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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.
| 551. |
Which of the two `: V (IV) ` or V (V) is paramagnetic ?Give reasons. At No ofV `=`23 |
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Answer» `._(23)V=[Ar]3d^(3)4s^(2) , V (IV)= [Ar]3d^(1) , V(V) = [Ar] 3d^(0)` Thus, V(IV) has one unpaired electron while V(V) has no unpaired electron. Hence, V(IV)is paramagnetic. |
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| 552. |
Which of the following is a diamagnetic ion?A. `Co^(2+)`B. `Cu^(2+)`C. `Mn^(2+)`D. `Sc^(3+)` |
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Answer» Correct Answer - D `Sc^(3+)` has `d^0` configuration , a diamagnetic species |
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| 553. |
The value of magnetic moment for `[Co(NH)_3 )_6 ]^(3+)` is zero, the unpaired electron would be |
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Answer» Correct Answer - A No unpaired `e^-` in 3d orbital `therefore` Magnetic moment will be zero. |
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| 554. |
The magnetic nature of `Zn ^(+2)`A. diamagneticB. paramagneticC. ferro magneticD. none |
| Answer» Correct Answer - A | |
| 555. |
The atomic radius of Zn is __ than that of Cu.A. lessB. greaterC. equalD. none |
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Answer» Correct Answer - B Atomic radii decreases from left to right |
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| 556. |
Oxidation state of +1 is possible withA. Cu and NiB. Cu and ZnC. Cu and CrD. Cr and Sc |
| Answer» Correct Answer - C | |
| 557. |
Which of the following atom would be rappled by magnetic filed ?A. TiB. CrC. NiD. Zn |
| Answer» Correct Answer - D | |
| 558. |
The element present immediately below Zn in the periodic table in the same column, has atomic number equal toA. 48B. 40C. 50D. 30 |
| Answer» Correct Answer - A | |
| 559. |
Which transition metal shows the highest oxidation state ?A. ScB. TiC. MnD. Zn |
| Answer» Correct Answer - C | |
| 560. |
The number of unpaired electrons in `Zn^(2+)` isA. 2B. 3C. 4D. 0 |
| Answer» Correct Answer - D | |
| 561. |
The common oxidation states of Ti areA. `+2`B. `+3`C. `+4`D. `+5` |
| Answer» Correct Answer - C | |
| 562. |
One of the following is diamagneticA. CuB. `Cu^(+)`C. `Cu^(2+)`D. all the above |
| Answer» Correct Answer - B | |
| 563. |
Which of the following would be diamagnetic?A. `Cu^(2+)`B. `Ni^(2+)`C. `Sc^(3+)`D. `Ti^(3+)` |
| Answer» Correct Answer - C | |
| 564. |
Assertion : Reduction potential of Mn ( + 3 to + 2) is more positive than that of Fe ( + 3 to + 2). Reason : Ionisation potential of Mn is more than that of Fe.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - C (c) Correct reason. `Mn^(2+)" "(3d^(5))` has more stable configuration than `Mn^(3+)" "(3d^(4))` while `Fe^(3+)" "(3d^(5))` has more stable configuration than `Fe^(2+)" "(3d^(6))`. Therefore, `E_(Mn^(3+)//Mn^(2+))^(@)` couple is more positive than `Fe^(3+)//Fe^(2+)` couple. |
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| 565. |
Why is `Cr^(3+)` reducing and `Mn^(3+)` oxidisin when both have `d^(4)` configuration ? |
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Answer» `Cr^(2+)` has the configuration `3d^(4)` . It can lose electron to form`Cr^(3+)` which has the stable `3d^(3)` configuration ( as it has half- filled `t_(2g)` level - explained in unit). Hence, it is reducing . On the other hand, `Mn^(3+)` also has `3d^(4)` configuration but it can gain electron to form `Mn^(2+)` which has stable `3d^(5)` configuration ( as it is exactly half- filled ). Hence, it is oxidizing. Alternatively. `E^(@)` value for `Cr^(3+) // Cr^(2+) ` is negative `( - 0.41V)` whereas `E^(@) `value for `Mn^(3+) // Mn^(3+)` is positive `( + 1.57V)` . Hence, `Cr^(2+)` ion can easily undergo oxidation to give `Cr^(3+)` ion and , therefore , acts as strong reducing agent where as `Mn^(3+)` can easily undergo reduction to give`Mn^(2+)` and hence acts as oxidizing agent. |
