This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Electron affinity of oxygen is less negative than sulphur. Justify this statement. |
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Answer» Solution :When an electron is added to oxygen, the added electron goes to the .L. SHELL (n=2). As the .L. shell POSSESS smaller region of space, the added electron feels SIGNIFICANT repulsion from the other ELECTRONS present in this level. E.A of O=-141 KJ `mol e^(-1)` E.A of S=-200 KJ `mol e^(-1)`. |
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| 2. |
Electron affinity of Fluorine is less than that of Chlorine because |
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Answer» Electronegativity of Fluorine is more |
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| 3. |
Electron affinity of chlorine is more than that of fluorine explain. |
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Answer» |
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| 4. |
Electronaffinityis positivewhen |
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Answer» O changesinto `O^(-)` |
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| 5. |
Electron affinity is measured in |
| Answer» Answer :D | |
| 6. |
Electron affinity is |
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Answer» Energy required to TAKE out an ELECTRON from an ISOLATED gaseous atom |
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| 8. |
Electromeric effect is a temporary effect. Assign reason. |
| Answer» Solution :The ELECTROMERIC effect is noticed in the organic compounds containing multiple bond i.e., DOUBLE bond `( gt C=C lt` or `gt C=O)` or triple blood `(-C equiv N)` under the influence of the attacking reagent. The moment the attacking reagent is REMOVED, pi `(pi)` electron pair comes BACK to its original position forming multiple bond again. Therefore, electromeric effect is a temporary effect. for details, consult section 12.22. | |
| 9. |
Electromagnetic radiationof wavelength242nm isjustsufficientto ionise thesodiumatom .Calculatethe ionisation energyofsodiumin Kj mol^(-1) |
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Answer» Solution :where `lambda= 242nm =242 xx 10^(9) m`wavelengthper photon`E= (hc )/(lambda ) ` permoleE = `(N_(A) hc )/(lambda )` `n=6.626 xx 10^(34)Js` `c=3.0xx 10^(8) m` `N_(A)= 6.022xx 10^(23)` permole `=(6.022 xx 10^(23) mol^(1)xx 6.626 xx 10^(34)js xx3.0 xx 10^(8) MS^(-1)J)/( (242 xx 10^(9) m))` `=494.7 xx KJ mol^(-1)495kJ mol^(1)` Ionisationenthalpy=495kJ `mol^(1)` |
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| 10. |
Electromeric effect involves the complete transfer of |
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Answer» `sigma `-ELECTRON |
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| 11. |
Electromeric effect can be relayed in |
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Answer» CARBON chain joined by SINGLE bond |
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| 12. |
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in kJ mol^(-1) |
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Answer» SOLUTION :`E = N hv = Nh (C)/(lamda) = ((6.02 xx 10^(23) mol^(-1)) xx (6.626 xx 10^(-34) JS) xx (3 xx 10^(8) ms^(-1)))/(242 xx 10^(-9) m)` `= 4.945 xx 10^(5) J mol^(-1) = 494.5 kJ mol^(-1)` |
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| 13. |
Electromagnetic radiation with minimum wavelength is |
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Answer» ULTRA violet |
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| 14. |
Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out |
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Answer» In the PRESENCE of `NaCl` |
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| 15. |
Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_(2)O_(2) The number of peroxybonds present in X and Y. Then x + y is? |
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Answer» |
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| 16. |
Electrolysis of X gives Y at anode. Vacuum distillation of Y gives H_(2)O_(2). The number of peroxy (O-O_ bonds present in X and Y respectively, are |
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Answer» 1,1 |
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| 17. |
Electrolysis of potassium salt of 3-methyl butanoic acid gives an alkane ‘A’. The name of alkylIodide which gives same alkane on treatment with sodium metal in dry ether solvent is |
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Answer» 1-Iodo-3-methyl butane |
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| 18. |
Electrolysis of molten saline hydride generally gives |
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Answer» HYDROGEN at CATHODE |
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| 19. |
Electrolysis of KCl. MgCl_2 . 6H_2O gives: potassium only magnesium only magnesium and chlorine potassium and magnesium. |
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Answer» POTASSIUM only |
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| 20. |
Electrolysis of aqueous solution of sodium salt of Maleic acid and Fumaric acid gives |
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Answer» Ethylene and acetylene respectively |
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| 21. |
Electrolysis of aqueous solution of CH_(3)COOK gives |
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Answer» `CH_(4)` |
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| 22. |
Electrolysis of aqueous solution of ammonium sulphate and sulphuric acid at anode gives |
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Answer» `H_2S_2O_8` |
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| 23. |
Electrolysis of an aqueous solution of potassium acetate, the hydrocarbon obtained |
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Answer» Methane |
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| 24. |
Electrolysis of a concentrated aqueous solution of a compound gave C_2H_6 on anode.The compound is |
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Answer» `CH_3COOK` |
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| 25. |
Electrolysis of 50% H_(2)SO_(4) produces |
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Answer» `H_(2)S_(2)O_(8)` at ANODE |
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| 26. |
Electrolysis of 50% H.SO, produces |
