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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.
| 51. |
Name one example of reaction in which dihydrogen acts (i) as an oxidizing agent (ii) as a reducing agent. |
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Answer» (i) In the reaction of dihydrogen with metals to form metal hydride it acts as an oxidizing agent 2Na(s) +H2 (g) --------------> 2Na+H Here Na has oxidized to Na+ and Dihydrogen is reduced to hydride (H- ) ion (ii) In the reaction of heated cupric oxide with dihydrogen to form H2O and copper metal dihydrogen acts as a reducing agent CuO(s) +H2 (g) ------------> Cu(s) +H2O |
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| 52. |
The acidity of diprotic acids in aqueous solutions increases in the orderA. `H_(2)S lt H_(2)Se lt H_(2)Te `B. `H_(2)Se lt H_(2)S lt H_(2)Te`C. `H_(2)Te lt H_(2)S lt H_(2)Se`D. `H_(2)Se lt H_(2)Te lt H_(2)S` |
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Answer» Correct Answer - A As the size of the element (X) increases, the length of H-X bond increases and hence its bond strength decreases. As a result, it becomes easier to break the H-X bond to liberate `H^(+)` ions and hence the acidity increases in the order : `H_(2)S lt H_(2)Se lt H_(2)Te, ` i.e., option (a) is correct. |
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| 53. |
Melting point, enthalpy of vapourisation and viscosity data of H2O and D2O is given below :On the basis of this data explain in which of these liquids intermolecular forces are stronger? |
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Answer» The melting point, enthalpy of vapourisation and viscosity values of all these items depend upon the intermolecular forces of attraction. Since their values are higher for D2O as compared to those of H2O, therefore, intermolecular forces of attraction are stronger in D2O than in H2O. |
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| 54. |
Name one example of a reaction in which dihydrogen acts (i) as an oxidising agent (ii) as a reducing agent. |
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Answer» Oxidising agent : `2Na + H_(2) to 2Na^(+)H^(-)`. Here, Na is oxidised to `Na^(+)` Reducing agent `CuO + H_(2) overset("Heat")to Cu + H_(2)O`. Here, CuO is reduced to Cu. |
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| 55. |
Dihydrogen reacts with dioxygen (O2) to form water. Write the name and formula of the product when the isotope of hydrogen which has one proton and one neutron in its nucleus is treated with oxygen? Will the reactivity of both the isotope be the same towards oxygen? Justify your answer. |
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Answer» The isotope of hydrogen containing one proton and one neutron is deuterium (D). 2D2(g) + O2(g) → 2D2O Deuterium oxide Since D-D bond is stronger than H – H bond, therefore, D2 is less reactive towards oxygen than H2. |
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| 56. |
In which of the following reactions does dihydrogen act as oxidising agent ?A. `Ca+H_(2)to`B. `H_(2)+O_(2)to`C. `H_(2)+F_(2)to`D. `CuO+H_(2)to` |
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Answer» Correct Answer - A `Ca+H_(2)toCaH_(2)`. H changes oxidation number from 0 to -1. Hence it is oxidising `Ca^(0)` to `Ca^(2+)`. |
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| 57. |
Dihydrogen reacts with dioxygen `(O_(2))` to from water .Write the name and formule of the product when the isotope of hydrogen which has one proton and one neutron in its nucles is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen ? Justify your answer. |
| Answer» The isotope of hydrogen is deuterium (D) and the formula of its oxide is heavy water `(D_(2)O)`. The reactivy of `D_(2)` towards oxygen is less as compared to that `H_(2)` because bond dissociation enthalpy of `D_(2)(443.35kJ mol^(-1))` is more than that of `H_(2) (435.88kJ mol^(-1))` | |
| 58. |
On the basis of electron affinity, comment on the resemblance of hydrogen with halogens. |
