Q: The density (in g cm ^ -3 ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is . Use: Atomic mass of metal =105.6 amu and Avogadro's co

aligned & CCP : \\ & ( Z =4, a =400 pm , M =105.6 ~g / mol ) \\ & Density = Z M a ^3 N _ A \\ & = 4 105.6 (400 10^ -10 )^3 6 10^ 23 =11.00 ~g / cm ^3 aligned

Q: For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one face are shown for clarity), the packing fraction is approximately (packing fraction = p

Total number of atoms in a unit cell No. of x = 6 × 1 2 = 3 atom (at face centres) No. of y = 8 × 1 8 = 1 atom (at all corners) So, the formula of compound is x 3 y Now If radius of x is r and radius of y is R then side length a = 2 R and face diagonal 2 a = 2 R + 2 r r = 2 − 1 R Now Packing fraction = volume of atoms Volume of unit &#16

Q: The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s).

M 2 × 1 2 X 4 × 1 4 = MX d 2 = a 2 2 + a 2 2 d 2 = a 2 2 + a 2 4 = 3 a 2 4 d 2 a 2 = 3 4 d a = 3 2 = 1 . 732 2 = 0 . 866 r + r - = 0 . 732

Question 1

The density (in g cm ^ -3 ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is . Use: Atomic mass of metal =105.6 amu and Avogadro's co

Accepted Answer

aligned & CCP : \ & ( Z =4, a =400 pm , M =105.6 ~g / mol ) \ & Density = Z M a ^3 N _ A \ & = 4 105.6 (400 10^ -10 )^3 6 10^ 23 =11.00 ~g / cm ^3 aligned

Question 2

Atoms of metals x , &#160; y  and z  form face-centred cubic (fcc) unit cell of edge length L x , body-centred cubic (bcc) unit cell of edge length L y , and simple cubic unit cell of edge l

Accepted Answer

Metal x forms FCC (edge length L x ) Metal y forms BCC (edge length L y ) Metal z forms SC (edge length L z ) Given r z = 3 2 r y and r y = 8 3 r x &#8756; &#160;&#160; r z = 3 2 &#215; 8 3 r x = 4 r x M z = 3 2 M y &#160;&#160; M z = 3 M x &#8756; M y = 2 M x Packing efficiency FCC &#160; &#62; &#160; BCC &#160; &#62; &#160; SC Packing efficiency unit cell x &#62; y &#62; z In FCC unit cell:- atoms along the face diagonals are in contact. &#8756; 2 &#160; L x = 4 r x &#8658; L x = 2 2 r x In BC

Question 3

Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in % ) of the resultant solid is closest to

Accepted Answer

Atom X occupies FCC lattice sites as well as alternate tetrahedral voids of FCC. In FCC, tetrahedral voids are 8 (in a unit cell) Hence Atom X in a unit cell (FCC lattice sites) = 8 &#215; 1 8 + 6 &#215; 1 2 = 4 Atom X in a unit cell (in T.V.) = 1 2 &#215; 8 = 4 Total atom X  in one unit cell = 8 For relation between a and r, since Tetrahedral void  forms at 1 4 th of body diagonal, a 3 4 = 2 r a = 8 r 3 Packing efficiency = 8 &#215; 4 3 &#960;r 3 a 3 &#215; 100 = 8 &#215; 4 3 &#9

Question 4

For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one face are shown for clarity), the packing fraction is approximately (packing fraction  = p

Accepted Answer

Total number of atoms in a unit cell No. of  x   = 6 &#215; 1 2 = 3  atom (at face centres) No. of  y   = 8 &#215; 1 8 = 1  atom (at all corners) So, the formula of compound is  x 3 y Now If radius of  x  is  r  and radius of  y  is  R  then side length  a = 2 R and face diagonal  2 a = 2 R + 2 r r = 2 &#8722; 1 R Now Packing fraction  = volume &#160; of &#160; atoms &#160; Volume &#160; of &#160; unit &#16

Question 5

The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s).

Accepted Answer

M 2 &#215; 1 2 X 4 &#215; 1 4 = MX d 2 = a 2 2 + a 2 2 d 2 = a 2 2 + a 2 4 = 3 a 2 4 d 2 a 2 = 3 4 d a = 3 2 = 1 . 732 2 = 0 . 866 r + r - = 0 . 732

Solid State — Previous Year Questions & Solutions