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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.
| 701. |
f-block elements are known as `"…..............."` |
| Answer» Correct Answer - inner transition elements | |
| 702. |
Prussian blue is a complex with the formula `"…..............."`. |
| Answer» Correct Answer - `K overset(III)(Fe) [ overset(II) (Fe)(CN)_(6)]` | |
| 703. |
The green coloured salt formed on heating `(NH_(4))_(2) Cr_(2)O_(7)` is `"................."` |
| Answer» Correct Answer - `Cr_(2)O_(3)` | |
| 704. |
Complete the following reaction `: (NH_(4))_(2) Cr_(2)O_(7) overset("Heat")(rarr)` |
| Answer» `(NH_(4))_(2)Cr_(2)O_(7)(s)overset("Heat")(rarr)N_(2)(g)+4H_(2)O(l) + Cr_(2)O_(3)(s)` | |
| 705. |
Write ionic reaction for the reaction between `MnO_(4)^(-)` ions and oxalate ions at 333K. |
| Answer» `2MnO_(4)^(-)+16H^(+) + 5C_(2)O_(4)^(2-) overset("Heat")(rarr) 2Mn^(2+) +8 H_(2)O + 10 CO_(2)` | |
| 706. |
Misch metal isA. An alloy of aluminiumB. A mixture of chromium and lead chromateC. An alloy of lanthanoid metalsD. An alloy of copper |
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Answer» Correct Answer - c Misch metal is an ally of lanthanoid metal. |
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| 707. |
What would be energy order of d-orbitals of tetrahedral complexes when they undergo splitting ?A. `d_(xy) ~= d_(y_2) ~= d_(xz) lt d_(x^2-y^2) ~= d_z^2`B. `d_(x^2- y^2) ~= d_z^2 lt d_(xy) ~= d_(yz) ~= d_(xy)`C. `d_(xy) ~= d_(z^2) lt d_(yz) ~= d_(xz) ~= d_(x^2-y^2)`D. `d_(x^2-y^2) ~= d_(xz) lt d_(xy) ~= d_(yz) ~= d_z^2` |
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Answer» Correct Answer - B Tetrahedral complexes Splitting of d-orbitals as per CFT is `underset"eg"((d_(x^2-y^2)-=d_z^2)) lt underset(t_"2g")((d_"xy" -= d_"yz" -=d_"xz"))` |
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| 708. |
When a photographic film is exposed and developed, the image is due to the formation ofA. AgB. `Ag_(2)O`C. AgBrD. `[Ag(S_(2)O_(3))_(2)]^(3-)` |
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Answer» Correct Answer - A In an exposed film, we have exposed `AgBr`. This is fixed up by a developer known as quinol `C_(6)H_(4)(OH)_(2)`. The following reaction takes place `2AgBr+C_(6)H_(4)(OH)_(2)to2Ag+2HBr+C_(6)H_(4)O_(2)("quinone")` |
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| 709. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` Consider the following list of reagents `:` Acidified `K_(2)Cr_(2)_(7)` ,alkaline `KMnO_(4), CuSO_(4), H_(2)O_(2), Cl_(2),O_(3), FeCl_(2) , HNO_(3)` and`Na_(2)S_(2)O_(3)` . The total number of reagents that can oxidize aqueous iodide to iodine is |
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Answer» `6KI+ K_(2)Cr_(2)O_(7) +7H_(2)SO_(4) rarr 4K_(2)SO_(4) + Cr_(2)(SO_(4))_(3)+ 3I_(2) + 7H_(2)O` `2CuSO_(4)+ 4KI rarr Cu_(2)I_(2)+ 2K_(2)SO_(4)+ I_(2)` `H_(2)O_(2) +2KI rarr I_(2)+ 2KOH` `Cl_(2)+ 2KI rarr 2KCl + I_(2)` `O_(3)+ 2KI + H_(2)O rarr 2KOH + I_(2) O_(2)` `2FeCl_(3) + 2KI rarr 2FeCl_(2)+ 2KCl + I_(2)` `4HNO_(3) + 2KI rarr 2KNO_(3) + 2NO_(2)+ I_(2)+ 2H_(2)O` |
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| 710. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` The coefficient of `H_(2)SO_(4)` on balancing the equation `K_(2)Cr_(2)O_(7) + H_(2)SO_(4) + KI rarrK_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + H_(2)O+I_(2)` is |
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Answer» The balanced equation is `:` `K_(2)Cr_(2)O_(7) + 7H_(2)SO_(4) +6KI rarr 4K_(2)SO_(4)+ Cr_(2)(SO_(4))_(3)+ 3I_(2)+ 7H_(2)O` |
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| 711. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` The oxidation number of Mn in the porduct of alkanline oxidative fusion of `MnO_(2)` is |
