Analyze option A: $[\mathrm{FeCl}_4]^-$

• Oxidation state of Fe: $x + 4(-1) = -1 \Rightarrow x = +3$ So the complex is $[\mathrm{Fe}^{3+}\mathrm{Cl}_4]^-$.
• $\mathrm{Fe}^{3+}$ is a $d^5$ ion.
• $\mathrm{Cl}^-$ is a weak-field ligand.
• Four-coordinate complexes with weak ligands like $\mathrm{Cl}^-$ are generally tetrahedral rather than square planar.

Therefore, A is correct.

Analyze option B: $[\mathrm{Co}(en)(NH_3)_2Cl_2]^+$

• Here, $en$ is ethylenediamine, a bidentate ligand.
• Let oxidation state of Co be $x$: $x + 0 + 2(0) + 2(-1) = +1 \Rightarrow x = +3$ So cobalt is $\mathrm{Co}^{3+}$, i.e. $d^6$.
• Coordination number is 6, so geometry is octahedral.

Now count geometrical isomers:

• Since $en$ occupies two adjacent positions, the remaining four positions are occupied by $2NH_3$ and $2Cl$.
• In octahedral complexes of type $[M(AA)B_2C_2]$, there are 3 geometrical isomers:
  1. both $B$ ligands trans,
  2. both $C$ ligands trans,
  3. both pairs cis.

Hence $[\mathrm{Co}(en)(NH_3)_2Cl_2]^+$ has 3 geometrical isomers, not 2.

Therefore, B is incorrect.

Analyze option C: magnetic moments

For $[\mathrm{FeCl}_4]^-$

• $\mathrm{Fe}^{3+}$ is $d^5$.
• Tetrahedral with weak-field ligand $\mathrm{Cl}^-$ gives high-spin configuration.
• Number of unpaired electrons: $n = 5$
• Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35} \approx 5.92\,\text{BM}$

For $[\mathrm{Co}(en)(NH_3)_2Cl_2]^+$

• $\mathrm{Co}^{3+}$ is $d^6$.
• In octahedral $\mathrm{Co}^{3+}$ complexes, pairing energy is usually overcome; with ligands like $en$ and $NH_3$, the complex is generally low spin.
• Low-spin $d^6$ octahedral configuration: $t_{2g}^6 e_g^0$
• Number of unpaired electrons: $n = 0$
• So, $\mu = 0$

Therefore, $\mu([\mathrm{FeCl}_4]^-) > \mu([\mathrm{Co}(en)(NH_3)_2Cl_2]^+)$

Hence, C is correct.

Analyze option D: hybridisation of cobalt in $[\mathrm{Co}(en)(NH_3)_2Cl_2]^+$

• As discussed, Co is $\mathrm{Co}^{3+}$ ($d^6$).
• For low-spin octahedral $d^6$, the complex uses inner orbital hybridisation: $d^2sp^3$
• $sp^3d^2$ corresponds to outer orbital octahedral complexes.

Therefore, D is incorrect.

Final conclusion

Correct statements are: $\boxed{A,\ C}$

Comparison with stored correct answer

• Stored correct answer: A, C
• Derived answer: A, C

So, the derived answer agrees with the stored correct answer.