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| 566. |
Which out of the two ,`La(OH)_(3) and Lu(OH)_(3)`, is more basic and why? |
| Answer» `La(OH)_(3)` is more basic than `Lu(OH)_(3)` because the latter has greater covalent charcter as compared to the former on account of lanthanoid contraction. The release of `OH^(-)` ion from. `Lu(OH)_(3)` is more difficult and is less basic than `La(OH)_(3)`. | |
| 567. |
Write the electronic configuration of the element with atomic number `102`. |
| Answer» The configuration of the element (Z=102) is `[Rn]5f^(14)7s^(2)`. It is Nobellium (No). | |
| 568. |
The elements present in alnthanide series areA. 14B. 15C. 16D. 17 |
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Answer» Correct Answer - A We know that, lanthanide series contains 14 elements. This series started from Ce to Lu. |
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| 569. |
Actinides are known asA. 5-f series elementsB. `2^(nd)` inner transition elementsC. radio active elemenrtsD. all of these |
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Answer» Correct Answer - D We know that, all actinides are radioactive. The elements bismuth onwards are radioactive. |
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| 570. |
Lanthanides are known asA. rare earth elementsB. 4-f series elementsC. `1^(st)` inner transition elementsD. all of these |
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Answer» Correct Answer - D We know that, lanthanides are not widely spread in earth crust. Hence named as rare earth elements. |
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| 571. |
Zinc, cadmium and mercury are generally not considered as transition metals. Give reasons. |
| Answer» These elements in their most common oxidation state of `+2` have completely filled d-orbitals. Hence, d-d transition cannot occur. | |
| 572. |
Lanthandies and actinides areA. s-block elementsB. p-block elementsC. d-block elementsD. f-block elements |
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Answer» Correct Answer - A We known that, in lanthanides and actinides last electron is added in 4f and 5f orbitals, Hence these f-block elements. |
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| 573. |
How many elements are presents in the d-block of the periodic table ? |
| Answer» Correct Answer - 40 | |
| 574. |
How many transiiton series of elements are there in the periodic table ? Name them. |
| Answer» There are four main transition series of elements correspong to filling of 3d, 4d, 5d and 6d sublevels in 4th, 5th, 6th and7th periods. They are known as first transition series , second transition or 4d series, third transition or 5dseries and fourth transition or 6d series. | |
| 575. |
Why does a transition series contain 10 elements? |
| Answer» There are five d-orbitals in an energy level and each orbital can have two electrons. As we move from one element to the next, an electron is added and for complete filling of the five d orbitals , 10 electrons are required. | |
| 576. |
General electronic configuration of d-block elements isA. `(n-1)d^(1-10) ns^(1-2)`B. `ns^(2) np^(1-6)`C. `(n-2)f^(0-14) (n-1)^(1-2) ns^(2)`D. `(n-1)d^(1-5) ns^(1-2)` |
| Answer» Correct Answer - A | |
| 577. |
How many f-electrons are present in Tb when it show `+3` oxidation state ?A. 6B. 7C. 8D. 9 |
| Answer» Correct Answer - C | |
| 578. |
p-Amino-N,N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of aqueous solution of Y with reagent potassium hexacyanoferrate (II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, the treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown colouration due to the formation of Z. Q. Compound Z isA. `Mg_2[Fe(CN)_6]`B. `Fe[Fe(CN)_6]`C. `Fe_4[Fe(CN)_6]_3`D. `K_2Zn_3[Fe(CN)_6]_2` |
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Answer» Correct Answer - B `4FeCl_3+3K_4[Fe^(2+)(CN)_6]toFe_4[Fe(CN)_6]_3+2KCl` `FeCl_3+K_3[Fe^(3+)(CN)_6]toFe[Fe(CN)_6]+3KCl` |
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| 579. |
