To determine the correct statements for the complex $[Fe(CN)_6]^{3-}$, we need to analyze its oxidation state, electronic configuration, hybridization, and magnetic properties.

Determine Iron's Oxidation State in $[Fe(CN)_6]^{3-}$

First, let's find the oxidation state of the central metal ion, Iron (Fe). Let the oxidation state of Fe be '$x$'. The cyanide ligand ($CN^-$) has a charge of -1.

The sum of the oxidation states of all atoms in the complex must equal the overall charge of the complex, which is -3.

The equation is:

$x + 6 \times (-1) = -3$

$x - 6 = -3$

$x = -3 + 6$

$x = +3$

So, the oxidation state of Iron is +3.

Electronic Configuration of Iron

The atomic number of Iron (Fe) is 26.

• The electronic configuration of a neutral Fe atom is: [Ar] 3d6 4s2.
• For Fe³⁺, three electrons are lost: two from the 4s orbital and one from the 3d orbital.
• The electronic configuration of Fe³⁺ ion is: [Ar] 3d5.

Ligand Nature and Electron Pairing

The cyanide ligand ($CN^-$) is a strong field ligand. Strong field ligands cause significant crystal field splitting and promote the pairing of electrons in the d-orbitals of the metal ion.

Fe³⁺ has a 3d5 configuration. In an octahedral field, the d-orbitals split into two sets: t2g (lower energy) and eg (higher energy).

For a d5 configuration with a strong field ligand, the electrons fill the lower energy t2g orbitals first, and pairing occurs.

The distribution of 5 d-electrons is:

• t2g orbitals (3 orbitals): (↑↓) (↑↓) (↑)
• eg orbitals (2 orbitals): ( ) ( ) (empty)

This results in one unpaired electron in the t2g set.

Determine Hybridization and Magnetic Properties

For an octahedral complex with 6 ligands, the central metal ion needs 6 empty hybrid orbitals. Since $CN^-$ is a strong field ligand, it utilizes the inner 3d orbitals for hybridization, if available after pairing.

The available orbitals for Fe³⁺ are:

• 3d orbitals: t2g set has 5 electrons, leaving the eg set (two 3d orbitals) empty.
• 4s orbital: Empty.
• 4p orbitals: Empty.

To form 6 hybrid orbitals using inner d orbitals, one s orbital, and three p orbitals, the hybridization is d2sp3. This uses the two empty 3d orbitals (eg set), the one 4s orbital, and the three 4p orbitals.

Therefore, statement (D) d2sp3 hybridization is correct.

Statement (B) sp3d2 hybridization uses outer d orbitals and is incorrect for this complex with a strong field ligand.

Since there is one unpaired electron (as shown in the t2g configuration), the complex is paramagnetic.

Therefore, statement (A) Paramagnetic is correct.

Calculate Magnetic Moment

The magnetic moment ($\mu$) of a complex is calculated using the spin-only formula:

$\mu = \sqrt{n(n+2)}$ BM

where '$n$' is the number of unpaired electrons.

In $[Fe(CN)_6]^{3-}$, we found $n=1$.

Calculating the magnetic moment:

$\mu = \sqrt{1(1+2)} = \sqrt{1 \times 3} = \sqrt{3}$

$\mu \approx 1.732$ BM

Statement (C) states that the magnetic moment is 5.92 BM. This value corresponds to 5 unpaired electrons ($\sqrt{5(5+2)} = \sqrt{35} \approx 5.916$ BM). Since our calculated magnetic moment is approximately 1.732 BM, statement (C) is incorrect.

Conclusion on Correct Statements

Based on the analysis:

• Statement (A) Paramagnetic is correct.
• Statement (B) sp3d2 hybridization is incorrect.
• Statement (C) Magnetic moment = 5.92 BM is incorrect.
• Statement (D) d2sp3 hybridization is correct.

Therefore, the correct statements are (A) and (D) only.