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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.
| 751. |
The oxidation state of chromium in the final product formed by the reaction between Kl and acidified potassium dichromate solution is `:`A. `+3`B. `+2`C. `+6`D. `+4` |
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Answer» Correct Answer - A `Cr_(2)O_(7)^(2+)+14H^(+)+6l^(-)rarr2Cr^(3+)+7H_(2)O+3l_(2)` Thus , final state of Cr is `+3`. |
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| 752. |
Which solution is used by police to test that a person is drunk ? How is it done ? |
| Answer» Acidified potassium dichromate solution is used which has organge colour. The person is asked to breathe into the solution taken in a test tube. If the person has consumed alcohol, the organge colour will change into green colour due to oxidation of alcohol and reduction of acidified potassium dichromate to green coloured chromium sulphate. | |
| 753. |
Which alloy is generally used in making bullets, shells and lighter flints? What is its composition ? With which other element it is generally alloyed and why ? |
| Answer» The alloy used in known as mischmetal. For details refer to art. | |
| 754. |
Mercurous chloride black on adding `NH_(4)OH` to it. Explain. |
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Answer» `Hg_(2)+Cl_(2)+NH_(4)OHtounderset("Black")ubrace(Hg-NH_(2)Cl+Hg)` this is used to detect mercurous ions. |
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| 755. |
Permanent magnets are generally made of alloys ofA. FeB. CoC. NiD. Any one of these |
| Answer» Correct Answer - d | |
| 756. |
Which element is generally used in the X-rays tube for production of X-ray and why ? |
| Answer» Molybdenum is used because it is a heavy element. When cathode rays hit this element , X-rays are produced. | |
| 757. |
Complete and balance the following reactions: (i). `Mn^(2+)+PbO_2toMnO_4^(ɵ)+H_2O` (ii). `Ag^(o+)+AsH_3toH_3AsO_3+H^(o+)` |
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Answer» (i). `2Mn^(2+)+5PbO_2+4H^(o+)to2MnO_4^(ɵ)+5Pb^(2+)+2H_2O` (ii). `6Ag^(o+)+AsH_3+3H_2Oto6Agdarr+6H^(o+)+H_3AsO_3` |
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| 758. |
Write the balanced equations for extraction of silver from glance by cyanide process. |
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Answer» In case of silver metal, sulphide ore is treated with sodium cyanide. Sodium argentocyanide complex is formed which on treating with zinc metal gives silver. `Ag_2S+4NaCNto2Na[Ag(CN)_2]+Na_2S` `2Na[Ag(CN)_2]+ZntoNa_2[Na_2[Zn(CN)_4]+2AgIdarr` |
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| 759. |
Which of the following compounds will show magnetic moment of 1.72 BM ?A. `[Ni(CN)_(4)]^(2-)`B. `[CoCl_(6)]^(4-)`C. `[Cu(NH_(3))_(4)]^(2+)`D. `TiCl_(4)` |
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Answer» Correct Answer - C Cu has +2 oxidation state in `[Cu(NH_(3))_(4)]^(2+)` Electronic configuration of `Cu^(2+)(Z=29)` is `[Ar]^(18)3d^(9)4s^(0)` Number of unpaired electrons (n) = 1 Magnetic moment = `sqrt(n(n+2))BM=sqrt(1(1+2))=1.73BM` |
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| 760. |
Which element is used because for making filaments of electric bulbs and why? |
| Answer» Tungsten is used in making filaments of electric bulbs because if has very high melting point. | |
| 761. |
Answer the following questions briefly: (i). What is the actual reducing agent of haematite in blast furnace? (ii). Zinc, not copper is used for the recovery of metallic silver from the complex `[Ag(CN)_2]^(ɵ)`, explain. (iii). Why is chalcocite roasted and not calcinated during the recovery of copper? |
