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,b `sqrt(n(n+2))=2.2` We can see from the above calculation that the given value is closest to n=1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital. Hence, we can say that `CN^(−)` is a strong field ligand that causes the pairing of electrons (ii) `[Fe(H_(2)O)_(6)]^(2+)` `sqrt(n(n+2))=5.3` We can see from the above calculation that the given value is closest to n=4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d-orbital. Hence, we can say that `H_(2)O` is a weak field ligand and does not cause the pairing of electrons. (iii) `K_(2)[MnCl_(4)]` `sqrt(n(n+2))=5.9` We can see from the above calculation that the given value is closest to n=5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital. Hence, we can say that `Cl^(-)` is a weak field ligand and does not cause the pairing of electrons. |
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| 64. |
Choose the correct statement regarding bonding in `FeCl_(3)` (I) It contains `2c-2e^(-)` bond (II) It contains `3c-2e^(-)` bond (III) It contain co-ordinate bondA. (I),(II)B. (I),(III)C. (II),(III)D. (I),(II) &(III) |
| Answer» Correct Answer - B | |
| 65. |
How would you account for the increasing oxidising power in the series `VO_2^(o+)ltCr_2O_7^(2-)< MnO_4^(ɵ)`? |
| Answer» This is due to increasing stability of the lower species to which they are reduced. | |
| 66. |
Identify A, B and C and write chemical equtions. |
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Answer» `A to Na_(2)SO_(4)` `MnO_(2)+KOH to K_(2)MnO_(4)` (dark green compound) `K_(2)MnO_(4) + H_(2)SO_(4) to KMnO_(4)` (purple crystals) `B to MnSO_(4)` `{:(C to CH_(2)OH),(" "|),( " "CH_(2)OH):}` |
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| 67. |
Which of the following compounds is known as corrosive sublimate ?A. `ZnCI_(2)`B. `CdCI_(2)`C. `HgCI_(2)`D. `CuCI_(2)` |
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Answer» Correct Answer - C The commonest among mercuty (II) halide is mercury (II) chloride, which is known as corrosive sublimate. It may be prepared by heating mercuric sulphate with common salt in the presence of little manganese dioxide: `HgSO_(4) + 2NaCI` `overset (MnO_(2)) underset (Delta) rarr` `HgCI_(2) + Na_(2)SO_(4)` The function of `MnO_(2)` as a reducing is to prevent the formation of any mercurous chloride. Merciric chloride sublimes off and condenses of the coolar parts of the vessel. It can be prepared by dissolving mercuric oxide in hydrochloric acid or by dissolving Hg (or `Hg_(2) CI_(2)`) in aquaregia. (i) `HgO + 2HCI rarr HgCI_(2) + H_(2)O` (ii) `3HCI + HNO_(3) rarr NOCI + 2H_(2)O + CI_(2)` `Hg + 2CI rarr HgCI_(2)` `Hg_(2)CI_(2) + 2CI rarr 2HgCI_(2)` Mercuric chloride is commercially made by heating the metal mercury is a currenrt of chlorine: `Hg + CI_(2) rarr HgCI_(2)` Mercuric chloride forms white crystalline mass but form aqueous solution it crystallises into colorless needles. It is a covalent compound, readily soluble in water can be increased by the addition of chlotid ions, when a soluble tetrachlormercurate (II) complex ion is formed: `HgCI_(2) + 2CI^(-) rarr [Hg CI_(4)]^(2-)` It has corrosive action and is very poisonous (Its antidote is the white of an egg which eliminates it from the system in the form of a coagulated mass). On heating it sublimes. Therefore, it is known as corrosive sublimate. It has been used as an antiseptic since the middle ages. |
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| 68. |
Potassium iodide forms a scarlet precipitate of mercuric iodide when added to mercuric chloride solution. The precipitate of mercuric iodide dissolves in excess of potassium iodide forming a complexA. `K_(2)HgI_(4)`B. `K_(4)HgI_(6)`C. `K_(3)HgI_(3)`D. `KHgI_(2)` |
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Answer» Correct Answer - A Mercuric iodide is obtained as a scarlet prcipitate on addition of postassium iodide to a solution of mercuric chloride. It can also be obtaned by grinding mercury with appropriate amount of iodine. Mercuric iodide, although sparingly soluble in water, dissolves readily in a solution of potassium iodide due to th eformation of iodo complex: `HgCI_(2) + 2KI rarr 2KCI + Hf I_(2)` `HgI_(2) + 2KI rarr K_(2)[Hg I_(4)]` potassium tetraiodo-mercurate (II) The teraiodo complex forms light yellow crystals, `K_(2)[Hg I_(4)]`. `2H_(2)O`, freely soluble in water and alcohol. |
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| 69. |
Verdigris isA. Basic copper acetateB. Basic lead accetateC. Basic leadD. None of these |
