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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.
| 901. |
In which of the following complexes the metal ion is in zero oxidation state?A. `[Cu(NH_(3)_(4)]Cl_(2)]`B. `Zn_(2)[Fe(CN_(6)]`C. `Mn_(2)(CO)_(10)`D. `[Ag(NH_(3))_(2)]Cl` |
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Answer» Correct Answer - C Usually, metal carbonyls possesses the metals with lower oxidation state. Here, manganese have zero oxidation state. |
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| 902. |
`KMnO_(4)` in basic medium is used asA. strong oxidising agentB. strong reducing agentC. strong hydrogenating agentD. poor reducing agent |
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Answer» Correct Answer - A Due to the presence of nascent oxygen [O], `KMnO_(4)` (in basic medium) behaves like strong oxidising agent. `2KMnO_(4)+2KOHrarr2K_(2)MnO_(4)+H_(2)O+[O]` |
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| 903. |
The one which is not the characteristic property of transition element, isA. diamagnetic behaviourB. formation of complexesC. catalytic activityD. variable oxidation states |
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Answer» Correct Answer - A Since, the d-orbitals are partially filled in transition metal atoms/ions, they exhibit paramagnetism and not diamagnetism. |
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| 904. |
Statement-1: Zinc does nt show characteristic properties of transition metals. Statement-2: In zinc outermost shell is completely filled.A. If both Statement-I `&` Statement-II are True `&` the Statement-II is a correct explanation of the Statement-IB. If both Statement-I `&` Statement-II are True but Statement-II is not a correct explanation of the Statement-IC. If statement-I is True but the Statement-II is FalseD. If Statement-I is False but the but the Statement-II is True |
| Answer» Correct Answer - C | |
| 905. |
Find out the smallest ion from those given below.A. `Gd^(+3)`B. `Sm^(+3)`C. `Yb^(+3)`D. `Ce^(+3)` |
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Answer» Correct Answer - C From La to Lu , ionic radius decreases.Yb is the second last element of the 4f series. Therefore `Yb^(+3)` ion is the smallest ion. |
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| 906. |
The decrease in size of inner transition element isA. moreB. not regularC. lowD. unpredictable |
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Answer» Correct Answer - C Lanthanoid contraction - decreases in radius is very small |
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| 907. |
Which of the follow2ing show the oxidation state of `+4` ?A. PmB. EuC. SmD. Tb |
| Answer» Correct Answer - D | |
| 908. |
Which of the following transition metal cation has maxi- mum unpaired electrons?A. `Mn^(+2)`B. `Fe^(+2)`C. `Ni^(+3)`D. `Cu^(+1)` |
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Answer» Correct Answer - A `Mn^(+2) (3d^5 4s^0)` has five unpaired electron |
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| 909. |
The number of unpaired electrons in `Lu ^(3+)` are |
| Answer» Correct Answer - A | |
| 910. |
Assertion: `MnO` is basic whereas `Mn_2O_7` is acidic. Reason: Higher the oxidation state of a transition metal in its oxide, greater is the acidic character.A. If both (A) and (R) are correct and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect but (R) is correct. |
| Answer» Correct Answer - a | |
| 911. |
The number of unpaired electrons in cobalt atoms is (atomic number of Co= 27)A. 2B. 3C. 4D. 1 |
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Answer» Correct Answer - B Electronci configuration of `""_(27)Co.` `1s^(2) 2s^(2) sp^(6) 3p^(6) 3d^(7) 4s^(2),` hence unpaired `e^(-)=3` |
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| 912. |
An acidified solution of potassium permanganate oxidizesA. sulphatesB. sulphitesC. nitratesD. ferric salts |
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Answer» Correct Answer - B `{:(2MnO_4^(-),+6H^(+), +5" "SO_3^(2-) , to 2Mn^(2+) + 5SO_4^(2-) + 3H_2O),(,+1,+7 " "-8,):}` |
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| 913. |