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Answer» (i). For iron, first by calcination and roasting the ferrous oxide is oxidised to ferric. Then in blast furnace, Smelting is done where it is reduced to iron. `4FeO+O_2to2Fe_2O_3` `Fe_2O_3+COto2Fe_3O_4+CO_2` `Fe_2O_3+COto2FeO+CO_2` `FeO+COtoFe+CO_2` Carbon monoxide is the actual reducing agent of haematite in blast furnace. (ii). Zinc is more reducing and cheaper than copper. (iii). Chalcocite is a sulphide ore of copper therefore it must be roasted in the presence of air to get oxide. |
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| 762. |
On addition of `AgNO_(3)` to four different test tubes containing different solution, one of them gave a white precipitate . It may beA. `CHCl_(3)`B. `CaCL_(2)`C. `KNO_(3)`D. `"CCl"_(4)` |
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Answer» Correct Answer - B `AgNO_(3)+CaCl_(2)rarrunderset("White ppt.")(AgClldarr)` `CaCl_(2)` is ionic compound giving free `Cl^(-)` that gives white precipitate with `Ag^(+)` ion. `CHCl_(3),C Cl_(4)` are convalent and no `Cl^(-)` is formed. `KNO_(3)` does not react with `AgNO_(3)`. |
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| 763. |
Which of the following compounds will show magnetic moment of 1.72 BM ?A. `[Ni(CN)_(4)]^(2-)`B. `[CoCl_(6)]^(4-)`C. `[Cu(NH_(3)_(4)]^(2+)`D. `TiCl_(4)` |
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Answer» Correct Answer - C Cu has (+2) oxidation state in `[Cu(NH_(3))_(4)]^(2+)` Electronic configuration of `Cu^(2+)(Z=29),[Ar]^(18)3d^(9)4s^(0)` Magnetic moment (M) `=sqrt(n(n+2))BM=sqrt(1(1+2))=sqrt3=1.73BM` `[Cu(NH_(3)_(4)]^(2+)` shows the magnetic moment of 1.73BM. |
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| 764. |
Which element is used in making bone nails for surgery and why? |
| Answer» Tantalum is used because it is resistant to corresion. | |
| 765. |
In alkaline `H_(2)O_(2),Cr_(2)O_(7)^(2-)` changes to tetraperoxo species… having oxidation number of Cr as …A. `CrO_(4)^(2+),6`B. `Cro_(5),6`C. `CrO_(8)^(3-),5`D. `CrO_(8)^(3-),11` |
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Answer» Correct Answer - C Tera peroxo species `[Cr(O_(2))_(4)]^(3-)` oxidation number of `Cr=x-8=-3x=+5` |
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| 766. |
Explain the following facts: (a). Copper hydroxide is soluble in ammonium hydroxide but not in sodium hydeoxide. (b). Addition of an alkali to cuprous chloride solution gives a yellow precipitate which gradually changes to a red colour. |
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Answer» This is becaue `Cu^(2+)` can form many conplexes with `NH_4^(o+)` which are soluble in water. (b). `Cu_2Cl_2+2NaOHtoCu_2Odarr+2NaCl+H_2O` The precipitate changes colour from yellow to red due to formation of `Cu_2O`. |
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| 767. |
Write the balanced equations for the reaction occuring when gold is dissolved in aqua regia. |
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Answer» Gold dissolves in aqua regia due to the reaction with nascent chlorine. `[HNO_3+3HCltoNOCl+2H_2O+2Cl]xx3` `[Au+3CltoAuCl_3]xx2` `underline([AuCl_3+HCltoHAuCl_4]xx2)` `underline(2Au+3KNO_3+11HCl to2HAuCl_4+3NOCl+6H_2O)` |
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| 768. |
Write balanced equations for the reaction of alkaline perbromate with zinc giving tetrahydraoxozincate anion. |
| Answer» `BrO_4^(ɵ)+Zn+2OH^(ɵ)+H_2OtoBrO_3^(ɵ)+Zn(OH)_4^(2-)` | |
| 769. |
Mention the products formed in the following (i) Zinc oxide is treated with excess of sodium hydroxide solution (ii). Iodine is added to a solution of stannous chloride. |
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Answer» (i). Sodium zincate is formed. `ZnO+2NaOHtoNa_2ZnO_2+H_2O` (ii). Stannous chloride is a good reducing agent It reduces iodine to iodide. `SnCl_2+2HCl+I_2toSnCl_4+2HI` |
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| 770. |
