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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.
| 851. |
The maximum oxidation state of osmium isA. `+6`B. `+7`C. `+8`D. `+5` |
| Answer» Correct Answer - c | |
| 852. |
When `I^(Theta)` is oxidised by `MnO_(4)^(Theta)` in an alkaine medium, `I^(Theta)` converts intoA. `lO_(3)^(-)`B. `l_(2)`C. `lO_(4)^(-)`D. `lO^(-)` |
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Answer» Correct Answer - A When `l^(-)` is oxidised by `MnO_(4)^(-)`in alkaline medium`l^(-)`converts into`lO_(3)^(-)`. `{:(2KMnO_(4)+2KOHrarr2K_(2)MnO_(4)+H_(2)O+[O]),(ul(2K_(2)MnO_(4)+2H_(2)Orarr2MnO_(2)+4KOH+2[O])),(2KMnO_(4)+H_(2)O overset("alkaline")rarr2MnO_(2)+2KOH+3[O]),(Kl+3[O]rarrKlO_(3)),(Hence","),(bar(2KMnO_(4)+Kl+H_(2)Orarr2KOH+2MnO_(2)+KlO_(3))):}` |
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| 853. |
Using IUPAC norms write the formulas for the Potassium tetrachloropalladate (II) |
| Answer» Potassium tetrachloropalladate - `K_(2)[PtCl_(4)]` | |
| 854. |
Using IUPAC norms write the formulas for the Hexaamminecobalt (III) sulphate |
| Answer» Hexa ammie cobalt (III) sulphate - `[Co(NH_(3))_(6)]_(2)(SO_(4))_(3)` | |
| 855. |
Using IUPAC norms write the formulas for the Potassium tri(oxalato) chromate (III) |
| Answer» Potassium tri(oxalato) chromate (III) - `K_(3)[Cr(C_(2)O_(4))_(3)]` | |
| 856. |
lanthanoid contraction, the unique in the chemistry of lanthanoids, is basicallyA. the overall increases in atomic and ionic radii from La to LuB. the overall decreases in atomic and ionic radii from La to LuC. the overall increases in atomic radii only from La to LuD. the overall decreases in ionic radii only from La to Lu |
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Answer» Correct Answer - B Lanthanoid contraction , the unique feature in the chemistry of lanthanoids is basically the overall decease in atomic and ionic radii from La to Lu. |
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| 857. |
(a) What is Lanthanoids contraction ? (b) What are Interstial compounds ? Give one example (c) Calculate the spin only magnetic moment of `M^(2+)(aq)` ion [Z=29]. |
| Answer» (c) `M^(2+)(aq)` ion with Z=29 is `Cu^(2+)` ion. Its electronic configuration is `3d^(9) [uarrdarr uarr darr uarrdarruarr darruarr]`. With one unpaired electron, the magnetic moment can be calculated as `=sqrt(n(n+2))=sqrt(3)=1.732 BM` | |
| 858. |
Alloy forming tendency of transition element is due toA. large difference in atomic sizeB. small difference in atomic sizeC. more number of oxidation stateD. defect in their crystal lattice |
| Answer» Correct Answer - B | |
| 859. |
Colour of `Cu_(2)O` is due toA. d-d transitionB. charge transferC. f-f transitionD. geometry of `Cu_(2)O` |
| Answer» Correct Answer - B | |
| 860. |
Wh ich of the following is weak reducing agent ?A. NickelB. ScandiumC. CopperD. Titanium |
| Answer» Correct Answer - C | |
| 861. |
B.M is equal toA. `(eh)/(2pimc)`B. `(e h)/(4 pi m c)`C. `(eV)/(4 pi m c)`D. `(e V)/(2 pi m c)` |
| Answer» Correct Answer - B | |
| 862. |
Lower oxidation state in transition metals are stabilized byA. COB. OC. ID. Br |
| Answer» Correct Answer - A | |
| 863. |
The ions of `d-`block elements are mostly paramagnetic-A. Because their `d-`orbitals are completeB. Because they have mostly paired electronsC. Because they have mostly unpaired electronsD. Because they form coloured ions. |
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Answer» Correct Answer - C All atomic and molecular specie4s which contains one or more unpaired electrons are paramagnetic. Most of the ions of transition elements have unpaired electrons. Hence they are all paramagnetic. Some ions of `d-`block elements with no unpaired e,ectrpms are `Sc^(+3),Ti^(+4),Zn^(+2)` and `Cu^(+1)`. These ions are not paramagnetic and they are also without colour. |
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| 864. |
