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]

The correct answer is 1.

Isoelectronic species are atoms, molecules, or ions that have the same number of total electrons.

First, let us calculate the total number of electrons in $\mathrm{Ni}(\mathrm{CO}{)}_4$:
Atomic number of Nickel ($\mathrm{Ni}$) = 28
Number of electrons in one Carbon monoxide ($\mathrm{CO}$) molecule = 6 (for C) + 8 (for O) = 14 electrons.
Therefore, the total number of electrons in $\mathrm{Ni}(\mathrm{CO}{)}_4$ is:
$\text{Total electrons}=28+4\times 14=28+56=84$

Now, let us calculate the total number of electrons for each of the given species:

1. $V(\mathrm{CO}{)}_6$:
Total electrons = 23 (for V) + 6 × 14 = 23 + 84 = 107 electrons.

2. $\mathrm{Cr}(\mathrm{CO}{)}_5$:
Total electrons = 24 (for Cr) + 5 × 14 = 24 + 70 = 94 electrons.

3. $\mathrm{Cu}(\mathrm{CO}{)}_3$:
Total electrons = 29 (for Cu) + 3 × 14 = 29 + 42 = 71 electrons.

4. $\mathrm{Mn}(\mathrm{CO}{)}_5$:
Total electrons = 25 (for Mn) + 5 × 14 = 25 + 70 = 95 electrons.

5. $\mathrm{Fe}(\mathrm{CO}{)}_5$:
Total electrons = 26 (for Fe) + 5 × 14 = 26 + 70 = 96 electrons.

6. $\left[\mathrm{Co}\right(\mathrm{CO}{)}_3{]}^{3-}$:
Total electrons = 27 (for Co) + 3 × 14 + 3 (due to 3- charge) = 27 + 42 + 3 = 72 electrons.

7. $\left[\mathrm{Cr}\right(\mathrm{CO}{)}_4{]}^{4-}$:
Total electrons = 24 (for Cr) + 4 × 14 + 4 (due to 4- charge) = 24 + 56 + 4 = 84 electrons.

8. $\mathrm{Ir}(\mathrm{CO}{)}_3$:
Total electrons = 77 (for Ir) + 3 × 14 = 77 + 42 = 119 electrons.

Comparing the total number of electrons, only $\left[\mathrm{Cr}\right(\mathrm{CO}{)}_4{]}^{4-}$ has 84 electrons, which is equal to that of $\mathrm{Ni}(\mathrm{CO}{)}_4$.

Thus, there is only 1 species isoelectronic with $\mathrm{Ni}(\mathrm{CO}{)}_4$.