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Answer» `2MnO_(2)+ 4KOH + O_(2) rarr 2K_(2)MnO_(4) + 2H_(2)O` Ox. State of Mn in `overset(+1)(K_(2))overset(x)(M)noverset(-2)(O_(4)): 2+x-8=0` or `x= + 6` |
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| 712. |
`{:(,"Column I(Ion)",,"Column II ( Magnetic moment)"),((A),Cu^(2+),(p),2.9BM),((B),Fe^(2+),(q),4.0 BM),((C ),Ni^(2+),(r ),1.8 BM),((D),Co^(2+),(s),5.0 BM):}` |
| Answer» (a) `A-r,B-s,C-p,D-q` (b) `A-p,B-q,C-r,D-s`(c )`A-s,B-q,C-p,D-r` (d) `A-q,B-r,C-s,D-p` | |
| 713. |
`{:(,"Column I(Substance)",,"Column II(Eq. wt)"),((A),KMnO_(4)"in acidic solution",(p),"Mol."wt//6),((B),KMnO_(4)"in alkaline solution",(q),"Mol." wt//5),((C ),KMnO_(4)"in neutral solution",(r ),"Mol."wt//3),((D),K_(2)Cr_(2)O_(7)" in acidic medium ",(s ),"Mol. "wt//1):}` |
| Answer» (a) `A-s,B-r,C-q,D-p` (b) `A-q,B-s,C-r,D-p` (c )`A-r, B-s,C-p,D-q`(d ) `A-p, B-s,C-q,D-r` | |
| 714. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` The magnetic moment of a transition metal ion is foundto be 5.92 B.M. The number of unpaired electrons present in the species is |
| Answer» `mu = sqrt(n(n+2))` B.M. `mu 5.92 ` B.M. Only when n=5 | |
| 715. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` Th enumber of electrons present in the 4f-subshell ofTerbium ( atomic no. = 65) is |
| Answer» Electronic configuration of Terbium (Z= 65) is `[Xe]^(54) 4f^(9) 5d^(0)6s^(2)` | |
| 716. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` The highest oxidationstate shown by any transition element is |
| Answer» The highest oxidation state shown by the transition element is `+8` viz. Ru in 4d series and Os in 5d series. | |
| 717. |
The answer to each of the following questions is a single digit integer, ranging from 0 to 9. If the correct answers to the qustion numbers A,B,C and D (say) are 4,0,9 and 2 respectively, then the correct darkening of bubbles should be as shown on the side `:` In neutral or faintly alkaline solution , 8 moles of permanganate anion quantitatively oxidize thiosulphate anios to produce X moles ofsulphur containing product . The magnetic of X is |
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Answer» `8MnO_(4)^(-) + 3S_(2)O_(3)^(2-) +H_(2)O rarr 8 MnO_(2) + 6SO_(4)^(2-) + 2OH^(-)` `:. `Moles of `SO_(4)^(2-)` formed from 8 moles of `MnO_(4)^(-) = 6` |
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| 718. |
Choose the correct statement.A. First ionisation energy of Cu is higher than that of K.B. Second ionisation energy of Cu is lower than that of K.C. Thrid ionisation energy of Cu is higher than that of K.D. Third ionisation energy of Cu is lower than that of K. |
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Answer» Correct Answer - B::D (b,d) Both are correct. |
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| 719. |
Oxidation state of in `Fe_(3)O_(4)` isA. `(8)/(3)`B. `(3)/(4)`C. `(3)/(2)`D. `(1)/(2)` |
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Answer» Correct Answer - A Let x be the oxidation number of `Fe in Fe_(3)O_(4)`. therefore, `Fe_(3)O_(4),3X+4xx(-2)=0 or` `3x=8 or x=(8)/(3)` |
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| 720. |
Oxidation state of in `Fe_(3)O_(4)` isA. `+3`B. `+8`C. `+8//3`D. `+3//8` |
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Answer» Correct Answer - C `3(X) +4 (-2)=0 therefore X=+8//3=+2.6` |
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| 721. |
The compounds which are coloured only due to charge transfer are :A. `KMnO_(4)`B. `K_(2)Cr_(2)O_(7)`C. `Fe(OH)_(3)`D. `Cr(OH)_(3)`. |
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Answer» Correct Answer - A::B (a,b) Both are correct. |
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| 722. |
Which of the following compounds is not coloured ?A. `CuCl`B. `CuBr`C. `CuSO_(4)`D. `CuI` |
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Answer» Correct Answer - A CuBr is coloured due to Br, CuI is coloured due to I and `CuSO_(4)` is also coloured but due to `Cu^(2+)` ion. While `Cu^(+) and Cl^(-)` both are colourless. |
