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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. |
Which of the following d-block element has half - filled penultimate as well as valence subshell ?A. CuB. AuC. AgD. Cr |
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Answer» Correct Answer - D `_24Crto1s^2 2s^2 2p^6 3 s^2 3 p^6underset("half-filled")ubrace(4s^1 3d^5)` |
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| 2. |
General configuration of outermost and penultimate shell is `(n - 1) s^(2) (n - 1) p^(6) (n - 1)d^(x) ns^(2)`. If `n = 4` and `x = 5` then no. of protons in the nucleus will beA. `gt 25`B. `lt 24`C. 25D. 30 |
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Answer» Correct Answer - C Given `n = 4 x = 5` So `underset(3 s^(2))(4 - 1) s^(2) underset(3 p^(6))(4 - 1) p^(6) underset(3 d^(5))(4 - )d^(5) underset(4 s^(2))4 s^(2)` Total electron `= 2 + 6 + 2 = 15` Electron in `1 + 2` orbit `= 2 + 8 = 10` Total electron `= 10 + 15 = 25` No. of electron = No. proton So total proton = 25 |
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| 3. |
Statement-1: The highest oxidation state of chromium in its compound is +6. Statement-2: Chromium atom has only six electrons in ns and `(n-1)d` orbitals.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason not is the correct explanation of assertion.C. If assertion is trun but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 4. |
Assertion: Transitio metals show variable valence. Reason : Due to a large energy difference bteween the `ns^(2)` and `(n - 1)d` electrons.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 - C The assertion is correct but the reason is false. Actually transition metal show variable valency due to very samll difference between the `ns^(2)` and `(n + 1)d` electron, Therefore, assertion is correct but reason is false. |
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| 5. |
Oh heating pyrolusite with `KOH` in presence of air we getA. `KMnO_(4)`B. `K_(2) MnO_(4)`C. `Mn (OH)_(2)`D. `Mn_(3) O_(4)` |
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Answer» Correct Answer - B `underset("pyrollusite")MnO_(2) + KOH rarr K_(2) MnO_(4)` |
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| 6. |
The colour of `K_(2)Cr_(2)O_(7)` changes from red-orange to lemon-yellow on treatment with `KOH_((aq.))`, because of:A. The reduction of `Cr^(VI)` to `Cr^(III)`B. The formation of chromium hydroxidedC. The conversion of dichromate to chromateD. The oxidation of potassium hydroxide to potassium peroxide |
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Answer» Correct Answer - C The conversion of dichromate to chromate. `K_(2) Cr_(2) O_(7) + 2 KOH rarr underset(yellow) underbrace(2K_(2) Cr O_(4)) + H_(2) O` |
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| 7. |
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective verticalcolumns. Give special emphasis on the following points: (i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes. |
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Answer» (i) In the `1^(st), 2^(nd)` and `3^(rd)` transition series, the 3d, 4d and 5d orbitals are respectively filled. We know that elements in the same vertical column generally have similar electronic configurations. In the first transition series, two elements show unusual electronic configurations: `Cr(24)=3d^(5)4s^(1)` `Cu(29)=3d^(10)4s^(1)` Similarly, there are exceptions in the second transition series. These are: `Mo(42)=4d^(5)5s^(1)` `Tc(43)=4d^(6)5s^(1)` `Ru(44)=4d^(7)5s^(1)` `Rh(45)=4d^(8) 5s^(1)` `pb(46)=4d^(10)5 s^(0)` `Ag(47)=4d^(10)5s^(1)` There are some exceptions in the third transition series as well. These are: `W(74)=5d^(4)6s^(2)` `pt(78)=5d^(9)6s^(1)` `Au(79)=5d^(10)6s^(1)` As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar. (ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important. For example `[Foverset(II)e(Cn)_(6)]^(4-),[Coverset(III)(NH_(3))_(6)]^(3+),[Ti(H_(2)O)_(6)]^(3+)` are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example: `WCl_(6), ReF_(7), RuO_(4)`, etc. (iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the `1^(st)` transition series. (iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction. |