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Answer» `H_2S_2O_8` at ANODE |
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| 27. |
Electrolysis of 50%" "H_(2)SO_(4) produces |
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Answer» `H_(2)S_(2)O_(8)` at ANODE |
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| 28. |
Electro-osmosis was discovered by |
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Answer» Dorn |
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| 29. |
Electrinoic configuration of species M^(2+) is 1s^(2) 2s^22p^63s^23p^63d^6 and its atomic weight is 56.The number of neutrons in the nuclcus of species M Is………….. |
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Answer» 26 M:`Is^2``2s^2``2p^6``3s^2``3p^6``3d^8` ATOMIC number=26 Mass number=56 No.of netutrons=56-26=30 |
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| 30. |
Electriccookers have a coating of thatprotectes them against fire. |
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Answer» HEAVY lead |
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| 31. |
Electric cookers have a coating that protects them against firer. The coating is made of |
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Answer» Heavy LEAD |
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| 32. |
Elements with high electronegativity are generally |
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Answer» GOOD REDUCTANTS |
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| 33. |
Elaborate the term compound atoms as used by Dalton |
| Answer» Solution :The TERM compound atoms was used by Dalton to EXPRESS the substances obtained by the combination of atoms of different elements in a FIXED, simple and WHOLE number RATIO. | |
| 35. |
Eka-aluminium andeka- silicon were the namesgivenby Mendeleev for thethenunknowns elements. What isitsatomic number? Write its group number, electronic configuration, IUPAC and official names. |
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Answer» Solution :The element whichcomesafternercury in thesame group of theperiodictableis calledeka- mercury . Itsvariousparameters are : Z=80 + 32 = 112 IUPAC name : Uub OFFICIAL name : Cn ( coperniclum) E.C.=[RN]`5f^(14) 6d^(10) 7s^(2)` |
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| 36. |
Eka-aluminium and Eka-siliconare knownas |
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Answer» Galliumand GERMANIUM |
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| 37. |
Einstein was awarded Noble Prize for |
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Answer» General THEORY of relativity |
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| 38. |
Efficiency of energy conversion in nuclear fission of uranium and nuclear fusion of hydrogen are respectively: |
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Answer» `0.09%, 0.35%` |
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| 39. |
Efficiency of a Carnot's engine is 100% when |
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Answer» sing is at`0^(@)C` `T_(1) =` TEMPERATURE of sink, `T_(2) =` Temperature of source For`100%` efficiency , `eta= 1` HENCE, `1= 1- (T_(1))/(T_(2))` or `(T_(1))/(T_(2))=0` or `T_(1) = 0 K` |
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| 40. |
Effective magnetic moment of actinides with 'n' unpaired electrons can be calculated using following relation: |
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Answer» `g SQRT(J(J+1))BM` |
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| 41. |
Effect of fluorosis disease is...... |
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Answer» irritation in stomach. |
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| 42. |
Edge length of a cube is 400 pm , its body diagonal would be |
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Answer» 500 pm |
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| 43. |
Edge length of a cube is 400 pm, its body diagonal would be |
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Answer» 500 pm ` = 1.732 xx 400 "pm" = 692. 28"pm" = 693 "pm" ` |
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| 44. |
Ease of aromatic nucleophilic substitution among these compound will be in the order as : |
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Answer» `(V) gt (II) gt (III) gt (I) gt (IV)` |
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| 45. |
Ease of formation of anion is favoured by |
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Answer» LOWER VALUE of IONISATION energy |
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| 46. |
Earth's protective umbrella existing in the stratosphere is known as ______ |
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Answer» |
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| 47. |
E_(Al^(3+)//Al)^0 =- 1.66 V and E_(Tl^(3+)//Tl)^0 = 1.26 V. Then which of the following statements is correct? |
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Answer» ALUMINIUM has high TENDENCY to form `Al^(3+)` ions in aqueous solution. |
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| 48. |
Each unit cell of NaCl consists of 4 chlorine ions and: |
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Answer» 13 NA ATOMS |
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| 49. |
Each rubidium halide crystallising in the NaCl type lattice has a unit cell length 0.30 Å greater than for corresponding potassium salt (r_(K^+)=1.33 Å) of the same halogens . Hence, ionic radius of Rb^+ is |
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Answer» 1.03 Å Unit cell length of KX =`2(r_(K^+) + r_(X^-))` `therefore 2(r_(Rb^+)+r_(X^-))-2(r_(K^+)+r_(X^-))`=0.30 Å or `r_(Rb^+) - r_(K^+)`=0.15 Å or `r_(Rb^+) = r_(K^+)`+ 0.15 Å =1.33 +1.15=1.48 Å |
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| 50. |
Each rubidium halide crystallising in the NaCl type lattice has a unit cell length0.30 Ågreaterthan for corresponding potassium salt ( r_(K^(+))= 1.33 Å ) of the same halogen. Hence, ionic radius ofRb^(+) is |
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Answer» 1.03 Å Unit cell length of KX =`2 (r_(K^(+)) + r_(X^(-))` ` 2( r_(Rb^(+)) + r_(X^(-)))-2 (r_(K^(+)) + r_(X^(-))) = 0.30 Å` ` or r_(Rb^(+)) - r_(K^(+)) = 0.15 Å` ` orr_(Rb^(+) = r_(K^(+)) + 0.15 Å = 1.33 + 1.15 = 1.48 Å` |
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