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Q14P

Expert-verifiedFound in: Page 1273

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Assume that the total volume of a metal sample is the sum of the volume occupied by the metal ions making up the lattice and the (separate) volume occupied by the conduction electrons. The density and molar mass of sodium (a metal) are ${\mathbf{971}}{\mathbf{kg}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{3}}}$ and ${\mathbf{23}}{\mathbf{.}}{\mathbf{g}}{\mathbf{/}}{\mathbf{mol}}$, respectively; assume the radius of the Na+ ion is . (a) What percent of the volume of a sample of metallic sodium is occupied by its conduction electrons? (b) Carry out the same calculation for copper, which has density, molar mass, and ionic radius of 8960${\mathbf{kg}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{3}}}$, 63.5g/mol, and 135 pm, respectively. (c) For which of these metals do you think the conduction electrons behave more like a free-electron gas?**

- The percentage of the volume, the conduction electrons of a sample of metallic sodium occupy is 90 %.
- The percentage of the volume, the conduction electrons of a sample of metallic copper occupy is 12.4 %.

The metal whose conduction electrons behave more like a free-electron gas is sodium

- Density of sodium, ${\mathrm{d}}_{\mathrm{sodium}}=971\mathrm{lg}/{\mathrm{m}}^{3}$
- Molar mass of sodium, ${\mathrm{A}}_{\mathrm{sodium}}=23.0\mathrm{g}/\mathrm{mol}$
- Radius of sodium ion, ${\mathrm{R}}_{\mathrm{sodium}}=98\mathrm{pm}$
- Density, molar mass and ionic radius of copper are:

${\mathrm{d}}_{\mathrm{copper}}=8960\mathrm{kg}/{\mathrm{m}}^{3},{\mathrm{A}}_{\mathrm{copper}}=63.5\mathrm{g}/\mathrm{mol},{\mathrm{R}}_{\mathrm{copper}}=135\mathrm{pm}$

**Using the concept of molar mass, we can get the volume per cubic meter occupied by the conduction electrons in a metal. Now, using this value in the fractional formula, we can get the required ratio or percentage of the conduction electrons.**

Formulae:

The volume per cubic meter of a substance,

$\mathrm{V}\left(\mathrm{volume}/{\mathrm{cm}}^{3}\right)=\frac{{\mathrm{dN}}_{\mathrm{A}}}{\mathrm{A}}\left(\frac{4}{3}{\mathrm{\pi R}}^{3}\right),\mathrm{where}{\mathrm{N}}_{\mathrm{A}}=6.022\times {10}^{23}/\mathrm{mol}$ (i)

The fraction of volume available for conduction electrons, $\mathrm{Fraction}=1-\frac{{\mathrm{V}}_{\mathrm{substance}}}{1.00{\mathrm{m}}^{3}}$ (ii)

Using the given data in equation (i), the volume per cubic meter of sodium occupied by the sodium ions can be given as follows:

$\mathrm{V}\left(\mathrm{volume}/{\mathrm{cm}}^{3}\right)=\frac{\left(971\mathrm{kg}/{\mathrm{m}}^{3}\right)\left(6.022\times {10}^{23}/\mathrm{mol}\right)}{\left(23.0\times {10}^{-3}\mathrm{kg}/\mathrm{mol}\right)}\left(\frac{4}{3}\mathrm{\pi}{\left(98.0\times {10}^{-12}\mathrm{m}\right)}^{3}\right)\phantom{\rule{0ex}{0ex}}=0.100{\mathrm{m}}^{3}$

Now, the fraction of volume available for conduction electrons can be given using the above value in equation (ii) as follows:

$\mathrm{Fraction}=1-\frac{0.100{\mathrm{m}}^{3}}{1.00{\mathrm{m}}^{3}}\phantom{\rule{0ex}{0ex}}=0.900\mathrm{or}90.0\%$

Hence, the value of the occupancy percentage is 90.0 % .

Using the given data in equation (i), the volume per cubic meter of sodium occupied by the sodium ions can be given as follows:

$\mathrm{V}\left(\mathrm{volume}/{\mathrm{cm}}^{3}\right)=\frac{\left(8960\mathrm{kg}/{\mathrm{m}}^{3}\right)\left(6.022\times {10}^{23}/\mathrm{mol}\right)}{\left(63.5\times {10}^{-3}\mathrm{kg}/\mathrm{mol}\right)}\left(\frac{4}{3}\mathrm{\pi}{\left(135.0\times {10}^{-12}\mathrm{m}\right)}^{3}\right)\phantom{\rule{0ex}{0ex}}=0.1876{\mathrm{m}}^{3}$

Now, the fraction of volume available for conduction electrons can be given using the above value in equation (ii) as follows:

$\mathrm{Fraction}=1-\frac{0.1876{\mathrm{m}}^{3}}{1.00{\mathrm{m}}^{3}}\phantom{\rule{0ex}{0ex}}=0.124\mathrm{or}12.4\%$

Hence, the value of the occupancy percentage is 12.4 % .

From the above calculations, the metal whose conduction electrons are more like free-electron gas is sodium because the electrons occupy a greater portion of the space available.

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