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| 8. |
Describe the oxidising action of potassium dichromate and write the ionic equations of reaction with: (i). Iodide (ii). Iron (II) solution and (III). `H_2S` |
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Answer» `K_(2)Cr_(2)O_(7)` acts as a very strong oxidising agent in the acidic medium. `K_(2)Cr_(2)O_(7)+4H_(2)SO_(4) to K_(2)SO_(4)+CR_(2)(SO_(4))_(3)+4H_(2)O+3[O]` `K_(2)Cr_(2)O_(7)` takes up electrons to get reduced and acts as an oxidising agent. The reaction of `K_(2)Cr_(2)O_(7)` with other iodide, iron (II) solution, and `H_(2)S` are given below. (i) `K_(2)Cr_(2)O_(7)` oxidizes iodide to iodine. `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O),(" "2I^(-) to I_(2)+2e^(-)xx3),(underline(bar(Cr_(2)O_(7)^(2-)+6I^(-)+14H^(+) to 2Cr^(3+)+3I_(2)+7H_(2)O))):}` (ii) `K_(2)Cr_(2)O_(7)` oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions. `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O),(" "Fe^(2+) to Fe^(3+)+e^(-)xx6),(underline(bar(Cr_(2)O_(7)^(2-)+14H^(+)+6Fe^(2+) to 2Cr^(3+)+6Fe^(3+)+7H_(2)O))):}` (iii) `K_(2)Cr_(2)O_(7)` oxidizes H2S to sulphur. `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O),(" "H_(2)S to S+2H^(+)+2e^(-)xx3),(underline(bar(Cr_(2)O_(7)^(2-)+3H_(+)S+8H^(+) to 2Cr^(3+)+3S+7H_(2)O))):}` |
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| 9. |
Describe the preparation of potassium dichromate from iron chromite ore.What is the effect of increasing pH on a solution of potassium dichromate? |
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Answer» Potassium dichromate is prepared from chromite ore (`FeCr_(2)O_(4)`) in the following steps. Step (1): Preparation of sodium chromate `4FeCr_(2)O_(4)+16NaOH +7O_(2) to 8Na_(2)CrO_(4)+2Fe_(2)O_(3)+8H_(2)O` Step (2): Conversion of sodium chromate into sodium dichromate `2Na_(2)CrO_(4)+conc. H_(2)SO_(4) to Na_(2)Cr_(2)O_(7)+Na_(2)SO_(4)+8H_(2)O` Step(3): Conversion of sodium dichromate to potassium dichromate `Na_(2)Cr_(2)O_(7)+2KCl to K_(2)Cr_(2)O_(7)+2NaCl` Potassium dichromate being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration. The dichromate ion `(Cr_(2)O_(7)^(2-))` exists in equilibrium with chromate `(CrO_(4)^(2-))` ion at pH 4. However, by changing the pH, they can be interconverted. `2CrO42- to "Ac id "2HCrO_(4)- to "Acid "2Cr2O72-` `Cr2O72- to "Base "2HCrO42- to "Base "2CrO42-` |
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| 10. |
The transition elements have a general electronic configuration:A. `ns^(2) np^(6) nd^(1 - 10)`B. `(n - 1) d^(1 - 10) ns^(0 -2) np^(0 -6)`C. `(n - 1) d^(1 - 10) ns^(1 -2)`D. None of these |
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Answer» Correct Answer - C General electronic configuration of transition of element is [Noble gas] `(n - 1) d^(1 - 10) ns^(1 - 2)`. |
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| 11. |
Brass is an alloy ofA. Silver and copperB. Copper and zincC. Copper and tinD. Copper , zinc and tin |
| Answer» Correct Answer - B | |
| 12. |
Stainless steel is an alloy ofA. copperB. nickel and chromiumC. ManganeseD. zine |
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Answer» Correct Answer - B Stainless steel contains `11.5%Cr` and `2.0%Ni` with `Fe`. |
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| 13. |
Which of the following transition elements does not exhibit variable oxidation states ?A. ZnB. NiC. TiD. Sc |