| Answer» Halogens have electronic configuration `ns^(2)np^(5)`. So, they have strong tendency to gain one electron to attain stable nearest noble gas configuration. Similarly, hydrogen with `1s^(2)` electronic configuration has a tendency to gain one electron and form `H^(-)` ion and attain helium gas configuration. | |
| 59. |
In which of the properties listed below hydrogen does not show resemblance with halogens ? I Electropositive character II Electronegative character III Neutral nature of `H_(2)O` IV. AtomcityA. I and IIIB. I onlyC. II and IIID. III and IV |
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Answer» Correct Answer - A Halogens have electronegative character and `H_(2)O` does not resemble with oxides formed by halogens. |
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| 60. |
In which of the properties listed below hydrogen does not show resemblance with halogens ? I Electropositive character II Electronegative character III Neutral nature of `H_(2)O` IV. AtomicityA. I and IIIB. I onlyC. II and IIID. III and IV. |
| Answer» Correct Answer - A | |
| 61. |
To an aqueous solution of `AgNO_(3)` some NaOH(aq) is added, till a brown ppt. is obtained. To this `H_(2)O_(2)` is added dropwise. The ppt. turns black with the evolution of `O_(2)`. The black ppt. isA. `Ag_(2)O`B. `Ag_(2)O_(2)`C. AgOHD. None of these. |
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Answer» Correct Answer - D `2AgNO_(3)(aq)+2NaOH(aq)tounderset("Brown ppt.")(Ag_(2)O(s))+H_(2)O(l)+2NaNO_(3)(aq)` `Ag_(2)O(s)+H_(2)O_(2)(aq)toH_(2)O(l)+O_(2)(g)underset("Black ppt.")(+2Ag(s))` The finally divided Ag is black in colour. |
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| 62. |
Which of the following equation depicts reducing nature of `H_(2)O_(2)`?A. `2[Fe(CN)_(6)]^(4-)+2H^(+)+H_(2)O_(2)rarr2[Fe(CN)_(6)]^(3-)+2H_(2)O`B. `I_(2)+H_(2)O_(2)+2OH^(-)rarr 2I^(-)+2H_(2)O+O_(2)`C. `Mn^(2+)+H_(2)O_(2)rarr Mn^(4+)+2OH^(-)`D. `PbS+4H_(2)O_(2)rarr PbSO_(4)+4H_(2)O` |
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Answer» Correct Answer - B Reducing nature of `H_(2)O_(2)` means it reduces other substance and itself gets oxidised. In such reactions `O_(2)` is evolved. In (a), `Fe^(2+)` gets oxidised to `Fe^(3+)`. In (b), `I_(2)` gets reduced to `I^(-)`. In (c ), `Mn^(2+)` gets oxidised to `Mn^(4+)` In (d), `PbS^((2-))` gets gets oxided to `overset((+6))(PbSO_(4))`. |
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| 63. |
When a substance A reacts with water, it produces a combustible gas B and s solution of substance C in water. When another substance D reacts with this solution of C, it produces the same gas B on warming but D can produces gas B on reaction with dilute sulphuric acid at room temperature . A imparts a deep golden yellow colour to a smokeless flame of Bunsen. A,B,C and D respectively are:A. `Na,H_(2),NaOH, Zn`B. `K, H_(2),KOH, Al`C. `Ca,H_(2), Ca(OH)_(2), Sn`D. `CaC_(2), C_(2)H_(2),Ca(OH)_(2),Fe` |
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Answer» Correct Answer - A `underset(A)(2Na) + 2H_(2)O to underset(B)(H_(2))+2ZnO_(2) + underset(C)(2NaOH) and underset(D)(Zn)+underset(C)(NaOH) to Na_(2)ZnO_(2) + underset(B)(H_(2)) underset(D)Zn+ H_(2)SO_(4)("dil.") to ZnSO_(4) + underset(B)(H_(2))` Since Na gives golden yellow flame, K gives pink violet flame while Ca gives brick red flame, therefore, option (a) is correct. |
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| 64. |
When a substance A reacts with water it produces a combustible gas B and a solution of substance C in water. When another substance D reacts with this solution of C, it also produces the same gas B on warming but D can produce gas B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden yellow colour a smokeless flame to Bunsen burner. A,B,C, and D respectively are `:` |
| Answer» Correct Answer - Na, `H_(2),NaOH,Zn` | |
| 65. |
In the reaction, ` 2FeSO_(4) +H_(2)SO_(4)+H_(2)O_(2)rarr Fe_(2)(SO_(4))_(3)+2H_(2)O` The oxidising agent isA. `FeSO_(4)`B. `H_(2)SO_(4)`C. `H_(2)O_(2)`D. both `(b)` and `(c)` |
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Answer» Correct Answer - C `H_(2)O_(2)` is an oxidising agent, it is reduced to water. `overset(-1)(H_(2)O_(2)) rarr overset(-2)(H_(2)O)` Oxidation number of oxygen decreases. |
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| 66. |
