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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.
| 1. |
Lanthanide series start from the elements with atomic numberA. 27 to 71B. 58 to 71C. 56 to 70D. 60 to 74 |
| Answer» Correct Answer - B | |
| 2. |
The correct order of ionic radii of `Y^(3+), La ^(3+), Eu ^(3+) and Lu ^(3+)` isA. `Y^(3+) lt Lu ^(3+) lt Eu ^(3+) lt La ^(3+)`B. `La ^(3+) lt Eu ^(3+) lt Lu ^(3+) lt Y^(3+)`C. `Lu ^(3+), lt Eu ^(3+) lt La ^(3+) lt Y ^(3+)`D. `Y ^(3+) lt La ^(3+)lt Lu ^(3+) lt Eu^(3+)` |
| Answer» Correct Answer - A | |
| 3. |
Lanthanide contraction is due to increase inA. iionic radiiB. shielding of 4f electronsC. effective nuclear chargeD. size of 4f orbitals |
| Answer» Correct Answer - C | |
| 4. |
Lanthanide contraction is related toA. valence electronB. ionic radiiC. densitiesD. nuclear mass of various membere of the series |
| Answer» Correct Answer - B | |
| 5. |
The steady decrease in atomic size in lanthanide series is called lanthanide confraction and in all amounts toA. 150 PmB. `10 Pm`C. `20 Pm`D. `40 Pm` |
| Answer» Correct Answer - B | |
| 6. |
Lanthanide contraction is due to increase inA. shielding of 4f electronsB. effective nuclear chargeC. atomic numberD. size of 4f orbital |
| Answer» Correct Answer - B | |
| 7. |
Lanthanide contraction due toA. poor shielding of 5f orbitalsB. poor shielding of 3f orbitalsC. prerfect shielding of 4f orbitalsD. poor shielding of 4f orbitals |
| Answer» Correct Answer - D | |
| 8. |
The lanthanide contraction is responsible for the fact thatA. Zr and Yb have same sizeB. Zr and Nb have same oxidation sateC. Zr and Hf have same radiusD. Zr and Zn have same oxidation state |
| Answer» Correct Answer - C | |
| 9. |
The most characteristic oxidation state of lanthanide isA. `+2`B. `+3`C. `+4`D. `+6` |
| Answer» Correct Answer - B | |
| 10. |
Name a member of lanthanide series which is well known to exhibit `+4` oxidation state `,`A. CeB. LaC. LuD. Pr |
| Answer» Correct Answer - a | |
| 11. |
The steady decrease in ionic size in lanthanides due ot lanthanide contraction and in all amount toA. `10` pmB. `40` pmC. `18` pmD. `50` pm |
| Answer» Correct Answer - C | |
| 12. |
The pricipal oxidation state of lanthanide isA. `+2`B. `+3`C. `+4`D. `+5` |
| Answer» Correct Answer - B | |
| 13. |
Lightest and the heaviest transition metals respectively areA. Sc, OSB. Sc, FeC. OS, ScD. Cu, Fe |
| Answer» Correct Answer - A | |
| 14. |
In which of the lanthanide elements, 5d elements does not shift to 4f orbita ls ?A. Ce, Eu, YbB. La, Gd, LuC. Ce, Nd, DyD. Sm, Ho, Er |
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Answer» Correct Answer - B `La(57) to [Xe] 6s^2 5d^1 4f^0` Gd (64) `to [Xe] 6s^2 5d^1 4f^7` `Lu (71) to [Xe ] 6s^2 5d^1 4f^14` |
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| 15. |
Which of the following lanthanoids show `+2` oxidation state besides the characterisitc oxidation state `+3` lanthanoids ?A. CeB. E uC. YbD. Ho |
| Answer» Correct Answer - b,c | |
| 16. |
Which of the following lanthanoids show `+2` oxidation state besides the characteristic oxidation state +3 of lanthanoids?A. CeB. EuC. YbD. Ho |
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Answer» Correct Answer - B::C (a) Cerium (Z = 57) implies Electronic configuration = `[Xe]4f^(5)5d^(0)6s^(2)` Oxidation state of Ce = +3, +4 (b) Europium (Z = 63) implies Electronic configuration = `[Xe]4f^(7)5d^(0)6s^(2)` Oxidation state of Eu = +2, +3 (c ) Ytterbium (Z = 70) implies Electronic configuration = `[Xe]4f^(14)5d^(0)6s^(2)` Oxidation state of Yb = +2, +3 (d) Holmium (Z = 67) implies Electronic configuration = `[Xe]4f^(11)5d^(0)6s^(2)` Oxidation state of Ho = +3 |
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| 17. |
Which of the following ions show higher spin only magnetic moment value ?A. `Ti^(3+)`B. `Mn^(2+)`C. `Fe^(2+)`D. `Co^(3+)` |
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Answer» Correct Answer - B::C (b,c) are both correct |
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| 18. |
The outer expected electronic configuration `5f^(5),6d^(1), 7s^(2)` is of the elementA. AcB. AmC. NoD. Tb |
| Answer» Correct Answer - D | |
| 19. |
