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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.
| 401. |
Which of the following pairs has the same size ?A. `Zn^(2+), Hf^(4+)`B. `Fe^(2+),Ni^(2+)`C. `Zr^(4+),Ti^(4+)`D. `Zr^(4+),Hf^(4+)` |
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Answer» Correct Answer - D `Zr^(4+) and Hf^(4+)` have similar ionic radii due to lanthanoid contraction. |
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| 402. |
Given reasons for the following in one or two sentences: silver bromide is used in photography? |
| Answer» Silver bromide is used in photography because silver bromide is photosensitive. It decomposes and is converted into metallic silver grains when light is incident on it. | |
| 403. |
State the conditions under which the following preparations are carried out. Give the necessary equations which need not be balanced. Potassium permanganate from manganese dioxide. |
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Answer» Potassium permanganate can be prepared from manganese dioxide as follows. (i). `MnO_2+4KOH+O_2to2K_2MnO_4+2H_2O` (ii). `2K_2MnO_4+Cl_2to2KMnO_4+2KCl` The above reactionn can also be carried out with ozone or carbon dioxide, or electrolytically. |
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| 404. |
Which of the following statements concerning transition elements is false?A. They are all metalsB. They easily form complex coordination compoundsC. they from compounds containing unapaired electrons and their ions are mostly coloured electrons and there are mostly colouredD. They show multiple oxidation states always differing by units of two |
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Answer» Correct Answer - D The false statement about transition elements is that they show multiple oxidation states always differ by a unit of two, because they show variable oxidation state but not always have a difference of two unit. |
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| 405. |
Complete and balance the following reactions: (i). `Zn+NO_3^(ɵ)toZn^(2+)+NH_4^(o+)` (ii). `Cr_2O_7^(2-)+C_2H_4OtoC_2H_4O_2+Cr^(3+)` |
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Answer» (i). `4Zn+NO_3^(ɵ)+10H^(o+)to4Zn^(2+)+NH_4^(o+)+3H_2O` (ii). `Cr_2O_7^(2-)+3C_2H_4O+8H^(o+)to3C_2H_4O_2+2Cr^(3+)+4H_2O` |
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| 406. |
Which of the following is expected to be coloured?A. CuClB. `CuF_(2)`C. `Ag_(2)SO_(4)`D. `MgF_(2)` |
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Answer» Correct Answer - C In `CuF_(2), Cu` is present as `Cu^(2+)`.`Cu^(2)=[Ar]3d^(9),4s^(0)` Since , in it unpaired electron is present, it is coloured. |
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| 407. |
`(NH_4)_2Cr_2O_7` on heating gives a gas which is also given byA. heating `NH_4NO_2`B. Heating `NH_4NO_3`C. `Mg_3N_2+H_2O`D. `Na+H_2O_2` |
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Answer» Correct Answer - A `(NH_4)_2Cr_2O_7overset(Delta)toCr_2O_3+4H_2O+N_2uarr` `NH_4NO_2overset(Delta)to2H_2O+N_2uarr` |
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| 408. |
Which of the following compounds is expected to be coloured?A. `Ag_(2)SO_(4)`B. `CuF_(2)`C. `MgF_(2)`D. CuCl. |
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Answer» Correct Answer - B (b) `CuF_(2)` is coloured because `Cu^(2+)(3d^(9))` has unpaired electron. |
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| 409. |
The maximum number of transition metal elements that can be accommodated in any seriesA. 4B. 6C. 8D. 10 |
| Answer» Correct Answer - D | |
| 410. |
The properties of tungsten and molybdenum are sinmilar becauseA. both belong to d-blockB. both belong to same group of the periodic tableC. both have same number of electronsD. both have same radii |
| Answer» Correct Answer - D | |
| 411. |
The pair of atoms with anomalous (exceptional) electron congigurationsA. `Cu, Cr`B. `Cu, Zn`C. `Cr, Cd`D. `Cr, Zn` |
| Answer» Correct Answer - A | |
| 412. |
The properties of `3 ^(rd)` transition series elements and `2^(nd)` transition series elements are same because ofA. both belong to d-blockB. both belong to same group of the periodic tableC. both have same number of electronsD. both have same radii |
| Answer» Correct Answer - D | |
| 413. |
Transition elemtns are colouredA. Due to small sizeB. Due to metallic natureC. Due to unpaired d-electronsD. All of these |
