Supporting Information

Transcription

1 Supporting Information Heteroepitaxial Growth of High-Index-Faceted Palladium Nanoshells and Their Catalytic Performance Feng Wang, Chuanhao Li, Ling-Dong Sun,* Haoshuai Wu, Tian Ming, Jianfang Wang,* Jimmy C. Yu, and Chun-Hua Yan* Department of Physics, The Chinese University of Hong Kong, Shatin, Hong Kong SAR, China, State Key Lab of Rare Earth Materials Chemistry and Applications, Peking University, Beijing , China, and Department of Chemistry, The Chinese University of Hong Kong, Shatin, Hong Kong SAR, China (J.F.W.); (L.-D.S.); (C.-H.Y.) (I) Figures Figure S1. (a and b) Large-area SEM images of the THH Au nanocrystals at different orientations. S1

2 Figure S. SEM images of the products obtained from the growth on the elongated THH Au nanocrystals with varying volumes of the HPdCl4 solution: (a) 5 L; (b) 10 L; (c) 15 L; and (d) 0 L. Figure S3. TEM images of the products obtained from the growth on the elongated THH Au nanocrystals with varying volumes of the HPdCl4 solution: (a) 10 L; (b) 15 L; and (c) 0 L. S

3 Figure S4. (a) SEM image of the octahedral Au nanocrystals. (b) TEM image of the octahedral Au nanocrystals. Figure S5. SEM images of the products obtained from the growth on the TOH Au nanocrystals with varying volumes of the HPdCl4 solution: (a) 5 L; (b) 10 L; (c) 15 L; and (d) 0 L. S3

4 Figure S6. TEM images of the products obtained from the growth on the TOH Au nanocrystals with varying volumes of the H PdCl 4 solution: (a) 10 L; (b) 15 L; and (c) 0 L. Figure S7. (a) SEM image of the cubic Pd nanocrystals. (b) TEM image of the cubic Pd nanocrystals. Figure S8. (a) SEM image of the cubic Au@Pd nanocrystals. (b) TEM image of the cubic Au@Pd nanocrystals. S4

5 (II) Calculations of the total number of the Pd atoms on the surfaces and the total number of the surface Pd atoms at steps for the different metal nanocrystals (Tables S1 and S) (1) Pd nanocubes (1.1) Number of the Pd atoms per nanocube Scheme S1-1. Schematic showing a Pd nanocube. The Pd nanocube sample has an average edge length of 33 3 nm. Its volume is (33 nm) 3 = nm 3 Pd has the face-centered-cubic structure with a lattice constant of nm. The volume of a unit cell is (0.389 nm) 3 = nm 3 Each unit cell contains 4 Pd atoms. The number of Pd atoms in a single Pd nanocube is ( nm 3 ) / ( nm 3 ) 4 = (1.) Number of the surface Pd atoms on a single Pd nanocube Scheme S1-. Schematic showing the (100) facet of a Pd nanocube. A Pd nanocube is enclosed by 6 square facets. Its total surface area is S5

6 (33 nm) 6 = nm Each two-dimensional unit cell on the (100) facet contains two Pd atoms. Its area is (0.389 nm) = nm The number of the surface Pd atoms on a single Pd nanocube is ( nm ) / (0.151 nm ) = (1.3) Total number of the surface Pd atoms on the Pd nanocubes in the catalytic reaction solution The Pd nanocubes in the catalytic reaction solution were produced from ml of H PdCl 4 at 0.01 M. Because 65% of H PdCl 4 was consumed for the nanocube growth, the molar number of the consumed Pd atoms is mol. Therefore the total number of the consumed Pd atoms is ( mol) ( mol 1 ) = The number of the Pd nanocubes in the catalytic reaction solution is ( ) / ( ) = The total number of the surface Pd atoms on the Pd nanocubes in the catalytic reaction solution is ( ) ( ) = () Au@Pd nanocubes (.1) Number of the Pd atoms in a single Au@Pd nanocube Assume that the edge length of the Au octahedron is a. The distance between one apex to the opposite apex is therefore a a a. The volume of the octahedron is The average distance from one apex to the opposite apex of the Au octahedron sample is measured to be 17 1 nm. The volume of each Au octahedron is then (17 nm) 3 / 6 = 819 nm 3 S6