The general electronic configuration of 3d series is ?A. `[Ar] 3d^(1-10) 4s^(1 or 2)`B. `[Kr] 3d^(1-10) 4s^(1 or 2)`C. `[Ar] 3d^(0-10) 4s^(1 or 2)`D. `[Kr] 3d^(0-10) 4s^(1 or 2)` |
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Answer» Correct Answer - A 3d series Sc (21) to Zn (30), `[Ar] 4s^(1 or 2) 3d^(1-10)` |
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| 865. |
Which among the following is a transition element?A. AlB. CsC. PtD. S |
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Answer» Correct Answer - C Pt belongs to 5d series |
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| 866. |
Which one of the following constitutes a set of transition elements?A. Sn,Bi, MnB. Fe,Au, NiC. Na,Mg, AID. Hg, Pb, Cu |
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Answer» Correct Answer - B Fe-3d , Au-5d, Ni-3d series `therefore` Transition elements |
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| 867. |
Manganese belongs toA. 1st transition seriesB. 2nd transition seriesC. 3rd transition seriesD. 4th transition series |
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Answer» Correct Answer - A `Mn(25) 4s^2 3d^5` belong to 3d (1st) transition series |
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| 868. |
In the transition element the incoming electron occupies `[n-1]` d sublevel in preference toA. (n-1)s orbitalB. (n - 1)p orbitalC. np orbitalD. ns orbital |
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Answer» Correct Answer - C (n-1) d orbitals have lower energy than np orbitals |
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| 869. |
`Ce^(+ 4)` ion is isoelectronic withA. `La^(+3)`B. `Gd^(+3)`C. `Lu^(+3)`D. Both (a) and (c) |
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Answer» Correct Answer - A `Ce^(4+) to [Xe] 4f^0 5d^0 6s^0 , La^(3+) to [Xe] 4f^0 6d^0 6s^0` |
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| 870. |
`Zn^(+2)` ion is isoelectronic withA. `Fe^(+2)`B. `Cu^+`C. both a and bD. `Ni^(+2)` |
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Answer» Correct Answer - B `Zn^(+2)` ion is isoelectronic with `Cu^(+1)` as both have same number of `e^-` and hence same configuration `Zn^(2+) - 1s^2 2s^2 3s^2 3p^6 4s^0 3d^10` `Cu^(+) - 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^0` `therefore ` isoelectronic |
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| 871. |
Which of the following contains the maximum number of unpaired electrons ?A. `MnSO_4`B. `CuSO_4`C. `FeSO_4`D. `ZnSO_4` |
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Answer» Correct Answer - A Mn in `MnSO_4` contains maximum number of unpaired electron. Mn in `MnSO_4` is present in +2 oxidation state as `Mn^(+2)` ions. It has five unpaired electron `(3d^5 4s^0)` |
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| 872. |
`Ag^+` ion is isoelectronic with which of the following ion?A. `Cd^(+2)`B. `Au^(+3)`C. `Pt^(+1)`D. `Pt^(+3)` |
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Answer» Correct Answer - A `Ag^+` ion is isoelectronic with `Cd^(+2)` ions are both have same number of electrons and same electronic configuration . `Agj^(+) 5s^0 4d^10` `Cd^(+2) 5s^0 4d^10` |
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| 873. |
The electronic configuration of four elements are given below. Which element does not belong to the same family as others ?A. `[Xe] 4f^14 5d^10 6s^2`B. `[Kr] 4d^10 5s^2`C. `[Ar] 3d^10 4s^2`D. `[Ne] 3s^2 3p^5` |
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Answer» Correct Answer - D `[Ne] 3s^2 3p^5` belongs to group 17 (Halogen family) whereas others belong to group 12 (last group of transition series) |
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| 874. |
Which of the following statement is correct?A. Iron belongs to same transition series as silverB. Iron belongs to third transition series in the periodic tableC. Iron belongs to first transition seriesD. Iron belongs to 5d series |
