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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 :

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 [tex]g/cm^{3}[/tex]

              radius = 132 pm

Relation between edge length and volume is as follows.

             a = ∛V

                = ∛No. of atoms

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

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

                = 2.4 ∛No. of atoms

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

  • SC, 1 atom and r = [tex]\frac{a}{2}[/tex]

Here, a = 2.4, and r = 1.2 [tex]A^{o}[/tex] which is not right.

  • For BCC, there are two atoms and r = [tex]\frac{\sqrt{3}a}{4}[/tex]

So,         a = [tex]1.26 \times 2.4[/tex]

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

  • For FCC, there are 4 atoms and r = [tex]\frac{\sqrt{2}a}{4}[/tex]

So,       a = [tex]1.59 \times 2.4[/tex]

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

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