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Answer» Correct Answer - A Basic copper acetate. Verdigris `(CH_(3) COO)_(2) Cu. Cu (OH)_(2)` is a blue green power used in green pigment and in dyes, and also in the manufacture of inscticides and fungicides. |
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| 70. |
`AgCl` dissolves in a solution of `NH_(3)` but not in water becauseA. `NH_(3)` is a better solvent then `H_(2) O`B. `Ag^(+)` forms a complex ion with `NH_(3)`C. `NH_(3)` is a stronger base than `H_(2) O`D. The dipole moment of water is higher than `NH_(3)` |
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Answer» Correct Answer - B `Ag^(+)` forms a complex ion with `NH_(3)` `AgCl + 2 NH_(3) rarr [Ag (NH_(3))_(2)] Cl` |
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| 71. |
Assertion : `AgC1` dissolves in `NH_(4) OH` solution. Reason: Due to formation of a complex.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `AgC1` on adding to a solution of `NH_(4)OH` solution dissolves to form a complex diamine silver chloride. `AgC1 + 2NH_(4) Oh -lt Ag (NH_(2))_(2) C1 + 2H_(2) O` Therefore, both assertion and reaosn are ture and reason in a correct explanation of assertion. |
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| 72. |
Mercuric chloride reacts with aqueous ammonia to form a white precipitate ofA. `Hg (NH_(3))_(2)CI_(2)`B. `Hg (NH_(2))_(2)CI_(2)`C. `Hg (NH_(3))_(2)CI`D. `HG (NH_(2))CI` |
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Answer» Correct Answer - D When aqueous ammonia is added to the solution of mercuric chloride, a white precipitate) is formed: `HgCI_(2) + 2NH_(3) (aq) rarr Hg(NH_(2))CI + NH_(4)CI` The infusible white precipitate decomposes without melting on heating. On the other hand, gaseous ammonia or ammonium chloride produces a fusible white precipitate of diammine mercury (II) chloride. `HgCI_(2) + 2NH_(3) rarr Hg(NH_(3))_(2) CI_(2)` |
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| 73. |
Which of the statement is not correct?A. Potassium permanganate is powerful oxidising substaneB. Potassium is a weaker oxidising substnace than potassium dichromateC. Potassium permanganate is a stronger oxidising substance the potassium dichromateD. Potassium dichromate oxidises a scondary alcohol into a kenton |
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Answer» Correct Answer - B In acidic medium, `KMnO_(4)` gives 5 oxygen atoms while acidic `K_(2)Cr_(2)O_(7)` gives 3 oxygen atoms. |
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| 74. |
How many water molecules in blue vitriol are coordinated to the metal ?A. FiveB. FourC. ThreeD. Two |
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Answer» Correct Answer - B The crystalline copper (II) sulphate, `CuSO_(4)` . `5H_(2)O`, has the structure in which four water molecules are coordinated to the central copper catio9n at the centre of a square but the fifth water molecule is held by hydrogen bonds between a sulphate ion and a coordinated water molecule. The fifth hudrogen bonded water molecule is deeply embedded in the crystal lattice and hence not easily removed: `[Cu(H_(2)O)_(4)]^(2+) SO_(4)^(2-).H_(2)O` |
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| 75. |
Acidified solution of chromic acid on treatment with hydrogen peroxide yieldsA. `CrO_(3) + H_(2) O + O_(2)`B. `CrO_(5) + H_(2) O`C. `Cr_(2) O_(3) + H_(2) O + O_(2)`D. `H_(2) Cr_(2) O7 + H_(2) O + O_(2)` |
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Answer» Correct Answer - B `K_(2) Cr_(2)O_(7) + H_(2)SO_(4) + 4H_(2)O_(4) rarr K_(2)SO_(4) + 2CrO_(5) + 5H_(2) O` |
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| 76. |
On heating at `750^(@)C`, blue virtiol yieldsA. `Cu + SO_(2) + SO_(3)`B. `Cu_(2)O + SO_(3)`C. `CuO + SO_(3)`D. `Cu + S + H_(2)O` |
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Answer» Correct Answer - C Blue virtiol crystals effloresce on expouser and get converted into a pale blue powder, `CuSO_(4)` . `3H_(2)O` . When heated to `100^(@)C`, bluish white monohydrate `CuSO_(4)` . `H_(2)O` is formed. The monohydrate loses last molecule of water at `250^(@)` giving the anhydrous salt `CuSO_(4)`, which is white. These changes are reversible, the colorless anhydrous salt (white) regains its blue color when moistened with a drop of water (test of water due to the formation of blue virtiol, `CuSO_(4)` . `5H_(2)O`). If the anhydrous salt is heated at `750^(@)C`, it decomposes into cuopric oxide and sulphur trioxide: `CuSO_(4)` `overset (750^(@)C) rarr` `CuO + SO_(3)` |
