Rh has 9 valence electrons. Using the 18-electron rule for each metal (count each CO as a 2-electron donor in total), find the number of Rh-Rh bonds in the tetranuclear cluster $Rh X_{4} (CO) X_{12}$ (answer as an integer).

Numerical answer type, tolerance 0.5.

Answer: 6

Total valence electrons =

4 \times 9 + 12 \times 2 = 60

. Required for four 18-electron metals =

72

. Deficit = 12 electrons, and each M-M bond supplies 2 electrons to the cluster count, so the number of bonds is

12/2 = 6

(a tetrahedral

Rh X_{4}

core).