Triple point of water isA. `273 K`B. `373 K`C. `203 K`D. `193K` |
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Answer» Correct Answer - a The triple point of any substance is that temperature and pressure at which the material can exist ini all three phases `(` Solid, liquid and gas `)` in equilibrium specifically the tiple point of water is `273. 16K` at `611.2 Pa.` |
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| 67. |
Triple point of water isA. `273K`B. ` 373 K`C. `203 K`D. `193 K` |
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Answer» Correct Answer - a The triple point of any substance is that temperature and pressure at which the material can exist ini all three phases `(` Solid, liquid and gas `)` in equilibrium specifically the tiple point of water is `273. 16K` at `611.2 Pa.` |
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| 68. |
Two ice cubes are pressed over each other and unite to form one cube . Which force is responsible of holding them together ?A. Hydrogen bond formationB. Van der Waals forcesC. Covalent attractionD. Ionic interaction |
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Answer» Correct Answer - A Two ice cubes when pressed over each other unite due to hydrogen bond formation. |
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| 69. |
Water gas is produced byA. passing steam over the red hot cokeB. Passing steam and air over red hot cokeC. burning coke in excess of airD. burning coke in limited supply of air |
| Answer» Correct Answer - A | |
| 70. |
In context with the industrial preparation of hydrogen from water gas `(CO+H_(2))` , which of the following is the correct statement ?A. `CO` is oxidised to `CO_(2)` with steam in the presence of a catalyst followed by absorption of `CO_(2)` in alkaliB. `CO` and `H_(2)` are fractionally separated using difference in theire densitiesC. `CO` is removed by absorption in aqueous `Cu_(2)Cl_(2)`D. `CO` is removed by absorption in aqueous `Cu_(2)CL_(2)` |
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Answer» Correct Answer - a The water gas is mixed with excess of steam and passed `45^(@)C` over a heated catalyst `(Fe_(2)O_(3)+Cr_(2)O_(3)).CO` is mostly oxidised to `CO_(2)` and more `H_(2)` is set free. `CO_(2)` is absorbed in alkali. `underset("Water gas")(ubrace(CO+H_(2))) + underset("Steam")(H_(2)O) overset("Catalyst")(to) CO_(2) + 2H_(2)` |
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| 71. |
In context with the industrial preparation of hydrogen from water gas `(CO+H_(2))`. Which of the following is the correct statement?A. CO is removed by absorption in aqueous `Cu_(2)Cl_(2)` solutionB. `H_(2)` is removed through occulusion with PdC. CO is oxidized to `CO_(2)` with steam in the presence of a catalyst followed by absorption of `CO_(2)` in alkaliD. CO and `H_(2)` are fractionally separated using differences in their densities |
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Answer» Correct Answer - C Bosch process `: CO + H_(2) + underset("Stream")(H_(2)O) rarr CO_(2) + 2H_(2) overset("absorption of "CO_(2))rarr H_(2)` |
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| 72. |
In context with the industrial preparation of hydrogen from water gas `(CO+H_(2))` , which of the following is the correct statement ?A. CO is oxidised to `CO_(2)` with steam in the presence of a catalyst followed by absorption of `CO_(2)` in alkaliB. CO and `H_(2)` are fractionally separated using difference in their densitiesC. CO is removed by absorption in aqueous `Cu_(2)Cl_(2)` solutionD. `H_(2)` is removed through occlusion with Pd. |
| Answer» Correct Answer - A | |
| 73. |
In context with the industrial preparation of hydrogen from water gas `(CO+H_(2))`, which of the following is the correct statement ?A. CO is oxidising to `CO_(2)` with steam in the presence of a catalyst followed by absorption of `CO_(2)` in alkaliB. `CO and H_(2)` are fractionally separated using differences in their densitiesC. CO is removed by absorption in aqueous `Cu_(2)Cl_(2)` solutionD. `H_(2)` is removed through occlusion with Pd |
| Answer» Correct Answer - A | |
| 74. |