p-Amino-N,N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of aqueous solution of Y with reagent potassium hexacyanoferrate (II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, the treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown colouration due to the formation of Z. Q. Compound X isA. `NaNO_3`B. `NaCl`C. `Na_2SO_4`D. `Na_2S` |
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Answer» Correct Answer - D X is `Na_2S` . `Na_2S+2H^(o+)toH_2S+Na^(o+)` |
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| 580. |
What happens when (i) `(NH_(4))_(2)Cr_(2)O_(7) ` is heated ? (ii) `H_(3)PO_(3) ` is heated ? |
| Answer» (ii) `4H_(3)PO_(3)overset(473K) (rarr) 3H_(3)PO_(4)+ PH_(3) ` | |
| 581. |
The number of moles of `KMnO_(4)` reduced by `1 "mol of" KI` in alkaline medium is |
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Answer» Correct Answer - 2 `2KMnO_(4)+KI+H_(2)Orarr2KOH+2MnO_(2)+KIO_(3)` |
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| 582. |
Acidic `K_(2) Cr_(2)O_(7)` oxidises KI toA. KOIB. `KOI_(3)`C. HID. `I_(2)` |
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Answer» Correct Answer - D `K_(2)Cr_(2)O_(7)+ 7H_(2)SO_(4)+6KI to 4K_(2)SO_(4)+ Cr_(2) (SO_(4))_(3) +7H_(2)O+3I_(2)` |
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| 583. |
Acidic `K_(2)Cr _(2)O_(7)` oxidises `S_(2) ` to `K_(2)SO_(4)` and itself reduced intoA. `Cr_(2)(SO_(4))_(3)`B. `CrO_(3)`C. `H_(2)CrO_(4)`D. `KOH ` |
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Answer» Correct Answer - A Acidic `K_(2)Cr_(2) O_(7)` oxidises `SO_(2)` to `K_(2) SO_(4)` and itself reduced into chromic sulphate. |
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| 584. |
Which one of the following is reduced by hydrogen peroxide in acid medium?A. Potassium permagneteB. Potassium iodideC. Ferrous sulphateD. Potassium ferrocyanide |
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Answer» Correct Answer - A `H_(2)O_(2)` reduces acidified `KMnO_(4)` solution. As a result. The pink colour of `KMnO_(4)` is changed. |
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| 585. |
Which statement is not correctA. Potassium paramagnet is a powerful oxidising substanceB. Potassium paramagnet is a weakeer oxidising substance than potassium dichromateC. Potassium permanganate is a stronger oxidising substance than potassium dichromateD. Potassium dichromate oxidises a secondary alcohol into a ketone |
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Answer» Correct Answer - B In acidic medium, `KMnO_(4)` gives 5 oxygen while acidic `K_(2)Cr_(2)O_(7)` gives 3 oxygen. |
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| 586. |
Acidic `KMnO_(4)` oxidises `SO_(2)` toA. SB. `SO_(3)`C. `H_(2)SO_(4)`D. MnSO_(4)` |
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Answer» Correct Answer - C `2KMnO_(4)+2H_(2)O+5SO_(4)to 2MnSO_(4)+K_(2)SO_(4)+2H_(2) SO_(4)` |
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| 587. |
Which of the following compounds is formed when a mixture of `K_(2)Cr_(2) O_(7) and KCl` is heated with conc. `H_(2)SO_(4)` ?A. `CrO_(2)Cl_(2)`B. `CeCL_(3)`C. `CrOCl_(2)`D. `CrO_(2)Cl_(2)` |
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Answer» Correct Answer - A Red vapours of `CrO _(2) Cl_(2)` are obtained. `K_(2) Cr_(2) O_(7)+ 2H_(2)SO_(4)to KHSO_(4) +2CrO_(3) +H_(2)O` `KCl+H_(2) SO_(4) to KHSO_(4) +HCl` `{:(CrO_(3)+ 2HCl to " "CrO_(2)Cl_(2) " "+H_(2)O), (" ""Chromyl chloride"),(" "("Red oily vapours")):}` |
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| 588. |
The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium isA. `IO_(3)^(-), MnO_(2)`B. `I_(2), MnO_(4)^(2-)`C. `I_(2), MnO_(2)`D. `IO^(-), MnO_(4)^(2-)` |
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Answer» Correct Answer - A `KMnO_(4)+KI + H_(2)O to 2KOH+2MnO_(2)+ KIO_(3).` |
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| 589. |
The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium isA. `IO_3^-`B. `I_2`C. `IO^-`D. `IO_4^(-)` |
