Question

What is the distance between $N{{a}^{+}}$ and $C{{l}^{-}}$ in a $NaCl$ crystal if its density is 2.165 $g/c{{m}^{3}}$? $NaCl$ crystallizes in the fcc lattice?

Answer 1

The distance between the atoms in the solid structure can be calculated by the formula $\rho =\dfrac{Z\text{ x M}}{{{N}_{a}}\text{ x }{{\text{a}}^{3}}}$, where Z is the number of atoms in the unit cell, M is the molecular mass of the compound, a is the edge length of the unit cell, ${{N}_{a}}$ is the Avogadro’s number, and $\rho$ is the density of the substance.

Complete step by step answer:
- The distance between the atoms in the solid structure can be calculated by the formula $\rho =\dfrac{Z\text{ x M}}{{{N}_{a}}\text{ x }{{\text{a}}^{3}}}$, where Z is the number of atoms in the unit cell, M is the molecular mass of the compound, a is the edge length of the unit cell, ${{N}_{a}}$ is the Avogadro’s number, and $\rho$ is the density of the substance.
- The value of the Avogadro’s number is $6.023\text{ x 1}{{\text{0}}^{23}}$.
Z is the number of atoms in the unit cell and the $NaCl$ crystal has an fcc structure, and in the fcc structure, the number of atoms is 4, so Z will be 4.
The molecular mass of $NaCl$ is = 23 + 35.5 = 58.5
The given density is 2.156 $g/c{{m}^{3}}$.
- Now, putting all these values in the formula, we get:
$2.156 = \dfrac{\text{4 x 58}\text{.5}}{\text{6}\text{.023 x 1}{{\text{0}}^{23}}\text{ x }{{\text{a}}^{3}}}$
${{\text{a}}^{3}} = \dfrac{\text{4 x 58}\text{.5}}{\text{6}\text{.023 x 1}{{\text{0}}^{23}}\text{ x }2.156}$
On solving this, we get:
$a = 5.64\text{ x 1}{{\text{0}}^{-8}}$
We can also write this as:
$a = 5.64\overset{\circ }{\mathop{\text{A}}}\,$
This is the edge length of the sodium chloride crystal, to find the distance between the ions we have to divide it by 2:
$a=\dfrac{5.64\overset{\circ }{\mathop{\text{A}}}\,}{2} = 2.82\overset{\circ }{\mathop{\text{A}}}\,$
So, the distance will be $2.82\overset{\circ }{\mathop{\text{A}}}\,$.

Note: If the structure of the unit cell is simple then the number of atoms is 1, if the structure of the unit cell is a body-centered-cubic cell then the number of atoms will be 2, and if the structure of the unit cell is a face-centered-cubic cell then the number of atoms will be 4.