Excess of dilute sodium hydroxide solution is gradually added with shaking to an aqueous solution of zinc sulphate. What would you observe? |
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Answer» `ZnSO_4+2NaOHtoNa_2SO_4+Zn(OH)_2` (white) `overset(NaOH)toNa_1ZnO_2` (soluble and colourless) |
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| 771. |
Identify A, B, C, D and E. A while substance A reacts with dilute `H_2SO_4` to produce a colourless gas B and a colourless solution C. The reaction between B and acidified `K_2Cr_2O_7` solution produces a green solution and slightly coloured precipitate D. The substance D burns in air to produce a gas E which reacts with B to yield D and a colourless liquid. Anhydrous copper sulphate is turned blue on adiition of this colourless liquid. Addition of aqueous `NH_3` or `NaOH` to C produces first a precipitate |
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Answer» `ZnS(A)+dil.H_2SO_4rarrH_2S(B)+ZnSO_4(C)` `3H_2S+Cr_2O_7^(2-)+8H^(o+)rarr2Cr^(3+)(Green)+3S(D)+7H_2O` `S(D)+O_2rarrSO_2(E)` `SO_2(E)+H_2S(B)toS(D)+H_2O` (Colourless liquid) `Zn^(2+)+2OH^(ɵ)to2n(OH_2)darroverset(OH^(ɵ))toZn(OH)_4^(2-)` (Clear solution) (C) |
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| 772. |
On addition of small amount of `KMnO_(4)` to concentrated `H_(2)SO_(4)` , a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following `:`A. `Mn_(2)O_(7)`B. `MnO_(2)`C. `MnSO_(4)`D. `Mn_(2)O_(3)` |
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Answer» Correct Answer - a `2KMnO_(4) + 2H_(2)SO_(4) ( "conc.") rarr Mn_(2)O_(7) + 2KHSO_(4) + H_(2)O` |
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| 773. |
`CuSO_(4)` reacts with KCN to formA. `Cu(CN)_(2)`B. `CuCN`C. `K_(2)[Cu(CN)_(4)]`D. `K_(3)[Cu(CN)_(4)]`. |
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Answer» Correct Answer - D (d) `2CuSO_(4)+10KCNrarr2K_(3)[Cu(CN)_(4)]+2K_(2)SO_(4)+C_(2)N_(2)` |
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| 774. |
On strong heating, `AgNO_(3)` produces the gasesA. `N_(2)O" and "NO`B. `N_(2)O" and "O_(2)`C. `NO" and "N_(2)O`D. `N_(2)O" and "NO` |
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Answer» Correct Answer - B (b) `2AgNO_(3)rarr2Ag+2NO_(2)+O_(2)` |
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| 775. |
Explain briefly how `+2` state become more and stable in the first half of the first row transition elements with increasing atomic number? |
| Answer» In all the element listed, with the removal of valenece 4s electrons (+2 oxidation state ), the 3d-orbitals get gradually occupied. Since the number of empty d-orbitals decrease or the number of unpaired electrons in 3d orbitals increase, the stability of the cations `(M^(2+))` increase from `Sc^(2+)` to `Mn^(2+)` | |
| 776. |
The electronic configuration of the fourth transition element isA. `1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),4s^(2)`B. `1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(2),4s^(2)`C. `1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4s^(2),4p^(2)`D. `1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4s^(2),4p^(1)` |
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Answer» Correct Answer - B Transition elements are those elements which have partially filled d-subshell in their elementary from . Therefore, the general electronic configuration of d-block elements is `(n-1)d^(1-10)ns^(1-2)` |
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| 777. |
For `M^(2+) //M` and `M^(3+) //M^(2+)` system, `E^(@)` values for some metals are as follows `,` `Cr^(2) //Cr= - 0.9 V l, Cr^(3+) //Cr^(2) = - 0.4 V, Mn^(2+) //Mn = - 1.2 V, Mn^(3+) //Mn^(2+) = + 1.5 V, Fe^(2+) //e= - 0.4 V, Fe^(3+) //Fe^(2+) = + 0.8 V` Use this data to comment upon (i) the stability of `Fe^(3+)` in acid solution as compared to that of `Cr^(3+)` and `Mn^(3+)` (ii) the case with which iron can be oxidized as compared to the similar process for either Cr or Mn metals. |