In context with beryllium, which one of the following statements is incorrect ?A. It is rendered passive by nitric acidB. its forms `Be_2C`C. Its salts rarely hydrolyseD. Its hydride is electron-deficient and polymeric |
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Answer» Correct Answer - C Beryllium salts are covalent in nature because of very small size of `Be^(2+)` ion and its high polarising power, so it is easily hydrolysed. e.g.`BeCl_2+2H_2O to Be(OH)_2+2HCl` |
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| 75. |
What causes the temporary and permanent hardness of water ? |
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Answer» Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates `(MHCO_(3), "where" M = Mg, Ca)` in water. Permanent hardness of water is because of the presence of soluble salts of calcium and magnesium in the form of chlorides in water. |
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| 76. |
20 volume `H_(2)O_(2)` solution has a strength of aboutA. `30%`B. `6%`C. `3%`D. `10%`. |
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Answer» Correct Answer - B 20 volumes of `H_(2)O_(2)` means 20 litres of oxygen is obtained at STP by the decomposition of 1 litre `H_(2)O_(2)` solution `{:(2H_(2)O_(2),to,2H_(2)O+,O_(2)),(2 "mole",,,1 "mole"),(68 g,,,22*4 "litre at" STP):}` `22*4` litres oxygen is obtained from 68 g `H_(2)O_(2)` `:.` 20 litres oxygen is obtained from `=(68)/(22*4)xx20=60*7 g H_(2)O_(2)` `:.` 1000 ml of `H_(2)O` solution contains `60*7` g `H_(2)O_(2)` 100 ml `(~~100 g)` solution contains `(60*7)/(1000)xx100` `:.` Strength of `H_(2)O_(2)` solution `=6*07 g H_(2)O_(2)=6*07%` |
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| 77. |
Hydrogen peroxide is ……..A. an oxidising agentB. a reducing agentC. both an oxidising and a reducing agentD. neither oxidising nor reducing agent |
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Answer» Correct Answer - B Hydrogen peroxide acts as an oxiding as well as reducing agent in both acidic and alkaline media. |
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| 78. |
`H_(2)O_(2)` is manufactured these daysA. by the action of `H_(2)O_(2)` on `BaO_(2)`B. by the action of `H_(2)SO_(4)` on `Na_(2)O_(2)`C. by electrolysis of `50% H_(2)SO_(4)`D. by burning hydrogen in excess of oxygen. |
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Answer» Correct Answer - C Electrolysis of `50%` sulphuric acid gives per disulphuric acid `(H_(2)S_(2)O_(8))` which on distillation yields `30%` solution of hydrogen peroxide. |
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| 79. |
Why is sodium chloride less soluble in heavy water than in ordinary water ? |
| Answer» Due to lower dielectric constant of `D_(2)O` over `H_(2)O, NaCl` is less soluble in `D_(2)O` than in `H_(2)O`. | |
| 80. |
One part of heavy water is present in `X` parts of ordinary water. Here `X` isA. `10`B. `60`C. `6000`D. `60000` |
| Answer» Correct Answer - C | |
| 81. |
How is dihydrogen prepared from water by using a reducing agent? |
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Answer» Dihydrogen is prepared from water by the action of reactive metal like Na or K which is a strong reducing agent due to low ionisation enthalpy. `2Na+2H_(2)O to 2NaOH+H_(2)` `2K+2H_(2)O to 2KOH+H_(2)` |
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| 82. |
Which combination cannot be used for the preparation of hydrogen gas in the laboratory ? I. Zinc/conc. `H_(2)SO_(4)` II. Zinc/ `HNO_(3)` III. Pure zinc/dil. `H_(2)SO_(4)`A. I and IIB. I, II, IIIC. III onlyD. I and III |
| Answer» Correct Answer - B | |
| 83. |
Water cannot be used to extinguish petrol fire. Explain. |
| Answer» Petrol lighter than water and the two are immiscible. When water is sprayed over pertrol. The letter floats over water and its vapours still burn. | |
| 84. |
Hydrogen peroxide acts as:A. An oxidising agentB. a reducing agentC. An acidD. All the three. |
| Answer» (d) `H_(2)O_(2)` acts as all the three | |
| 85. |
Assertion. NaCl is less soluble in heavy water than in ordinary water. Reason. Dielevtric constant of ordinary water is more than that of heavy water. |
| Answer» Correct Answer - A | |
| 86. |
