the cluster is a closo polyhedron because , with . Other rules may be considered when predicting the structure of clusters: For clusters consisting mostly of transition metals, any main group elements present are often best counted as ligands or interstitial atoms, rather than vertices. Larger and more electropositive atoms tend to occupy vertices of high connectivity and smaller more electronegative atoms tend to occupy vertices of low connectivity. In the special case of boron hydride clusters, each boron atom connected to 3 or more vertices has one terminal hydride, while a boron atom connected to two other vertices has two terminal hydrogen atoms. If more hydrogen atoms are present, they are placed in open face positions to even out the coordination number of the vertices. For the special case of transition metal clusters, ligands are added to the metal centers to give the metals reasonable coordination numbers, and if any hydrogen atoms are present they are placed in bridging positions to even out the coordination numbers of the vertices. In general, closo structures with n vertices are n-vertex polyhedra. To predict the structure of a nido cluster, the closo cluster with n + 1 vertices is used as a starting point; if the cluster is composed of small atoms a high connectivity vertex is removed, while if the cluster is composed of large atoms a low connectivity vertex is removed. To predict the structure of an arachno cluster, the closo polyhedron with n + 2 vertices is used as the starting point, and the n + 1 vertex nido complex is generated by following the rule above; a second vertex adjacent to the first is removed if the cluster is composed of mostly small atoms, a second vertex not adjacent to the first is removed if the cluster is composed mostly of large atoms. Example: Electron count: 10 × Pb + 2 (for the negative charge) = 10 × 4 + 2 = 42 electrons. Since n = 10, 4n + 2 = 42, so the cluster is a closo bicapped square antiprism. Example: Electron count: 4 × S – 2 (for the positive charge) = 4 × 6 – 2 = 22 electrons. Since n = 4, 4n + 6 = 22, so the cluster is arachno. Starting from an octahedron, a vertex of high connectivity is removed, and then a non-adjacent vertex is removed. Example: Os6(CO)18 Electron