Question 11

Isoelectronic species in the context of metal carbonyls refers to species having the same total valence electron count (18-electron rule).

1. $Ni(CO)_4$: $Ni$ (Group 10) has 10 valence electrons. $4 \times CO$ contribute $4 \times 2 = 8$ electrons. Total = $10 + 8 = 18$.
2. Evaluate the given species:

• $V(CO)_6$: $5 + 6(2) = 17$ (No)
• $Cr(CO)_5$: $6 + 5(2) = 16$ (No)
• $Cu(CO)_3$: $11 + 3(2) = 17$ (No)
• $Mn(CO)_5$: $7 + 5(2) = 17$ (No)
• $Fe(CO)_5$: $8 + 5(2) = 18$ (Yes)
• $[Co(CO)_3]^{3-}$: $9 + 3(2) + 3 = 18$ (Yes)
• $[Cr(CO)_4]^{4-}$: $6 + 4(2) + 4 = 18$ (Yes)
• $Ir(CO)_3$: $9 + 3(2) = 15$ (No)

3. Total number of species with 18 valence electrons is 3 ($Fe(CO)_5$, $[Co(CO)_3]^{3-}$, and $[Cr(CO)_4]^{4-}$).