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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

Contributor-Level 10

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

Contributor-Level 10

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

Contributor-Level 10

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

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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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11 months ago

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alok kumar singh

Contributor-Level 10

20. Option (i)

Explanation: 2H2O + Cl→2OH+ H2 + Cl2

The ΔG?  is +422 kJ for this reaction. We get ΔE? = −2.2 V, when we convert it to ΔE?   (using G? = E? F). It will, of course, require an external e.m.f. greater than 2.2 V. However, electrolysis necessitates an excess potential in order to overcome some other impeding reactions. Thus,  Cl2 is obtained through electrolysis, which produces H2 and aqueous NaOH as byproducts. In addition, molten NaCl is electrolyzed. However, in that case, Na metal is formed rat

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alok kumar singh

Contributor-Level 10

19. The metal gold is a nonreactive metal. It is found as a native metal in its free state.

So, it is not compulsory to separate it by a chemical process. Thus, gold is leached to get it in its pure form. In the metallurgy of gold (Au), the respective metal is leached with a dilute mixture of NaCN or KCN in the presence of air where Gold (Au) is oxidized by oxygen of the air to Au+cation which then merges with ions to form a soluble compound.

4Au (s)+8CN (aq)+O2 (g)+2H2O (aq)→4 [Au (CN)2] (aq)+4OH (aq)

Gold is then obtained from this soluble compound x by replacement method by using a much more electropositive Z

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