Question

The number of total valence electrons in the following complex is \([Os(N)Br_2(PMe_3)(NMe_2)]^-\)

Hint:

Be careful with the nitrido (\(N^{3-}\)) and amido (\(NR_2^-\)) ligands. In neutral counting, Nitrido is 3e and Amido is 1e.

Answer 1

{Step 1: Understanding the Concept:}
The 18-electron rule is evaluated by counting valence electrons from the metal and those donated by ligands.
{Step 2: Detailed Explanation:}
Using the neutral atom counting method:
- Osmium (Os): Group 8 element $\rightarrow$ 8e.
- Nitrido (N): Terminal triply bonded ligand $\rightarrow$ 3e.
- Bromine (\(Br_2\)): 2 halogen atoms $\rightarrow$ \(2 \times 1 = 2e\).
- Phosphine (\(PMe_3\)): Neutral 2-electron donor $\rightarrow$ 2e.
- Amido (\(NMe_2\)): Radical amido ligand $\rightarrow$ 1e.
- Negative charge: Adds 1e.
Total = \(8 + 3 + 2 + 2 + 1 + 1 = 17\) electrons.
{Step 3: Final Answer:}
The total valence electron count is 17.