Among the oxides, `Mn_(2)O_(7)(I), V_(2)O_(3) (II), V_(2)O_(5) (III),CrO(IV)` and `Cr_(2)O_(3)(V)`, the basic oxides areA. I and IIB. II and IIIC. III and IVD. II and IV |
| Answer» Generally, the oxides in the lower oxidation states of metals are basic. | |
| 914. |
Basic properties of `TiO_(2), ZrO_(2)` and `HfO_(2)` are in the order `:`A. `TiO_(2) lt ZrO_(2) lt HfO_(2)`B. `ZrO_(2)ltHfO_(2) lt TiO_(2)`C. `HfO_(2) ltTiO_(2)lt ZrO_(2)`D. `TiO_(2) lt HfO_(2) lt ZrO_(2)` |
| Answer» Ti, Zr and Hf lie in Group 4. The metal oxides are basic and their basic character increases down the group | |
| 915. |
Which of the following is amphoteric oxide? `Mn_(2)O_(7),CrO_(3),Cr_(2)O_(3),CrO,V_(2)O_(5),V_(2)O_(4)`A. `V_(2)O_(5),Cr_(2)O_(3)`B. `Mn_(2)O_(7), CrO_(3)`C. `CrO, V_(2)O_(5)`D. `V_(2)O_(5), V_(2)O_(4)` |
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Answer» Correct Answer - A `V_(2)O_(5)` and `Cr_(2)O_(3)` are amphoteric oxide because both react with alkalies as well as acids. Note In lower oxides, the basic character is predominant while in higher oxides, the acidic character is predominant. |
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| 916. |
`KMnO_(4)` acts as an oxidizing agent in acidic medium. The number of moles of `KMnO_(4)` that will be needed to react with one mole of sulphide ions in acidic solution isA. `(2)/(5)`B. `(3)/(5)`C. `(4)/(5)`D. `(1)/(5)` |
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Answer» Correct Answer - a `2KMnO_(4)+ 3H_(2)SO_(4) rarr K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5(O)` `H_(2)S+ (O) rarr H_(2)OI +S ] xx 5` `bar ( 2KMnO_(4) + 3H_(2)SO_(4) + 5H_(2)Srarr K_(2)SO_(4) + 8 H_(2)O 5S)` 1 mole of `S^(2-)` ions will required`KMnO_(4) = ( 2)/( 5) `moles |
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| 917. |
Using IUPAC norms write the systematic names of the `[Co(NH_(3))_(6)]Cl_(3)` |
| Answer» Hexa ammine cobalt (III) chloride | |
| 918. |
Using IUPAC norms write the systematic names of the `[Ti(H_(2)O)_(6)]^(3+)` |
| Answer» Hexa aquo titanium (III) ion | |
| 919. |
Which of the following statements is not correct?A. Copper liberates hydrogen from acidsB. In its higher oxidation states, manganese forms stable compounds with oxygen and fluorineC. `Mn^(3+)` and `Co^(3+)` are oxidising agents in aqueous solutionD. `Ti^(2+)` and `Cr^(2+)` are reducing agents in aqueous solution |
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Answer» Correct Answer - A Copper lies below hydrogen in the electrochemical series and hence does not liberate `H_(2)` from acids. Therefore, option (a) is not correct. Other three options (b,c,d) are correct. |
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| 920. |
Cerium `(Z= 58)` is an important member of the lanthanoids . Which of the following statements about cerium is incorrect ?A. The common oxidation states of cerium are `+3` and `+4`B. The `+3` oxidation state of cerium is more stable than `+4` oxidation state.C. The `+4` oxidation state of cerium is not known in solutionD. Cerium (IV) acts as an oxidaizing agent. |
| Answer» `Ce^(4+)` exist in solution though it tends to change to more stable `Ce^(3+)` | |
| 921. |
Name of gas that can readily decolorise acidified `KMnO_(4)` solutionA. `CO_(2)`B. `SO_(2)`C. `NO_(2)`D. `P_(2)O_(5)` |
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Answer» Correct Answer - B `SO_(2)` readily decolourises acidified `KMnO_(4)` solution. |
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| 922. |
Identify the incorrect statement among the following.A. 4f and 5f orbitals are equally shieldedB. d-block elements show irregular and erratic chemical properties among themselvesC. La and Lu have partially filled d-orbitals and no other partially filled orbitals.D. The chemistry of various lanthanoids is quite similar. |
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Answer» Correct Answer - A 4f orbitals are more shielded than 5f orbitals due to their greater penetration in the nucleus. |
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| 923. |
