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

$$[Mn(NH_3)_6]^{3+}$$, $$[MnCl_6]^{3-}$$, $$[FeF_6]^{3-}$$, $$[CoF_6]^{3-}$$, $$[Fe(NH_3)_6]^{3+}$$ and $$[Co(en)_3]^{3+}$$

[Given, atomic number: Mn = 25, Fe = 26, Co = 27; en = $$H_2NCH_2CH_2NH_2$$]

For an octahedral complex, whether it is diamagnetic (all electrons paired) or paramagnetic (one or more unpaired electrons) depends on

(i) the metal-ion oxidation state  →  its $$d^{\,n}$$ configuration, and
(ii) the ligand field strength  →  high-spin (weak field) or low-spin (strong field) arrangement.

Weak-field ligands: $$Cl^-,$$ $$F^-$$
Intermediate/strong ligands: $$NH_3,$$ en ($$H_2NCH_2CH_2NH_2$$). For $$Co^{3+},$$ en is strong enough to give a low-spin complex.

Ligands neutral, so oxidation state $$Mn^{3+}$$. Mn (Z = 25) gives $$d^4$$: $$[Ar]\,3d^4$$.
$$NH_3$$ is not strong enough to pair electrons in $$Mn^{3+}$$, hence high-spin $$t_{2g}^{3}e_g^{1}$$  →  4 unpaired electrons. Paramagnetic.

Again $$Mn^{3+} : d^4$$. $$Cl^-$$ is a weak ligand, so high-spin $$t_{2g}^{3}e_g^{1}$$ with 4 unpaired electrons. Paramagnetic.

$$Fe^{3+} : d^5$$. $$F^-$$ is weak, giving high-spin $$t_{2g}^{3}e_g^{2}$$ with 5 unpaired electrons. Paramagnetic.

$$Co^{3+} : d^6$$. $$F^-$$ weak  →  high-spin $$t_{2g}^{4}e_g^{2}$$ with 4 unpaired electrons. Paramagnetic.

$$Fe^{3+} : d^5$$. $$NH_3$$ is not strong enough to force pairing in $$Fe^{3+}$$, so high-spin $$t_{2g}^{3}e_g^{2}$$ with 5 unpaired electrons. Paramagnetic.

$$en$$ is a bidentate strong-field ligand. For $$Co^{3+} : d^6$$ the crystal-field splitting $$\Delta_o$$ is large enough to cause pairing  →  low-spin $$t_{2g}^{6}e_g^{0}$$. All electrons are paired. Diamagnetic.

Counting the diamagnetic species, only Case 6 qualifies.

Total number of diamagnetic complexes = 1.