General electronic configurationof actinoids is `(n-2)f^(1-14) ( n-1)d^(0-2) ns^(2)` . Which of the following actinoids have one electron in 6d orbital ?A. U ( Atomic no. 92)B. Np ( Atomic no . 93)C. Pu ( Atomic no. 94)D. Am ( Atomic no. 95) |
| Answer» Correct Answer - a,b | |
| 20. |
Which of the following lanthanoids show `+2` oxidation state besides the characteristic oxidation state +3 of lanthanoids?A. CeB. EuC. YpD. Ho |
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Answer» Correct Answer - B::C (b,c) Both Europian and Ytterbium show +2 oxidation ste in addition to +3 oxidation state. |
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| 21. |
General electronic configuration of actinoids is `(n-2)f^(1-14)(n-1)d^(0-2)ns^(2)`. Which of the following actinoids have one electron in 6d orbital?A. U (Atomic number. 92)B. Np (Atomic number. 93)C. Pu (Atomic number. 94)D. Am (Atomic number. 95) |
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Answer» Correct Answer - A::B General electronic configuration of actinoids is `(n-2)f^(1-14)(n-1)d^(0-2)ns^(2)`. U and Np each have one electron in 6d orbital. |
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| 22. |
General electronic configuration of actinoids is `(n-2)f^(1-14)(n-1)d^(0-2)ns^(2)`. Which of the following actinoids have one electron in 6d orbital?A. U (atomic no. 92)B. Np(atomic no. 93)C. Pu (atomic no. 94)D. Am (atomic no. 95) |
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Answer» Correct Answer - A::B (a,b) Both uranium are Neptunium have one electron in 6d orbital. |
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| 23. |
In the form of dichromate, Cr(VI) is a strong oxidising agent in acidic medium but Mo(VI) in `Mo0_(3)` and W(VI) in `W0_(3)` are not becauseA. Cr(VI) is more stable than Mo(VI) and W(VI)B. Mo(VI) and W(VI) are more stable than Cr(VI).C. higher oxidation states of heavier members of group-6 of transition sereis are more stable.D. lower oxidation states of heavier members of group-6 of transition sereis are more stable. |
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Answer» Correct Answer - B::C (b,c) are the correct answers. |
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| 24. |
Which of the following actinoids show oxidation states upto +7?A. AmB. PuC. UD. Np |
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Answer» Correct Answer - B::D The oxidation states of the following actinoids are (a) Americium (Z = 95), Electronic configuration = `[Rn]5f^(7)6d^(0)7s^(2)` Oxidation states shown by Am = + 3, + 4, + 5, + 6. (b) Plutonium (Z = 94), Electronic configuration = `[Rn]5f^(6)6d^(0)7s^(2)` Oxidation states shown by Pu = + 3, + 4, + 5, + 6, 7. (c ) Uranium (Z = 92), Electronic configuration = `[Rn]5f^(3)6d^(1)7s^(2)` Oxidation states shown by U = + 3, + 4, + 5, + 6. (d) Neptunium (Z = 93), Electronic configuration = `[Rn]5f^(4)6d^(1)7s^(2)` Oxidation states shown by Np = + 3, + 4, + 5, + 6, + 7. |
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| 25. |
Which of the following actinoids show oxidation states upto `+7` ?A. AmB. PuC. UD. Np |
| Answer» Correct Answer - b,d | |
| 26. |
Most common oxidation state of actinoids is ......... .A. `+4`B. `+5`C. `+2`D. `+3` |
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Answer» Correct Answer - D Common oxidation state of actinide is +3 which shows increasing stability for heavier elements |
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| 27. |
The highest magnetic moments is shown by the transition metal ion with the outermost electronic configuration is:A. `3d ^(2)`B. `3d ^(5)`C. `3d^(7)`D. `3d^(9)` |
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Answer» Correct Answer - B `3d^(6)` as this configuration corresponds to maximum number of unpaired electrons. |
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| 28. |
The highest magnetic moments is shown by the transition metal ion with the outermost electronic configuration is:A. `3d^2`B. `3d^5`C. `3d^7`D. `3d^9` |
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Answer» Correct Answer - B `3d^5` have five unpaired `e^-` in 3d orbitals (which is maximum ) |
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| 29. |
