The user wants to find the number of species from a given list that are isoelectronic with $\mathrm{Ni}(\mathrm{CO})_4$.

Step 1: Define 'isoelectronic' and calculate the electron count for the reference species.

Two species are considered isoelectronic if they have the same total number of electrons. First, we need to calculate the total number of electrons in the reference species, $\mathrm{Ni}(\mathrm{CO})_4$.

• The atomic number (Z) of Nickel (Ni) is 28. A neutral Ni atom has 28 electrons.
• The carbonyl ligand (CO) consists of a Carbon atom (Z=6, 6 electrons) and an Oxygen atom (Z=8, 8 electrons). So, each CO molecule has $6 + 8 = 14$ electrons.
• The complex $\mathrm{Ni}(\mathrm{CO})_4$ has one Ni atom and four CO ligands.
• The total number of electrons in $\mathrm{Ni}(\mathrm{CO})_4$ is: $\text{Total electrons} = (\text{electrons from Ni}) + 4 \times (\text{electrons from CO})$ $\text{Total electrons} = 28 + 4 \times 14 = 28 + 56 = 84 \text{ electrons}$

So, we need to find the species from the list that also have a total of 84 electrons.

Step 2: Calculate the total number of electrons for each species in the list.

We will calculate the total electron count for each species using their atomic numbers and charges.

1. $\mathrm{V}(\mathrm{CO})_6$

   • Atomic number of V = 23.
   • Total electrons = $23 + 6 \times 14 = 23 + 84 = 107$. Not isoelectronic.

2. $\mathrm{Cr}(\mathrm{CO})_5$

   • Atomic number of Cr = 24.
   • Total electrons = $24 + 5 \times 14 = 24 + 70 = 94$. Not isoelectronic.

3. $\mathrm{Cu}(\mathrm{CO})_3$

   • Atomic number of Cu = 29.
   • Total electrons = $29 + 3 \times 14 = 29 + 42 = 71$. Not isoelectronic.

4. $\mathrm{Mn}(\mathrm{CO})_5$

   • Atomic number of Mn = 25.
   • Total electrons = $25 + 5 \times 14 = 25 + 70 = 95$. Not isoelectronic.

5. $\mathrm{Fe}(\mathrm{CO})_5$

   • Atomic number of Fe = 26.
   • Total electrons = $26 + 5 \times 14 = 26 + 70 = 96$. Not isoelectronic.

6. $\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-}$

   • Atomic number of Co = 27.
   • The charge is 3-, so we add 3 electrons to the total count.
   • Total electrons = $27 + 3 \times 14 + 3 = 27 + 42 + 3 = 72$. Not isoelectronic.

7. $\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-}$

   • Atomic number of Cr = 24.
   • The charge is 4-, so we add 4 electrons to the total count.
   • Total electrons = $24 + 4 \times 14 + 4 = 24 + 56 + 4 = 84$. This is isoelectronic with $\mathrm{Ni}(\mathrm{CO})_4$.

8. $\operatorname{Ir}(\mathrm{CO})_3$

   • Atomic number of Ir = 77.
   • Total electrons = $77 + 3 \times 14 = 77 + 42 = 119$. Not isoelectronic.

Step 3: Count the isoelectronic species.

From the calculations above, only one species, $\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-}$, has the same total number of electrons (84) as $\mathrm{Ni}(\mathrm{CO})_4$.

Therefore, the total number of isoelectronic species is 1.

Note: One might be tempted to use the 18-electron rule (valence electron count). For $\mathrm{Ni}(\mathrm{CO})_4$, the valence electron count is $10 + 4\times2 = 18$. Species like $\mathrm{Fe}(\mathrm{CO})_5$ ($8 + 5\times2 = 18$) also have 18 valence electrons. However, 'isoelectronic' strictly means the same total number of electrons, not just valence electrons. $\mathrm{Fe}(\mathrm{CO})_5$ has a total of 96 electrons, which is different from 84.