Question

A certain metal fluoride crystallizes in such a way that the fluoride ions occupy simple cubic lattice sites, while the metal ions occupy the body centers of half the unit cells. The formula for the metal fluoride is

Answer 1

Answer:

$MF_2$

Explanation:

In a simple cubic lattice lattice, the atoms are present at the eight corners of the cibe.

Since it is mentioned that the fluorine is present at the corners and also 1 corners are shared by 8 unit cells. So, share of atom in one unit cell is:- $8\times \frac{1}{8}=1$

Also, the metal, M occupy half of the body centre. A cube has only one body centre. So, share of M in each unit cell:- $\frac{1}{2}$

Thus, the formula is:-

$M_{\frac{1}{2}}F$ Or simplifying $MF_2$

Answer 2

Answer:

• The formula of the metal fluoride is = $MF_2$

Explanation:

given that fluoride ions occupy simple cubic lattice and metal ions occupy the body center of half the cube

therefore,

The number of $F^-$ ions per unit cell = $\frac{1}{8}*1$

$= \frac{1}{8}$

There is only one body center per unit cell

The total number of metal ions per unit cell = $\frac{1}{2}*1$

$= \frac{1}{2}$

Therefore the formula for metal fluoride

$= MF_2$

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