The number of geometrical isomers found in the metal complexes $$[PtCl_2(NH_3)_2]$$, $$[Ni(CO)_4]$$, $$[Ru(H_2O)_3Cl_3]$$ and $$[CoCl_2(NH_3)_4]^+$$ respectively, are:

We begin with the first complex $$[PtCl_2(NH_3)_2]$$. Platinum in the +2 oxidation state has a $$d^8$$ electronic configuration, and the well-known rule is that $$d^8$$ metal ions of the second and third transition series generally adopt a square-planar geometry. In a square-planar arrangement an $$MA_2B_2$$ type molecule can place the two identical ligands either next to each other (cis) or opposite to each other (trans). Therefore for $$[PtCl_2(NH_3)_2]$$ we obtain two distinct geometrical isomers, one cis and one trans.

Next we consider $$[Ni(CO)_4]$$. The oxidation state of nickel here is zero because carbon monoxide is a neutral ligand. The familiar 18-electron rule tells us that $$Ni(0)$$ prefers a tetrahedral coordination when it binds four ligands. In a perfect tetrahedron all four positions are equivalent; rotating the molecule can superimpose any arrangement of identical ligands on any other. Hence there is no way to create a second, nonsuperimposable arrangement simply by exchanging ligand positions. Consequently, the number of geometrical isomers is zero.

Now we study $$[Ru(H_2O)_3Cl_3]$$. Both $$H_2O$$ and $$Cl^-$$ are monodentate. With six monodentate ligands the usual geometry is octahedral. The complex is of the $$MA_3B_3$$ form. For an octahedron of the $$MA_3B_3$$ type there are exactly two geometrical possibilities: (i) the three identical ligands occupying one face of the octahedron (facial, abbreviated fac) and (ii) the three identical ligands arranged around the meridian of the octahedron (meridional, abbreviated mer). Therefore this complex possesses two geometrical isomers.

Finally we turn to $$[CoCl_2(NH_3)_4]^+$$. The cobalt is in the +3 oxidation state, which is well known to give an octahedral geometry with six coordination sites. The formula places it in the $$MA_4B_2$$ category, where $$A = NH_3$$ and $$B = Cl^-$$. In an octahedron of $$MA_4B_2$$ type the two $$B$$ ligands may lie opposite each other (trans) or adjacent to each other (cis). These two possibilities cannot be interconverted without breaking bonds, so they are distinct geometrical isomers, giving a total of two.

Collecting our results, the counts of geometrical isomers are:

$$[PtCl_2(NH_3)_2] : 2$$

$$[Ni(CO)_4] : 0$$

$$[Ru(H_2O)_3Cl_3] : 2$$

$$[CoCl_2(NH_3)_4]^+ : 2$$

So the required sequence is $$2,\,0,\,2,\,2$$.

Comparing with the options, this matches Option B.

Hence, the correct answer is Option B.