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Answer» Correct Answer - D Scandium always exists in the oxidation state (+3), and occurs as `Sc^(3+)` ion. The formation of `Sc^(3+)` requires the removal of the two 4s and one d-d specturm is impossible. Consequently the ion and its compounds are coplorless and dimagnetic. Of the simple ions of the 3d-serios, only `Sc^(3+)` is known to have `d^(0)` configuration. Zn (+2) is the only state known for Zn. However the answer of this question is (4), because Zn is a d-block element but not a transition element. The reduced tendency of higher oxidation states towards the end of the series is due to steady increase in the effective nuclear charge along the series. This pulls the (n-1) d orbitals into the electron core and hence they are not readily available for bonding. For instance, the only oxidation state for Zn is Zn(+2) where no (n-1) d electron is involved. On the other hand, early in the series, it is difficult to form species that do not utillise the (n-1) d electron, e.g., Sc(+2) is virually unknown while Ti(+4) is more stable than Ti(+2). |
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| 14. |
Name the transition element which does not exhibit variable oxidation states . |
| Answer» Sc(Z=21) does not exhibit variable oxidation states . | |
| 15. |
The transition metals have a strong tendency to form complaexes because of (i) smaller sizes of the metal ions (ii) variable oxidation states (iii) high ionic charges of metal ions (iv) availability of vacant d-orbitals for bond formation.A. (i), (ii), (iii), (iv)B. (i), (iii), (iv)C. (iv) onlyD. (i), (iii) |
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Answer» Correct Answer - B Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties. The transition metals form a large number of complex compounds due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of empty d-orbitals for bond formation with the ligands. |
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| 16. |
Assertion : The ability of oxygen to stabilize high oxidation states exceeds that of fluorine. Reason : The highest oxidation number in the oxides coincides with the group numberA. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason not is the correct explanation of assertion.C. If assertion is trun but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Oxygen can stabilize high oxidation states because of its ability to form multiple bonds to metals. |
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| 17. |
STATEMENT-1: `VO_(4)^(-3)` is coloured due to charge transfer . STATEMENT-2: Colour due to charge transfer is highly intense colour STATEMENT-3: Fe may form Fe-Fe bondA. FTFB. FTTC. TTFD. TTT |
| Answer» Correct Answer - 4 | |
| 18. |
Which of the following arrangements represent the correct order of the property stated against it ? (i) `Ni^(2+) lt Co^(2+) lt Fe^(2+) lt Mn^(2+)` : ionic size (ii) `Co^(3+) lt Fe^(3+) lt Cr^(3+) lt Sc^(3+)` : stability in aqueious solution (iii) `Sc lt Ti lt Cr lt Mn` : number of oxidation states (iv) `V^(2+) lt Cr^(2+) lt Mn^(2+) lt Fe^(2+)` : paramagnetic behaviourA. Sc gt Ti gt Cr lt Mn : number of oxidation statesB. `V^(2+) ltCr^(2+) ltMn^(2+) ltFe^(2+)` : paramagnetic behaviourC. `Ni^(2+) lt Co^(2+) ltFe^(2+) ltMn^(2+)` : ionic sizeD. `Co^(3+) lt Fe^(3+) lt Cr^(3+) lt Sc^(3+)` : stability in aqueous solution. |
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Answer» Correct Answer - B Number of unpaired electrons in `Fe^(2+)` is less than `Mn^(2+)`, so `Fe^(2+)` is less paramagnetic than `Mn^(2+)`. |
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| 19. |
Transition metals make the most efficient catalysts because of their ability toA. adopt multiple oxidation states and to form omplexesB. form coloured ionsC. show paramagnetism due to upaired electronsD. form a large number of oxides. |
| Answer» Correct Answer - A | |
