Class 12th
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New answer posted
a year agoContributor-Level 10
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.
New answer posted
a year agoContributor-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.
New answer posted
a year agoContributor-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.
New answer posted
a year agoContributor-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.
New answer posted
a year agoContributor-Level 10
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.
New answer posted
a year agoContributor-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.
New answer posted
a year agoContributor-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
New answer posted
a year agoContributor-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.
New answer posted
a year agoContributor-Level 10
36. Option (i) (A) and (B) is correct since AgBr shows both Frenkel as well as Schottky defects.
New answer posted
a year agoContributor-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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