Question

NH4Br has a face-centered cubic unit cell in which the Br- anions occupy corners and face centers, while the cations fit into the hole between adjacent anions. What is the radius of Br- if the ionic radius of NH4 is 138.3 pm and the density of NH4Br is 2.429 g/cm3

Answer 1

The radius of Br- in NH4Br is 69.15 pm.

Step-by-step explanation:

In NH4Br, the Br- anions occupy the corners and face centers of a face-centered cubic (FCC) unit cell, while the NH4+ cations fit into the octahedral holes between adjacent anions. The edge length of the unit cell is twice the radius of the anions. Therefore, the radius of Br- can be calculated by dividing the edge length of the unit cell by 2.

Given that the ionic radius of NH4+ is 138.3 pm, we know that the edge length of the unit cell is twice this value.

So, the radius of Br- is 138.3 pm ÷ 2 = 69.15 pm.

Answer 2

Ionic radius of the bromide ion is 227.9 pm.

Step-by-step explanation:

Number of atom in FCC unit cell = Z = 4

Density of ammonium bromide =
`2.429 g/cm^3`

Edge length of cubic unit cell= a= ?

Molar mass of ammonium bromide = 98 g/mol

Formula used :

`\rho=(Z* M)/(N_(A)* a^(3))`

where,

`\rho` = density

Z = number of atom in unit cell

M = atomic mass

`(N_(A))` = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'

`2.429 g/cm^3=(4* 98 g/mol)/(6.022* 10^(23) mol^(-1)* a^3)`

`a^3=(4* 98 g/mol)/(6.022* 10^(23) mol^(-1)* 2.429 g/cm^3)`

`a=6.447* 10^(-8) cm=6.447* 10^(-8) cm* 10^(10) pm=644.7 pm`

Ionic radius of bromide ion = r

To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:

`644.7 pm=2√(2)r`

Putting values in above equation, we get:

`r=(644.7 pm)/(2√(2))=227.9 pm`

Ionic radius of the bromide ion is 227.9 pm.