in the case of chromium(II) acetate. Organic molecules Organic molecules are often cyclic compounds containing one or more rings, such as benzene, and are often made up of many sigma bonds along with pi bonds. According to the sigma bond rule, the number of sigma bonds in a molecule is equivalent to the number of atoms plus the number of rings minus one. Nσ = Natoms + Nrings − 1 This rule is a special-case application of the Euler characteristic of the graph which represents the molecule. A molecule with no rings can be represented as a tree with a number of bonds equal to the number of atoms minus one (as in dihydrogen, H2, with only one sigma bond, or ammonia, NH3, with 3 sigma bonds). There is no more than 1 sigma bond between any two atoms. Molecules with rings have additional sigma bonds, such as benzene rings, which have 6 C−C sigma bonds within the ring for 6 carbon atoms. The anthracene molecule, C14H10, has three rings so that the rule gives the number of sigma bonds as 24 + 3 − 1 = 26. In this case there are 16 C−C sigma bonds and 10 C−H bonds. This rule fails in the case of molecules which, when drawn flat on paper, have a different number of rings than the molecule actually has - for example, Buckminsterfullerene, C60, which has 32 rings, 60 atoms, and 90 sigma bonds, one for each pair of bonded atoms; however, 60 + 32 - 1 = 91, not 90. This is because the sigma rule is a special case of the Euler characteristic, where each ring is considered a face, each sigma bond is an edge, and each atom is a vertex. Ordinarily, one extra face is assigned to the space not inside any ring, but when Buckminsterfullerene is drawn flat without any crossings, one of the rings makes up the outer pentagon; the inside of that ring is the outside of the graph. This rule fails further when considering other shapes - toroidal fullerenes will obey the rule that the number of sigma bonds in a molecule is exactly the number of atoms plus the number of rings, as will nanotubes - which, when drawn flat as if looking through one from the end, will have a face in the middle, corresponding to the far end of the nanotube, which