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Answer» Correct Answer - C Due to unpaired d-electrons. |
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| 414. |
Which of the following `M^(2+)` ions have stable configuration ?A. `Eu^(2+)` onlyB. `Eu^(2+)and Yb ^(2+) ` onlyC. `Yb^(2+)` onlyD. `Yb^(2+)and Sm^(2+)` only |
| Answer» Correct Answer - B | |
| 415. |
Among the following pairs of ions the lower oxidation state in aqueous solution is more stable than the other inA. `Ti^(+), Ti^(3+)`B. `Cu^(+), Cu^(2+)`C. `Cr^(2+), Cu^(3+)`D. `V^(2+),VO^(2+)` |
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Answer» Correct Answer - A `Ti^(+)` is more stable than `Ti^(3+)` because only valence `6p^(1)` is to be removed and valance `6s^(2)` electrons impart stability to the ions on account of inert pair effect. |
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| 416. |
Spin only of `Mn^(2+)` isA. 7 B. MB. 6 B.MC. 1.5 B.MD. 5.9 B.M |
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Answer» Correct Answer - D Spin only of `Mn^(2+)` is Electronic configuration of Mn and `Mn^(2+)` is , `""_(25)Mn=[Ar] 3d^(5),4s^(2)` `""_(25)Mn^(2+)=[Ar]3d^(5), 4s^(0)` `mu _(eff)=sqrt(n(n+2))B.M` `=sqrt(5(5+2))=sqrt35=5.9B.M.` |
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| 417. |
The metals present in insulin and haemoglobin are respectively:A. FeB. CoC. ZnD. Au |
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Answer» Correct Answer - C Insulin contain Zn in trace amount |
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| 418. |
Which of the following alloys contain (s) Cu and Zn?A. BronzeB. SteelC. AlnicoD. Brass |
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Answer» Correct Answer - D Their percentage compositions are Brass: `Cu =80, Zn= 20,` |
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| 419. |
Which of the following elements forms stable di-nuclear ions?A. ZnB. CdC. HgD. Fe |
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Answer» Correct Answer - C Hg (I) compounds contain mercurous ions `(Hg-Hg)^(2+) and not Hg^(+).` The two Hg atoms are bonded together using 6s-orbital. The only other metals which forms di nuclear ions are `(Zn-Zn)^(2+) and (Cd-Cd)^(2+).` But, these ions are unstable and have been only detected in melts of `Zn//ZnCl_(2) and Cd//Cd Cl_(2)` spectroscopically. |
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| 420. |
Transition metals are often paramagnetic due toA. Their high m.p. and b.p.B. The presence of vacant orbitalsC. The presence of one or more upaired electrons in the systemD. Their being less electropoitive than the elements of groups I-A and II-A |
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Answer» Correct Answer - C The presence of one or more unpaired electrons in the system. |
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| 421. |
Transition metals are often paramagnetic due toA. high m.p. and b.p.B. presence of unpaired electronsC. malleability and ductilityD. low m.p. and b.p. |
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Answer» Correct Answer - B Paramagnetic property is due to presence of unpaired `e^-` in (n-1)d orbitals |
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| 422. |
The oxidation number of Mn in the product of alkaline oxidative fusion of `MnO_(2)` isA. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - D `2MnO_(2)+4KOH+O_(2)overset("Fusion"" ")(rarr)2K_(2)MnO_(4)+2H_(2)O` Oxidation number of `Mn "in" K_(2)MnO_(4)` is `2xx(1)+X+4(-2)=0 implies " "X=+6` |
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| 423. |
Find the oxidation number of `Mn` in the product of alkaline oxidative fusion of `MnO_(2)`. |
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Answer» Correct Answer - 6 `2MnO_(2)+4KOH+O_(2)rarr2K_(2)MnO_(4)+2H_(2)O.` The oxidation state of Mn in `K_(2)MnO_(4)` formed in the reaction is +6.` |
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| 424. |
Rusting of iron in moist air involves-A. Loss of electrons by ironB. Gain of electrons by ironC. Neither gain nor less of electronsD. Hydration of iron |
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Answer» Correct Answer - A Rusting of iron in moist air involves the loss of electrons by iron. The oxidation of iron to ferrous ion takes place and a brown deposit takes place on the surface of iron. `Fe to Fe^(+2)+2e` Rusting is an electrochemical process. |