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| 723. |
Which of the following statement is not correct about transition metalsA. They have low M.PB. They shows variable O.SC. They tend to adopt closely packed structuresD. They have high lattice energy |
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Answer» Correct Answer - A All transition metals have very high melting point. |
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| 724. |
The valence shell electronic configuration of `Cr^(2+)` ion isA. `4s ^(0) 3d ^(4)`B. `4s^(2)3d^(2)`C. `4s^(2)3d^(0)`D. `3p^(6)4s^(2)` |
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Answer» Correct Answer - A Electronic configuration of chromium `Cr to [Ar] 3d^(5) 4s^(1)` `Cr ^(2+)to [Ar]3d^(4)4s ^(0).` |
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| 725. |
The valence shell of transition element consists ofA. n d orbitalsB. (n - 1) d orbitalsC. ns nd orbitalsD. (n - 1) d ns orbitals |
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Answer» Correct Answer - D (n-1) d and ns orbitals |
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| 726. |
Which of the following f-block elements has half filled electrocin configuration having at. No. ?A. 95,96,102 and 103B. 70,71, 102 and 103C. 63,64,95 and 96D. 63,64,70 and 71 |
| Answer» Correct Answer - C | |
| 727. |
Which of the following electron configurations is correct for iron,(atomic number26)?A. `1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6`B. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5`C. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7`D. `1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6` |
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Answer» Correct Answer - D Fe (26) -`1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6` |
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| 728. |
The f-block element having at n. 95 has electronic configurationA. `[Xe] 5f^(7) 6d^(0) 7s^(2)`B. `[Rn] 5f^(6) 6d^(1) 4s^(2)`C. `[Rn] 5f^(7) 6d^(0) 7s^(2)`D. `[Xe] 4f^(2) 5d^(0) 6s^(2)` |
| Answer» Correct Answer - C | |
| 729. |
The electronic configuration of copper isA. `[Ar] 4s^2 3d^9`B. `[Ar]4s^1 3d^10`C. `[Kr] 5s^2 4d^9`D. `[Kr] 5s^1 4d^10` |
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Answer» Correct Answer - B Anamalous configuration of Cu , Z=29 . `[Ar] 4s^1 3d^10` |
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| 730. |
Which among the following forms coloured salts?A. metalsB. non- metalsC. transition elementsD. s - block elements |
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Answer» Correct Answer - C Unpaired electrons in d-orbitals and d-d transition impart colour. |
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| 731. |
The 3d-series elements rangesA. Z = 21 to 30B. Z = 22 to 30C. Z = 20 to 30D. Z = 31 to 40 |
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Answer» Correct Answer - A Sc (21) - `4s^2 3d^1` and Zn (30) - `4s^2 3d^6` |
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| 732. |
Why do the transition metal ions have high enthalpy of hydration ? |
| Answer» This is due to their small and large nuclear charge. This is so because when we move along any transition series, the nuclear charge increases and size decreases. | |
| 733. |
The ability of d-block elements to form complexes is due to:A. Small and gighly charged ionsB. Vacant low energy orbitals to accept lone pair of electrons from ligandsC. Low polarising power of cationD. None is correct |
| Answer» Correct Answer - a,b | |
| 734. |
Select the colured compound amongst the following :A. `TiCl_(4)`B. `CrCl_(3)`C. `ZnCl_(2)`D. `CuCl`. |