| 20. |
Which of the following compounds is not coloured ?A. `Na[CuCl_4]`B. `Na_2[CdCl_4]`C. `K_4[Fe(CN)]_6`D. `K_3[Fe(CN)_6]` |
| Answer» Correct Answer - B | |
| 21. |
The ions from among the following which are colourless are:A. (i) and (ii) onlyB. (i),(ii) and (iii)C. (iii) and (iv)D. (ii) and (iii) |
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Answer» Correct Answer - A (i) Valence shell electron configuration of `Ti^(4+)` is `3d^(0) 4 s^(0)`. As there is no unparied electrons for d-d trastition, the solution of ions will be colourless (ii) Valence shell electron configuration of `Cu^(+)` is `3 d^(10) 4 s^(0)`. As all electons are unparied, there is no d-d trasition, so the solution of ions will be colourless. (iii) Valence shell electron configuration of `Co^(3 +)` is `3 d^(6) 4s^(0)`. As there are 4 upaired electrons, there is d-d transition of electron, so the solution of ions will be coloured. (iv) Valenec shell electron configuration of `Fe^(2 +)` is `3 d^(6) 4 s^(0)`. As there are 4 upaired electrons. there is d-d transition of electron, so the solution of ions will be coloured. |
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| 22. |
Which of the following transition metal ions is colourless ?A. `V^(2+)`B. `Cr^(3+)`C. `Zn^(2+)`D. `Ti^(3+)` |
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Answer» Correct Answer - C `Zn%(2+)to 3d^10` has no unpaired electrons for excitation. |
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| 23. |
Colour of transition metal ions are due to absorption of some wavelength. This results inA. d-s transitionB. s-s transitionC. s-d transitionD. d-d transition |
| Answer» Correct Answer - D | |
| 24. |
In which of the following ions, the colour is not due to d-d transition?A. `[Ti(H_2O)_6]^(3+)`B. `[Cu(NH_3)_4]^(2+)`C. `[CoF_6]^(3-)`D. `CrO_4^(2-)` |
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Answer» Correct Answer - D In `CrO_4^(2-)` Cr is in +6 oxidation state and has `d^0` configuration. The Colour is due to charge transfer and not due to d-d transition. |
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| 25. |
Why `IE_(3)` of Mn is high ? |
| Answer» Due to half filled electronic configuration of `Mn^(2+)` | |
| 26. |
In 3d series which element have highest `IE_(3)` ? |
| Answer» Zn due to full filled `d^(10) ` E.C. | |
| 27. |
Choose the pair in which `IE_(1)` of first element is greater than `IE_(1)` of second element but in case of `IE_(2)` order is/are reversedA. `Mn gt Cr `B. `Mn gt Fe `C. `Zn gt Cu `D. All of these |
| Answer» Correct Answer - D | |
| 28. |
That the electronic configuration of ytterbium `(Z = 70)` is `4f&(14) 5s^(2)` and of lutetium `(Z) = 71)` is `4 f^(14) 5 d^(1) 6 s^(2)` can be explained on the basis ofA. the extra stability of the half-filled orbitals conesB. the extra stability of the completely filled orbitalsC. the usual rules for the arrangement of electron in their orbitsD. None of these |
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Answer» Correct Answer - C Ytterbium and luteium follow the usual rules for the arrangement in their oribts. |
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| 29. |
Which of the following can be employed for the conversion of potassium manganate to potassium permanganate?A. `Cl_(2)`B. `O_(3)`C. `SO_(2)`D. `KNO_(3)` |
| Answer» Correct Answer - A::B | |
| 30. |
A solution of `KMnO_4` is reduced to various products depending upon its pH. At pH lt 7 it is reduced to a colourless solution (A), at pH = 7 it forms a brown precipitate (B) and at pH gt 7 it gives a green solution ( C), (A),(B) and( C) areA. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_2,MnO_4^(2-)):}`B. `{:("(A)","(B)","(C)"),(MnO_2,mn^(2+),MnO_4^(2-)):}`C. `{:("(A)","(B)","(C)"),(Mn^(2+),MnO_4^(2-),MnO_2):}`D. `{:("(A)","(B)","(C)"),(MnO_4^(2-),Mn^(2+),MnO_2):}` |