7 The average edge length of the nanocube is 47 nm. Its volume is (47 nm) 3 = nm 3 The volume of the Pd shell is then calculated to be ( nm 3 ) (819 nm 3 ) = nm 3 The number of the Pd atoms in the shell of a single Au@Pd nanocube is ( nm 3 ) / ( nm 3 ) 4 = Scheme S-1. Schematic showing (a) a Au octahedron and (b) a Au@Pd nanocube. (.) Number of the surface Pd atoms on a single Au@Pd nanocube The total surface area of a Au@Pd nanocube is (47 nm) 6 = nm The number of the surface Pd atoms on a single Au@Pd nanocube is ( nm ) / (0.151 nm ) = (.3) Total number of the surface Pd atoms on the Au@Pd nanocubes in the catalytic reaction solution The Pd shells of the Au@Pd nanocubes in the catalytic reaction solution were grown with ml of H PdCl 4 at 0.01 M. Because 73% of H PdCl 4 was consumed for the shell growth, the molar number of the consumed Pd atoms is mol. Therefore the total number of the consumed Pd atoms is ( mol) ( mol 1 ) = The number of the Au@Pd nanocubes in the catalytic reaction solution is then S7

8 ( ) / ( ) = The total number of the surface Pd atoms on the Au@Pd nanocubes in the catalytic reaction solution is ( ) ( ) = (3) THH Au nanocrystals (3.1) Volume of a single THH Au nanocrystal Scheme S3-1. Schematic showing (a) a THH Au nanocrystal and (b) its projection. The nanocrystal is viewed along the [00 1] axis. The THH Au nanocrystal sample has an average thickness of 71 8 nm and length of 15 1 nm. Since the THH nanocrystal is enclosed with {730} facets, the length ratio between the two perpendicular sides of the right triangle indicated in red in Scheme S3-1 is 7:3. Suppose that one side is S8

9 7x. The other side is then 3x. The third side is calculated to be 58 x. The value of x can be calculated from the nanocrystal thickness to be x = (71 nm) / ( ) = 3.55 nm Other related lengths are also calculated and shown along with the projected model in Scheme S3-1. A THH nanocrystal can be decomposed into a cuboid, two square pyramids, and four elongated pyramids. The elongated pyramid can further be decomposed into a square pyramid and a triangular prism, as shown in Scheme S3-. The square pyramid has a base side length of 14x. Its height is 3x. Therefore, the volume of the square pyramid is x 3x = 196x 3 = 196 (3.55 nm) 3 = nm 3 For the triangular prism, the base is an isosceles. Its base length and height are 14x and 3x, respectively. The height of the prism is 54 nm (Scheme S3-1). The volume of the triangular prism is then 1 (14x)(3x)(54 nm) = 1x (54 nm) = 1 (3.55 nm) (54 nm) = nm 3 The cuboid has a square base with a side length of 14x. Its height is nm (Scheme S3-1). Therefore, the volume of the cuboid is (14x) (103.7 nm) = 196x (103.7 nm) = 196 (3.55 nm) (103.7 nm) = nm 3 By combining together all of the above components, the volume of a THH Au nanocrystal is calculated to be ( nm 3 ) + ( nm 3 ) + 4[( nm 3 ) + ( nm 3 )] = nm 3 S9

10 Scheme S3-. Schematic showing (a) the decomposition of a THH Au nanocrystal into a cuboid, two square pyramids, and four elongated pyramids, and (b) the decomposition of an elongated pyramid into a square pyramid and a triangular prism. (4) THH Au@Pd nanocrystals (4.1) Volume of a single THH Au@Pd nanocrystal The THH Au@Pd nanocrystal sample has an average thickness of 77 7 nm and length of nm. The volume of a THH Au@Pd nanocrystal is calculated in a similar way to that of a THH Au nanocrystal. For a THH Au@Pd nanocrystal, x = (77 nm) / ( ) = 3.85 nm The total volume of a THH Au@Pd nanocrystal is nm 3 (4.) Number of the Pd atoms in a THH Au@Pd nanocrystal The volume of the Pd shell in a THH Au@Pd nanocrystal is ( nm 3 ) ( nm 3 ) = nm 3 The number of the Pd atoms in the shell in a THH Au@Pd nanocrystal is calculated according to the unit cell volume to be ( nm 3 ) / ( nm 3 ) 4 = (4.3) Numbers of the surface Pd atoms and the surface Pd atoms at steps on a single THH Au@Pd nanocrystal S10

