The total number of unpaired electrons present in the complex K$$_3$$[Cr(oxalate)$$_3$$] is ________.

In the complex $$\text{K}_3[\text{Cr(oxalate)}_3]$$, we first determine the oxidation state of chromium. Each oxalate ion ($$\text{C}_2\text{O}_4^{2-}$$) carries a charge of $$-2$$, and there are three of them, giving a total negative charge of $$-6$$. The three $$\text{K}^+$$ ions contribute $$+3$$ charge. For the complex to be neutral: $$+3 + x + 3(-2) = 0$$, so $$x = +3$$. Chromium is in the $$+3$$ oxidation state.

The electronic configuration of $$\text{Cr}^{3+}$$ is $$[\text{Ar}]\, 3d^3$$. Now we need to determine whether the electrons are paired or unpaired. Oxalate is a bidentate ligand, and three oxalate ions create an octahedral environment around Cr. Oxalate is a weak-field ligand (it falls on the weaker side of the spectrochemical series), so it does not cause pairing of the $$d$$-electrons.

With three electrons in the $$d$$-orbitals and a weak field (high-spin configuration), all three electrons occupy separate $$t_{2g}$$ orbitals with parallel spins according to Hund's rule: $$t_{2g}^3 \, e_g^0$$. This is the same arrangement regardless of whether the field is strong or weak, since three electrons in three degenerate $$t_{2g}$$ orbitals naturally remain unpaired.

Therefore the total number of unpaired electrons is $$3$$.