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| 425. |
In which of the following d-d transition involves absorption in the ultraviolent region ? |
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Answer» Correct Answer - D The wavelenght of light of light absorbed by `[Co(CN)_(6)]^(3-)` is 310 pm. This corresponds to the U.V region and not to the visible region. |
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| 426. |
How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide?A. 1B. 2C. 3D. 4 |
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Answer» `K_(2)Cr_(2)O_(7) + 4H_(2)SO_(4) rarrK_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 4H_(2)O+ 3(O)` `ul2KI + H_(2)SO_(4) + (O) rarrK_(2)SO_(4) + I_(2) + H_(2) O ] xx3)` `K_(2)Cr_(2)O_(7) + 7H_(2)SO_(4)+ 6KI rarr 4K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 7GH_(2)O + 3I_(2)` |
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| 427. |
The coloured spot of`KMnO_(4)` on any article can be bleached byA. `SO_(2)+H^(+)`B. `C_(2)O_(4)^(2-)+H^(+)`C. `H_(2)O_(2)+H^(+)`D. all of these |
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Answer» Correct Answer - D `KMnO_(4)` in presence of `H^(+)` ionscan be reduced to colourless `Mn^(2+)` by all the given chemicals. |
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| 428. |
Which of the following is not used as a catalyst ?A. FeB. NiC. PtD. `Cl_(2)` |
| Answer» Correct Answer - D | |
| 429. |
the catalyst used in the hydrogenation of oils is :A. ZnB. NiC. MoD. Fe |
| Answer» Correct Answer - B | |
| 430. |
The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted toA. `Mn_2O_3`B. `MnO_2`C. `MnO_4^(-)`D. `MnO_4^(2-)` |
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Answer» Correct Answer - B `{:(MnSO_4 to ,MnO_2 ),(+2,+4),(Mn^(2+)to ,Mn^(4+) + 2e^(-)):}` `therefore` Equivalent weight of `MnSO_4 = "Molecular weight"/2` |
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| 431. |
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples. |
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Answer» In transition elements the oxidation states vary by unity (due to incomplete filling of d-orbitals) Eg: Mn exhibits +2, +3, +4, +5, +6 and +7 all differing by 1. In non-transition element, this variation is selective, always differing by 2. Eg : S exhibits 2, 4, 6 oxidation states. N exhibits 3, 5 etc. |
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| 432. |
How is the variability of oxidation states of the transition elements different from that of the non-transition elements ? Illustrate with examples. |
| Answer» The variability in oxidation states of transition metals is due to the incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., `Fe^(2+)` and `Fe^(3+), Cu^(+)` and `Cu^(2+)` and `Cu^(2+)` etc. In case of non-transition elements, the oxidation states differ by units of two , e.g., `Pb^(2+)` and `Pb^(4+), Sn^(2+)` and `Sn^(4+)` etc. Moreover , in transition elements, the higher oxidation states are more stable for heavier elements in a group. For example , in group 6, Mo(VI) and W(VI) are more stable than Cr(VI) . In p-block elements , the lower oxidation states are more stable for heavier members due to inert pair effect, e.g., in group 16, Pb (II) is more stable than Pb(IV). | |
| 433. |
Knowing that the chemistry of lanthanoids (Ln) is dominated by its `+3` oxidation state, which of the following statement is incorrect?A. The ionic sizes of Ln (III) decreases in general with increasing atomic numberB. Ln (III) compounds are generally colourlessC. Ln (III) hydroxides are mainly basic in characterD. Because of the large size of the Ln(III) ions the bonding in its compounds is predominently ionic in character |
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Answer» Correct Answer - B `(Xe)6s^(2)6sd^(1)` `Ln^(+3)=(Xe)` diamagnetic species. |
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| 434. |
Knowing that the chemistry of lanthanoids (Ln) is dominated by its `+3` oxidation state, which of the following statement is incorrect?A. Because of the large size of the Ln(III) ions the bonding in its compounds is predominantly inoni in character.B. The ionic sizes of Ln(III) decrease in general with increasing atomic numberC. Ln(III) compounds are generally colourlessD. Ln(III) hydroxides are mainly basic in character. |