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Answer» (i) `Cr^(3+) //Cr^(2+)` has a negative reduction potential. Hence, `Cr^(3+)` cannot be reduced to `Cr^(2+) ` ,i.e, `Cr^(3+)` is most stable . `Mn^(3+) //Mn^(2+)` has large positive `E^(@)` value. Hence, `Mn^(3+)` can be easily reduced to `Mn^(2+)` , i.e., `Mn^(3+)` is least stable. `E^(@)` value for`Fe^(3+)//Fe^(2+)` is positive but small. Hence, `Fe^(3+) ` si more stable than `Mn^(3+)` but less stable than `Cr^(3+)` (ii) Oxidation potentials for the given pairs will be `+ 0.9 V , + 1.2 V` and `+ 0.4 ` volt . Thus, the order of their getting oxidized will be in the order `Mn gt Cr gt Fe` |
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| 778. |
Which of the following statements(s) is `//` are correctly applicable to all the transition elements with atomic number 21-29 ? 1. The lowest oxidation states is `+1`.2. The 4s orbital is completely filled in the group state. 3. The 3d orbital is incompletely filled in the group state. 4. The ion in the `+2` oxidation state is paramagnetic.A. 1,2 and 3B. 3 and 4 onlyC. 4 onlyD. 3only |
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Answer» Lowest oxidation state is `+1` only in case of Cu. Hence, (1) is wrong. 4s orbital is not completely filled in Cu and Cr. Hence, (2) is wrong. 3d orbital is not incompletely filled in cae of Cu. Hence, (3) is wrong. Ions is `+2` oxidation state of all elements ( 21-29) hav unpaired electrons and, therefore, paramagnetic . Hence, ( 4) is correct. |
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| 779. |
Among the following compound s that is both paramagnetic and coloured isA. `K_(2)Cr_(2)O_(7)`B. `(NH_(4))_(2)[TiCl_(6)]`C. `VOSO_(4)`D. `K_(3)[Cu(CN)_(4)]` |
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Answer» For a compoun to be coloured, the central atom should have incompletely filled d-orbital and to be a pramagnetic , there should be at least one unpaired electron Here, we have `{:("Compound","Central atom","Electronic config."),(K_(2)Cr_(2)O_(7),._(24)Cr(VI),3s^(2)3p^(6)),((NH_(4))_(2)[TiCl_(6)],._(22)Ti(IV),3s^(2)3p^(6)),(VOSO_(4),._(23)V(IV),3s^(2)3p^(6)3d^(1)),(K_(3)[Cu(CN)_(4)],._(29)Cu(I),3s^(2)3p^(6)3d^(10)):}` |
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| 780. |
In what way is the electronic configuration of trantision elements different from that of the non-transition elements ? |
| Answer» Transition elements contain incompletely filled d-subshell, i.e., their electronic configuration is `( n-1) d^(1-10) ns^(0-2)` whereas non- transition elements have no d-subshell or their d-subshell is completely filled and have `ns^(1-2) ` or `ns^(2) np ^(1-6)` in their outermost shell. | |
| 781. |
Why are `Mn^(2+)` compounds more stable than `Fe^(2+)` comounds towards oxidation to their `+3` state ? |
| Answer» Electronic configuration of `Mn^(2+)` is `3d^(5)` which is half filled and hence stable. Therefore, 3rd ionizatio enthalpy is very high, i.e., 3rd electron cannot be lost easily. In case of `Fe^(2+)` , electronic configuration is `3d^(6)` . Hence, it can lose one electron easily to give the stable configuration `3d^(5)` | |
| 782. |
In What way is the electronic configuration of the transition elements different from the of the non-transition elements? |
| Answer» Electronic configuration of transition element : `(n-1)d^(1-10)ns^(1-2)`. Electronic configuration of non- transition element `:ns^(1-2) " or " ns^(2)np^(1-6)`. From comparison, it is quite evident that the transition elements have incomplete d- orbitals (s-orbitals in some cases ) while teh non-transition elements have no d-orbitals present in the valence shells of the atoms. This is responsible for the difference in the characteristics of the element belonging to these calssess of elements. | |
