A complex is diamagnetic if it contains no unpaired electrons. To determine the magnetic properties of the given complexes, we need to find the number of unpaired electrons in the central metal ion for each complex using Crystal Field Theory.

We will analyze each complex step-by-step.

1. $\left[\mathrm{Mn}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

• Oxidation state of Mn: Let the oxidation state be $x$. $\mathrm{NH}_3$ is a neutral ligand. So, $x + 6(0) = +3 \implies x = +3$. The metal ion is $\mathrm{Mn}^{3+}$.
• Electronic configuration: The atomic number of Mn is 25. The configuration of Mn is $[\mathrm{Ar}] 3d^5 4s^2$. For $\mathrm{Mn}^{3+}$, the configuration is $[\mathrm{Ar}] 3d^4$.
• Ligand and Geometry: $\mathrm{NH}_3$ is a strong field ligand. The coordination number is 6, so the geometry is octahedral. For a $d^4$ system with a strong field ligand, a low-spin complex is formed.
• Electron filling: The four electrons fill the lower energy $t_{2g}$ orbitals. The configuration is $t_{2g}^4 e_g^0$. This results in one pair of electrons and two unpaired electrons.
• Magnetic property: Since there are unpaired electrons, the complex is paramagnetic.

2. $\left[\mathrm{MnCl}_6\right]^{3-}$

• Oxidation state of Mn: Let the oxidation state be $x$. $\mathrm{Cl}^-$ has a charge of -1. So, $x + 6(-1) = -3 \implies x = +3$. The metal ion is $\mathrm{Mn}^{3+}$.
• Electronic configuration: The configuration of $\mathrm{Mn}^{3+}$ is $[\mathrm{Ar}] 3d^4$.
• Ligand and Geometry: $\mathrm{Cl}^-$ is a weak field ligand. The geometry is octahedral. For a $d^4$ system with a weak field ligand, a high-spin complex is formed.
• Electron filling: The electrons occupy orbitals to maximize spin. The configuration is $t_{2g}^3 e_g^1$. This results in four unpaired electrons.
• Magnetic property: The complex is paramagnetic.

3. $\left[\mathrm{FeF}_6\right]^{3-}$

• Oxidation state of Fe: Let the oxidation state be $x$. $\mathrm{F}^-$ has a charge of -1. So, $x + 6(-1) = -3 \implies x = +3$. The metal ion is $\mathrm{Fe}^{3+}$.
• Electronic configuration: The atomic number of Fe is 26. The configuration of Fe is $[\mathrm{Ar}] 3d^6 4s^2$. For $\mathrm{Fe}^{3+}$, the configuration is $[\mathrm{Ar}] 3d^5$.
• Ligand and Geometry: $\mathrm{F}^-$ is a weak field ligand. The geometry is octahedral. For a $d^5$ system with a weak field ligand, a high-spin complex is formed.
• Electron filling: The configuration is $t_{2g}^3 e_g^2$. This results in five unpaired electrons.
• Magnetic property: The complex is paramagnetic.

4. $\left[\mathrm{CoF}_6\right]^{3-}$

• Oxidation state of Co: Let the oxidation state be $x$. $\mathrm{F}^-$ has a charge of -1. So, $x + 6(-1) = -3 \implies x = +3$. The metal ion is $\mathrm{Co}^{3+}$.
• Electronic configuration: The atomic number of Co is 27. The configuration of Co is $[\mathrm{Ar}] 3d^7 4s^2$. For $\mathrm{Co}^{3+}$, the configuration is $[\mathrm{Ar}] 3d^6$.
• Ligand and Geometry: $\mathrm{F}^-$ is a weak field ligand. The geometry is octahedral. For a $d^6$ system with a weak field ligand, a high-spin complex is formed.
• Electron filling: The configuration is $t_{2g}^4 e_g^2$. This results in four unpaired electrons.
• Magnetic property: The complex is paramagnetic.

5. $\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_6\right]^{3+}$

• Oxidation state of Fe: Let the oxidation state be $x$. $\mathrm{NH}_3$ is neutral. So, $x + 6(0) = +3 \implies x = +3$. The metal ion is $\mathrm{Fe}^{3+}$.
• Electronic configuration: The configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] 3d^5$.
• Ligand and Geometry: $\mathrm{NH}_3$ is a strong field ligand. The geometry is octahedral. For a $d^5$ system with a strong field ligand, a low-spin complex is formed.
• Electron filling: The configuration is $t_{2g}^5 e_g^0$. This results in one unpaired electron.
• Magnetic property: The complex is paramagnetic.

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

• Oxidation state of Co: Let the oxidation state be $x$. Ethylenediamine (en) is a neutral ligand. So, $x + 3(0) = +3 \implies x = +3$. The metal ion is $\mathrm{Co}^{3+}$.
• Electronic configuration: The configuration of $\mathrm{Co}^{3+}$ is $[\mathrm{Ar}] 3d^6$.
• Ligand and Geometry: en is a strong field ligand. The geometry is octahedral. For a $d^6$ system with a strong field ligand, a low-spin complex is formed.
• Electron filling: The six electrons pair up in the lower energy $t_{2g}$ orbitals. The configuration is $t_{2g}^6 e_g^0$. This results in zero unpaired electrons.
• Magnetic property: Since there are no unpaired electrons, the complex is diamagnetic.

Conclusion: Out of the six given complexes, only one, $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$, is diamagnetic. Therefore, the total number of diamagnetic species is 1.