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Answer» Correct Answer - A In alkaline medium the stable oxidation state of Mn is +6 . Thus `MnO_4^-` is reduced to `MnO_4^(2-)` and `I^-` is oxidised to `IO_3^(-)` `6MnO_4^(-) + I^(-) + 6OH^(-) to 6MnO_4^(2-) + IO_3^(-) + 3H_2O` |
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| 590. |
On adding `KMnO_(4)` to cold conc. `H_(2)SO_(4),` it givesA. `MnO_(2)`B. `Mn_(2)O_(3)`C. `Mn_(2)O_(7)`D. `MnO_(3).` |
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Answer» Correct Answer - C `2K MnO_(4)+ 2H_(2) SO_(4)to 2KHSO_(4)+ underset(("Green oily liquid"))(Mn_(2) O_(7)+H_(2)O)` `Mn_(2)O_(7)` explodes on heating. `2Mn_(2)O_(7)overset(Delta)to4MnO_(2)+3O_(2)` |
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| 591. |
Pick out the incorrect statement for transition metals,A. They from alloysB. transition metals do not exhibit variable oxidation statesC. transition metal ions are collouredD. transition metals and majority of their compounds are paramagnetic. |
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Answer» Correct Answer - B a) Transition metals exhibit variable oxidation states, because the energies of `( n-1)` d and nsorbitals have comparable values, due to which `(n-1)` d-orbital electrons alos participate in bond formation with ns electrons. a) `Cu^(+)` is not a transition metal ion, because it has fully-filled d-orbitals. c) Transition metal ion are coloured, because of d-d transitions of electron by the absorption of radiations of electron by the absorption of radiation of visible range. Therefore, the trsnmitted is possible, if d-orbitals are incompletely filled. d) Trnsition metals and majority of their compound are paramagnetic, because they have unpaired electrons in the `(n-1)` d-orbitals. have unpaired electrons in the `(n-1)` d-orbitals. |
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| 592. |
Which one of the following statements is not true?A. Transition metals form alloysB. Transition metals from complexesC. Zn, Cd and Hg are transition metalsD. `K_(2) [PtCl_(6)]` is a well known compound, but corresponding nickel compound is not known |
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Answer» Correct Answer - C Zn, Cd and Hg are not regarded transition metals, because the netural atoms as well as stable ion have fully-filled d-orbitals. a) Because transition metals have similar atomic sizes, therefore, in the crystal lattice, one transition metal atoms can be easily replaced by the atoms of another transition metal to form an alloy. b) Transition metals form complexes, because they form small highly charged cations, which have vacant orbitals of suitable energy to accept electron pairs donated by ligands. d) Because the sum of first four inization energies of Pt is less than the sum of first four ionization energies of Ni, so `K_(2)[PtCl_(6)]` is more stable than the corresponding nickel compound. |
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| 593. |
`MnO_(4)^(-)` ions can be reduced in strongly alkaline medium to giveA. `MnO_(2)`B. `Mn^(1+)`C. `MnO _(4) ^(2-)`D. `MnO_(3)^(-)` |
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Answer» Correct Answer - C `underset(("Purple"))(MnO_(4)^(-))+e^(-) to underset(("Green "))(MnO_(4)^(2-))` |
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| 594. |
Pick out the incorrect statement for transition metalsA. they have low melting and boiling points (or low enthalpies of atomization)B. 5d-elements have higher energies than 3d or 4d-elementsC. Zr and Hf have almost identical atomic and ionic radiiD. thtey form interstitial compounds. |
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Answer» Correct Answer - A a) Transition metals have high enthalpies of atmization, because they form strong metallic bonds, due to participation of `(n-1)` d-electrons along with ns-electrons in metallic bonding. b) This is due to the greater reffective nuclear charge acting on outer valence electrons, because of weak shielding of nucleus by 4f-electrons. c) The normal size in increase from Zr to Hf is almost exactly balanced by lanthanide contraction. d) Because small size, atom of H, He, C, N occupy the vacant spaces in the crystal lattice of transition metal. |