Explain the following: (i) Temporary hard water becomes soft on boiling. (ii) Water can extinguish most fires but not petrol fire. (iii) Hard water is softened before being used in boilers. |
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Answer» (I) Upon boiling, the hydrogen carbonates of calcium and magnesium decompose to form corresponding carbonates. They, being water insoluble, get precipated and can be removed by filtration. Thus, temporary hard water becomes soft on boiling `Ca(HCO_(3))_(2) overset("Boil")to underset("ppt")(CaCO_(3))+H_(2)O+CO_(2)` `Mg(HCO_(3))_(2) overset("Boil")to underset("ppt")(MgCO_(3))+H_(2)O+CO_(2)` (ii) When water is added to petrol fire in order to extinguish it, petrol lighter floats over water. As a result, the fire flares up rather than getting extinguished. (iii) When hard water slowly form an insoluble deposit all along insdie the boiler. This is called boiler scale. As the scale is poor conductor of heat, there is wastage of fuel. Morever, it also redcues the life of the boiler. Therefore, hard water has to be softened before being used in boilers. |
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| 87. |
Assertion : When sodium hydride in fused state is electrolysed, hydrogen is dicharged at anode. Reason : Sodium hydride is an electrovalent compound in which hydrogen is present as cation.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - C Sodium hydride is an electrovalent compound in which hydrogen is present as an anion, `H^(-)` which is discharged as `H_(2)` at anode on electrolysis. |
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| 88. |
Account for the following in the preparation of `H_(2)O_(2)` from barium perioxide `(BaO_(2))` and dilute `H_(2)SO_(4)`. (a) A thin paste of hydrated barium perioxde is used instead of nahydrous perioxide. (b) The temperature of the reaction mixture is kept at `0^(@)C`. (c ) The final solution must be kept slightly acidic. |
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Answer» (a) Anhydrous `BaO_(2)` is not used becasuse the reaction with dilute `H_(2)SO_(4)` is highly exothermic and will accelerate the decomposition of `H_(2)O_(2)` into `H_(2)O and O_(2)`. Moreever, in this case a layer of `BaSO_(4)` initially formed in the reaction gets deposited over barium and any futher raction with acid does not take place. Keeping this in mind. hydrated barium peroxide `(BaO_(2).8H_(2)O)` is used (b) The temperature of the reaction mixture is kept at `0^(@)C` becasue higher temperature will cause the decomposition of the hydrogen peroxide. (c ) The traces of `H^(+)` ions present in the final solution will act as inhibitor and will retard the decomposition of `H_(2)O_(2)`. |
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| 89. |
Hydrocarbon perioxide is a thich syrup liquid with a better taste. It can be prepared by a number of methods both in the laboratory as well as commercially. However, the concentrations of hydrogen peroxide is a big problem since it readily decomposes when heated under normal conditions of temperature and pressure. (i) How is the strength of `H_(2)O_(2)` generally expressed? (II) What is the meaning of 20 volume `H_(2)O_(2)` solution? (iii) What are the values associated with the use of hydrogen perioxde? |
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Answer» (i) Strength of `H_(2)O_(2)` is generally expressed in terms of volume i.e. as 10 volume, 15 volume, 20 volumes etc. (ii) A 20 volume of `H_(2)O_(2)` solution means that 1 mL of `H_(2)O_(2)` evolves 20mL of `O_(2)` under N.T.P. conditions (iii) Hydrogen proxide is useful in a number of ways. It acts as a bleachin agent for delicate materials like, silk, wool nad a constitutent of hair dyes. It is also reagarded as a better bleaching agent in the laundary than chlorine since it does not release any poisnous vapoours unlike chlorine and thus, promotes Green Chemistry. It is used as an antiseptic fore washing fresh wounds. It is also as an oxidant in rocket fuels. |
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| 90. |
Permanent hardness can be removed by addingA. `Cl_(2)`B. `Na_(2)CO_(3)`C. `Ca(O Cl)Cl`D. K. |