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| 77. |
Addition of NaOH on `Zn^(2+)` ion gives a white ppt. which on addition of excess of NaOH which dissolves . In this solution Zn exists inA. `Zn^(2+)` formB. Cationic formC. Anionic formD. Both (2) & (3) |
| Answer» Correct Answer - C | |
| 78. |
The colours of ppt. A,B and C respectively areA. Black , yellow , deep yellowB. Black , red , yellowC. Brown, red , whiteD. Black , white , red |
| Answer» Correct Answer - B | |
| 79. |
Each coinage metal hasA. 8B. 2C. 18D. 32 |
| Answer» Correct Answer - C | |
| 80. |
The products (A) and (B) are respectivelyA. `KI,I_(2)`B. `I_(2),KIO_(3)`C. `KIO_(3),KIO_(4)`D. `I_(2),I_(2)` |
| Answer» Correct Answer - B | |
| 81. |
The pair having similar magnetic moment isA. `Ti^(2+),V^(3+)`B. `Cr^(3+),Mn^(2+)`C. `Mn^(2+),Fe^(3+)`D. `Fe^(2+),Mn^(2+)` |
| Answer» Correct Answer - C | |
| 82. |
Due to lanthanoid contraction which of the following properties is not expected to be similar in the sme vertical columns of second and third row transition elements ?A. Atomic radiiB. Ionisation elergiesC. Magnetic momentsD. Lattice energies |
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Answer» Correct Answer - C Magnetic moments depend upon the number of unpaired electrons. |
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| 83. |
STATEMENT-1 : In zinc, outermost shell is completely filled and STATEMENT-2: Zn does not much resemblance with transition metals.A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - D | |
| 84. |
Europium isA. s-block elementB. p-block elementC. f-block elementD. d-block element |
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Answer» Correct Answer - C Europium is a `f`-block elements as ti follows the general electronic configuration of the `f`-block elements `(4 f^(1 -14) 5 d^(0,1) 6s^(2))` `Eu = [Xe] 4 f^(7) 6s^(2)` |
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| 85. |
Which of the following statements is incorrect ?A. Magnetic susceptibility of transition metal compounds is experimentally measured by Gouy method.B. The unit of magnetic moment is Bohr magneton, `mu_(B)`.C. The value of Bohr magneton is given by the formula, `mu_(B) = eh//4pi m_(e)`D. One Bohr magneton is equal to `9.27xx10^(-23) jT^(-1)` |
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Answer» Correct Answer - D `mu_(B) = (eh)/(4pi m_(e)) = 9.27xx10^(-24)JT^(-1)` |
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| 86. |
Which of the following is not ferromagnetic ?A. `Mn^(4+)`B. `Ti^(4+)`C. CoD. both (1) and (2) |
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Answer» Correct Answer - D The metals iron, cobalt, nickel, and certain alloys are vastly more magnetic than any other known substance: these metals are said to be ferromagnetic. Ferromagnetism is due to unbalanced electron spin in the inner electron orbits of the elements concerned which gives the atom a resultant magnetic moment. The ionic spacing in ferromagnetic crystals is such that very large forces, called exchange force, cause the alignment of all the individual magnetic moments of large groups of atoms to given highly magnetic domains. In an unmagetized piece of ferromagnatic material such as iron, these domains are orignment at random, their magnetic axes pointing in all direction. The application of an extrenal field serves to line up the domain axes, giving rise to the observed magnetism. Ferromagnetic substance have very large magnetic permeabilities, which vary with the strength of the applied field. A given ferromagnetic substance loses its ferromagnetic properties at a certain critical temperature, the curie temperature for the substance. `Mn^(4+) (3d^(3))` is paramagnetic due to unpaired electrons while `Ti^(4+) (3d^(0))` is diamagnetic due to paired spins. |
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| 87. |
On the basis of position in the electrolchemical series, the metal which does not displace `H_(2)` from water and acid is `:`A. `Ba`B. `Al`C. `Pb`D. `Hg` |
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Answer» Correct Answer - D Hg because it lies below, `H_(2)` in electrochemical series and thus cannot reduce it |
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| 88. |
Which of the following series of d-block elements exhibits the minimum deviation in electronic configuration ?A. 3d seriesB. 4d seriesC. 5d seriesD. 6d series |