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| 595. |
Pick out the incorrect statement.A. `MnO_(4)^(2-)` is quite strongly oxidizing and stable only in very strong alkalies. In dilute alkali, water or acidic solutions , it disproportionates.B. In acidic solutions, `MnO_(4)^(-)` is reduced to `Mn ^(2+)` and thus, `KMnO_(4)` is widely used as oxidizing agentC. `KMnO_(4)` does not act as oxidizing agent in alkaline mediumD. `KMnO_(4)` is manufactured by the fusion of pyrolusite ore with KOH in presence of air or `KNO_(3).` followed by electrolytic oxidation in alkaline solution. |
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Answer» Correct Answer - C Alkaline `KMnO_(4)` also acts as oxidizing agent. `MnO_(4)^(-) +2H_(2) O+3e^(-)to MnO_(2)+ 4OH^(-)` a) `MnO_(4)^(2-)` disproportionate in water, dilute alkali or acidic solution. `overset((+6))(MnO_(4)^(2-))+4H^(+)to overset((+7))(2MnO_(4)^(-))+overset((+4))(+MnO_(2))+2H_(2)O` b) `MnO_(4)^(-)` is reduced to `Mn^(2+)` involbing a change of five electrons. `MnO_(4)^(-)+8H^(+)+5e^(-)to Mn ^(2+) +4H_(2)O` d) `MnO_(2) +2KOH +1/2O_(2) to K_(2)MnO_(4) +H_(2) O` Anode : `MnO_(4)^(2-)overset("Alkaline")underset("Solution")toMnO_(4)^(-)+e^(-)` Cathode : `2H^(+)+2e^(-) to H_(2)` |
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| 596. |
Pick out incorrect statemment about `K_(2) Cr_(2)O_(7).`A. It oxidazes KI to `I_(2)`B. It oxidizes Ki to `I_(2)`C. It oxidizes `SO_(2)` to ` K)_(2)SO_(4)`D. It oxidizes `SO_(2)` to S |
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Answer» Correct Answer - D It oxidises `SO_(2)` to `K_(2)SO_(4)` |
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| 597. |
Pick out the incorrect statement about `K_(2)Cr_(2)O_(7)`A. It act as oxidising agent in basic mediumB. It dissolves in alkali to form chromateC. It oxidizes `FeSO_(4)` solution to `Fe_(2)(SO_(4))_(3)`D. It is used as cleansing agent for glassware etc. when mixzed with cold conc `H_(2)SO_(4).` |
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Answer» Correct Answer - A a) It decomposes on heating to a white heat. `4K_(2) Cr_(2) O_(7)overset(Delta)to 4K_(2) CrO_(4)+2Cr _(2)O _(3)+ 3O_(2)` b) `underset(("Orange-red"))(K_(2)Cr_(2)O_(7))+ 2KOH to underset(("Yellow"))(2K_(2)CrO_(4))+ H_(2)O` c) `K_(2) Cr_(2) O_(7)+ underset(("dil))(4H_(2)SO_(4))toK_(2)SO_(4)+ Cr_(2)(SO_(4))_(3) +4H_(2) O +3O` `(2FeSO_(4) +H_(2)SO_(4) + O to Fe_(2) (SO_(4))_(3)+ +H_(2) Oxx3)/(K_(2) Cr_(2) O_(7)+ 7H_(2) SO_(4)+ 6FeSO_(4)to3Fe_(2)(SO_(4))_(3)+ K_(2)SO_(4)+Cr_(2) (SO_(4))_(3)+ 7H_(2)O)` d) With cold and concentrated `H_(2)SO_(4),` it forms `CrO_(3),` which forms `H_(2)CrO_(4)` with water, which has cleansing action. `K_(2) Cr_(2) +2H_(2) SO_(4) to 2Cr O_(3) +2KHSO_(4)+ H_(2)O` |
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| 598. |
Pick out the incorrect statement:A. `MnO_(4)^(2-)` is quite strongly oxidizing and stable only in very strong alkalies, In dilute alkali, neutral solutions, it disproportionatesB. In acidic solution, `MnO_(4)^(-)` is reduced to `Mn^(2+)` and thus, `KMnO_(4)` is widely used as oxidising agentC. `KMnO_(4)` does not acts as oxidising agent in alkaline mediumD. `KmnO_(4)` is manufactured by the fusion of pyrolusite ore with `KOH` in presence of air or `KNO_(3)`, followed by electrolytic oxidation in alkaline solution |
| Answer» Correct Answer - C | |
| 599. |
Which ion is colourless in water ?A. `Ti^(+3)`B. `Sc^(+3)`C. `Cr^(+3)`D. `V^(+3)` |
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Answer» Correct Answer - B `Sc^(+3) (3d^0 4s^0)` ion is colourless in water as it has no unpaired electrons |
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| 600. |
Which of the following ion is colourless in aqueous solution ?A. `Fe^(2+)`B. `Mn^(2+)`C. `Ti^(3+)`D. `Sc^(3+)` |
| Answer» Correct Answer - d | |