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Answer» Correct Answer - B In `CrCl_(3)` , `Cr^(3+) (Z = 24)` is coloured due to the presence of three unpaired d-electrons. All other ions have paired electrons in d-orbitals and are colourless. |
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| 735. |
The ability of d-block elements to form complexes is due toA. large size and high nuclear chargeB. small size and low nuclearC. small size and high nuclear chargeD. none of these. |
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Answer» Correct Answer - C It is the correct answer. |
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| 736. |
Which one of these is not know?A. `CuCl_(2)`B. `CuI_(2)`C. `CuF_(2)`D. `CuBr_(2)` |
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Answer» Correct Answer - B `CuI_(2)` is an unstable compound. It decompose to cuprous iodide and liberates iodine. `2CuI_(2) to Cu_(2)I_(2) + I_(2)` |
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| 737. |
Which is not true for describing the catalyti activity of transition metals?A. Their ability to adopt multiple oxidation statesB. Their ability to from bonds between reactant molecule and atoms of the surface of catalysts.C. Increasing the concentration of reactiants at the catalyst surfaceD. Strenghtening the bonds in the reacting moleucles. |
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Answer» Correct Answer - D Because of the formation of bonds between the reactant molecules and active centres on the catalyst surface, there is the weakeing of the bonds between the reactant molecules. |
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| 738. |
The correct statemet isA. the earlier members of lanthanoid series resemble calcium in their chemical properties.B. the extent of actinoid contraction is almost the same as lanthanoid contrction.C. in general, Inathnoid and actinoids and actinoids do not show variable oxidation states.D. `Ce^(4+)` in aquesous soltuion is not known. |
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Answer» Correct Answer - A Lanthanoids are placaed in the same group (2) in which the element Ca occupies a position. Therefore, the earlier members of the series resemble Ca in their chemical properties. |
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| 739. |
Give reasons for the following : Variation in the radii of transition elements are not as pronounced as those of representative elements. |
| Answer» As we proceed along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the d-subshell increases the screening effect which tends to counterbalance the effect of the increased nuclear charge. | |
| 740. |
The colours of lanthanide ions arise due to ______transition.A. p - fB. d - dC. f - dD. f - f |
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Answer» Correct Answer - D Unpaired `e^-` in 4f orbitals leading to f-f transition giving colour of lanthanoid ions . |
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| 741. |
From La to Lu basicity of hydoxidesA. increasesB. remains sameC. decreasesD. cannot be predicted |
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Answer» Correct Answer - C Basicity decreases from La to Lu as atomic radii decreases from La to Lu |
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| 742. |
Which of the following has lowest basicity ?A. `Ce (OH)_(3)`B. `Bk(OH)_(3)`C. `Md(OH)_(3)`D. `No (OH)_(3)` |
| Answer» Correct Answer - A | |
| 743. |
Which is not the chracteristics of the transition metals ?A. They show variable oxidation statesB. They form coloured compoundsC. They have low enthalpy of atomisationD. They acts as good catalysts |
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Answer» Correct Answer - C Transition metals have high enthalpy of atomisation. |
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| 744. |
Which of the following ion acts as an oxidising agent?A. `Eu^(2+)`B. `Np^(4+)`C. `Sm^(2+)`D. `Yb^(2+)` |
