Question

Solid vanadium crystallizes in a body-centered cubic structure and has a density of 6.00 g/cm3. First determine the number of atoms of vanadium in a unit cell. Then, calculate the atomic radius of vanadium.

Answer 1

The given question is incomplete. The complete question is as follows.

Solid vanadium crystallizes in a body-centered cubic structure and has a density of 6.00 g/cm3. Assuming the vanadium atomic radius is 132 pm, is the vanadium unit cell primitive cubic, body centered cubic, or face centered cubic.

Explanation:

The given data is as follows.

            Density = 6.00 $g/cm^{3}$

              radius = 132 pm

Relation between edge length and volume is as follows.

             a = ∛V

                = ∛No. of atoms

                = $\sqrt[3]{\frac{mass}{density \times N_{A}}}$

                = $\sqrt[3]{\frac{50.941 g/mol}{6.0 g/cm^{3} \times 6.022 \times 10^{23}}}$

                = 2.4 ∛No. of atoms

So, there will be three possibilities which are as follows.

• SC, 1 atom and r = $\frac{a}{2}$

Here, a = 2.4, and r = 1.2 $A^{o}$ which is not right.

• For BCC, there are two atoms and r = $\frac{\sqrt{3}a}{4}$

So,         a = $1.26 \times 2.4$

                = 3.02 and, r = 1.31 $A^{o}$ which is a good fit to the measured radius.

• For FCC, there are 4 atoms and r = $\frac{\sqrt{2}a}{4}$

So,       a = $1.59 \times 2.4$

              = 3.81 and r = 1.35 $A^{o}$ which will not fit to the measured radius as well as BCC.