Calculating number of M-M bonds in Ir4(CO)12

Ir4(CO)12 is a cluster compound of iridium metal and carbon monoxide ligands. The structure of the compound is tetrahedral with four iridium atoms at the corners and twelve carbon monoxide ligands at the edges.

Step 1: Determine the oxidation state of the metal atoms

In Ir4(CO)12, each iridium atom has a zero oxidation state. This is because it is a metal cluster compound where the metal atoms are in a zero oxidation state.

Step 2: Calculate the total number of valence electrons

The total number of valence electrons can be calculated by adding the valence electrons of each atom in the compound.

4 × (number of valence electrons of Ir) + 12 × (number of valence electrons of CO)

4 × 9 + 12 × 2 = 48

Step 3: Count the number of electrons used in bonding

Each carbon monoxide ligand uses two electrons to form a bond with the iridium atom.

12 × 2 = 24

Step 4: Calculate the number of remaining electrons

The remaining electrons can be calculated by subtracting the number of electrons used in bonding from the total number of valence electrons.

48 - 24 = 24

Step 5: Determine the number of M-M bonds

The number of M-M bonds can be calculated by dividing the number of remaining electrons by two, since each bond involves two electrons.

24 / 2 = 12

However, since each bond involves two metal atoms, we need to divide the total number of bonds by two to get the number of M-M bonds.

12 / 2 = 6

Therefore, Ir4(CO)12 has six M-M bonds.