Respuesta :
Answer:
Ionic radius of the bromide ion is 227.9 pm.
Explanation:
Number of atom in FCC unit cell = Z = 4
Density of ammonium bromide = [tex]2.429 g/cm^3[/tex]
Edge length of cubic unit cell= a= ?
Molar mass of ammonium bromide = 98 g/mol
Formula used :
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho [/tex] = density
Z = number of atom in unit cell
M = atomic mass
[tex](N_{A})[/tex] = Avogadro's number
a = edge length of unit cell
On substituting all the given values , we will get the value of 'a'
[tex]2.429 g/cm^3=\frac{4\times 98 g/mol}{6.022\times 10^{23} mol^{-1}\times a^3}[/tex]
[tex]a^3=\frac{4\times 98 g/mol}{6.022\times 10^{23} mol^{-1}\times 2.429 g/cm^3}[/tex]
[tex]a=6.447\times 10^{-8} cm=6.447\times 10^{-8} cm\times 10^{10} pm=644.7 pm[/tex]
Ionic radius of bromide ion = r
To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:
[tex]644.7 pm=2\sqrt{2}r[/tex]
Putting values in above equation, we get:
[tex]r=\frac{644.7 pm}{2\sqrt{2}}=227.9 pm[/tex]
Ionic radius of the bromide ion is 227.9 pm.