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Answer» Correct Answer - A At pH lt 7, in acidic medium `MnO_4^(-)+8H^(+)+5e^(-) to underset(("brown precipitate"))(MnO_2)+4OH^(-)` At pH gt 7, in alkaline medium `MnO_4^(-)+e^(-)to underset(("green"))(MnO_4^(2-))` |
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| 31. |
Explain why does colour of `KMnO_(4)` disappear when oxalic acid is added to its solution in acidic medium?A. the pH of the solution changes on adding oxalic acid, hence `KMnO_4` is decolourisedB. `KMnO_4` oxidises oxalic acid to `CO_2` and itself changes to `Mn^(2+)` ions which are colourlessC. `KMnO_2` is oxidised to potassium sulphate which is colourlessD. on exposure ot air the acidic solution of `KMnO_4` becomes colourless. |
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Answer» Correct Answer - B `5C_2O_4^(2-)+underset(("coloured"))(2MnO_4^(-))+16H^(+)tounderset(("colourless"))(2Mn^(2+))+8H_2O+10CO_2` |
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| 32. |
The oxidation number of `+6` thermodynamically preferred forA. CrB. MoC. WD. both (2) and (3) |
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Answer» Correct Answer - D Group 6 consists of the elements Cr, Mo, W and Sg which are silvery, lustrous, hard but fairly soft (when pure) and have very high melting points and low volatility. The melting point of W is the next highest to C. The group oxidation state of `+6` is exhibited by all the three elements (Cr, Mo and W) in their oxoanions. For Mo and W, the O.N. of `+6` is thermodynamically preferred . However, for Cr, the `+6` state is highly oxidizing, the O.N. of `+3` is most stable for Cr in aqueous solution and in complexes. `Cr(+2)` is strongly reducing. Despite their thermodynamic instability, kinetic factors enable serveral `Cr(+6)` compounds to exist. The most important of these are the chromates and dichromates. Chromium is used in leather industry and in the manufacture of stainless steel and chrome plating (Cr provides a shiny protective coating for Fe and steel surface. Cr metal is not inert in itself, instead, it has a very thin, tough, transparent oxide coating that confers the protection). Mo is used in X- ray tubes. Compounds of Mo and W are employed as cataysts. |
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| 33. |
The most common and most stable oxidation state for all the elements of titanium family isA. `+2`B. `+3`C. `+4`D. `+1` |
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Answer» Correct Answer - C Group 4 consist of elements-titanium (Z = 22), zirconium (Z = 40), hafnium (Z = 72) and rutherfordium (Z = 104). They are lustrous silvery metals with high melting points. These elements arer relatively electropositive but less so than those of Group 3. If heated, they react directly with most non-metals, particularly with oxygen, hydrogen (reversibly) and in case of Ti, with nitrogen also. With the exception of hydrofluoric acid, minerals acid have little effect unless hot. This unreactivity results from a thin impermeble oxide film which form which forms on the surface and prevents further attack. The best solvent for all metals is HF, because they form hexafluoro complexes. `Ti + 6HF rarr H_(2) [TiF_(6)] + 2H_(2)` The most stable oxidation state is `+4`. The lower oxidation states are not known for heaver elements. Even for Ti, they are easily oxidized to `+4` state. |
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| 34. |
The number of unpaired electrons in gaseous species of `Mn^(3+), Cr^(3+)` and `V^(3+)` respectively are………….and most stable species is…………..A. 4,3 and 2, `V^(3+)`B. 3,3 and 2, `Cr^(3+)`C. 4,3 and 2, `Cr^(3+)`D. 3,3 and 3, `Mn^(3+)` |
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Answer» Correct Answer - C `Mn^(3+)=3d^4=4` unpaired electrons, `Cr^(3+)=3d^3=3` unpaired electrons, `V^(3+)=3d^2=2` unpaired electrons. `Cr^(3+)` is most stable out of these in aqueous solution because it has half-filled `t_(2g)`level (i.e., `t_(2g)^(3))`. |