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Answer» Correct Answer - C `Ln^(3+)` compounds are generally coloured in the solid state as well as in presence of f-electrons which undergo f-f transitions. |
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| 435. |
How would you account for the following ? Among lanthanoids, Ln (III) compounds are predominant . However, occasionally in solutions or in the solid compounds , `+2` and `+4 ` ions are alos obtained. |
| Answer» `+2` and `+4` oxidation states are shown by those elements which by losing 2 or 4 electrons acquire a stable configureation of `f^(0), f^(7) ` or `f^(14)` , e.g., `Eu^(2+)` is `[Xe] 4f^(7), Yb^(2+)` is `[Xe4f^(14) , Ce^(4+) ` is `[Xe]4f^(0)` and `Tb^(4+)` is `[Xe]4f^(7)` | |
| 436. |
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spinn only magnetic moment value of `Cr^(3+)` ion is `"…..........."`A. 2.87 B.M.B. 3.87 B.M.C. 3.47 B.M.D. 3.57 B.M. |
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Answer» Correct Answer - b `Cr^(3+) = 3d^(3)` . Hence, `mu = sqrt(n( n+2)) B.M. = sqrt( 3(3+2)) = sqrt( 15) = 3.87 B.M` |
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| 437. |
The color of `KMnO_(4)` is due to :A. `L to M` charge transfer transitionB. `sigma-sigma^(**)` transitionC. `M to L` charge transfer transitionD. `d-d` transition |
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Answer» Correct Answer - A The colour of `KMnO_(4)` is due to ligands `(L)to` metal charge transfer transition. Due to high value of positive charge on `Mn` |
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| 438. |
The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of `Cr^(3+)` ion isA. 2.87 BMB. 3.87 BMC. 3.47 BMD. 3.57 BM |
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Answer» Correct Answer - B The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of `Cr^(3+)` ion is `3d^(3)` Hence, magnetic moment `(mu) = sqrt(n(n+2))BM` `" "=sqrt(3(3+2))=sqrt(15)` `" "=3.87 BM` |
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| 439. |
The spin only magnetic moment of `Fe^(3+)` ion (inBM) is approximatelyA. 4B. 7C. 5D. 6 |
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Answer» Correct Answer - D The electronic configuration of is `._(26)Fe=[Ar]3d^(6),4s^(2)` `Fe^(3+)=[Ar]3d^(5)` Thus , five unpaired electrons are present. :. Spin only magnetic moment, `mu=sqrt(n(n+2))BM` `n=5("unpaired electrons")` `mu=sqrt(5(5+2))BM=sqrt35=5.91~~6BM` |
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| 440. |
Which one of the following arrangements does not represent the correct order of the property stated against it?A. `Ni^(2+)ltCo^(2+)ltFe^(2+)ltMn^(2+):` ionic sizeB. `Co^(3+)ltFe^(3+)ltSc^(3+):` stability in aqueous solutionC. `ScltTiltCrltMn:` number of oxidation statesD. `V^(2+)ltCr^(2+)ltMn^(2+)ltFe^(2+)`: paramagnetic behaviour |
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Answer» Correct Answer - B::D On the basis of `CFSE`, stability of `Co^(3+)` is highest and stability of `Sc^(3+)` is minimum among `Co^(+3), Fe^(+3),Cr^(+3), Sc^(+3)` `{:("Paramagnetic behaviour"),(V^(2+)ltCr^(2+)ltFe^(2+) lt Mn^(2+)):}` |
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| 441. |
Which of the following arrangements does not represent the correct order of the property stated against it?A. `V^(2+)ltCr^(2+)ltMn^(2+)ltFe^(3+)` paramagnetic behaviourB. `Ni^(2+)ltCo^(2+)ltFe^(2+)ltMn^(2+)` ionic sizeC. `Co^(3+)ltFe^(3+)ltCr^(3+)ltSc^(3+)` stability in aqueous solutionD. `Sc lt Ti lt Cr lt Mn ` number of oxidation states |
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Answer» Correct Answer - A (a)`V^(2+)=3` unpaired electrons `Cr^(2+)=4` unpaired electrons `Mn^(2+)=5` unpaired electrons `Fe^(3+)=4` unpaired electrons Hence, the order of paramagnetic behaviour should be `V^(2+) lt Cr^(2+)=Fe^(2+) lt Mn^(2+)`(b) Ionic size decreases from left to right in the same period. (c)`Co^(3+)//Co^(2+)=1.97Fe^(3+)//Fe^(2+)=0.77` `Cr^(3+)//Cu^(2+)=-0.41` Hence, stability increases as `Co^(3+) lt Fe^(3+) lt Cr^(3=) lt Sc^(3+)`. (d) The oxidation states increases as we go from group 3 to group 7 in the same period . |
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| 442. |
Which of the following arrangements does not represent the correct order of the property stated against it ?A. `Sc ltTi lt Cr lt Mn :` number of oxidation statesB. `V^(2+) ltCr^(2+) lt Mn^(2+) lt Fe^(2+) :` paramagnetic behaviourC. `Ni^(2+) lt Co^(2+) lt Fe^(2+) lt Mn^(2+) :` ionic sizeD. `Co^(3+) lt Fe^(3+) lt Cr^(3+) lt Sc^(3+) :` stability in aqueous solution. |
