Which of the following is diamagnetic?

We have to decide which of the four octahedral coordination entities is diamagnetic, that is, which one possesses no unpaired electron. A complex is diamagnetic when the total number of unpaired electrons, $$n_u,$$ is zero. The standard plan is to find, for every complex,

1. the oxidation state of the central metal ion,
2. its corresponding ground-state $$d$$ electron count,
3. whether the ligand set produces a high-spin (weak-field) or low-spin (strong-field) splitting,
4. the distribution of those $$d$$ electrons into the octahedral levels $$t_{2g}$$ and $$e_g,$$ and finally
5. the value of $$n_u.$$

First recall the octahedral crystal-field diagram: in octahedral geometry the five $$d$$ orbitals split into the lower-energy triply degenerate $$t_{2g}$$ set and the higher-energy doubly degenerate $$e_g$$ set. When the ligands are strong-field (large $$\Delta_0$$), electrons pair inside $$t_{2g}$$ first (low-spin); when the ligands are weak-field (small $$\Delta_0$$), Hund’s rule dominates and electrons spread into $$e_g$$ before pairing (high-spin).

With that background we examine each option one by one.

Option A : $$[Fe(CN)_6]^{3-}$$

We first compute the oxidation state of iron. The charge on each $$CN^-$$ ligand is $$-1,$$ so for six ligands the total ligand charge is $$6(-1)=-6.$$ The overall complex carries $$-3.$$ Writing the balance, $$ x + (-6)= -3 \;\Longrightarrow\; x = +3. $$ Hence the central ion is $$Fe^{3+}.$$

The atomic number of iron is $$26,$$ so the neutral configuration is $$[Ar]\,3d^6 4s^2.$$ Removing three electrons for $$Fe^{3+}$$ leaves $$ Fe^{3+} : 3d^5. $$

The ligand $$CN^-$$ is a classic strong-field ligand (spectrochemical series). Therefore the complex is low-spin. In an octahedral low-spin arrangement five electrons fill the lower $$t_{2g}$$ orbitals as $$ t_{2g}^5\,e_g^0. $$ The orbital diagram shows four paired electrons and one left unpaired, so $$ n_u =1 \;\neq 0. $$ Hence the complex is paramagnetic, not diamagnetic.

Option B : $$[Co(ox)_3]^{3-}$$   (here $$ox^{2-}=C_2O_4^{2-}$$)

Let us again evaluate the oxidation state. Each oxalate carries $$-2,$$ so $$ x + 3(-2) = -3 \;\Longrightarrow\; x = +3. $$ Therefore we have $$Co^{3+}.$$

Cobalt’s atomic number is $$27,$$ giving the neutral configuration $$[Ar]\,3d^7 4s^2.$$ Removal of three electrons yields $$ Co^{3+} : 3d^6. $$

Now we decide whether the complex is high-spin or low-spin. Oxalate is a bidentate ligand that lies in the mid-to-strong field region; more importantly, the central ion bears a high charge $$+3,$$ which enhances the crystal-field splitting. Experimentally the majority of $$Co^{3+}$$ octahedral chelates, including $$[Co(ox)_3]^{3-},$$ are low-spin. We shall therefore fill the six $$d$$ electrons into the low-spin pattern: $$ t_{2g}^6\,e_g^0. $$ All six electrons are now paired inside the lower level, so $$ n_u = 0. $$ Because no unpaired electrons remain, the complex is diamagnetic.

Option C : $$[FeF_6]^{3-}$$

The oxidation-state calculation is identical to Option A, because each $$F^-$$ contributes $$-1.$$ We again find $$ Fe^{3+} : 3d^5. $$

The ligand $$F^-$$ is a very weak-field ligand. Hence the complex will be high-spin. In an octahedral high-spin $$d^5$$ configuration the electrons distribute as $$ t_{2g}^3\,e_g^2, $$ one in each orbital, giving $$ n_u = 5. $$ Five unpaired electrons make the substance strongly paramagnetic, so it is not diamagnetic.

Option D : $$[CoF_6]^{3-}$$

Again taking six $$F^-$$ ligands, we find $$ Co^{3+} : 3d^6. $$

Because $$F^-$$ is a weak-field ligand, the splitting is small and the complex remains high-spin. For a high-spin $$d^6$$ ion we obtain $$ t_{2g}^4\,e_g^2, $$ with the four $$t_{2g}$$ electrons arranged as three unpaired and one paired, plus two unpaired in $$e_g,$$ giving $$ n_u = 4. $$ Therefore the complex is paramagnetic, not diamagnetic.

Summarising the counts:

$$[Fe(CN)_6]^{3-}: n_u = 1 \quad (\text{paramagnetic})$$
$$[Co(ox)_3]^{3-}: n_u = 0 \quad (\text{diamagnetic})$$
$$[FeF_6]^{3-}: n_u = 5 \quad (\text{paramagnetic})$$
$$[CoF_6]^{3-}: n_u = 4 \quad (\text{paramagnetic})$$

Only the second complex, $$[Co(ox)_3]^{3-},$$ has zero unpaired electrons. Hence, the correct answer is Option B.