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]

To decide which of the given carbonyl species are isoelectronic with $$Ni(CO)_4$$ we must compare the total number of electrons (all valence and core electrons of the metal, all electrons of the ligands, and any extra electrons arising from charge) present in each species.

1. Compute the total electrons in $$Ni(CO)_4$$.
  • Atomic electrons in Ni : $$28$$ (atomic number).
  • Each CO molecule contains $$14$$ electrons.
  • Four CO ligands contribute $$4 \times 14 = 56$$ electrons.
  • Charge on the complex is zero, so no extra electrons.
Hence, the total electron count is
$$28 + 56 = 84$$ electrons.

2. Evaluate every listed species in the same way.

Case 1: $$V(CO)_6$$   23 (V) $$+ 6 \times 14 = 23 + 84 = 107$$ electrons. Case 2: $$Cr(CO)_5$$   24 (Cr) $$+ 5 \times 14 = 24 + 70 = 94$$ electrons. Case 3: $$Cu(CO)_3$$   29 (Cu) $$+ 3 \times 14 = 29 + 42 = 71$$ electrons. Case 4: $$Mn(CO)_5$$   25 (Mn) $$+ 5 \times 14 = 25 + 70 = 95$$ electrons. Case 5: $$Fe(CO)_5$$   26 (Fe) $$+ 5 \times 14 = 26 + 70 = 96$$ electrons. Case 6: $$[Co(CO)_3]^{3-}$$   27 (Co) $$+ 3 \times 14 = 27 + 42 = 69$$;
  extra electrons due to charge $$= 3$$.
  Total $$= 69 + 3 = 72$$ electrons. Case 7: $$[Cr(CO)_4]^{4-}$$   24 (Cr) $$+ 4 \times 14 = 24 + 56 = 80$$;
  extra electrons due to charge $$= 4$$.
  Total $$= 80 + 4 = 84$$ electrons. Case 8: $$Ir(CO)_3$$   77 (Ir) $$+ 3 \times 14 = 77 + 42 = 119$$ electrons.

3. Compare each total with the 84 electrons of $$Ni(CO)_4$$.
Only $$[Cr(CO)_4]^{4-}$$ matches exactly.

Therefore, the number of species that are isoelectronic with $$Ni(CO)_4$$ is 1.