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The Solid State

Short Answer Type

161. A metal (at. mass = 50) has a bcc crystal structure. The density of the metal is 5.96 g cm^–3. Find the volume of its unit cell?

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162. Calculate the density of silver which crystallizes in a face centred cubic lattice with unit cell length 0.4086 nm (At. mass of Ag = 107.88)

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Long Answer Type

163. An element crystallizes in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 200 g of this element contains 24 x 10²³ atoms.

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164. An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm^–3. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.

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Short Answer Type

165. An atom has fcc crystal whose density is 10 gm^–3 and cell edge is 100 pm. How many atoms are present in its 100 g?

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Long Answer Type

166. Analysis shows that a metal oxide has the empirical formula M_0.98 O_1.00. Calculate the percentage of M²⁺ and M³⁺ ions in this crystal.

Or

In an ionic compound the anion (N) form cubic close type of packing. While the cation (M⁺) ions occupy one third of the tetrahedal voids. Deduce the empirical formula of the compound and the coordination number of (M) ions.

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167. An element E crystallizes in body centred cubic structure. If the edge length of the cell is 1.469 x 10^–10 m and the density is 19.3 g cm^–3, calculate the atomic mass of this element. Also calculate the radius of this element.

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168. Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body- centred cubic
(iii) face - centred cubic . (With the assumptions that atoms are touching each other).

(i) Simple Cubic: A simple cubic unit cell has one sphere (or atom) per unit cell. If r is the radius of the sphere, then volume occupied by one sphere present in unit cell = $4 over 3 πr cubed$
Edge length of unit cell (a) = r + r = 2r
Volume of cubic $left parenthesis straight a cubed right parenthesis space equals space left parenthesis 2 straight r right parenthesis cubed space equals space 8 straight r cubed$
Volume of occupied by sphere = $4 over 3 πr cubed$
Percentage volume occupied = percentage of efficiency of packing
$equals space fraction numerator Volume space of space sphere over denominator Volume space of space cube space end fraction cross times 100 equals space fraction numerator begin display style 4 over 3 end style πr cubed over denominator 8 straight r cubed end fraction cross times 100 equals space 1 over 6 cross times 3.143 cross times 100 space equals space 52.4 %$
For simple cubic metal crystal the efficiency of packing = 52.4%.
(b) Body centred cubic: From the figure it is clear that the atom at the centre will be in touch with other two atoms diagonally arranged and shown with solid boundaries.
$In space increment EFD space space space space space space space space space straight b squared space equals space straight a squared plus straight a squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 straight a squared space space space space space space space space space space space space space space space space space space space space space space space space straight b space equals space square root of 2 straight a end root Now space in space increment AFD space space space space space space space space space space space space space space space space space space space space space space space space straight c squared space equals space straight a squared plus space straight b squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight a squared plus 2 straight a squared space equals space 3 straight a squared space space space space space space space space space space space space space space space space space space space space space space space space straight c space equals space square root of 3 straight a end root$
The length of the body diagonal c is equal to 4r where r is the radius of the sphere (atom). But c = 4r, as all the three spheres along the diagonal touch each other
∴ $square root of 3 straight a end root space equals space 4 straight r$
or $straight a equals fraction numerator 4 straight r over denominator square root of 3 end fraction$
or $straight r equals fraction numerator square root of 3 over denominator 4 end fraction straight a.$
(i) Simple Cubic: A simple cubic unit cell has one sphere (or atom)
Fig. BCC unit cell. In this sort of structure total number of atoms is two and their volume is $2 cross times open parentheses 4 over 3 close parentheses πr cubed.$
Volume of the cube, a³ will be equal to
$open parentheses fraction numerator 4 over denominator square root of 3 end fraction straight r close parentheses space or space straight a cubed space equals space open parentheses fraction numerator 4 over denominator square root of 3 end fraction straight r close parentheses cubed.$
Therefore, Percentage of efficiency Volume occupied by four - spheres
$equals space fraction numerator in space the space unit space cell over denominator Total space volume space of space the space unit space cell end fraction cross times 100 % equals space fraction numerator 2 cross times open parentheses begin display style 4 over 3 end style close parentheses πr cubed cross times 100 over denominator left parenthesis 4 divided by square root of 3 straight r right parenthesis cubed end fraction % equals space fraction numerator left parenthesis 8 divided by 3 right parenthesis space πr cubed space cross times space 100 over denominator left square bracket 64 divided by left parenthesis 3 square root of 3 right parenthesis straight r cubed right square bracket end fraction % space equals space 68 %$

(c) Face centred cubic: A face centred cubic cell (fcc) contains four spheres (or atoms) per unit volume occupied by one sphere of radius
r = 4/3 $straight pi$ r³Volume occupied by four spheres present in the unit cell
r = 4/3 $straight pi$ r³ x 4 = 16/3 $straight pi$ r³Fig. Face centred cubic unit cell. Figure indicates that spheres placed at the corners touches a face centred sphere. Length of the face diagonal
= r + 2r + r = 4r
(i) Simple Cubic: A simple cubic unit cell has one sphere (or atom)
Hence, for face centred cubic, efficiency of packing = 74%.