Identify the diamagnetic octahedral complex ions from below:
A. $$[Mn(CN)_6]^{3-}$$
B. $$[Co(NH_3)_6]^{3+}$$
C. $$[Fe(CN)_6]^{4-}$$
D. $$[Co(H_2O)_3F_3]$$
Choose the correct answer from the options given below: 

The complexes are octahedral, so their magnetic behaviour depends on

• oxidation state of the metal → number of $$d$$ electrons
• strength of the ligands (spectro-chemical series)
• whether the complex is high-spin or low-spin → number of unpaired electrons

Complex $$[Mn(CN)_6]^{3-}$$

Oxidation state of Mn: $$x-6 = -3 \;\Rightarrow\; x = +3$$, so Mn is $$Mn^{3+}$$.

Electronic configuration of $$Mn^{3+}: [Ar]\,3d^4$$.

$$CN^-$$ is a strong-field ligand, so the complex is low-spin. For a d4 ion in a low-spin octahedral field we fill $$t_{2g}$$ first: $$t_{2g}^4 e_g^0$$. This arrangement has $$2$$ unpaired electrons, hence the complex is paramagnetic, not diamagnetic.

Complex $$[Co(NH_3)_6]^{3+}$$

Oxidation state of Co: $$x = +3$$ ⇒ $$Co^{3+}$$.

Electronic configuration of $$Co^{3+}: [Ar]\,3d^6$$.

$$NH_3$$ is a borderline/medium-field ligand, but with the highly charged $$Co^{3+}$$ centre the complex is low-spin. Thus the six electrons occupy $$t_{2g}$$ only: $$t_{2g}^6 e_g^0$$. All electrons are paired ⇒ diamagnetic.

Complex $$[Fe(CN)_6]^{4-}$$

Oxidation state of Fe: $$x-6 = -4 \;\Rightarrow\; x = +2$$, so Fe is $$Fe^{2+}$$.

Electronic configuration of $$Fe^{2+}: [Ar]\,3d^6$$.

$$CN^-$$ is a strong-field ligand giving a low-spin complex. For d6: $$t_{2g}^6 e_g^0$$, all paired ⇒ diamagnetic.

Complex $$[Co(H_2O)_3F_3]$$ (overall neutral)

Let oxidation state of Co be $$x$$:
$$x + 0\,(3H_2O) - 1\,(3F^-) = 0 \;\Rightarrow\; x = +3$$, so $$Co^{3+}$$ again (d6).

However, $$H_2O$$ and $$F^-$$ are weak-field ligands, so the complex is high-spin. Electron distribution: $$t_{2g}^4 e_g^2$$ with $$4$$ unpaired electrons ⇒ paramagnetic.

Summary:
A ⇒ paramagnetic
B ⇒ diamagnetic
C ⇒ diamagnetic
D ⇒ paramagnetic

Therefore the diamagnetic octahedral complexes are B and C only.

Correct option: Option D (B and C Only).