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| 35. |
When `KMnO_(4)` reacts with acidified `FeSO_(4)`A. Only `FeSO_(4)` is oxidisedB. Only `KMnO_(4)` is oxidisedC. `FeSO_(4)` is oxidised `KMnO_(4)` and is reducedD. None of these |
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Answer» Correct Answer - C `FeSO_(4)`is oxidised and `KMnO_(4)` is reduced. `2KMnO_(4) + 3H_(2) SO_(4) rarr K_(2) SO_(4) + 2MNSO_(4)` `[2FeSO_(4) + H_(2)SO_(4) rarr Fe_(2) (SO_(4))_(3) + 2H] xx 5` `[2H + [O] rarr H_(2) O] xx 5` `bar(2KMnO_(4) + 8H_(2) SO_(4) + 10 FeSO_(4) rarr)` `K_(2) SO_(4) + 2MnSO_(4) + 5Fe_(2) (SO_(4))_(3) + 8H_(2)O` In this reaction, oxidation state of `Mn` is changing from `+ 7` to `+ 2` while oxidation state of `Fe` is changing from `+2` to `+3`. |
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| 36. |
Which one of the following statements related to lanthanons is incorrect ?A. Europium shows `+2` oxidation stateB. The basicity decreases as the ionic radius decreases form Pr to Lu.C. All the lanthanons are much more reactive than aluminiumD. `Ce(+4)` solution are widely used as oxidizing agent in volumetric analysis. |
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Answer» Correct Answer - B The basicity decreases as the ionic radius decreases form `Ce^(3+)` to `Lu^(3+)` . Thus `Ce(OH)_(3)` is the strongest base while `Lu(OH)_(3)` is the weakest base. |
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| 37. |
`F_(2)` is formed by reacting `K_(2) MnF_(6)` withA. `SbF_(5)`B. `MnF_(3)`C. `KSvF_(6)`D. `MnF_(4)` |
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Answer» Correct Answer - A `K_(2)MnF_(6) + 2 SbF_(5) rarr 2KSbF_(6) + MnF_(3) + (1)/(2) F_(2)` In this reaction, the stronger Lewis acid `SbF_(5)` displaces the weaker one, `MnF_(4)` from its salt. `MnF_(4)` and fluorine. |
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| 38. |
What is the oxidation state of iron in haemoglobin ? |
| Answer» Correct Answer - 2 | |
| 39. |
What should be the oxidation state of iron for maximum magnetic moment ? |
| Answer» Correct Answer - 3 | |
| 40. |
The correct statement about interstitial compound isA. Interstitial compound are formed by small atoms such as `C,N,H,B` etc .B. These compounds are nonstoichiometricC. They show difference in physical properties but similarity in chemical propertiesD. All of these |
| Answer» Correct Answer - D | |
| 41. |
White vitriol, and blue vitriol are respectivelyA. `ZnSO_(4).7H_(2)O,CuSO_(4).5H_(2)O`B. `FeSO_(4).7H_(2)O, ZnSO_(4) . 7H_(2)O`C. `ZnSO_(4).7H_(2)O,FeSO_(4).7H_(2)O`D. `ZnSO_(4).7H_(2)O,CuSO_(4)` |
| Answer» Correct Answer - A | |
| 42. |
The most common oxosalt of copper (II) is blue virtiol. It has the compositionA. `CuSO_(4) . 4H_(2)O`B. `CuSO_(4) . 5H_(2)O`C. `CuSO_(4) . 3H_(2)O`D. `CuSO_(4) . H_(2)O` |
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Answer» Correct Answer - B It is made commercially by spraying hot dilute sulphuric acid on scrap copper in the presence of a current of air, when the metal slowly dissolves: `2Cu + 2H_(2)SO_(4) + O_(2) rarr 2CUSO_(4) + 2H_(2)O` The crude copper (II) sulphate obtains iron (II) sulphate as impurity. Dilute nitric acid is added to oxidise iron (II) to iron (III) sulphate, which remains in solution after srystallization. The pentahydrate, `CuSO_(4)` . `5H_(2)O` crystallizes out. In the laboratory it is prepared by dissolving copper oxide in dilute sulphuric acid and concentrating the solution, when deep blue crystals of the pentahydrate, known as blue virtiol, are obtained: `CuO + H_(2) SO_(4) rarr CuSO_(4) + H_(2)O` Crystalline copper sulphate is readily soluble in water. The salt, `CuSO_(4)` . `5H_(2)O`, is insoluble in a alchol and is precipitated on adding alcohol to the aqueous solution. |