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Answer» Correct Answer - B The ions whose oxidatio state change from +4 to +3 act as an oxidising agent while those ions whose oxidation state change from +2 to +3 tends to act as reducing agents. Therefore, `Np^(4+)` is an oxidising agent. |
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| 745. |
Lanthanoid contraction is caused due to:A. the appreciable shielding on outer electrons by 4f-electrons from the nuclear chargeB. the appreciable shielding on outer electrons by 5d-electrons from the nuclear chargeC. the same effective nuclear charge from Ce to LuD. the imperfect shielding on outer electrons by 4f-electrons from the nuclear charge |
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Answer» Correct Answer - D Lanthanoid contraction is due ineffctive shielding produced by larger f-subshell. |
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| 746. |
Amount of oxalic acid present in a solution can be determined by its titration with `KMnO_(4)` solution in the presence of `H_(2)SO_(4)` . The titration gives unsatisfactory rasult when carried out the presence of HCl because HClA. reducs permanganate to `Mn^(2+)`B. oxidises oxalic acid to carbon dioxide and waterC. gets oxidized by oxalic acid to chlorineD. furnishes `H^(+)` ions in addition to those from oxalic acid. |
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Answer» Correct Answer - A In presence of `H_(2)SO_(4), KMnO_(4)` oxidises oxalic acid to `CO_(2)` .In presence of HCl, `KMnO_(4)` not only oxidises oxalic acid but also oxidises HCl to `Cl_(2)` and itself it is reduced to `Mn^(2+)` |
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| 747. |
A student accidently added conc. `H_(2)SO_(4)` to potassium permanganateand itexplodeddue to the formation of a explosive which isA. MnOB. `Mn_(2)O_(3)`C. `Mn_(2)O_(5)`D. `Mn_(2)O_(7)` |
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Answer» Correct Answer - D `Mn_(2)O_(7)` is formed which decomposes explosively on heating. |
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| 748. |
`KMnO_(4)`gets reduced toA. `K_(2)MnO_(4)` in neutral mediumB. `MnO_(2)` in acidic mediumC. `Mn^(2+)` in alkaline mediumD. `MnO_(2)` in neutral medium |
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Answer» Correct Answer - D `KMnO_(4)` is reduced to `MnO_(2)` in neutral medium `2KMnO_(4) + H_(2)O rarr 2MnO_(2)+ 2KOH +3(O)` or `MnO_(4)^(-) + 2H_(2)O+ 3e^(-) rarr MnO_(2) + 4OH^(-)` In alkaline medium, it is reduced to `K_(2)MnO_(4)` or `MnO_(2)` . In acidic medium, it is reduced to `Mn^(2+)` |
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| 749. |
The colour of `KMnO_(4)` is due toA. `L rarr M` charge transfer transitionB. `sigma-sigma^(**)` transitionC. `M rarrL` charge transfer transitionD. d-d transition |
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Answer» Correct Answer - A The deep purple colour of `KMnO_(4)` is not due to d-d transition but due to charge transfer from O to Mn, i.e., from Ligand (L) to Metal (M) which reduces the oxidation state of Mn from `+7` to `+6` momentarily. |
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| 750. |
In which of the pairs of ions given, there is an ion that forms a co-ordination compound both aqueous sodium hydroxide and ammonia and another ion that forms a co-ordination compound only with aqueous sodium hydroxide ?A. `Pb^(2+),Cu^(2+)`B. `Zn^(2+),Al^(3+)`C. `Cu^(2+),Zn^(2+)`D. `Al^(3+),Cu^(2+)` |
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Answer» Correct Answer - B `Zn^(2+),Al^(3+)` (I)`Zn(HO)_(2)+2NaOHrarr[Zn(OH)_(4)]^(2-)+2Na^(+)` `underset("Anhydrous")(ZnCl_(2))+4NH_(3)rarrunderset("zinc chloride")underset("Tetraamine")(Zn(NH_(3))_(4))Cl_(2)` (II) `Al(OH)_(3)+NaOHrarr[Al(OH)_(4)]^(-)+Na^(+)` `Al^(3+)+3NH_(3)rarrAl(NH_(3))_(3)` |
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