The B4C crystal structure has rhombohedral symmetry
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Author : LZH
Update time : 2023-07-05 20:01:40
What is Boron carbide? Lightweight B6O has properties similar to those of B4C, combining great hardness with low mass density, high thermal conductivity, high chemical inertness, and excellent wear resistance. However, nanoindentation experiments observed similar amorphous shear band formation along the (011̅1̅) plane in B6O, our QM shearing studies of single crystal B6O along the same slip system (011̅1̅)/⟨1̅101⟩ show an unusual structural recovery without fracture of icosahedra. We conclude that this difference between the QM study and experiment arises from the difference in loading conditions. In experimental indentation conditions, a component of compressive stress always accompanies the shear components. The deformation mechanism deduced from our QM simulations suggests that incorporating B6O into B4C might dramatically improve the flexibility of these superhard materials.
The B4C crystal structure has rhombohedral symmetry Experimental and QM studies agree that the B4C crystal structure has rhombohedral symmetry with the R3̅m space group. The unit cell of B4C contains one 12-atom icosahedron located at the apexes of the primitive cell with one 3-atom linear chain aligned along the [111]r direction. Each icosahedron involves two crystallographic sites: six polar sites (p) connecting directly to other icosahedra and six equatorial sites (denoted as e) connecting with the 3-atom chains. The B4C structure was originally proposed to consist of one B12 icosahedron and one C–C–C chain, denoted as B12(CCC). Later, X-ray diffraction (XRD) determined that the chain is C–B–C, so the icosahedron must be B11C, but whether the C is at the polar or equatorial site is not determined experimentally. QM calculations predicted that (B11Cp)(CBC) is the ground-state structure, with (B11Ce)(CBC) 0.29 eV/unit cell higher in energy. One can think of this (B11Cp)(CBC) structure as having a B+ in the C–B–C chain making strong σ bonds to the two C of the linear C–B–C chain, transferring one electron to the icosahedron to provide the 26 skeleton electrons within the icosahedron needed to satisfy Wade’s rule. Thus, one can write the structure for B4C as (B11Cp)1–(CBC)1+.
The crystal structure of B6O is similar to that of B4C The crystal structure of B6O is similar to that of B4C, leading to a rhombohedral lattice with the R3̅m space group. The unit cell is composed of one B12 icosahedral cluster at the apexes of the rhombohedral unit cell and one O–O chain along the trigonal axis. (7, 16, 34) However, each oxygen atom bonds to three icosahedra through B–O single bonds with 1.50 Å bond length, while the distance between two chain oxygen atoms is 3.07 Å, precluding direct O–O bonding. (35) Because each O atom makes three σ bonds in a plane, it must have one π-like lone pair, making it a formal O+. Thus, again the B12 icosahedron will gain 2 electrons to provide the 26 skeleton electrons needed to satisfy Wade’s rule. This leads to a representation as (B12)2–(O)+(O)+ for B6O.
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