1. Find the oxidation state of chromium in K_3[Cr(oxalate)_3].

   • The complex ion is $[Cr(ox)_3]^{3-}$ because there are $3K^+$ outside.
   • Each oxalate ligand $(ox)$ has charge $-2$.

   Let oxidation state of Cr be $x$: $x + 3(-2) = -3$ $x - 6 = -3$ $x = +3$

   So, chromium is $Cr^{3+}$.

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

   • Atomic number of Cr = 24
   • Neutral Cr: $[Ar]3d^5 4s^1$
   • For $Cr^{3+}$, remove electrons first from $4s$ and then $3d$: $Cr^{3+} = [Ar]3d^3$

3. Nature of ligand and geometry.

   • Oxalate is a bidentate ligand.
   • Three oxalate ligands give coordination number 6, so the complex is octahedral.

4. Distribution of $d^3$ electrons in octahedral field.

   In an octahedral crystal field: $t_{2g} < e_g$

   For $d^3$ configuration, the electrons occupy: $t_{2g}^3 e_g^0$

   By Hund's rule, all three electrons remain unpaired in the three $t_{2g}$ orbitals.

5. Count unpaired electrons.

   Number of unpaired electrons = $3$.

Therefore, the total number of unpaired electrons present in $K_3[Cr(oxalate)_3]$ is: $\boxed{3}$