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

Step 1: Understanding Isoelectronic

Isoelectronic species have the same number of valence electrons. The number of valence electrons in metal carbonyls is determined by:

$$\text{Total valence electrons} = \text{Atomic number of metal} - \text{Oxidation state} + 2 \times \text{Number of CO ligands}$$

For Ni(CO):

$$28 + 4 \times 2 = 36 \text{ electrons}$$

Step 2: Checking Other Species

• Cr(CO): $24 + 5 \times 2 = 34$ (Not isoelectronic)
• Cu(CO): $29 + 3 \times 2 = 35$ (Not isoelectronic)
• Mn(CO): $25 + 5 \times 2 = 35$ (Not isoelectronic)
• Fe(CO): $26 + 5 \times 2 = 36$ (Isoelectronic ✅)
• [Co(CO)]₃: $27 + 3 + 3 \times 2 = 36$ (Isoelectronic ✅)
• [Cr(CO)]: $24 + 4 + 4 \times 2 = 36$ (Isoelectronic ✅)
• Ir(CO): $77 + 3 \times 2 = 83$ (Not isoelectronic)

Step 3: Conclusion

The three species isoelectronic with Ni(CO) are:

• Fe(CO)
• [Co(CO)]₃
• [Cr(CO)]

Total count = 3.