Evaluate Properties and Isomerism of Coordination Complexes

To determine the correct statement(s), we analyze each option individually:

Statement (A): $[\text{FeCl}_4]^-$ exhibits tetrahedral geometry.

1. Oxidation State of Iron ($\text{Fe}$): $\text{Oxidation state of Fe} + 4(-1) = -1 \implies \text{Fe} \text{ is in } +3 \text{ oxidation state ($\text{Fe}^{3+}$).}$

2. Electronic Configuration:

   • Free atom $\text{Fe}$: $[\text{Ar}]\,3d^6 4s^2$
   • $\text{Fe}^{3+}$ ion: $[\text{Ar}]\,3d^5$

3. Geometry and Hybridization:

   • $\text{Cl}^-$ is a weak field ligand.
   • For a 4-coordinate complex of $\text{Fe}^{3+}$ ($3d^5$) with weak-field ligands, pairing of electrons does not occur, and steric repulsion leads to $sp^3$ hybridization.
   • Thus, $[\text{FeCl}_4]^-$ adopts a tetrahedral geometry.

Therefore, Statement (A) is correct.

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Statement (B): $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$ possesses two geometrical isomers.

1. Complex Type: This is an octahedral complex of the type $[\text{M(en)a}_2\text{b}_2]^+$, where $\text{en}$ is a symmetrical bidentate ligand, $\text{a} = \text{NH}_3$, and $\text{b} = \text{Cl}^-$.

2. Geometrical Isomers: Since the bidentate ligand $\text{en}$ must occupy adjacent (cis) coordination sites, the remaining four sites are occupied by two $\text{NH}_3$ and two $\text{Cl}^-$ ligands. The possible arrangements are:

   • Isomer 1: $\text{Cl}^-$ ligands are trans to each other (which forces $\text{NH}_3$ ligands to be cis to each other).
   • Isomer 2: $\text{NH}_3$ ligands are trans to each other (which forces $\text{Cl}^-$ ligands to be cis to each other).
   • Isomer 3: Both $\text{Cl}^-$ ligands are cis to each other AND both $\text{NH}_3$ ligands are cis to each other.

3. Thus, the complex has 3 geometrical isomers (one of which is optically active, existing as a pair of enantiomers).

Therefore, Statement (B) is incorrect.

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Statement (C): $[\text{FeCl}_4]^-$ displays a higher spin-only magnetic moment than $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$.

1. Magnetic Moment of $[\text{FeCl}_4]^-$:

   • $\text{Fe}^{3+}$ ($3d^5$) in a tetrahedral field (weak field) has the configuration $e^2 t_2^3$.
   • Number of unpaired electrons, $n = 5$.
   • Spin-only magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\text{ BM}$

2. Magnetic Moment of $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$:

   • Oxidation state of Cobalt: $x + 0 + 2(0) + 2(-1) = +1 \implies x = +3\ (\text{Co}^{3+})$
   • $\text{Co}^{3+}$ has a $3d^6$ electronic configuration.
   • Strong-field ligands ($\text{en}$ and $\text{NH}_3$) cause maximum pairing of $d$-electrons, yielding a low-spin octahedral complex with configuration $t_{2g}^6 e_g^0$.
   • Number of unpaired electrons, $n = 0$.
   • Spin-only magnetic moment: $\mu = 0\text{ BM}$

Since $5.92\text{ BM} > 0\text{ BM}$, $[\text{FeCl}_4]^-$ has a higher spin-only magnetic moment than $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$.

Therefore, Statement (C) is correct.

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Statement (D): The cobalt ion in $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$ features $sp^3d^2$ hybridization.

1. In $[\text{Co(en)}(\text{NH}_3)_2\text{Cl}_2]^+$, $\text{Co}^{3+}$ ($3d^6$) is surrounded predominantly by strong-field chelating and amine ligands ($\text{en}$ and $\text{NH}_3$).
2. Due to strong crystal field splitting ($\Delta_o > P$), electrons pair up in the inner $3d$ orbitals ($t_{2g}^6$), leaving two vacant $3d$ orbitals.
3. The complex utilizes inner $3d$ orbitals along with $4s$ and $4p$ orbitals, forming an inner orbital octahedral complex with $d^2sp^3$ hybridization (not $sp^3d^2$).

Therefore, Statement (D) is incorrect.

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Conclusion:

The correct statements are A and C.