The number of paramagnetic metal complex species among $$[Co(NH_3)_6]^{3+}$$, $$[Co(C_2O_4)_3]^{3-}$$, $$[MnCl_6]^{3-}$$, $$[Mn(CN)_6]^{3-}$$, $$[CoF_6]^{3-}$$, $$[Fe(CN)_6]^{3-}$$ and $$[FeF_6]^{3-}$$ with same number of unpaired electrons is _____.

First find the oxidation state of the metal ion and its $$d$$-electron count in every complex. Then decide whether the ligand present is a strong-field (low-spin) or weak-field (high-spin) ligand. This gives the number of unpaired electrons, $$n_{u}$$, in each complex.

Case 1: $$[Co(NH_3)_6]^{3+}$$
Co: atomic number $$27 \Rightarrow [Ar]\,3d^7\,4s^2$$.
Oxidation state $$=+3 \Rightarrow Co^{3+}: d^{6}$$.
$$NH_3$$ is an intermediate-strong field ligand for $$Co^{3+}$$, so the complex is low spin:
$$t_{2g}^{6}e_g^{0} \Rightarrow n_u = 0$$ (diamagnetic).

Case 2: $$[Co(C_2O_4)_3]^{3-}$$
$$C_2O_4^{2-}$$ is a bidentate, moderate field ligand. For $$Co^{3+}(d^{6})$$ its crystal-field splitting is still large enough to give low spin:
$$t_{2g}^{6}e_g^{0} \Rightarrow n_u = 0$$ (diamagnetic).

Case 3: $$[MnCl_6]^{3-}$$
Let the oxidation number of Mn be $$x$$:
$$x + 6(-1) = -3 \Rightarrow x = +3 \Rightarrow Mn^{3+}: d^{4}$$.
$$Cl^-$$ is a weak-field ligand ⇒ high spin:
$$t_{2g}^{3}e_g^{1}$$ has $$4$$ unpaired electrons, $$n_u = 4$$.

Case 4: $$[Mn(CN)_6]^{3-}$$
Same oxidation calculation gives $$Mn^{3+}: d^{4}$$.
$$CN^-$$ is a strong-field ligand ⇒ low spin:
$$t_{2g}^{4}e_g^{0}$$ (two orbitals singly-filled, one paired) gives $$n_u = 2$$.

Case 5: $$[CoF_6]^{3-}$$
Charge balance: $$Co^{3+}: d^{6}$$.
$$F^-$$ is weak-field ⇒ high spin for $$d^{6}$$:
$$t_{2g}^{4}e_g^{2}$$ gives $$n_u = 4$$.

Case 6: $$[Fe(CN)_6]^{3-}$$
$$Fe^{3+}: d^{5}$$ (since $$x+6(-1)=-3 \Rightarrow x=+3$$).
Strong-field $$CN^-$$ ⇒ low spin:
$$t_{2g}^{5}e_g^{0}$$ gives $$n_u = 1$$.

Case 7: $$[FeF_6]^{3-}$$
$$Fe^{3+}: d^{5}$$.
Weak-field $$F^-$$ ⇒ high spin:
$$t_{2g}^{3}e_g^{2}$$ gives $$n_u = 5$$.

Collect the results:
$$\begin{array}{lcl} [Co(NH_3)_6]^{3+} &:& n_u = 0 \\ [Co(C_2O_4)_3]^{3-} &:& n_u = 0 \\ [MnCl_6]^{3-} &:& n_u = 4 \\ [Mn(CN)_6]^{3-} &:& n_u = 2 \\ [CoF_6]^{3-} &:& n_u = 4 \\ [Fe(CN)_6]^{3-} &:& n_u = 1 \\ [FeF_6]^{3-} &:& n_u = 5 \end{array}$$

Remove the diamagnetic complexes ($$n_u = 0$$). The remaining paramagnetic ones and their $$n_u$$ values are:

$$[MnCl_6]^{3-}: 4,\; [Mn(CN)_6]^{3-}: 2,\; [CoF_6]^{3-}: 4,\; [Fe(CN)_6]^{3-}: 1,\; [FeF_6]^{3-}: 5$$.

Only $$[MnCl_6]^{3-}$$ and $$[CoF_6]^{3-}$$ share the same number of unpaired electrons ($$n_u = 4$$).

Hence, the number of paramagnetic complexes having an identical count of unpaired electrons is $$\boxed{2}$$.