1. Find oxidation state and electronic configuration of cobalt in both complexes

In both complexes, cobalt is in the oxidation state $+3$.

• For $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$: $x+6(0)=+3 \Rightarrow x=+3$
• For $\left[\mathrm{CoF}_6\right]^{3-}$: $x+6(-1)=-3 \Rightarrow x=+3$

So in both cases, the metal ion is $\mathrm{Co}^{3+}$.

Cobalt has atomic number $27$: $\mathrm{Co}: [Ar] \, 3d^7 4s^2$

Therefore, $\mathrm{Co}^{3+}: [Ar] \, 3d^6$

So both are $d^6$ octahedral complexes.

2. Consider ligand strength

• $\mathrm{NH}_3$ is a stronger field ligand.
• $\mathrm{F}^-$ is a weak field ligand.

For an octahedral $d^6$ ion:

• Strong field ligand $\Rightarrow$ low spin $\Rightarrow$ electrons pair up in $t_{2g}$.
• Weak field ligand $\Rightarrow$ high spin $\Rightarrow$ electrons occupy higher orbitals before pairing.

3. For $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$

Since $\mathrm{NH}_3$ is strong field, this is a low-spin octahedral complex.

Electronic arrangement: $t_{2g}^6 e_g^0$

All electrons are paired, so it is a spin-paired complex.

Also, because pairing occurs in 3d orbitals, two 3d orbitals become vacant and hybridisation is: $d^2sp^3$ So it is also an inner orbital complex.

4. For $\left[\mathrm{CoF}_6\right]^{3-}$

Since $\mathrm{F}^-$ is weak field, this is a high-spin octahedral complex.

Electronic arrangement: $t_{2g}^4 e_g^2$

There are unpaired electrons, so it is a spin-free complex.

Its hybridisation is: $sp^3d^2$ So it is an outer orbital complex.

5. Match with options

• $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$: spin paired complex
• $\left[\mathrm{CoF}_6\right]^{3-}$: spin free complex

Therefore, the correct option is: $\boxed{\text{B}}$

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They match.