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| 43. |
When a solution of mercuric chloride treated with an excess of stannic chloride it produces, eventuallyA. `Hg_(2)CI_(2)`B. `Hg_(2)SnCI_(6)`C. HgD. `HgCI_(4)` |
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Answer» Correct Answer - C Mercury (II) chloride solution is reduced by many reducing agents, e.g., formaldehyde, tin (II) chloride, sulphur diooxider, etc., precipitating with mercuryu (I) chloride first, which with excess of reducing agent turns black owing to the formation of metallic mercury. `HgCl_(2)overset(SnCl_(2))tounderset("White")(Hg_(2)Cl_(2))overset(SnCl_(2))to underset("black")(Hg)` |
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| 44. |
Postassium permanganate is manufactured a large scale by fusing - with KOH and `KNO_(3)` to form `K_(2)MnO_(4)`, which is than electrolytically oxidized in an alkaline medium.A. manganite, `Mn_(2)O_(3)` . `H_(2)O`B. hausmannite, `Mn_(3)O_(4)`C. pyrolusite, `MnO_(2)`D. braunite, `Mn_(2)O_(3)` |
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Answer» Correct Answer - C `MnO_(2)` `overset ("fuse with KOH") underset ("oxidize with" KNO_(3)) rarr` `MnO_(4)^(2-)` `overset ("electrolytic oxidation") underset ("in alkaline solution") rarr` `MnO_(4)^(-)` The function of `KNO_(3)` is to provide `O_(2)` to oxidize `MnO_(2)` to `K_(2)MnO_(4)` : `2MnO_(2) + 4KOH + O_(2) rarr 2KMnO_(4) + 2H_(2)O` On passing current, the oxygen evolved at anode oxidises manganese into permangnate. `2K_(2)MnO_(4) + H_(2)O + O rarr 2KMnO_(4) + 2KOH` To oxidize manganate to permanganate, we can also use `CO_(2), CI_(2)` or `O_(3)` but not `H_(2)O_(2)` as it is relatively a weak oxidizing agent. |
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| 45. |
In the reaction, `KNmO_(4) + 16 HC1 rarr 5C1_(2) + 2KC1 + 8H_(2) O` the reduction product isA. `C1_(2)`B. `MnC1_(2)`C. `H_(2) O`D. `KC1` |
| Answer» Correct Answer - B | |
| 46. |
On what ground can you say that scandium `(Z=21)` is a transition element but zinc `(Z=30)` is not? |
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Answer» E.C of `""_(21)Sc=1s^(2)2s^(2)2p^(6)3s^(2)3p^(2)3d^(1)4s^(2)`. In this E.C 3d orbital have one electron in its ground state . That is why it is regarded as a transition element. E.C of Zn=`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(10)4s^(2)` Zn does not have partly filled orbital in its ground state or in its excited state that is why it is not considered as a typical transition element . |
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| 47. |
Arrange the following in increasing value of magnetic moments. (i) `[Fe(Cn)_6]^(4-)` (ii) ` [Fe(CN)_6]^(3-)` (iii) `[Cr(NH_3)_6]^(3+)` (iv) `[Ni(H_2O)_4]^(2+)`A. `ilt ii lt iii lt iv`B. `iltiiltivltiii`C. `iiltiii ltiltiv`D. `iii ltilt ii ltiv` |
| Answer» Correct Answer - B | |
| 48. |
Hybridisation of chromium ions in chromate and dichromate ions is respectivelyA. `sp^(2) and sp^(2)`B. `sp^(2) and sp^(3)`C. `sp^(3) and sp^(2)`D. `sp^(3) and sp^(3)` |
| Answer» Correct Answer - D | |
| 49. |
`V_2O_5` reacts with alkalies as well as acids to giveA. `VO_4^(3-) and VO^(2+)`B. `VO^(2+) and VO_4^(+)`C. `VO_2^(+) and VO^(2+)`D. `VO_4^(3-) and VO_4^(+)` |
| Answer» Correct Answer - D | |
| 50. |
Which of the following reactions do not result in the preparation of potassium dichromate? (I) `4FeCr_2O_4+8 Na_2CO_3+7O_2to` (II) `Na_2CrO_4+H_2SO_4 to ` (III) `Na_2Cr_2 O_7 + 2KCl to `A. I and IIB. II and IIIC. I and IIID. I, II, and III |
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Answer» Correct Answer - A (I) `4FeCr_2O_4+8 Na_2CO_3+7O_2to 8Na_2CrO_4+2Fe_2O_3+8CO_2` (II) `2Na_2CrO_4+H_2SO_4toNa_2Cr_2O_7+Na_2SO_4+H_2O` (III) `Na_2Cr_2O_7+2KCl to K_2Cr_2O_7+2NaCl` |
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