11 Scheme S4-1. Schematic showing the high-index (730) facet. (a) Stereoview. (b) Two-dimensional projection along the [001] axis. A THH nanocrystal is enclosed by 4 {730} facets. Among them, 8 are trapezoids and 16 are triangles. The height of both the trapezoid and triangle is composed of periodic segments from bottom to top. Scheme S4-1 shows one of these segments. The number of the surface atoms and that of the surface atoms at steps can be determined from the perspective of the surface area. Scheme S4-1b shows that the total number of the surface atoms on the high-index {730} facets is equal to that of the surface atoms on the {010} planes that are projected from the {730} facets. In addition, the number ratio of the surface atoms at steps and all of the surface atoms on the {730} facets is found from Scheme S4-1b to be Consider the cuboid enclosed with the {010} planes for a THH Au@Pd nanocrystal. Its length is (13 nm) 6 (3.85 nm) = nm S11

12 The base of the cuboid is a square. The side length of the base is (77 nm) 6 (3.85 nm) = nm The total surface area of the cuboid is (53.90 nm) + 4 ( nm) (53.90 nm) = nm The volume of a two-dimensional unit cell on the {010} planes is nm. The total number of the surface Pd atoms on the {730} facets of a single THH Au@Pd nanocrystal is then ( nm ) / (0.151 nm ) = The total number of the surface Pd atoms at steps on the {730} facets of a single THH Au@Pd nanocrystal is = (4.4) Total numbers of the surface Pd atoms and the surface Pd atoms at steps on the THH Au@Pd nanocrystals in the catalytic reaction solution The Pd shells of the THH Au@Pd nanocrystals in the catalytic reaction solution were grown with ml of H PdCl 4 at 0.01 M. Because 73% of H PdCl 4 was consumed for the shell growth, the molar number of the consumed Pd atoms is mol. Therefore the total number of the consumed Pd atoms is ( mol) ( mol 1 ) = Because each THH Au@Pd nanocrystal contains Pd atoms, the number of the THH Au@Pd nanocrystals in the catalytic reaction solution is then ( ) / ( ) = The total number of the surface Pd atoms on the THH Au@Pd nanocrystals in the catalytic reaction solution is ( ) ( ) = S1

13 The total number of the surface Pd atoms at steps on the THH nanocrystals in the catalytic reaction solution is ( ) ( ) = (5) TOH Au nanocrystals (5.1) Volume of a single TOH Au nanocrystal A TOH nanocrystal can be created by pulling out the centers of the eight triangular faces of an octahedron. Therefore a TOH nanocrystal can be decomposed into one octahedron and eight triangular pyramids. Scheme S5-1. Schematic showing the decomposition of a TOH Au nanocrystal into an octahedron and eight triangular pyramids. The base side length of each triangular pyramid is assumed to be y (Scheme S5-a). Each triangular pyramid has three side faces. The length of the two equal sides is assumed to be x (Scheme S5-b). We need to find out x because y can be measured. The blue line in Scheme S5-a is the projection of the (1) facet. Its length is then x y. In the right triangle with the blue line as the hypotenuse, the length ratio of the three sides is determined according to the projected atomic model of the (1) facet shown in Scheme S5-c to be 3 : : 1. Then, on the basis of the red right triangle, x can be determined by x y x y 3 3 x y y S13

14 The result is x y Scheme S5-. Schematic showing the models of a TOH nanocrystal. (a) Entire TOH nanocrystal projected along the [ 110] axis. (b) One side face of the triangular pyramid. (c) Two-dimensional projection of the high-index (1) facet along the [ 110] axis. The average distance from one apex to the opposite apex is measured to be 67 3 nm for the TOH Au nanocrystal sample. Therefore, y = 3.7 nm y = 67 nm. We obtain For the triangular pyramids, its base is an equilateral triangle with a side length of y = 47.4 nm. The height of the triangular pyramid is 6 3 y = 19.4 nm. The volume of the triangular pyramid is then 1 3 (47.4 nm) (19.4 nm) = nm The volume of the octahedron is S14

