The total number of possible isomers for square planar [Pt(Cl)(NO$$_2$$)(NO$$_3$$)(SCN)]$$^{2-}$$ is:

We have the square-planar complex $$[\,\text{Pt}(\text{Cl})(\text{NO}_2)(\text{NO}_3)(\text{SCN})\,]^{2-}.$$ Its central metal ion $$\text{Pt}^{\;II}$$ invariably adopts a square-planar geometry, so four ligands occupy the four corners of a square in the same plane.

Step 1 — geometric arrangements: For an $$\text{MA B C D}$$ square-planar species in which all four ligands A, B, C, D are different, the total number of possible geometric (cis-trans) arrangements is obtained as follows.

We first place one ligand (say $$\text{Cl}$$) at any corner. Because a square can be rotated in the plane through $$90^{\circ}, 180^{\circ}$$ or $$270^{\circ}$$ without producing a new arrangement, rotation eliminates the apparent fourfold multiplicity. What distinguishes the isomers is therefore only which ligand is placed trans (opposite) to the fixed ligand. Three other ligands are available, so

$$N_{\text{geometric}} = 3.$$

Step 2 — linkage possibilities: Two of the remaining ligands are ambidentate, i.e. they can coordinate through either of two different donor atoms.

• $$\text{NO}_2^-$$ may bind through N to give a nitro ligand or through O to give a nitrito ligand – 2 linkage forms.
• $$\text{SCN}^-$$ may bind through S to give a thiocyanato ligand or through N to give an isothiocyanato ligand – 2 linkage forms.

The nitrate ion $$\text{NO}_3^-$$, when monodentate, can coordinate only through an oxygen atom, so it contributes no extra linkage variation.

Because the choices of linkage for $$\text{NO}_2^-$$ and $$\text{SCN}^-$$ are independent, the total number of linkage combinations for each geometric arrangement is

$$N_{\text{linkage per geometry}} = 2 \times 2 = 4.$$

Step 3 — overall count: Multiplying the number of geometric isomers by the number of linkage variants associated with each one, we obtain

$$N_{\text{total isomers}} = N_{\text{geometric}} \times N_{\text{linkage per geometry}} = 3 \times 4 = 12.$$

Hence, the correct answer is Option B.