hydrogens at the vertices of a triangle, and tetrahedral, in which the carbon is in the center of a regular tetrahedron and hydrogens are at each vertex The orbital description of carbon, being ls 2s 2p, would suggest that, in order to form four bonds, a 2s electron must be promoted to the empty p-orbital to provide an electronic structure as shown below Recalling the geometry of the three p-orbitals, the pyramidal geometry might seem to be favored, since the three p-orbitals would form bonds at 90 angles with the s-orbital on the opposite face. This can easily be shown to be incorrect, however, by simply examining the number of molecules having the molecular formula CHCl, As shown below, if carbon was square-planar, there should be two forms of the molecule CH,Cl,(molecules having the same molecular formula, but different structures are termed isomers), one in which the two chlorines are on the same face of the square, and one in whi they are in opposite corners. If carbon was pyramidal, again two isomers would be predicted, one in which both chlorines are in the trigonal plane (with an angle of 120 ) and one in which one chlorine is in the apical position, with an angle of For a tetrahedron, however, since all bond angles are equal at 109.5,both chlorines would always occupy one ad jacent face of the tetrahedron (no matter which face you choose, as shown below) and only one isomer would be predicted.hydrogens at the vertices of a triangle, and, • tetrahedral, in which the carbon is in the center of a regular tetrahedron and hydrogens are at each vertex. The orbital description of carbon, being 1s2 2s2 2p2 , would suggest that, in order to form four bonds, a 2s electron must be promoted to the empty p-orbital to provide an electronic structure as shown below: 2s 2px 2py 2pz Recalling the geometry of the three p-orbitals, the pyramidal geometry might seem to be favored, since the three p-orbitals would form bonds at 90 angles, with the s-orbital on the opposite face. This can easily be shown to be incorrect, however, by simply examining the number of molecules having the molecular formula CH2Cl2. As shown below, if carbon was square-planar, there should be two forms of the molecule CH2Cl2 (molecules having the same molecular formula, but different structures are termed isomers), one in which the two chlorines are on the same face of the square, and one in which they are in opposite corners. If carbon was pyramidal, again two isomers would be predicted, one in which both chlorines are in the trigonal plane (with an angle of 120 ) and one in which one chlorine is in the apical position, with an angle of 90 . For a tetrahedron, however, since all bond angles are equal at 109.5 , both chlorines would always occupy one adjacent face of the tetrahedron (no matter which face you choose, as shown below) and only one isomer would be predicted