Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \AA=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm} \mathrm{Pt}\) sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one \(\mathrm{Pt}\) atom can be estimated from its atomic diameter of \(2.8 \AA\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

First, we will find the volume of the 2 nm sphere using the formula for the volume of a sphere, which is \((4/3)\pi r^3\). To do this, we need to convert the diameter to radius and the units to meters. Radius of the sphere (in meters): \(r = 1.0 \times 10^{-9}\) The sphere's volume: \(V_\text{sphere} = (4/3) \pi r^3 = (4/3) \pi (1.0 \times 10^{-9})^3\) Next, we will find the volume of one platinum atom. As platinum has face-centered cubic (fcc) arrangement, we start by calculating the volume of the unit cell, since each unit cell contains an equivalent of 4 atoms in an fcc lattice. Conversion of edge length to meters: Edge length of Pt unit cell: \(a = 3.924 \times 10^{-10} \mathrm{~m}\) The volume of the unit cell: \(V_\text{unit cell} = a^3 = (3.924 \times 10^{-10})^3\) As each unit cell contains 4 Pt atoms, the volume of a single Pt atom: \(V_\text{Pt atom} = \frac{V_\text{unit cell}}{4}\) Now, we can estimate the number of Pt atoms in the sphere by dividing the sphere's volume by the volume of one Pt atom: Total number of Pt atoms: \(N_\text{total} = \frac{V_\text{sphere}}{V_\text{Pt atom}}\)

We will be using the sphere's surface area formula, which is \(4\pi r^2\), and an estimation of the "footprint" of one Pt atom (atomic diameter of 2.8 Å). First, determine the total surface area of the sphere: Surface area of the sphere: \(A_\text{sphere} = 4\pi r^2 = 4\pi(1.0\times 10^{-9})^2\) Next, find the area of one Pt atom's "footprint": Pt atom's diameter (in meters): \(d = 2.8 \times 10^{-10}\) Area of one Pt atom: \(A_\text{Pt atom} = \pi (\frac{d}{2})^2 = \pi (1.4 \times 10^{-10})^2\) We can estimate the number of Pt atoms on the surface by dividing the surface area of the sphere by the area of one Pt atom: Number of Pt atoms on the surface: \(N_\text{surface} = \frac{A_\text{sphere}}{A_\text{Pt atom}}\)

To find the percentage of the Pt atoms on the surface of the nanoparticle, divide the number of surface Pt atoms by the total Pt atoms and multiply by 100: Percentage of surface Pt atoms: \(\frac{N_\text{surface}}{N_\text{total}} \times 100\%\) Calculate the values for each step mentioned above to find the percentage of Pt atoms on the surface.

Similar to the calculations made in part (a)–(c) above, determine the following values for the 5 nm nanoparticle: 1. Total number of Pt atoms in a 5 nm sphere. 2. Number of Pt atoms on the surface of a 5 nm Pt sphere. 3. The percentage of Pt atoms on the surface of a 5 nm nanoparticle.

Based on the results obtained in part (c) and (d), compare the percentage of surface Pt atoms to surface Pt atoms in the 2 nm and 5 nm nanoparticles. The nanoparticles with a higher percentage of surface atoms will interact more with the surroundings (higher surface area to volume ratio) and are expected to have greater catalytic activity.