The total number of isomers for a square planar complex: $$[MCl(F)(NO_2)(SCN)]$$ is:

We have the square-planar complex $$[\,MCl(F)(NO_2)(SCN)\,]\,.$$

A square-planar species possesses four positions at the corners of a square. Each position is cis (adjacent) to the two neighbouring positions and trans (opposite) to the position across the square. Because free rotation of the whole molecule does not create a new isomer, the only stereochemical issue is which ligand is opposite (trans to) which other ligand.

First we find the geometrical (cis-trans) possibilities. For a formula of type $$MA\,B\,C\,D$$ with four different monodentate ligands the number of different trans pairs equals the number of ways of dividing the four ligands into two unordered pairs. The known combinatorial result is

Hence we can list the three distinct geometrical isomers as follows (showing only the two trans pairs each time):

So, three geometrical isomers exist.

Next we examine possible linkage isomerism. Both $$NO_2^{-}$$ and $$SCN^{-}$$ are ambidentate:

• $$NO_2^{-}$$ can bind through nitrogen to give a nitro ligand $$(-NO_2)$$ or through oxygen to give a nitrito ligand $$(-ONO).$$ Hence, $$NO_2^{-}$$ offers $$2$$ linkage modes.
• $$SCN^{-}$$ can bind through sulfur to give thiocyanato $$(-SCN)$$ or through nitrogen to give isothiocyanato $$(-NCS).$$ Hence, $$SCN^{-}$$ also offers $$2$$ linkage modes.

The chloride and fluoride ligands are not ambidentate, so no extra choices arise from them. Because the choice of donor atom for $$NO_2^{-}$$ is independent of the choice for $$SCN^{-},$$ the total multiplication factor due to linkage isomerism is

Now we combine the two kinds of isomerism. For each of the three geometrical arrangements we may realise the complex in four different linkage forms, giving

No optical isomerism occurs, because every square-planar complex is achiral (it lies in a plane and has an internal mirror plane).

So, the grand total of distinguishable isomers is $$12.$$

Hence, the correct answer is Option A.