The oxidation state of chrominium in the final product formed in the reaction between `KI` and acidified potassium dichromate soluttion isA. `+4`B. `+6`C. `+2`D. `+3` |
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Answer» Correct Answer - D `overset(+6)(Cr_(2))O_(7)^(2-)+14H^(+)+6I^(-)3//4®" " overset(+3)(2Cr)+7H_(2)O+3I_(2)` |
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| 924. |
The equilibrium `2 CuIhArr Cu+Cu^(II)` In aqueous medium at `25^@C` shifts towards the left in the presence ofA. `NO^(ɵ)`B. `Cl^(ɵ)`C. `SCN^(ɵ)`D. `CN^(ɵ)` |
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Answer» Correct Answer - B::C::D `(b,c,d) 2Cu^(o+)hArr Cu+Cu^(2+)` `Cl^(ɵ)` and `CN^(ɵ)` both make ppt. with `Cu^(o+)` and hence drive the reaction to the left. |
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| 925. |
Steel containsA. Fe+C + MnB. Fe+ C+NiC. Fe+ MnD. Fe+ Mn +Cr |
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Answer» Correct Answer - A Steel contains Fe, C , Mn |
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| 926. |
Percentage of silver in German silver isA. `1.5%`B. `5%`C. `10%`D. zero% |
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Answer» Correct Answer - D German silver does not contain silver |
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| 927. |
Alloy is an example ofA. GelB. AerosolC. Solid solutionD. Emulsion |
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Answer» Correct Answer - C Alloy is mixture of two or more metals forming solid solution. (One metal dispersed in another metal) |
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| 928. |
`Ce^(4+)` is stable .This is because ofA. half-filled d-orbitalB. all paired electrons in d-orbitalC. empty orbitalD. fully filled d-orbital |
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Answer» Correct Answer - C The electronic configuration of Ce is `Ce_(58)=[Xe]4f^(1)5d^(1)6s^(2)` (predicted) or `=[Xe]4f^(2)5d^(0)6s^(2)` (observed) `Ce^(4+)=[Xe]4f^(0)5d^(0)6s^(0)` Since, in `+4` oxidation state, all 4f,5d and 6s orbitals are empty and Ce gains the stable configuration of nearest inert gas . Hence, `Ce^(4+)` is most stable . |
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| 929. |
Which gives +7 oxidation state ?A. Mn(25)B. Cr(24)C. Cu(29)D. Fe(26) |
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Answer» Correct Answer - A For d-block elements, oxidation state=ns electron+(n-1)d electrons (unpaired) (a) `Mn(25)=[Ar]3d^(5)4s^(2),O.S.=5+2=7` (b) `Cr(24)=[Ar]3d^(5)4s^(1),O.S.=5+1=6` `(c ) Cu(29)=[Ar]3d^(10)4s^(1),O.S.=0+1=1` (d)` Fe(25)=[Ar]3d^(6)4s^(2),O.S.=+2 and +3` Thus , only Mn exhibits `+7` oxidation state among the given elements . |
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| 930. |
Element with maximum atomic number isA. LanthanumB. ActiniumC. ScandiumD. Halfnium |
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Answer» Correct Answer - B La-57, Ac-89, Sc-21 , Hf-72 `therefore` Ac-89 |
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| 931. |
In lanthanides, last electrons enters into (n - 2) r subshell. What is the value of n ?A. 4B. 6C. 7D. 8 |
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Answer» Correct Answer - B Lanthanide `to [Xe] 4f^(1-14) 5d^(0//1)6s^2 therefore` n=6, n-2= 4 |
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| 932. |
Assertion : An aqueous solution of ferric chloride is acidic due to hydrolysis. Reason : Ferric chloride is a covalent compound and exists as a dimer.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 - B (b) Correct explanation. Upon hydrolysis, ferric chloride forms ferric hydroxide (weak base) and hydrochloric acid (strong acid). |
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| 933. |
The correct option(s) to distinguish nitrate salts of `Mn^(2+)" and Cu^(2+)` taken separately is (are)A. `Mn^(2+)` show the characteristic green colour in the flame testB. Only `Cu^(2+)` show the formation of precipitate by passing `H_(2)S` in acidic mediumC. Only `Mn^(2+)` show the formation of precipitate by passing `H_(2)S` in faintly basic mediumD. `Cu^(2+)//Cu` has higher reduction potential than `Mn^(2+)//Mn` (measured under similar conditions). |