15 1 3 3 y y = y y = 3 (47.4 nm) (3.7 nm) = nm 3 The total volume of a single TOH Au nanocrystal is calculated to be ( nm 3 ) + 8( nm 3 ) = nm 3 (6) TOH Au@Pd nanocrystals (6.1) Volume of a single TOH Au@Pd nanocrystal The average distance from one apex to the opposite apex is 78 5 nm for the TOH Au@Pd nanocrystal sample. Therefore, y = 7.6 nm. The volume of a single TOH Au@Pd nanocrystal is y y y y y y y nm (6.) Number of the Pd atoms in a TOH Au@Pd nanocrystal The volume of the Pd shell in a TOH Au@Pd nanocrystal is ( nm 3 ) ( nm 3 ) = nm 3 The number of the Pd atoms in the shell in a TOH Au@Pd nanocrystal is calculated according to the unit cell volume to be ( nm 3 ) / ( nm 3 ) 4 = (6.3) Numbers of the surface Pd atoms and the surface Pd atoms at steps on a single TOH Au@Pd nanocrystal S15

16 Scheme S6-1. Schematic showing the high-index (1) facet. (a) Viewed along the [ 110] axis. (b) Viewed along [1] axis. A TOH nanocrystal is enclosed by 4 {1} triangular facets. The two sides of each triangular facets are equal to each other. Scheme S5-c shows a segment of the projection of one (1) facet along the [ 110] axis. The hypotenuse of the projection corresponds to one segment of the height of one triangular facet. There are 9 exposed atoms along the hypotenuse of the projection. The unit cell length of Pd is nm. The length of the hypotenuse is 9 (0.389 nm) =.476 nm The height of each triangular facet is x y = y = 7.6 nm Therefore, the total number of the exposed atoms along the height direction of each triangular facet is (7.6 nm) / (.476 nm) 9 = 100 The Pd atoms are close-packed along the base direction of each triangular facet. The distance between two neighboring atoms along the base is (0.389 nm) = nm The base length of each triangular facet is 55. nm. the number of the Pd atoms along the base is (55. nm) / (0.751 nm) = 01 The number of each triangular facet is then calculated to be = The total number of the surface Pd atoms on the {1} facets of a single TOH Au@Pd nanocrystal is then = S16

17 According to Scheme S5-c, one third of the surface Pd atoms are at steps on each {1} facet. Therefore, the total number of the surface Pd atoms at steps on the {1} facets of a single TOH Au@Pd nanocrystal is (6.4) Total numbers of the surface Pd atoms and the surface Pd atoms at steps on the TOH Au@Pd nanocrystals in the catalytic reaction solution The Pd shells of the TOH Au@Pd nanocrystals in the catalytic reaction solution were grown with ml of H PdCl 4 at 0.01 M. Because 73% of H PdCl 4 was consumed for the shell growth, the molar number of the consumed Pd atoms is mol. Therefore the total number of the consumed Pd atoms is ( mol) ( mol 1 ) = Because each TOH Au@Pd nanocrystal contains Pd atoms, the number of the TOH Au@Pd nanocrystals in the catalytic reaction solution is then ( ) / ( ) = The total number of the surface Pd atoms on the TOH Au@Pd nanocrystals in the catalytic reaction solution is ( ) ( ) = The total number of the surface Pd atoms at steps on the TOH Au@Pd nanocrystals in the catalytic reaction solution is ( ) ( ) = S17

18 (III) Tables Table S1. Numbers of the Surface Pd Atoms and All of the Pd Atoms per Nanocrystal statistics cubic Pd cubic THH TOH number of the surface atoms number of the Pd atoms Table S. Numbers of the Surface Pd Atoms on the Metal Nanocrystals in the Catalytic Reaction Solutions statistics cubic Pd cubic Au@Pd THH Au@Pd TOH Au@Pd molar number of the Pd atoms number of the nanocrystals total number of the surface atoms total number of the surface atoms at steps S18