
Calcium crystallizes in a cubic unit cell with density 3.2g/cc. Edge-length of the unit cell is 437pm.The number of nearest neighbors of a Ca atom is
A) 4
B) 6
C) 8
D) 12
Answer
554.4k+ views
Hint: The equation relating density, edge length and number of atoms in a unit cell is as follows,
$d=\dfrac{M\times Z}{{{N}_{A}}\times {{a}^{3}}}$
The atomic weight of calcium is 40g
Complete step by step solution:
So in the question the density of crystal with cubic unit cell with Ca is given and the edge length of the unit cell is also given and from this we have to find the number of atoms present in the unit cell.
From the number of atoms present in the unit cell we could say about the type of unit cell and finally we could say about the coordination number i.e. the number of nearest atoms with respect to Ca can be found out.
Let’s move to the mathematical solution part of this problem,
The density equation for the crystal is,
$d=\dfrac{M\times Z}{{{N}_{A}}\times {{a}^{3}}}$
Where d = density of the crystal
Z is the number of atoms in unit cell
M is the atomic weight of the given atom
${{N}_{A}}$=Avogadro number =$6.022\times {{10}^{23}}$
a =edge length
And we have to find the number of atoms in the unit cell.
So we alter the equation of density as,
$z=d\times {{N}_{A}}\times \dfrac{{{a}^{3}}}{M}$
Density =3.2g/cc
Edge-length= 437pm =$437\times {{10}^{-10}}$cm
Mass of Ca is 40
\[z=\dfrac{\left( 3.2 \right)\times 6.023\times {{10}^{23}}\times {{\left( 437\times {{10}^{-10}} \right)}^{3}}}{40}\]
Z=4.021$\approx $4
As z=4, then it comes under the FCC type of unit cell ie face centered cubic.
In FCC unit cells the neighboring atoms with respect to the central atom, here it is Ca is 12.The coordination number is 12 so the correct answer for this question is also 12.There are 12 neighboring atoms.
The correct option for this question is option (D).
Note: For BCC-body centered cubic units, the net amount contribution of atoms is 2 and the coordination number is 8.
The HCP-hexagonal closest packed unit cell contains 4 atoms and the coordination number is 12 for this unit cell.
So for different unit cells the value of Z and coordination number changes and the density of the crystal changes with respect to the unit cell present in the entire crystal.
$d=\dfrac{M\times Z}{{{N}_{A}}\times {{a}^{3}}}$
The atomic weight of calcium is 40g
Complete step by step solution:
So in the question the density of crystal with cubic unit cell with Ca is given and the edge length of the unit cell is also given and from this we have to find the number of atoms present in the unit cell.
From the number of atoms present in the unit cell we could say about the type of unit cell and finally we could say about the coordination number i.e. the number of nearest atoms with respect to Ca can be found out.
Let’s move to the mathematical solution part of this problem,
The density equation for the crystal is,
$d=\dfrac{M\times Z}{{{N}_{A}}\times {{a}^{3}}}$
Where d = density of the crystal
Z is the number of atoms in unit cell
M is the atomic weight of the given atom
${{N}_{A}}$=Avogadro number =$6.022\times {{10}^{23}}$
a =edge length
And we have to find the number of atoms in the unit cell.
So we alter the equation of density as,
$z=d\times {{N}_{A}}\times \dfrac{{{a}^{3}}}{M}$
Density =3.2g/cc
Edge-length= 437pm =$437\times {{10}^{-10}}$cm
Mass of Ca is 40
\[z=\dfrac{\left( 3.2 \right)\times 6.023\times {{10}^{23}}\times {{\left( 437\times {{10}^{-10}} \right)}^{3}}}{40}\]
Z=4.021$\approx $4
As z=4, then it comes under the FCC type of unit cell ie face centered cubic.
In FCC unit cells the neighboring atoms with respect to the central atom, here it is Ca is 12.The coordination number is 12 so the correct answer for this question is also 12.There are 12 neighboring atoms.
The correct option for this question is option (D).
Note: For BCC-body centered cubic units, the net amount contribution of atoms is 2 and the coordination number is 8.
The HCP-hexagonal closest packed unit cell contains 4 atoms and the coordination number is 12 for this unit cell.
So for different unit cells the value of Z and coordination number changes and the density of the crystal changes with respect to the unit cell present in the entire crystal.
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