Hybridization and shape of $Cs[FeCl_4]$ is:

$sp^3$, tetrahedral

The correct answer is Option (2) → $sp^3$, tetrahedral

Let us go through the details of the hybridization and shape of \( \text{Cs[FeCl}_4] \) step by step.

Determining the Oxidation State of Iron
In \( \text{Cs[FeCl}_4] \):

Cesium (Cs) has a known oxidation state of +1.

Chloride ions (Cl) each have an oxidation state of -1.

Setting up the equation based on the charges, we have:

\(\text{Fe} + 4(-1) = 0\)

From this, we can solve for the oxidation state of iron:

\(\text{Fe} - 4 = 0 \implies \text{Fe} = +4\)

Thus, iron is in the +4 oxidation state in this complex.

The coordination number refers to the number of ligand atoms that are directly bonded to the central metal atom (iron in this case). Here, iron is coordinated by four chloride ions (\( \text{Cl}^- \)), which gives it a coordination number of 4.

The hybridization can be determined based on the coordination number:

For a coordination number of 4, two common types of hybridization exist:

\( \text{sp}^3 \): This hybridization leads to a tetrahedral geometry.

\( \text{dsp}^2 \): This hybridization leads to a square planar geometry, typically seen with transition metals that have d-orbitals involved.

However, in the case of \( \text{Cs[FeCl}_4] \):

The \( \text{Fe}^{4+} \) ion has a configuration of \( [Ar] \, 3d^6 \, 4s^0 \) (as it loses the 4s electrons before the 3d).

Since it uses the 4s and 3p orbitals to form bonds with four chloride ions, it undergoes \( \text{sp}^3 \) hybridization.

Geometry

With \( \text{sp}^3 \) hybridization and four bonding pairs of electrons (from the four \( \text{Cl}^- \) ligands), the molecule adopts a tetrahedral geometry.

In a tetrahedral arrangement, the four ligands are positioned at the corners of a tetrahedron, maximizing the distance between them to minimize electron repulsion.

The structure can be visualized as having four chloride ions situated around the iron atom in a three-dimensional tetrahedral shape

Conclusion

Putting all of this together, we find that the hybridization of \( \text{Cs[FeCl}_4] \) is \( \text{sp}^3 \), leading to a tetrahedral geometry.

Thus, the final answer is sp³, tetrahedral.