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Answer» Correct Answer - B::D (b,d) Both these are correct. (b) In weakly acidic medium, only `Cu^(2+)` ions form black precipitate of CuS. (d) `E_(Cu^(2+)//Cu)^(@)(+0.34V)` is higher than that of `E_(Mn^(2+)//Mn)^(@)(-1.18V)` The other two options are not correct. (a) `Cu^(2+)" and not "Mn^(2+)` shows characteristic green colour in the flame test. (c) Both `Cu^(2+)" and "Mn^(2+)` show the formation of precipitate (CuS-Black) & MnS-Brown) by passing `H_(2)S` in faintly basic medium. |
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| 934. |
Which one of the following does not correctly represent the correct order of the property indicated against itA. `Ti lt V lt Cr lt Mn`, Increasing number of oxidation statesB. `Ti^(3+) lt V^(3+) lt Cr^(3+) lt Mn^(3+)`: increasing magnetic momentC. `Ti lt V lt Cr lt Mn`: increasing melting pointsD. `Ti lt V lt Mn lt Cr`: increasing 2nd ionization enthalpy |
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Answer» Correct Answer - C The correct increasing order of m.p is `Mn lt Ti lt Cr lt V` `(1245^(@)C) (1668^(@)C)(1875^(@)C)(1900^(@)C)` |
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| 935. |
Transition metals with `4S^(1)` configuration are :A. CrB. MnC. NiD. Cu |
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Answer» Correct Answer - A::D (a,d) Both these options are correct. |
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| 936. |
In the following sequence in aqueous solution, the species X, Y and Z, respectively, are `S_(2)O_(3)^(2-) overset(Ag^(+))rarrunderset(("Clear solution"))X overset (Ag^(+))rarr underset(("White ppt."))Y overset("With time")rarr underset(("Black ppt."))Z`A. `[Ag(S_(2)O_(3))_(2)]^(3-),Ag_(2)S_(2)O_(3),Ag_(2)S`B. `[Ag(S_(2)O_(3))_(3)]^(5-),Ag_(2)SO_(3),Ag_(2)S`C. `[Ag(SO_(3))_(2)]^(3-),Ag_(2)S_(2)O_(3),Ag`D. `[Ag(SO_(3))_(3)]^(3-),Ag_(2)SO_(4),Ag`. |
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Answer» Correct Answer - A (a) `Ag^(+)+S_(2)O_(3)^(2-)rarrunderset(darrAg^(+))underset("Clear solution (X)")([Ag(S_(2)O_(3))_(2)]^(3-))` `underset("Black ppt. (Z)")underset()(Ag_(2)S)overset("CaO")larrunderset("White ppt. (Y)")underset()(Ag_(2)S_(2)O_(3))` |
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| 937. |
Among these, identify the species with an atom in `+6` oxidation state: .A. `KMnO_(4)`B. `K_(2)MnO_(4)`C. `CrO_(2)Cl_(2)`D. `K_(3)Cr(CN)_(6)` |
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Answer» Correct Answer - B::C (b,c) Oxidation state of Mn in `K_(2)MnO_(4)` and of cr in `CrO_(2)Cl_(2)` are +6 |
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| 938. |
Which of the following combination will produce `H_(2)` gas ?A. Fe metal and Conc. `HNO_(3)`B. Cal metal and Conc. `HNO_(3)`C. Au metal and NaCN(aq) in the presence of airD. Zn metal and NaOH(aq) |
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Answer» Correct Answer - D (d) It is the correct answer. `Zn+2NaOHrarrNa_(2)ZnO_(2)+H_(2)` Fe becomes passive with conc. `HNO_(3)` Cu liberates `NO_(2)` gas with conc. `HNO_(3)` Gold forms a soluble complex with NaCl in the presence of air. |
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| 939. |
The catalyst used in the manufacture of `H_(2)SO_(4)` by contact process isA. NO(g)B. `V_(2)O_(5)`C. MoD. Platinised asbestos. |
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Answer» Correct Answer - B::D (b,d) Both are correct. |
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| 940. |
`uarr Y(g)overset(KI)(larr)CuSO_(4)overset(dilH_(2)SO_(4))(rarr)X(`Bluecolour`),X` and `Y` areA. `X=I_(2),Y=[Cu(H_(2)O)_(4)]^(2+`B. `X=[Cu(H_(2)O)_(4)]^(2+),Y=I_(2)`C. `X=[Cu(H_(2)O)_(4)]^(+),Y=I_(2)`D. `X=[Cu(H_(2)O)_(5)]^(2+),Y=I_(2)` |
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Answer» Correct Answer - B `{:(CuSO_(4)+KItoCu_(2)I_(2)+K_(2)SO_(4)+I_(2)),(" "darr),([Cu(H_(2)O)_(4)]^(2+)),(("Blue coloe")):}` |
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| 941. |
What is a double salt? Give example. |
