
Silver has an atomic radius of $144pm$ and the density of silver is $10.6gc{m^{ - 3}}$ .To which type of cubic crystal does silver belong?
(A) HCP
(B) BCC
(C) FCC
(D) SC
Answer
550.5k+ views
Hint:Before solving the numerical it is important to understand the concept of cubic crystal. Unit cells occur in many different varieties. As one example, the cubic crystal system is composed of three different types of unit cells: (1) simple cubic , (2) face-centered cubic , and (3)body-centered cubic .
Complete step-by-step solution:For calculating the type of packing the following formula is used:
$\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$
Here, $\rho $ = density; and the given value is:
${N_A}$ = Avogadro’s number; and its value is:
$M$ = Atomic mass; and the given value is: $M = 107.87g/mol$
$a$ = edge length of unit cell; and the given value we need to evaluate out with the help of $r$ which is the radius of silver; and here the given value is: $r = 144pm$
= number of atom in unit cell
Case 1: When silver has simple cubic unit cell:
In the simple cubic structure there is only one lattice point at each corner of the cube-shaped unit cell. They mark the position of either a single atom, or the same group of atoms, known as the motif, which is repeated across the lattice
We have: $Z = 1$
$a = 2r $
$\Rightarrow a = 2 \times 144pm$
$\Rightarrow a = 288pm $
$\Rightarrow a = 288 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {2.88 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} \ne 7.49g/c{m^3}$
Case 2: when silver has simple face centered cubic unit cell:
We have: $Z = 4$
$a = 2.828r $
$\Rightarrow a = 2.828 \times 144pm $
$\Rightarrow a = 4.07 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {4.07 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} = 10.6g/c{m^3}$
Case 3: when silver has simple body centered cubic unit cell:
Body-centered cubic (BCC) is the name given to a type of atom arrangement found in nature. A body-centered cubic unit cell structure consists of atoms arranged in a cube where each corner of the cube shares an atom and with one atom positioned at the center. The center of the unit cell consists of 1 full atom, therefore the total number of atoms in the BCC unit cell structure is 2; one at the center plus eight one-eighth atoms at the corners.
We have: $Z = 2$
$a = 2.309r $
$\Rightarrow a = 2.309 \times 144pm $
$\Rightarrow a = 3.32 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {3.32 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} \ne 9.78g/c{m^3}$
Hence silver belongs to a face centered cubic unit cell.
Hence the correct option is (C).
Note: Unit cell is defined as the smallest repeating unit having the full symmetry of the crystal structure. A simple cubic unit cell is the simplest repeating unit in a simple cubic structure.The relationship between atomic radius and edge length in a unit cell depends on the type of the cell.
Complete step-by-step solution:For calculating the type of packing the following formula is used:
$\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}$
Here, $\rho $ = density; and the given value is:
${N_A}$ = Avogadro’s number; and its value is:
$M$ = Atomic mass; and the given value is: $M = 107.87g/mol$
$a$ = edge length of unit cell; and the given value we need to evaluate out with the help of $r$ which is the radius of silver; and here the given value is: $r = 144pm$
= number of atom in unit cell
Case 1: When silver has simple cubic unit cell:
In the simple cubic structure there is only one lattice point at each corner of the cube-shaped unit cell. They mark the position of either a single atom, or the same group of atoms, known as the motif, which is repeated across the lattice
We have: $Z = 1$
$a = 2r $
$\Rightarrow a = 2 \times 144pm$
$\Rightarrow a = 288pm $
$\Rightarrow a = 288 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {2.88 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} \ne 7.49g/c{m^3}$
Case 2: when silver has simple face centered cubic unit cell:
We have: $Z = 4$
$a = 2.828r $
$\Rightarrow a = 2.828 \times 144pm $
$\Rightarrow a = 4.07 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {4.07 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} = 10.6g/c{m^3}$
Case 3: when silver has simple body centered cubic unit cell:
Body-centered cubic (BCC) is the name given to a type of atom arrangement found in nature. A body-centered cubic unit cell structure consists of atoms arranged in a cube where each corner of the cube shares an atom and with one atom positioned at the center. The center of the unit cell consists of 1 full atom, therefore the total number of atoms in the BCC unit cell structure is 2; one at the center plus eight one-eighth atoms at the corners.
We have: $Z = 2$
$a = 2.309r $
$\Rightarrow a = 2.309 \times 144pm $
$\Rightarrow a = 3.32 \times {10^{ - 8}}cm$
$10.6g/c{m^3} = \dfrac{{1 \times 107.87g/mol}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}} \times {{\left( {3.32 \times {{10}^{ - 8}}cm} \right)}^3}}}$
$10.6g/c{m^3} \ne 9.78g/c{m^3}$
Hence silver belongs to a face centered cubic unit cell.
Hence the correct option is (C).
Note: Unit cell is defined as the smallest repeating unit having the full symmetry of the crystal structure. A simple cubic unit cell is the simplest repeating unit in a simple cubic structure.The relationship between atomic radius and edge length in a unit cell depends on the type of the cell.
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