Question

Among V(CO)₆, Cr(CO)₅, Cu(CO)₃, Mn(CO)₅, Fe(CO)₅, [Co(CO)₃]^3-, [Cr(CO)₄]^4-, and Ir(CO)₃, the total number of species isoelectronic with Ni(CO)₄ is _________.
[Given, atomic number: V = 23, Cr=24, Mn 25, Fe 26, Co 27, Ni 28, Cu = 29, Ir = 77]

Answer 1

Correct Answer: 1

To solve the problem, we need to find the species that are isoelectronic with Ni(CO)₄.

1. Determining the Number of Electrons in Ni(CO)₄:
Ni(CO)₄ has the following number of electrons:

$ \text{Ni} = 28, \, \text{CO} = 14, \, \text{so} \, 28 + 4(14) = 28 + 56 = 84 \, \text{electrons}$.

2. Checking Other Species:
Now, we check other species to see if they have 84 electrons, considering both the atomic numbers and any charges.

3. Species and Their Electron Counts:

• V(CO)₆: $ \text{V} = 23, \, \text{CO} = 14, \, \text{so} \, 23 + 6(14) = 23 + 84 = 107 \, \text{electrons}$
• V(CO)₆^-: $ 23 + 6(14) + 1 = 23 + 84 + 1 = 108 \, \text{electrons}$
• Cr(CO)₅: $ \text{Cr} = 24, \, \text{CO} = 14, \, \text{so} \, 24 + 5(14) = 24 + 70 = 94 \, \text{electrons}$
• Cu(CO)₃: $ \text{Cu} = 29, \, \text{CO} = 14, \, \text{so} \, 29 + 3(14) = 29 + 42 = 71 \, \text{electrons}$
• Mn(CO)₅: $ \text{Mn} = 25, \, \text{CO} = 14, \, \text{so} \, 25 + 5(14) = 25 + 70 = 95 \, \text{electrons}$
• Fe(CO)₅: $ \text{Fe} = 26, \, \text{CO} = 14, \, \text{so} \, 26 + 5(14) = 26 + 70 = 96 \, \text{electrons}$
• [Co(CO)₃]^3-: $ \text{Co} = 27, \, \text{CO} = 14, \, \text{so} \, 27 + 3(14) + 3 = 27 + 42 + 3 = 72 \, \text{electrons}$
• [Cr(CO)₄]^4-: $ \text{Cr} = 24, \, \text{CO} = 14, \, \text{so} \, 24 + 4(14) + 4 = 24 + 56 + 4 = 84 \, \text{electrons}$
• Ir(CO)₃: $ \text{Ir} = 77, \, \text{CO} = 14, \, \text{so} \, 77 + 3(14) = 77 + 42 = 119 \, \text{electrons}$

4. Conclusion:
Only [Cr(CO)₄]^4- has 84 electrons and is isoelectronic with Ni(CO)₄.

Final Answer:
The final answer is $\boxed{1}$.