Select the correct statements for [Fe(CN)₆]³⁻

(A) Paramagnetic

(B) sp³d² hybridization

(C) Magnetic moment = 5.92 BM

(D) d²sp³ hybridization

Choose the correct answer from the options given below:

The correct option is (A) and (D) only.

Let us analyze the coordination entity $[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6{]}^{3-}$ step-by-step to understand why statements (A) and (D) are correct:

1. Oxidation State of the Central Metal Ion:

Let the oxidation state of iron (Fe) be $x$.

Since cyanide (${\mathrm{CN}}^{-}$) is a monodentate ligand with a charge of $-1$:

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

$x-6=-3$

$x=+3$

Thus, the iron ion is in the $+3$ oxidation state (${\mathrm{Fe}}^{3+}$).

2. Electronic Configuration:

The ground state electronic configuration of neutral iron ($\mathrm{Fe}$, atomic number $26$) is:

$\left[\mathrm{Ar}\right]3d^{6}4s^2$

For the ${\mathrm{Fe}}^{3+}$ ion, three electrons are removed (two from the $4s$ orbital and one from the $3d$ orbital):

${\mathrm{Fe}}^{3+}=\left[\mathrm{Ar}\right]3d^5$

3. Nature of Ligands and Hybridization:

Cyanide (${\mathrm{CN}}^{-}$) is a strong field ligand. According to Crystal Field Theory and Valence Bond Theory, a strong field ligand causes pairing of the electrons in the $3d$ orbitals.

The five electrons in the $3d$ subshell of ${\mathrm{Fe}}^{3+}$ pair up as much as possible:

Before pairing (free ion): Five unpaired electrons ($\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$ $\uparrow$)

After pairing (due to ${\mathrm{CN}}^{-}$): Two pairs and one unpaired electron ($\uparrow \downarrow$ $\uparrow \downarrow$ $\uparrow$) in the $3d$ subshell.

This pairing leaves two internal $3d$ orbitals vacant. These two vacant $3d$ orbitals, along with one $4s$ orbital and three $4p$ orbitals, hybridize to form six $d^{2}s{p}^3$ hybrid orbitals.

Therefore, the hybridization is $d^{2}s{p}^3$ (inner orbital octahedral complex). This makes statement (D) correct and statement (B) incorrect.

4. Magnetic Behavior:

Since there is one unpaired electron remaining in the $3d$ orbitals, the complex is paramagnetic. This makes statement (A) correct.

We can calculate the spin-only magnetic moment ($\mu$) using the formula:

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

where $n$ is the number of unpaired electrons ($n=1$):

$\mu =\sqrt{1(1+2)}=\sqrt{3}\approx 1.73\text{BM}$

Since the magnetic moment is $1.73\text{BM}$ and not $5.92\text{BM}$, statement (C) is incorrect.

Conclusion:

Only statements (A) Paramagnetic and (D) $d^{2}s{p}^3$ hybridization are correct.