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Answer» Correct Answer - B Permanent hardness is removed by precipitating carbonates of `Ca^(+2)` and `Mg^(+2)` `CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)darr+2NaCl` |
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| 91. |
Assertion. `H_(2)O_(2)` liberates `O_(2)` when it reacts with acidified `KMnO_(4)` solution. Reason. `KMnO_(4)` oxidises `H_(2)O_(2)` to `O_(2)` |
| Answer» Correct Answer - A | |
| 92. |
A `5.0 cm^(3)` solution of `H_(2)O_(2)` liberates `1.27 g` of iodine from an acidified `KI` solution. The precentage strength of `H_(2)O_(2)` isA. `11.2`B. `5.6`C. `1.7`D. `3.4` |
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Answer» Correct Answer - D `H_(2)O_(2)=KI=I_(2)` mEq of `H_(2)O_(2)=5xxN` mEq of `I_(2)=("Weight")/(EW)xx1000` `=1.27/127xx1000=10 mEq` `I_(2)=MW=254/2=127` `:.` mEq of `I_(2)` `5xxN=10 :. N=2` `1N` of `H_(2)O_(2)=1.7%` of `H_(2)O_(2)` `2N` of `H_(2)O_(2)=3.4% H_(2)O_(2)` |
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| 93. |
Hydrogen perioxide is used to restore the colour of old oil painting lead oxide. Explain. |
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Answer» When the old oil painting are kept in the atmostphere for a long period, the traces of `H_(2)S` gas present in the atmosphere converts lead oxide (PbO) into lead sulphide (PbS) which is black is colour. Therefore, these paintings get transhed. The original whiteness can be restored by keeping then in hydrogen peroxide solution for sometime as a result of which lead sulphide is oxidised to lead sulphate. `underset("Black")(PbS)+4H_(2)O_(2) to underset("white")(PbSO_(4))+4H_(2)O` |
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| 94. |
Statement-1. Electrolysis of NaH in the fused state liberates `H_(2)` at the anode. Statement -2. NaH contains `H^(-)` ions.A. Statement-1 is true, Statement-2 is True , Statement-2 is a correct explanation for statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for statement-1C. Statement-1 is True, Statement -2 is FalseD. Statement -1 is False , Statement-2 is True. |
| Answer» Correct Answer - A | |
| 95. |
Which of the following statements concerning protium, and tritium is not true ?A. They are isotopes of each otherB. They have similar electronic configurationsC. They exist in the nature in the ratio of 1:2:3D. Their mass numbers are in the ratio of 1:2:3 |
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Answer» Correct Answer - C Actually these exist in the ratio `{:("Protium",:,"Deuterium",:,"Tritium"),(1,:,1.56xx10^(-2),:,1xx10^(-17)),(,,,,):}` |
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| 96. |
H2O2 is used to restore the colour of old paintings containing lead sulphide. Write a balanced equation for the reaction that takes place in this process. |
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Answer» PbS(Black) + 4H2O2 → PbSO4(White) + 4H2O |
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| 97. |
Statement-1: `H_(2)O_(2)` liberates `O_(2)` when it reacts with acidified `KMnO_(4)` solution Statement-2: `KMnO_(4)` oxidised `H_(2)O_(2)` to `O_(2)`.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
| Answer» Correct Answer - A | |
| 98. |
Old paintings of lead are generally washed with dilute solution of hydrogen peroxide in order to regain its colour. Why? |
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Answer» Black colour in paintings is due to lead sulphide (PbS). This is oxidized to lead sulphate by hydrogen peroxide. |
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| 99. |
Size of neucleus increases from protium to tritium so in `H_(2),D_(2)&T_(2)` area off overlapping also increases in the same order. Q. Which overlapping is responsible for bond formation in `H_(2),D_(2),T_(2)` respectively?A. 1s-1s in eachB. 2s-2s in eachC. 1s-2s in eachD. 1s-2s I `T_(2)`, 1s-1s in rest |
| Answer» Correct Answer - A | |
| 100. |
Size of neucleus increases from protium to tritium so in `H_(2),D_(2)&T_(2)` area off overlapping also increases in the same order. Q. `H_(2),D_(2)&T_(2)` show their bond-enthalpies asA. `H_(2)=D_(2)=T_(2)`B. `H_(2) gt D_(2) gt T_(2)`C. `H_(2) lt D_(2) lt T_(2)`D. `D_(2) lt H_(2) lt T_(2)` |
| Answer» Correct Answer - C | |