Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

41.

Frenkel and Schottky defects are

  • nucleus defects

  • non-crystal defects

  • crystal defects

  • nuclear defects


42.

In the crystal of CsCl, the co-ordination number changes from 8:8 to 6:6 by:

  • applying high pressure

  • applying low temperature

  • applying high temperature

  • applying low pressure


43.

A compound CuCl has face-centred cubic structure. Its density is 3.4 g cm-3 , The length of unit cell is:

  • 8.734 A0

  • 5.834 A0

  • 5.783 A0

  • 8.153 A0


44.

If the positions of Na+ and Cl- are interchanged in NaCl , the crystal lattice with respect to Na+ and CI- is:

  • bcc and fcc

  • fcc and bcc

  • both fcc

  • both bcc


Advertisement
Advertisement

45.

The crystal with metal deficiency defect is :

  • NaCl

  • FeO

  • KCl

  • ZnO


B.

FeO

The metal deficiency defect arises due to the missing of a cation from its lattice site and presence of the cation with higher charge in the adjacent site. Hence, the metals showing variable valency like transition metal show this defect , e.g. , FeO.


Advertisement
46.

ZnO is white when cold and yellow when heated. It is due to the development of

  • Frenkel defect

  • metal excess defect

  • Schottky defect

  • metal deficiency defect.


47.

Among the following, the compound that is both paramagnetic and coloured is

  • K2Cr2O7

  • (NH4)2[TiCl6]

  • VOSO4

  • K3[Cu(CN)4]


48.

If an atom crystallises in bee lattice with r= 4 Å then the edge length will be

  • 2 Å

  • 8 Å

  • 2.39 Å

  • 9.23 Å


Advertisement
49.

A mineral having the formula AB2 crystallizes in the cubic close-packed lattice , with the A atoms occupying the lattice points. The fraction of the tetrahedral sites occupied by B atoms is :

  • 20 %

  • 40 %

  • 60 %

  • 100 %


50.

A forms hcp lattice and B are occupying 1/3rd of tetrahedral voids, then the formula of compound is

  • AB

  • A3B2

  • A2B3

  • AB4


Advertisement