Four successive members of the first row transition elements are listed below with their atomic number. Which one of them is expected to have the highest third ionisation enthalpy ?A. Vanadium (Z=23)B. Chromium (Z=24)C. Iron(Z=26)D. Manganese (Z=25) |
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Answer» Correct Answer - D In `._(23)V=1s^(2),2s^(2),2p^(6),3s^(2)3p^(6)3d^(3),4s^(2)` Third electron which is removed in third ionisation potential belongs to `3d^(3)` -subshell. `._(24)Cr=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6)3d^(5),4s^(1)` third electron which is removed in thired ionisation potential belongs to `3d^(3)`-subshell. `._(26)Fe=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(6),4s^(2)` Third electron which is removed in third ionisation potantial belong to `3d^(6)`-subshell `._(25)Mn=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(5),4s^(2)` Third electron which is removed in third ionisation potential belongs to `3d^(5)`-subshell. In all elements shell and subshell are same . Required amount of energy (enthalpy) is based upon the stability in an ion because it is half-filled subshell (while other are incomplete). So , Mn shows highest third ionisation potential or enthalpy. |
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| 30. |
Write balanced equation for the follownig `:` (i) Potassium permanganate is added to hot solution of manganous sulphate. (ii) Nitrogen is obtained in the reaction of aqueous ammonia with potassium permanganate. |
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Answer» (i)` 2KMnO_(4) +H_(2) rarr 2KOH + 2MnO_(2)+ 3(O)` `MnSO_(4) + H_(2)O+ (O) rarr MnO_(2)+ H_(2)SO_(4) ] xx 3` `2KOH + H_(2)SO_(4) rarr K_(2)SO_(4) + 2H_(2)O` `bar(2KMnO_(4)+ 3MnSO_(4) + 2H_(2)Orarr 5MnO_(2) + K_(2)SO_(4)+ 2H_(2)SO_(4))` (ii) `2KMnO_(4) + H_(2)O rarr 2MnO_(2) + 2KOH + 3(O)` `2NH_(3) + 3(O) rarr N_(2)+3H_(2)O` `bar(2KMnO_(4) + 2NH_(3) rarr 2MnO_(2) + 2KOH + 2H_(2)O+ N_(2))` |
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| 31. |
Which of the following does not considered as transition element ?A. AuB. HgC. LaD. Pt |
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Answer» Correct Answer - B Hg has completely filled 6s and 5d orbitals `6s^2 5d^10` `therefore` Not transition element |
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| 32. |
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in the respective vertical columns. Give special emphasis on the following point : (i) electronic configuration (ii) oxidation states (iii) inonisation enthalpies (iv) atomic sizes. |
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Answer» (i) Electronif configuration. There are some exceptions in the electronic configurations. In all the three series. (ii) Oxidation state. The elements belonging to the different series but present in the same group have similar electronic configuration and therefore, exhibit almost same variable oxidation state. In general, these are maximum in the middle of the series while minimum towards the end. (iii) Ionisation enthalpies. In general, the ionisation enthalpies in all the three transition series increase from left to the right. However, the gaps in the two successive elements in a particular series are small and are also not regular. The first three ionisation enthalpies of the elements present in the first transition series are given in the text part. The `Delta_(i)H^(1)` values of the elements belonging to 5d series an higher as compared to those belonging to 3d to 4d series in the same group because of poor shelding by intervening 4f electrons present. (iv) Atomic size. In all the three transition series, the atomic as well as ionic radii of the elements increase from left to the right. The values for 3d series are given in the text part. However, the increase in their values are not as much as expected since the shielding by (n-1)d electrons in not as much as expected. In a particular group, the atomic radius of the elements belonging to 4d series is more than the element in the 3d series. However, the gaps in the elements belonging to 4d and 5d series are negligible on account of lanthanoid contraction which the elements of 5d experience. |
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| 33. |
