Thomas McCullough, C.S.C. st. Edward's Un~versity Austin, Texos
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s i d e hkulation of the Tetrahedral Bond Angle
Very early in the typical elementary organic chemistry course, the student is introduced to the tetrahedral structure of the aliphatic carbon atom and the famous value of the angle between any two bonds, 109' 28'. At this point the matter of bond angle is dropped and the more enquiring student is left to ponder the source of such an unlikely number. Unless well versed in the science of solid geometry, the student will not likely proceed beyond wonderment as very few organic texts include the calculation of tetrahedral bond angle; I have heard of just one which utilizes solid geometry. However, a simple calculation can be made using plane geometry and the knowledge that in methane the permanent dipole moment is zero (see figure). A plane is passed through carbon atom C and two of the hydrogens, X and H, of the methane molecule. Hydrogen atom Y will extend in front of the plane and hydrogen atom Z will extend behind. By a simple rotation of the molecule about the H-C axis, Y and then Z can be placed in the position now occupied by X as would be expected from the equality of angles between any two bonds. The dipole force along CH is exactly balanced by three opposing components of vectors of equal magnitude acting through C X , C Y , and CZ. Setting the
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476 / Journal o f Chemical Education
force up equal to the total force down we find that CH = 3CX sin a However, since a11 four hydrogen atoms are equivalent, the dipole force along CH is equal to that along C X . Cancelling CX and CH from the above equation results in the simple relationship sin a =
From trigonometry tables we find ar to be 19' 28'. The tetrahedral bond angle is thus 90' plus 19' 28'.
