硅-铝催化剂酸中心形成及其结构

Formation and Structural Characteristics of Acidic Centers of Silica-Alumina Catalyst

含Si、Al元素的固体酸催化剂,其活性与其酸性质密切相关。催化剂酸性来源于不同氧配位数的Al原子,而与Si原子无关。催化剂的酸类型、酸强弱取决于Al原子的氧配位数,即与Al原子形成配位的氧原子的个数。Al原子的氧配位数与其外层轨道的杂化方式有关。13Al外层电子排布是3s23p1d0,其外层轨道有sp3、sp3d和sp3d2杂化3种杂化方式。Al原子外层轨道经sp3杂化形成4个sp3杂化轨道,其中有3个未成键的sp3杂化轨道和1个空的sp3杂化轨道。如果1个Al原子与2个O原子(sp3杂化)形成共价键,其结构为[AlO2],这种Al原子外层还有1个价电子和1个空轨道,空轨道可吸引孤对电子,[AlO2]是二配位L酸中心;如果1个Al原子与3个O原子形成共价键,其结构为[AlO3],则Al原子外层有1个空轨道,[AlO3]是三配位L酸中心。如果1个Al原子与4个O原子和1个H原子形成共价键,其结构为四配位的H[AlO4]正四面体;Al原子外层轨道经sp3d杂化,可形成5个sp3d杂化轨道,有3个未成键的sp3d杂化轨道和2个空的sp3d杂化轨道,1个Al原子可与5个O原子和2个H原子形成共价键,其结构为五配位的H2[AlO5]正四方锥或三角双锥;Al原子外层轨道经sp3d2杂化,可形成6个sp3d2杂化轨道,有3个未成键的sp3d2轨道和3个空的sp3d2杂化轨道,1个Al原子可与6个O原子和3个H原子形成共价键,其结构为六配位的H3[AlO6]正八面体。H[AlO4]有提供1个H+的可能,是四配位B酸中心,H2[AlO5]有提供2个H+的可能,是五配位B酸中心,H3[AlO6]有提供3个H+的可能,是六配位B酸中心。

The catalytic activity of silica-alumina catalyst heavily depends on its acidic properties.The acidity of silica-alumina catalyst comes from Al atoms which have different oxygen coordination,but not from Si atoms.It is proposed that the type of acid center and acidic strength of silica-alumina catalyst are determined by coordination number of Al atom,i.e.the number of oxygen atoms coordinated with Al atom,which relates to the hybrid manner of Al atom outer orbital.13Al atom takes the outer orbital distribution of 3s23p1d0,which has three kinds of orbital hybrid,i.e.sp3,sp3d,and sp3d2.Four hybrid orbitals will be created if Al atom takes sp3 hybridization.Among the four sp3,three orbitals are unbonded,and the other is unoccupied.One Al atom can bond with two O atoms,forming [AlO2] a unit.There is a valence electron in the outer orbital of Al atom.At the same time,Al atom in [AlO2] unit still has an unoccupied orbital,which can accept electrons,so the [AlO2] is a two-coordination Lewis acid center.Meanwhile,one Al atom may bond with three O atoms,forming a [AlO3] unit.In [AlO3] unit,Al atom has an unoccupied orbital,so the [AlO3] unit is a three-coordination Lewis acid center.If one Al atom bonds with four O atoms and one H[AlO4] atom,forming a four-coordination unit H,which form is regular tetrahedron.Otherwise,Al atom can take sp3d hybridization,making the Al atom have three unbonded hybrid orbitals and two unoccupied orbitals,then one Al atom can bond with five O atoms and two H atoms,forming a five-coordination unit H2[AlO5],which form is tetragonal pyramid or trigonal bipyramid.Likely,through sp3d2 hybridization,six sp3d2 hybrid orbitals will be created around Al atom,among which three are unbonded orbitals and the other three are unoccupied orbitals.In this case,one Al atom can bond with six O atoms and three H atoms,forming a six-coordination unit H3[AlO6],which form is regular hexahedron.One H[AlO4] can provide one H＋,being a four-coordination BrΦnsted acid center.Meanwhile,one H2[AlO5] can provide two H＋,being a five-coordination BrΦnsted acid center,and H3[AlO6] can provide three H＋,being a six-coordination BrΦnsted acid center.