Potassium dichromate is usedA. in electroplatingB. as a reducing agentC. oxidise ferrouse ion into ferric ions in acidic media asD. oxidise ferrous ions into ferric ions in acidic media as an oxidsing agent |
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Answer» Correct Answer - C Potassium dichromate is an oxidising agent, thus it oxidises ferrous ion into ferric ion in acidic medium. `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O),(" "[Fe^(2+)rarrFe^(3+)+e^(-)]),(bar(Cr_(2)O_(7)^(2-)+6Fe^(2+)+14 H^(+) rarr 6Fe^(3+)+2Cr^(3+)+7H_(2)O)):}` |
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| 34. |
Which of the following compounds is used as the starting material for the preparation of potassium dichromate?A. `K_2SO_4.Cr_2(SO_4)`B. `PbCrO_4` (chrome yellow)C. `FeCr_(2)O_4` (chromite)D. `PbCrO_4_4PbO` (chrome red) |
| Answer» Correct Answer - c | |
| 35. |
Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on solution of potassium dichromate ? |
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Answer» Preparation from chromite: For answer, consult Section 8. Effect to increasing pH: The solution of potassium dichromate `(K_(2)Cr_(2)O_(7))`in water is orange in colour. Increasing the pH i.e. on adding the base, the potassium dichromate changes to potassium chromate `(K_(2)Cr_(2)O_(4))` which yellow in colour. Thus, on increasing the pH, the colour of the solution changes from orange to yellow. `underset("(Orange)")(K_(2)Cr_(2)O_(7)(aq))+2KOH(aq) to underset("(Yellow)")(2K_(2)CrO_(4)(aq))+H_(2)O(l)` |
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| 36. |
Silver atom has completely filled `d` orbitals `(4d^(10))` in its ground state. How can you say it is a transition element? |
| Answer» Silver `(Z=47)` belong to group 11 of d-block ( Cu, Ag, Au) and its outer electronic configuration is `4d^(10)5s^(1)`. It shows +1 oxidation state ( `4d^(10)` configuration) in silver halides ( e.g. AgCl). However, it can also exhibit +2 oxidation state ( `4d^(9)` configuration) in compound like `AgF_(2)` and AgO. Due to the presence of half filled d-orbital, silver is a transition metal | |
| 37. |
Silver atom has completely filled d-orbitals `( 4d^(10))` in its ground state. How can you say that it is a transition element ? |
| Answer» The outer electronic configuration of `Ag(Z = 47)` is `4d^(10) 5s^(1)` .In addition to `+1` , it shows an oxidation state of `+2` ( e.g., `AgO` and `AgF_(2)` exist ) . In `+2` oxidation state, the configuration is `d^(9)` , i.e., the d-subshell is incompletely filled. Hence, it is a transition element. | |
| 38. |
Which of the following compounds is used as the starting material for the preparation of potassium dichromate?A. `K_(2)SP_(4). Cr_(2)(SO_(4))_(3). 24H_(2)O`B. `PbCrO_(4)`C. `FeCr_(2)O_(7)`D. `PbCrO_(4). PbO` |
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Answer» Correct Answer - C `FeCr_(2)O_(4)` is used as the starting material for `K_(2)Cr_(2)O_(7)`. |
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| 39. |
In general, the Transition elements exhibit their highest oxidation states in their compounds with elements like:A. CB. SC. S and PD. F and O |
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Answer» Correct Answer - d It is because these have small size, can have highest coordination number. |
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| 40. |
Though copper , silver and gold have completely filled sets of d-orbitals yet they are considered as transition metals. Why? |
| Answer» These metals in their common oxidation states have incompletely filled d-orbitals, e.g., `Cu^(2+)` has `3d^(9)` and `Au^(3+)` has `5d^(8)` configuration. | |
| 41. |
Consider the following statements for transition elements. (I) form sets of compounds which disply different oxidation states of the metal. (II) from coloured ions in solution. (III) burn vigorously in presence of oxygen. (IV) replace `H_(2)` from dilute acids.A. I,II III are correctB. II,III,Iv are correctC. I,II are correctD. All are correct |
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Answer» Correct Answer - C (I) True, due to involvement of ns electrons in bonding. (II) True, due to unpaired electrons in d-orbitals. (III) Since, `E_((M^(2+)//M))^(@)=+ve` `E_((M//M^(2+)))^(@)=-ve ` Thus oxidation of the metal is not easy. Hence, false. (Iv) They do not displeace `H_(2)` with dilute acid as per their `E^(o)` values. Hence, false. only I nad II are true statements . |