| 783. |
Why does not copper liberate hydrogen from acids. ? |
| Answer» Copper placed below hydrogen in the electrochemical series. It is not in a position to release electrons to `H^(+)` ions of the acid. Therefore, hydrogen gas is not evolved. | |
| 784. |
Although `+3` is the characteristic oxidation state for langthanoids but certain also shows `+4` oxidation state because `"…...................."` |
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Answer» Correct Answer - b,c `._(58) Ce=[Xe]^(54) 4f^(2) 5d^(0)6s^(2) `. Hence, `Ce^(4+) = [Xe]^(54) 4f^(0) `. Hence, (b) and (c ) are true. |
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| 785. |
Assertion: `E^(ɵ)` for `(Mn^(3+))/(Mn^(2+))` is more positive than for `(Cr^(3+))/(Cr^(2+))`. Reason: The third ionisation energy of Mn is larger than that of Cr.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 | |
| 786. |
Which of the following pair of elements has `(n-1) d^10 ns^2` electronic configuration ?A. Fe, Co, NiB. Cu, Ag, AuC. Zn, Cd, HgD. Sc, Y, La |
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Answer» Correct Answer - C `Zn-3d^10 4s^2 , Cd-4d^(10) 5s^2 , Hg- 5d^10 6s^2` `therefore (n-1) d^10 ns^2` |
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| 787. |
The colours of the transition metal are due to d-d excitation. The energy required for d-d electron axcitation is available in the visible range. Transition metal ions have the tendency to absorb certain rediations from the visible region and exhibit the complementary colour. The transition metal ions which have completely filled d-orbitals are colourless as the excitation of electron or electrons is not possible within d-orbitals. The transition metal ions which have completely empthy d-orbitals are also colourless. In `KMnO_4` and `K_2Cr_2O_7`, there are no unpaired electrons at the central atom but they are dep in colour. The colour of these compounds is due to charge transfer spectrum. for example in `MnO_4` electron is momentrily transferred from O to the metal and thuys oxygen changes from `O^(2-)` and `O^(ɵ)` maganese from `Mn^(7+)` to `Mn^(6+)`. Q. Which of the following is paramagnetic as well as coloured ion?A. `Cu^(o+)`B. `Cu^(2+)`C. `Sc^(3+)`D. `Zn^(2+)` |
| Answer» Correct Answer - b | |
| 788. |
Select the correct order of sizes for the following d-block elements.A. `Zr=Hf,Nb=Ta,Fe~=Co~=Ni`B. `Zr^(4+) lt Zr, Nb^(3+) lt Ta^(3+), Fe^(3+) lt Fe^(2+) lt Fe`C. `Zr^(4+)=Hf^(4+),Nb^(3+)=Ta^(3+),Fe lt Co lt Ni`D. `Zr^(4+) lt Hf^(4+),Nb^(3+)=Ta^(3+),Ni lt Cu lt Co` |
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Answer» Correct Answer - C `Zr=Hf,Nb=Ta ` due to lanthanide contraction. and `Fe=[Ar]3d^(6)4s^(2)` `Co=[Ar]3d^(7)4s^(2)` `Ni=[Ar]3d^(8)4s^(2)` |
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| 789. |
`(n-1)d^(10)ns^(2)`is the electronic configuration ofA. ZnB. CdC. HgD. All of these |
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Answer» Correct Answer - A The general electronic configuration of group 12 elements (Zn, Cd and Hg) are represented by the general formula `(n-1)d^(10)ns^(2)` |
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| 790. |
When an electron from a lower energy d-orbital is excited to a higher energy d-orbital (i)The energy of excitation corresponds to the frequencey of light absorbed. (ii) This frequency generally lines in the visible region. (iii) The colour observed corresponds to the compementary colour of the light absorbed. (iv) The frequencey of the light absorbed is determined by the nature of the ligand. Which one of the above mentioned statements are correct?A. I,II and IvB. I,II and IIIC. I,II,III,and IvD. II and iv |
