Chemistry

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Payal Gupta

Contributor-Level 10

45. Option (iv) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids is not true since the occupation of the voids depends upon the stoichiometry of the compound instead of the radii of the ions. 

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Payal Gupta

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44. Option (ii) Ferromagnetic substances cannot be magnetised permanently is correct since ferromagnetic substances are those which are very strongly attracted by the magnetic field and can be magnetized permanently.

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Payal Gupta

Contributor-Level 10

43. Option (ii) n-type semiconductor is correct since silicon is doped with electron rich impurities like Phosphorus or Arsenic and this leads to the increase in the conductivity due to the negatively charged electron . Hence, silicon doped with electron rich impurity is termed as n -type semiconductor. 

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Payal Gupta

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42. Option (iv) Electronic defects is correct since doping can be done with an impurity which is electron rich or electron deficient as compared to the intrinsic semiconductor like Si or Ge . Such impurities lead to the electronic defects in them. 

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Payal Gupta

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41. Option (iii) 4 is correct since in  2-D scope structure each sphere is in contact with four of its neighbours. Thus, its coordination number is found to be 4. 

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Payal Gupta

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40. Option (i) Cl-ion form fcc lattice and Na+ ions occupy all octahedral voids of the unit cell is correct since In NaCl,  chloride ion form fcc lattice and sodium ions occupy all the octahedral voids of the unit cell. The coordination number of both the cation and anion is found to be 6.

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Payal Gupta

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39. Option (iv) since in this arrangement the spheres of fourth layer are exactly aligned with those of the first layer is correct since in hexagonal close packed structure there is ABAB.type arrangement which means that in this case spheres of third layer are exactly aligned with that of the first layer. Hence, the pattern of spheres is repeated in alternate layers. Thus, the statement at (iv) is not true about hexagonal close packing. 

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Payal Gupta

Contributor-Level 10

38. Option (iii) 32 is correct  since the packing efficiency in bcc arrangement is 68%, therefore the percentage empty space  can be calculated as – 100 - 68 = 32 

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Payal Gupta

Contributor-Level 10

37. Option (ii) hcp and ccp is correct since in hcp and ccp the packing efficiency is 74% ( maximum ), hence this pair of packing is considered as the most efficient packing. 

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Payal Gupta

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36. Option (i) (A) and (B) is correct since AgBr shows both Frenkel as well as Schottky defects. 

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