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Answer» Correct Answer - C `Fe(26)to [Ar] 4s^2 3d^6` , belongs to 3d - series which is first transition series |
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| 875. |
The outer electronic configuration of Ag is `4d^10 5s^1`, it belongs toA. 5th period, group 4B. 4th period, group 5C. 5th period, group 11D. 4th period, group 11 |
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Answer» Correct Answer - C A outer electronic configuration of Ag is `4d^10 5s^1` it belongs to 5th period (`5s^1` means outermost shell is 5), group 11 (`4d^10 5s^1` means it belongs to 2nd last group of 4d series) |
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| 876. |
No is the member of 5f series. The other members of this group areA. Th and PmB. Cm and SmC. Tm and ThD. Am and Fm |
| Answer» Correct Answer - D | |
| 877. |
The last ement in actinide series isA. NoB. BkC. LrD. Fm |
| Answer» Correct Answer - C | |
| 878. |
ActinidesA. are all synthetic elementsB. including elements of atomic number 104C. have only short lived isotopesD. have variable valency |
| Answer» Correct Answer - D | |
| 879. |
Californium is the member ofA. 4f seriesB. 5f seriesC. p-blockD. d-block |
| Answer» Correct Answer - B | |
| 880. |
Which of the following element has maximum, first ionisation potential?A. VB. TiC. CrD. Mn |
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Answer» Correct Answer - D The first inization energies of Ti, V, Cr and Mn are 656, 650, 652 and 717 kJ/mole respectively. I.E. increase in a period from left to right hence, magnese has maximum first ionisation potential. |
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| 881. |
Match list I with list II and select the correct answer using the codes given below the lists `{:("List I", "List II"), ("Metal Ions", "Magnetic moment (B,M)"),(A.Cr^(3+), 1. sqrt35),(B.Fe ^(2+), 2.sqrt30),(C. Ni^(2+), 3. sqrt24), (D. Mn^(2+), 4. sqrt15),(, 5.sqrt8):}`A. `A-1, B-3, C-5 D-4`B. `A-2, B-3, C-5, D-1`C. `A-4, B-3, C-5, D-1`D. `A-4, B-5, C-3, D-1` |
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Answer» Correct Answer - C Magnetic moment, `musqrt(n(n+2))` Where n = number of unpaired electrons For `Cr^(3+)(3s^(2)3p^(6)3d^(3)),mu=sqrt(3(3+2))=sqrt15 B.M. ` For `Fe ^(2+)(3s^(2)3p^(6)3d^(6)), mu= sqrt(4(4+2))= sqrt24 B.M.` For `Ni^(2+)(2s^(2)3p^(6)3d^(8)), mu =sqrt(2(2+2))=sqrt8 B.M.` `Mn^(2+)(3x^(2)3p ^(6) 3d^(5), mu =sqrt(5(5+2)))= sqrt35B.M` |
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| 882. |
Match list I and II and select the correct answer using the codes given below the lists `{:("List-I", "List-II"), ("At. no. of elements ","Names of the elements "),(A.96,1."Berkelium"), (B.97,2."Curium"),(C. 98, 3."Californium"),(D.99,4."Einsteinium"):}`A. `{:(A,B,C,D),(1,2,1,1):}`B. `{:(A,B,C,D),(4,3,2,3):}`C. `{:(A,B,C,D),(4,3,1,3):}`D. `{:(A,B,C,D),(2,1,3,4):}` |
| Answer» Correct Answer - C | |
| 883. |
Thr outer electronic configuration of lawrencium isA. `7s^(2), 5f^(14), 6d^(1)`B. `4s^(2), 5f^(13) , 6d^(2)`C. `7s^(2), 5f^(14), 7p^(2)`D. `7s^(2), 5f^(14), 7p^(1)` |
| Answer» Correct Answer - A | |
| 884. |
Give the geometrical shapes of the following complex entities `[Co(NH_(3))_(6)]^(3+)` |
| Answer» Geometrical shape of `[Co(NH_(3))_(6)]^(3+)` is octahedral | |
| 885. |
Give the geometrical shapes of the following complex entities `[PtCl_(4)]^(2-)` |
| Answer» Geometrical shape of `[PtCl_(4)]^(2-)` is square planar | |
| 886. |
Give the geometrical shapes of the following complex entities `[Ni(CO)_(4)]` |
| Answer» Geometrical shape of `[Ni(CO)_(4)]` is tetrahedral | |
| 887. |
Account for the zero oxidation staie of Ni and Fe in `[Ni(CO)_(4)] and [Fe(CO)_(5)]` respectively. |
| Answer» In `[Ni(CO)_(4)] and [Fe(CO)_(5)]` the oxidation state of Ni and Fe is zero. These low oxidation states found when the complex compound has ligaqds capable of `pi`-acceptor character in addition to-the `sigma`-bonding. | |