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Answer» Correct Answer - D It has only one deviation. The expected configuration of Rg is [Rn] `6d^(9) 7 s^(2)` but its actual configuration is [Rn] `6d^(10) 7s^(1)`. The maximum deviation occur in 4d series: Nb, Mo, Tc, Ru, Rh, Pd and Ag. On the other hand, both 3d series and 5d series each have two deviations `[Cr, Cu and Pt. Au]` respectively. |
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| 89. |
`3 d^(10) 4 s^(0)` electronic configuration exhibitsA. `Hg^(++)`B. `Cu^(++)`C. `Cd^(++)`D. `Zn^(++)` |
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Answer» Correct Answer - D `Zn - 3 d^(10) 4 s^(2)` `Zn^(+ + ) - 3 d^(10) 4 s^(0)` |
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| 90. |
The inner transition element that is radioactive isA. PmB. GdC. LuD. Sm |
| Answer» Correct Answer - A | |
| 91. |
Transition metal with low oxidation state will act as:A. a baseB. an acidC. both (a) and (b)D. None of these |
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Answer» Correct Answer - B Transition metal ions in lower oxidation states form cations. These cations act as Lewis acids e.g., `Fe rarr Fe^(2 +)` (Lewis acid) `+ 2 e^(-e)`. |
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| 92. |
The names transition and inner transition metals are used to indicate the element of:A. d-block elements onlyB. f-block elements onlyC. p-and-d-blocks element respectivelyD. d-and f-blocks elements respectively |
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Answer» Correct Answer - D In electron configuration of d-block element last electron filled in `(n - 1) d` subshell & these element placed between s-block and p-block element in periodic table. In f-block element, last electron filled in `(n - 2)` f-subshell and these elements are placed in third `B` group in transition element. These element known as inner transition metal elements. |
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| 93. |
Which one of the following is reduced by hydrogen peroxide in acid medium?A. Potassium permanganateB. Potassium iodideC. Ferrous sulpateD. Potassium ferrocyanide |
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Answer» Correct Answer - A `H_(2) O_(2)` reduces acidified `KMnO_(4)` solution. As a result. The pind colour of `KMnO_(4)` is changed. |
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| 94. |
Which of the following represents the electronic configuration of a transition element?A. `1 s^(2), 2 s^(2) p^(6)`…………………..`n s^(2) p^(3)`B. `1 s^(2), 2 s^(2) p^(6)`…………………..`n s^(2) p^(6)`C. `1 s^(2), 2 s^(2) p^(6)`…………………..`n s^(2) p^(3) d^(10), (n + 1) s^(2) p^(1)`D. `1 s^(2), 2 s^(2) p^(6)`…………………..`n s^(2) p^(3) d^(3), (n + 1) s^(2)` |
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Answer» Correct Answer - D `1s^(2), 2s^(2), p^(6) ………..ns^(2) p^(6) d^(3), (n + 1) s^(2)` as last electron enters d-subshell. |
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| 95. |
Which of the following will have standard oxidation potential less than `SHE`?A. `Zn`B. `Fe`C. `Cu`D. `Ni` |
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Answer» Correct Answer - C `Cu` as it comes after `H` in electrochemical series. |
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| 96. |
Which of the following is not an amphoteric ion?A. `Al^(3+)`B. `Cr^(3+)`C. `Fe^(2+)`D. `Zn^(2+)` |
| Answer» Correct Answer - C | |
| 97. |
Which of the following oxides of manganes is amphoteric?A. `MnO_(2)`B. `Mn_(2)O_(3)`C. `Mn_(2)O_(7)`D. `MnO` |
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Answer» Correct Answer - A `MnO_(2)` forms amphoteric oxide to intermediate oxidation state. |
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| 98. |
Which of the following ions is not amphoteric?A. `A1^(3+)`B. `Fe^(3+)`C. `Cr^(3+)`D. `Zn^(2+)` |
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Answer» Correct Answer - B All the oxides of `Fe (FeO, Fe_(2) O_(3)` and `Fe_(3) O_(4))` are basic in nature. |
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| 99. |
Which one of the following is an example of non-typical transition elements?A. `Li, K, Na`B. `Zn, Cd, Hg`C. `Be, A1, Pb`D. `Ba, Ca, Sr` |
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Answer» Correct Answer - B `Zn` Cd and Hg are not-typical transition elements because they have complete d-orbits. |
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| 100. |
Which of the following is an amphoteric oxide?A. `V_2O_5,Cr_2O_3`B. `Mn_2O_7,CrO_3`C. `CrO,V_2O_5`D. `V_2O_5,V_2O_4` |
| Answer» Correct Answer - A | |