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| 42. |
`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media. Q. The molarity `(M)` of `KMnO_4` solution in the acidic medium isA. 0.2 MB. 0.02MC. 0.4 MD. 0.04 M |
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Answer» Correct Answer - b `MnO_4^(ɵ)-=S_2O_3^(2-)(2S_2O_3^(2-)toS_4O_6^(2-)+2e^(ɵ))` `(n=(2)/(2)=1)` `100mLxxN-=100mLxx0.1xx1` (n factor) `N=0.1` `M=(0.1)/(n "factor")=(0.1)/(5)=0.02M` |
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| 43. |
Iron, once dipped in concentrated `H_2SO_4`, does not displace copper from copper sulphate solution, becauseA. It is lessreactive than copperB. A layer of sulphate is deposited on itC. An inert layer of iron oxide s deposited on itD. All valence electrons of iron are consumed |
| Answer» Correct Answer - c | |
| 44. |
Write balanced equations for "the extraction of copper from pyrites by self-reduction". |
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Answer» Roasting: `2CuFeS_2+O_2overset(Delta)toCu_2S+2Fes+SO_2` `2Cu_2S+3O_2to2Cu_2O+3SO_2` `2FeS+3O_2to2FeO+2SO_2` Smelting: `FeO+SiO_2toFeSiO_3` (Slag) `Cu_2O+FeStoCu_2S+FeO` Bessemerisation: `Cu_2S+2Cu_2Oto6Cu+SO_2` |
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| 45. |
In self-reduction, the reducing species isA. SB. `O^(2-)`C. `S^(2-)`D. `SO_2` |
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Answer» Correct Answer - C Self-reduction. `Cu_2S+2Cu_2Oto6Cu+SO_2` the oxidation number of sulphur increases from `-2` to `+4`, therefore, `S^(2-)` Is the reducing species. |
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| 46. |
When a metal rod M is dipped into an aqueous colourless concetrated solution of compound N, the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aqueous `NH_3` dissolves O and gives an intense blue solution. Q. The compound N isA. `AgNO_3`B. `Zn(NO_3)_2`C. `Al(NO_3)_3`D. `Ph(NO_3)_2` |
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Answer» Correct Answer - A White ppt. of `O` Is of `AgCl` Since, `AgNO_3` is completely soluble silver compound, N is `AgNO_3`. |
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| 47. |
When a metal rod M is dipped into an aqueous colourless concetrated solution of compound N, the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aqueous `NH_3` dissolves O and gives an intense blue solution. Q. The final solution contains.A. `[Pb(NH_(3))_(4)]^(2+)" and "[CoCl_(4)]^(2-)`B. `[Al(NH_(3))_(4)]^(3+)" and "[Cu(NH_(3))_(4)]^(2+)`C. `[Ag(NH_(3))_(2)]^(+)" and "[Cu(NH_(3))_(4)]^(2+)`D. `[Ag(NH_(3))_(2)]^(+)" and "[Ni(NH_(3))_(6)]^(2+)` |
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Answer» Correct Answer - C (c) `AgNO_(3)(aq)+NaCl(aq)rarrunderset("white ppt.")underset()(AgCl(s))+NaNO_(3)(aq)` `AgCl(s)+2NH_(3)(aq)rarr[Ag(NH_(3))_(2)]^(+)Cl^(-)` `Cu(NO_(3))_(2)(aq)+4NH_(4)OH(aq)rarr[Cu(NH_(3))_(4)]^(2+)(aq)` |
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| 48. |
Complete the following reactions : (i) `Cr_(2)O_(7)^(2-)+Sn^(2+)+H^(+) to ` (ii) `MnO_(4)^(-)+Fe^(2+)+H^(+) to ` |
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Answer» (i) `Cr_(2)O_(7)^(2-)+14H^(+)+3Sn^(2+) to 2Cr^(3+)+3Sn^(4+)+7H_(2)O` (ii) `2MnO_(4)^(-) +16H^(+)+10Fe^(2+) to 2Mn^(2+)+8H_(2)O+10Fe^(3+)` |
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| 49. |
Irregular trend in the standard reduction potential value of the first row transition elements is due toA. regular variation of first and second row enthalpiesB. irregular variation of sublimation enthalpiesC. regular variation of sublimation enthalpiesD. increase in number of unpaired electrons |
| Answer» Correct Answer - b | |
| 50. |
In 4-f series penultimate shell contain how many electrons ?A. `d ^(0"to"1)`B. `d ^(1"to"10)`C. `d ^(0)`D. `d ^(10"to"0)` |
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Answer» Correct Answer - A We know that, in 4f series penultimate subshell is d. Which contains 0 to 1 electrons |
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