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Answer» The salts which contains two cations and one anion are called double salts. `rarr` These dissociates into simple ions completely when dissolved in water. Eg : `KCl.MgCl_(2).6H_(2)O` -Carnallite. |
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| 942. |
Magnetic moment 2.84 BM is given byA. `Cr^(2+)`B. `Co^(2+)`C. `Ni^(2+)`D. `Ti^(3+)` |
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Answer» Correct Answer - C `Cr^(2+)-3d^(4):4 "unpaired electron"` `Co^(2+)-3d^(7),3:4 "unpaired electron"` `Ni^(2+)-3d^(8):2 "unpaired electron"` `Ti^(3+)-3d^(1):1 "unpaired electron"` Magnetic moment `(mu)=sqrt(n(n+2))=sqrt(2(2+2))=2.86BM` It shows that `Ni^(2+)` with 2 unpaired electrons matches the value of magnetic moment. |
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| 943. |
Write the formulae for the follow Co-ordination compounds Amminebromidochloridonitrito-N-platinate (II) |
| Answer» `[Pt(NH_(3))BrCl(NO_(2))]^(-)` | |
| 944. |
What is the effect of increasing pH on a solution of potassium dichromate ? |
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Answer» On increasing pH of `K_(2)Cr_(2)O_(7)` (orange) it changes into `K_(2)CrO_(4)` (yellow) `underset("Orange")(Cr_(2)O_(7)^(-2))+2OH^(-) rarr underset("Yellow")(2CrO_(4)^(-2))+H_(2)O` |
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| 945. |
Correct statement `(s)"is"//"are"`A. an acidified solution of `K_(2)Cr_(2)O_(7)` liberates iodine from `KI`B. `K_(2)Cr_(2)O_(7)` is used as a standard solution for estimation of `Fe^(2+)` ionsC. in acidic medium, `M=N//6` for `K_(2)Cr_(2)O_(7)`D. `(NH_(4))_(2)Cr_(2)O_(7)` on heating decomposes to yield `Cr_(2)O_(3)` through an exothermic reaction |
| Answer» Correct Answer - A::B::C | |
| 946. |
Write the formulae for the follow Co-ordination compounds Potassium tetracyanonickelate (II) |
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Answer» `("Potassium")/("Counterion")("Tetracyano")/("Ligand")("Nickelate (II)")/("Metal")` `overset(II)(K_(x)[Ni(CN)_(4)]` To find the value of x, find the charge on the complex. `[Ni(CN)_(4)]x^(-)` (as K + positive) `+2+(-1)xx4=-x` `-x=-2" or "x=+2` So, the formula of the complex is `K_(2)[Ni(CN)_(4)]`. Similarly |
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| 947. |
Lantanide series arises due toA. stability of 6s-orbitalsB. stability of 5d-orbitalsC. stability of 4f-orbitalsD. equal stability of 5d and 4f-orbitals |
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Answer» Correct Answer - C We know that, in `6^(th)` row after 6s-orbitals. At this stage 5d-orbitals stop its spliting and 4f-orbitals, split into 7 orbitals, and become lower energy than, 5d-orbitals. Hence 4f-orbital are progressively filled and 14 new elements are added called as lanthanides. |
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| 948. |
Actinide series is arises due toA. stability of 5f-orbitalsB. stability of 7s-orbitalsC. stability of 6d-orbitalsD. equal stability of 6d and 5f-orbitals |
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Answer» Correct Answer - A We know that, in `7 ^(th)` row after 7s-orbitals the next electrons enters into 6d-orbitals. At this stage 6d-orbitals stop its spliting and 5f-orbitals, split into 7 orbitals, and become lower energy than, 6d-orbitals. Hence 5f-orbitals are progressively filled and 14 new elements are added called as actinides. |
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| 949. |
Zr and Hf have identical sizes. Elaborate. |
| Answer» Both Zr and Hf are present in group 4 (d-block). The atomic radius of Zr (160 pm) is quite close to that of Hf (159 pm). Actually, Hf experiences lanthanoid contraction which causes a decrease in its atomic size and atomic radius. Therefore, the two elements have identical sizes. | |
| 950. |
`Zn^(2+)` salt are white while `Cu^(2+)` salts are blue. Assign reason. |
| Answer» `Cu^(2+)` ion `(3d^(9))` has one filled d-orbitals. Therefore , it has an urge to take part in d-d transition. The compounds containing `Cu^(2+)` ion (e.g., `CuSO_(4).5H_(2)O`) are coloured. `Zn^(2+)` ion `(3d^(10))` has completely filled d-orbitals and there is ano scope for any electron transition. Therefore, `Zn^(2+)` salt are white and not coloured. | |