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Answer» Correct Answer - D When an electron from a lower energy d-orbital is excitated to a higher energy d-orbital, the energy of excitation corresponds to the frequancy of light absorbed. This frequency generally lines in the visible region. The coloue observed corresponds to the complementary colour of the light absorbed. The frequency of light absorbed is determined by the nature of the ligand. |
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| 791. |
In the standardization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by iodometry, th equivalent weight of `K_(2)Cr_(2)O` isA. `"Molecular weight"/2`B. `"Molecular weight"/6`C. `"Molecular weight"/3`D. same as the molecular weight |
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Answer» Correct Answer - B In iodometry, Kl reacts with `K_2Cr_2O_7` in acidic medium to give `I_2` which is titrated against `Na_2S_2O_3`. `underset"1 M"(" "K_2Cr_2O_7) + 7H_2SO_4 + 6KI to 4K_2SO_4 + Cr_2(SO_4)_3+ underset"6 x 127 parts"(7H_2O + 3I_2)` `underset"2 M"(" "2Na_2S_2O_3) +I_2 to underset"2 x 147 parts"(Na_2S_4O_6+2NaI)` For 1 M `Na_2S_2O_3` , 127 parts of `I_2` is required which is obtained from 1/6 M of `K_2Cr_2O_7`. Hence, eq. wt. of `K_2Cr_2 O_7` is Mol. wt./2 |
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| 792. |
How to standardise `Na_(2)S_(2)O_(3)` solution in iodometry? |
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Answer» `K_(2)Cr_(2)O_(7)` is primary standard `rArr" "` strength is know by weighing the salt in chemical balance and dissolving in measured amount of water. Then in acidic solution add. `KI` `Cr_(2)O_(7)""^(2-)+14H^(+)+6I^(-)to2Cr^(3+)+3I_(2)+7H_(2)O` This `I_(2)` is liberated can be estimated with `S_(2)O_(3)""^(2-)`. |
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| 793. |
What is the oxidation state of `Fe` and `NO` in nitroprusside ion. How can it be determined? |
| Answer» The nitroprusside ion is `[Fe(CN)_(5)NO]^(2-)`. This can be established by measuring the magnetic moment of the solid compound which should correspond to `(Fe^(2+)=3d^(6))`, that is, four unpaired electrons. In this ion, iron is in `Fe^(2+)` and `NO` is `NO^(+)`. | |
| 794. |
In laboratory `K_(2)Cr_(2)O_(7)` is used mainly not `Na_(2)Cr_(2)O_(7).` Why? |
| Answer» `Na_(2)Cr_(2)O_(7)` is deliquescent enough and changes its concetration and can not be taken as primary standard solution whereas `K_(2)Cr_(2)O_(7)` has no water of crystallisation and not deliquescent. | |
| 795. |
In the equation: `M+8CN^(-)+2H_(2)O+O_(2)to4[M(CN)_(2)]^(-)+4OH^(-)` metal M is:A. AgB. AuC. Cu^(+)`D. Hg |
| Answer» Correct Answer - A::B | |
| 796. |
Match the column-A. `{:(Column-I,Column-II),("Highest density",(P) Os):}`B. `{:(Column-I,Column-II),("Colourless salts",(Q) Os):}`C. `{:(Column-I,Column-II),("Maximum magnetic moment",(R) Cr):}`D. `{:(Column-I,Column-II),("Variable oxidation state",(S) Mn):}` |
| Answer» Correct Answer - A::B::C::D | |
| 797. |
Even after removal of magnetic field substance does not cease to exhibit magnetic character, the phenomenon isA. DiamagnetismB. ParamagnetismC. FerromagnetismD. Ferro-electroism |
| Answer» Correct Answer - C | |
| 798. |
The common oxidation states of gold are-A. 1,2 and 3B. 1,3C. 2 and 3D. 3,4 |
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Answer» The common oxidation states of gold are `+1` and `+3` The `+3` state is more stable. This is explained on the basic that `+1` ion gets disproportinate. `3 Au^(+1) to Au^(+3)+2Au` |
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| 799. |
Silver and gold atoms have nearly the same atomic radii due to `"…..........."` |
| Answer» Correct Answer - lanthanoid contraction | |
| 800. |
Which one of the following metal will not form amalgam ?A. GoldB. ZincC. MercuryD. Tin |
| Answer» Correct Answer - C | |