Choose the correct statement(s) among the following:

Let us examine each statement one by one.

Statement A
For $$\left[FeCl_{4}\right]^{-}$$ :
Fe has atomic number 26. In the complex each $$Cl^{-}$$ is -1, so $$x+4(-1)=-1 \;\Rightarrow\; x=+3$$. Hence the ion is $$Fe^{3+}$$ (configuration $$[Ar]\,3d^{5}$$).
With four ligating atoms the possibilities are tetrahedral or square-planar. Square-planar geometry is obtained only for strong-field, low-spin $$d^{8}$$ or $$d^{9}$$ systems (e.g. $$Ni^{2+}, \;Pd^{2+}$$ etc.). Chloride is a weak-field ligand and the $$d^{5}$$ ion does not favour square-planar arrangement. Therefore $$\left[FeCl_{4}\right]^{-}$$ is tetrahedral.
Statement A is correct.

Statement B
$$\left[Co(en)(NH_{3})_{2}Cl_{2}\right]^{+}$$ is octahedral with a symmetric bidentate ligand en (ethylenediamine) and two pairs of identical monodentate ligands ($$NH_{3},Cl^{-}$$). Treating en as occupying two adjacent positions, four positions remain for the pair $$NH_{3}$$ and the pair $$Cl^{-}$$. Two distinct arrangements exist:

1. $$NH_{3}$$ ligands mutually trans (hence $$Cl^{-}$$ cis).
2. $$Cl^{-}$$ ligands mutually trans (hence $$NH_{3}$$ cis).

Any other trial turns out to be superposable on one of the above by a symmetry operation of the octahedron, so only 2 geometrical isomers are possible.
Statement B is correct.

Statement C
Spin-only magnetic moments are given by $$\mu=\sqrt{n(n+2)}\,\text{BM}$$ where $$n$$ is the number of unpaired electrons.

For $$\left[FeCl_{4}\right]^{-}$$ : high-spin $$d^{5}$$ ⇒ $$n=5$$ $$\mu=\sqrt{5(5+2)}=\sqrt{35}\approx5.92\;\text{BM}$$.

For $$\left[Co(en)(NH_{3})_{2}Cl_{2}\right]^{+}$$ : Co is +3 ( $$d^{6}$$ ). en and $$NH_{3}$$ are strong-field ligands and even two $$Cl^{-}$$ ions cannot reduce the crystal-field splitting below the pairing energy for the highly charged $$Co^{3+}$$ centre. The complex is therefore low-spin $$t_{2g}^{6}e_{g}^{0}$$, i.e. $$n=0$$ unpaired electrons, $$\mu=0\;\text{BM}$$.

Thus $$\mu\bigl([FeCl_{4}]^{-}\bigr)\gt\mu\bigl([Co(en)(NH_{3})_{2}Cl_{2}]^{+}\bigr)$$. Statement C is correct.

Statement D
A low-spin octahedral $$Co^{3+}$$ complex utilises the two vacant inner (n-1)$$d$$ orbitals together with one $$s$$ and three $$p$$ orbitals, giving $$d^{2}sp^{3}$$ hybridisation (inner-orbital complex). The outer-orbital hybridisation $$sp^{3}d^{2}$$ would correspond to a high-spin case, which is not applicable here.
Statement D is incorrect.

Therefore the correct statements are:
Option A (tetrahedral geometry of $$[FeCl_{4}]^{-}$$),
Option B (two geometrical isomers),
Option C (larger magnetic moment for $$[FeCl_{4}]^{-}$$).

Answer: Option A, Option B, Option C.