1. Find oxidation state of Fe in $K_3[Fe(CN)_6]$

   Let oxidation state of Fe be $x$.

   In the complex ion $[Fe(CN)_6]^{3-}$:

   • Each $CN^-$ ligand has charge $-1$
   • Total ligand charge $= 6(-1) = -6$

   So, $x - 6 = -3$ $x = +3$

   Hence, iron is $Fe^{3+}$.

2. Write electronic configuration of $Fe^{3+}$

   Atomic number of Fe = 26.

   Neutral Fe: $[Ar]3d^6 4s^2$

   For $Fe^{3+}$, remove two electrons from $4s$ and one from $3d$: $Fe^{3+} = [Ar]3d^5$

3. Nature of ligand $CN^-$

   $CN^-$ is a strong field ligand, so it causes pairing of electrons in the $d$-orbitals.

   For an octahedral complex with $d^5$ configuration and strong field ligands, we get low-spin arrangement: $t_{2g}^5 e_g^0$

4. Count number of unpaired electrons

   In $t_{2g}^5$:

   • two orbitals are paired
   • one orbital has one unpaired electron

   Therefore, number of unpaired electrons: $n = 1$

5. Calculate spin-only magnetic moment

   Formula: $\mu_s = \sqrt{n(n+2)}\;\text{B.M.}$

   Substituting $n=1$: $\mu_s = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\;\text{B.M.}$

6. Nearest integer

   $1.73 \approx 2$

Final Answer

Spin-only magnetic moment of $K_3[Fe(CN)_6]$ is: $\boxed{2}$