Among the following complexes, the total number of diamagnetic species is ______.

${\left[\mathrm{M}\mathrm{n}{\left({\mathrm{N}\mathrm{H}}_3\right)}_6\right]}^{3+},{\left[{\mathrm{M}\mathrm{n}\mathrm{C}\mathrm{l}}_6\right]}^{3-<!--\; -\; -->},$ ${\left[{\mathrm{F}\mathrm{e}\mathrm{F}}_6\right]}^{3-<!--\; -\; -->},{\left[{\mathrm{C}\mathrm{o}\mathrm{F}}_6\right]}^{3-<!--\; -\; -->},{\left[\mathrm{F}\mathrm{e}{\left({\mathrm{N}\mathrm{H}}_3\right)}_6\right]}^{3+}and{[\mathrm{C}\mathrm{o}(\mathrm{e}\mathrm{n}{)}_3]}^{3+}$

[Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = H₂NCH₂CH₂NH₂]

The correct answer is 1.

To determine the total number of diamagnetic species among the given coordination complexes, we must analyze the oxidation state, electronic configuration, and the nature of the ligands (whether strong-field or weak-field) for the central metal ion in each complex.

Recall that:
- A species is paramagnetic if it contains one or more unpaired electrons.
- A species is diamagnetic if all of its electrons are paired (zero unpaired electrons).

Let's analyze each complex step-by-step:

1. ${\left[Mn{\left({NH}_3\right)}_6\right]}^{3+}$
- Oxidation state of Manganese: The neutral ligand ${\text{NH}}_3$ gives Mn an oxidation state of +3.
- Electronic configuration of Mn (Z = 25) is $\left[\text{Ar}\right]3d^{5}4s^2$. Thus, ${\text{Mn}}^{3+}$ has a $3d^4$ configuration.
- In octahedral geometry, the d-orbitals split into $t_{2g}$ and $e_g$ sets. Whether low-spin ($t_{2g}^4{e}_g^0$, with 2 unpaired electrons) or high-spin ($t_{2g}^3{e}_g^1$, with 4 unpaired electrons), a $d^4$ system will always have unpaired electrons.
- Therefore, this complex is paramagnetic.

2. ${\left[{MnCl}_6\right]}^{3-}$
- Oxidation state of Manganese: Since ${\text{Cl}}^{-}$ carries a -1 charge, the oxidation state of Mn is +3 ($3d^4$ configuration).
- Chlorine (${\text{Cl}}^{-}$) is a weak-field ligand, resulting in a high-spin configuration of $t_{2g}^3{e}_g^1$ (4 unpaired electrons).
- Therefore, this complex is paramagnetic.

3. ${\left[{FeF}_6\right]}^{3-}$
- Oxidation state of Iron: Fluoride (${\text{F}}^{-}$) is a monodentate anionic ligand, giving Fe an oxidation state of +3.
- Electronic configuration of Fe (Z = 26) is $\left[\text{Ar}\right]3d^{6}4s^2$. Thus, ${\text{Fe}}^{3+}$ has a $3d^5$ configuration.
- Fluoride is a weak-field ligand, yielding a high-spin configuration of $t_{2g}^3{e}_g^2$ (5 unpaired electrons).
- Therefore, this complex is paramagnetic.

4. ${\left[{CoF}_6\right]}^{3-}$
- Oxidation state of Cobalt: ${\text{Co}}^{3+}$.
- Electronic configuration of Co (Z = 27) is $\left[\text{Ar}\right]3d^{7}4s^2$. Thus, ${\text{Co}}^{3+}$ has a $3d^6$ configuration.
- Fluoride is a weak-field ligand, giving a high-spin configuration of $t_{2g}^4{e}_g^2$ (4 unpaired electrons).
- Therefore, this complex is paramagnetic.

5. ${\left[Fe{\left({NH}_3\right)}_6\right]}^{3+}$
- Oxidation state of Iron: ${\text{Fe}}^{3+}$ ($3d^5$ configuration).
- For ${\text{Fe}}^{3+}$, ${\text{NH}}_3$ behaves as a weak-field ligand (pairing energy is higher than crystal field splitting energy).
- Even if it were a strong-field ligand causing a low-spin configuration, $d^5$ in low-spin is $t_{2g}^5{e}_g^0$, which still has 1 unpaired electron.
- Therefore, this complex is paramagnetic.

6. ${\left[Co{\left(\text{en}\right)}_3\right]}^{3+}$
- Oxidation state of Cobalt: Since ethylenediamine (en) is a neutral bidentate ligand, Cobalt is in the +3 oxidation state.
- Electronic configuration of ${\text{Co}}^{3+}$ is $3d^6$.
- Ethylenediamine is a strong-field ligand, which causes pairing of electrons, leading to a low-spin configuration.
- All 6 d-electrons populate the lower energy $t_{2g}$ orbitals: $t_{2g}^6{e}_g^0$.
- Since all electrons are fully paired, the number of unpaired electrons is 0.
- Therefore, this complex is diamagnetic.

Conclusion:
Only 1 species (${\left[Co{\left(\text{en}\right)}_3\right]}^{3+}$) is diamagnetic.