1. Find oxidation state and d-electron count of the metal in each complex

• $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$: NH$_3$ is neutral, so Co is $+3$.

  Co: $Z=27$, so $\mathrm{Co}^{3+} = 3d^6$.

• $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$: oxalate is $-2$, so $x+3(-2)=-3 \implies x=+3$ Hence $\mathrm{Co}^{3+}=3d^6$.

• $[\mathrm{MnCl}_6]^{3-}$: Cl$^-$ is $-1$, so $x+6(-1)=-3 \implies x=+3$ Hence $\mathrm{Mn}^{3+}=3d^4$.

• $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$: CN$^-$ is $-1$, so Mn is $+3$. Thus $\mathrm{Mn}^{3+}=3d^4$.

• $[\mathrm{CoF}_6]^{3-}$: F$^-$ is $-1$, so Co is $+3$. Thus $\mathrm{Co}^{3+}=3d^6$.

• $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$: Fe is $+3$. Thus $\mathrm{Fe}^{3+}=3d^5$.

• $[\mathrm{FeF}_6]^{3-}$: Fe is $+3$. Thus $\mathrm{Fe}^{3+}=3d^5$.

2. Decide strong-field / weak-field and number of unpaired electrons

Use:

• Strong field ligands: CN$^-$, generally NH$_3$ with Co(III), and Co(III) complexes are usually low spin.
• Weak field ligands: F$^-$, Cl$^-$.
• Oxalate $(\mathrm{C}_2\mathrm{O}_4)^{2-}$ is not as strong as CN$^-$, but with Co(III) ($\Delta_o$ large), $[\mathrm{Co}(\mathrm{ox})_3]^{3-}$ is low spin.

Now evaluate each:

(i) $[\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}$

• Metal ion: $d^6$
• Co(III) gives large $\Delta_o$
• Low spin octahedral: $t_{2g}^6 e_g^0$
• Unpaired electrons $=0$
• Diamagnetic

(ii) $[\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_3]^{3-}$

• Metal ion: $d^6$
• Co(III) octahedral, low spin
• Configuration: $t_{2g}^6 e_g^0$
• Unpaired electrons $=0$
• Diamagnetic

(iii) $[\mathrm{MnCl}_6]^{3-}$

• Metal ion: $d^4$
• Cl$^-$ is weak field
• High spin octahedral: $t_{2g}^3 e_g^1$
• Unpaired electrons $=4$
• Paramagnetic

(iv) $[\mathrm{Mn}(\mathrm{CN})_6]^{3-}$

• Metal ion: $d^4$
• CN$^-$ is strong field
• Low spin octahedral: $t_{2g}^4 e_g^0$
• Unpaired electrons $=2$
• Paramagnetic

(v) $[\mathrm{CoF}_6]^{3-}$

• Metal ion: $d^6$
• F$^-$ is weak field
• High spin octahedral: $t_{2g}^4 e_g^2$
• Unpaired electrons $=4$
• Paramagnetic

(vi) $[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$

• Metal ion: $d^5$
• CN$^-$ is strong field
• Low spin octahedral: $t_{2g}^5 e_g^0$
• Unpaired electrons $=1$
• Paramagnetic

(vii) $[\mathrm{FeF}_6]^{3-}$

• Metal ion: $d^5$
• F$^-$ is weak field
• High spin octahedral: $t_{2g}^3 e_g^2$
• Unpaired electrons $=5$
• Paramagnetic

3. List only the paramagnetic species and their unpaired electrons

• $[\mathrm{MnCl}_6]^{3-} \rightarrow 4$
• $[\mathrm{Mn}(\mathrm{CN})_6]^{3-} \rightarrow 2$
• $[\mathrm{CoF}_6]^{3-} \rightarrow 4$
• $[\mathrm{Fe}(\mathrm{CN})_6]^{3-} \rightarrow 1$
• $[\mathrm{FeF}_6]^{3-} \rightarrow 5$

Among these, the species having the same number of unpaired electrons are:

• $[\mathrm{MnCl}_6]^{3-}$ and $[\mathrm{CoF}_6]^{3-}$, both with $4$ unpaired electrons.

So, the number of paramagnetic metal complex species with same number of unpaired electrons is $2$

4. Comparison with stored answer

Stored correct answer = $2$.

This matches the derived answer.