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Answer» (a) Number of oxidation states increases along a period.Hence, (a) is correct. (b) `V=3d^(3) 4s^(2) , V^(2+) = 3d^(3)= 3` unpaired `e^(-)` `Cr= 3d^(5) 4s^(1) , Cr^(2+)= 3d^(4) = 4` unpaired `e^(-)` `Mn= 3d^(5)4s^(2), Mn^(2+)= 3d^(2+)= 5` unpaired `e^(-)` `Fe=3d^(6) 4s^(2), Fe^(2+)= 3d^(6) = 4` unpaired`e^(-)` Hence, correct order of paramagnetic behaviour will be `V^(2+) lt Cr^(2+) = Fe^(2) lt Mn^(2+)`. Thus, `(b)` is incorrect.(c ) Ionic radii `(M^(2+))` increase along the period . Hence, (c ) is correct. (d) Stability of `M^(3+)` ion is aqueous soltuion increases as `E_(M^(3+) //M^(2+))^(@)` values decrease. Hence, (d) is correct |
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| 443. |
Which one of the following arrangements does not represent the correct order of the property stated against it?A. `Ni^(2+)ltCo^(2+)ltFe^(2+)ltMn^(2+)`: ionic sizeB. `Co^(3+)ltFe^(3+)ltCr^(3+)ltSc^(3+)`: Stability in aqueous solution.C. `ScltTiltCrltMn`: number of oxidation states.D. `V^(2+)ltCr^(2+)ltMn^(2+)ltFe^(2+)`: paramagnetic behaviour |
| Answer» Correct Answer - B::D | |
| 444. |
The spin only magnetic tnoment of `[MnBr_(4)]^(2-)` is 5.9 BM. Predict the geometry of the complex ion ? |
| Answer» Since the Co-ordination number of `Mn^(2+)` ion in the complex ion is 4, it will be either tetrahedral (`sp^(3)` hybridisation) or square planar (`dsp^(2)` hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d-orbitals. | |
| 445. |
The elements of the three transition series of the d-block are given below `:` `{:(Sc,Ti,V,Cr,Mn,Fe,Co, Ni,Cu,Zn),(Y,Zr,Nb,Mo,Tc,Ru,Rh,Rd,Ag,Cd),(La,Hf,Ta,W,Re,Os,Ir,Pt,Au,Hg):}` In any transition, as we move from left to right,the d-orbital sare progressively filled and their properties vary accordingly . Which element do you expect to have the smallest atomic radius ?A. ScB. ZnC. LaD. Hg |
| Answer» Atomic size decreases along the transition series from left to right though at the end of the series, there is a slight increase. Down the group, the size increases through elements of the second and third series have nearly the same size due to lanthanide contraction. | |
| 446. |
The elements of the three transition series of the d-block are given below `:` `{:(Sc,Ti,V,Cr,Mn,Fe,Co, Ni,Cu,Zn),(Y,Zr,Nb,Mo,Tc,Ru,Rh,Rd,Ag,Cd),(La,Hf,Ta,W,Re,Os,Ir,Pt,Au,Hg):}` In any transition, as we move from left to right,the d-orbital sare progressively filled and their properties vary accordingly . Which element do you expect to have the highest melting point ?A. LaB. WC. OsD. Pt |
| Answer» W has the highest melting point due to maximum number of unpaired electrons and strongest intermetallic bonds. | |
| 447. |
Aqueous `Cu^(2+)` ions are blue in colour, where as Aqueous `Zn^(2+)` ions are colourless. Why ? |
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Answer» `rarr` electronic configuration of `Cu^(+2)` ion is `[Ar]4s^(0)3d^(9)`. It contains one unpaired electron due to presence of this unpaired electron aq. `Cu^(+2)` ions are blue in colour. `rarr` Electronic configuration of `Zn^(+2)` ion is `[Ar] 4s^(0)3d^(10).` It contains no unpaired electrons, due to absence of unpaired electrons aq. `Zn^(+2)` ions are colourless. |
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| 448. |
Why `Zn^(2+)` is diamagnetic whereas `Mn^(2+)` is paramagnetic ? |
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Answer» `rarr Zn^(+2)` electronic configuration is `[Ar]4s^(0)3d^(10)`. It has no unpaired electrons. So it is dia magnetic. `rarr Mn^(2+)` electronic configuration is `[Ar]4s^(0)3d^(5).` It has five unpaired electrons so it is paramagnetic. |
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| 449. |
van-Arkel method is based onA. cupellation methodB. furnace refining methodC. poling methodD. None of the above |
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Answer» Correct Answer - D van-Arkel method is based on vapour phase refining. |
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| 450. |
Since, lanthanides relased electron from 4f, 5d and 6s orbitals, they showA. fixed oxidation stateB. variable oxidation stateC. two oxidation stateD. one oxidation state |
| Answer» Correct Answer - B | |