| 888. |
Write the IUPAC names of the follow coordination compounds. `[Co(NH_(3))_(5)(CO_(3))]Cl` |
| Answer» Pentaamminecarbonatocobalt (III). chloride | |
| 889. |
Write the IUPAC names of the follow coordination compounds. `K_(3)[Cr(C_(2)O_(4))_(3)]` |
| Answer» Potassium trioxalatochromate (III) | |
| 890. |
Although Cr, Mo and W belong to the same group (group 6) Cr (VI) is a strong oxidizing agent while Mo (VI) and W (VI) are not. Why ? |
| Answer» In group 6, Mo (VI) and W (VI) are more stable than Cr (VI). Thus Cr (VI) in the form of dichromate in acidic medium is a strong oxidising agent. Where as `MoO_(3) and WO_(3)` are not. | |
| 891. |
The point of dissimilarity between lanthanides and actinic isA. three outermost shells are partially filledB. they show oxidation state of `+3` (common)C. they are called inner transition elementsD. they are radioactive |
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Answer» Correct Answer - D Lanthanides and actinides both are not radioactive Except promethium , all lanthanides are non-radioactive While all actinides are radioactive. |
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| 892. |
Which of the following lanthanodie ion used as oxidizing agent ?A. only `Tb ^(3+)`B. only `Ce ^(2+)`C. only `Ce ^(4+) and Tb ^(4+)`D. only `Tb^(2+) and Dy_(3+)` |
| Answer» Correct Answer - C | |
| 893. |
f-block elements are called asA. Alkaline earth elementB. Earth metalC. Rare earth elementD. Representative element |
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Answer» Correct Answer - C Earlier these elements were obtained very rarely due to difficulties in their extraction method . |
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| 894. |
The least basic hydroxide isA. `La (OH)_(3)`B. `Ce (OH)_(3)`C. `Lu (OH)_(3)`D. `Pm (OH)_(3)` |
| Answer» Correct Answer - C | |
| 895. |
Which of the following hydroxide is more ionic and more basic in nature ?A. `Ce (OH)_(3)`B. `Lu (OH)_(3)`C. `Dy (OH)_(3)`D. `Tm(OH)_(3)` |
| Answer» Correct Answer - A | |
| 896. |
The most basic hydroxide isA. `Lu(OH)_(3)`B. `Ce (OH)_(3)`C. `La(OH)_(3)`D. `Yb (OH)_(3)` |
| Answer» Correct Answer - C | |
| 897. |
Most basic hydroxide among the following isA. `Lu(OH)_(3)`B. `Eu(OH)_(3)`C. `Yb(OH)_(3)`D. `Ce(OH)_(3)` |
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Answer» Correct Answer - D Due to lanthanide contraction,the size of `M^(3+)` ion (ie,`Lu^(3+),Eu^(3+),Yb^(3+) and Ce^(3+)`) decreases and thus, the basic strenth of their hydroxides decreases. The order of size of given `M^(3+)` ions is `Ce^(3+)gtEu^(3+)gtYb^(3+)gtLu^(3+)` `therefore` The order of basic strength of hydroxides is `Ce(OH)_(3)gtEu(OH)_(3)gtYb(OH)_(3)gtLU(OH)_(3)` |
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| 898. |
Most basic hydroxide isA. `Ce(OH)_3`B. `Pr(OH)_3`C. `Nb(OH)_3`D. `Sm(OH)_3` |
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Answer» Correct Answer - A Basicity of hydroxide decreases from La to Lu as the size increases `Ce(OH)_3 gt Pr(OH)_3 gt Nd(OH)_3 gt Sm(OH)_3` |
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| 899. |
Give reasons for following in one or two sentences: `CrO_3` is an acid anhydride? |
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Answer» `CrO_3` is an acid anhydride When it is added to water, it forms chromic acid. So it is the anhydride of chromic acid. `CrO_3+H_2OtoH_2CrO_4` |
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| 900. |
An important oxide ore if iron isA. SmaltiteB. GarnieriteC. PentalanditeD. Haematite |
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Answer» Correct Answer - D Haematite `(Fe_(2)O_(3))` Garnierite is hydrated double silicate of Mg and Si having 5-8% Ni. Pentalandite (NiS) ore associated with copper and iron pyrite. |
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