The correct statement on the isomerism associated with the following complex ions.
(a) $$[Ni(H_2O)_5NH_3]^{2+}$$
(b) $$[Ni(H_2O)_4(NH_3)_2]^{2+}$$ and
(c) $$[Ni(H_2O)_3(NH_3)_3]^{2+}$$ is:

First of all we recall that the nickel(II) ion, $$Ni^{2+}$$, generally forms octahedral complexes. In an octahedral field there are six positions arranged at the corners of a regular octahedron round the metal. When the ligands are monodentate (here $$H_2O$$ and $$NH_3$$), the type of isomerism that can appear depends only on how many ligands of each kind are present.

An octahedral complex $$[MA_6]$$ containing one kind of ligand has only one possible arrangement, so no isomerism is possible. As soon as two kinds of ligands are present, we label the possible stoichiometries and examine them one by one.

We need two facts.

• Fact 1 - Geometrical (cis-trans / fac-mer) isomerism.
  • $$MA_5B$$ has only one possible arrangement, so it cannot show geometrical isomerism.
  • $$MA_4B_2$$ can have the two $$B$$ ligands opposite each other (trans) or adjacent (cis), so it does show geometrical isomerism.
  • $$MA_3B_3$$ can exist as fac (three identical ligands occupy one triangular face) and mer (three identical ligands lie on a meridian), so it also does show geometrical isomerism.

• Fact 2 - Optical isomerism.
For octahedral complexes that contain only monodentate ligands, a necessary condition for optical activity is the absence of any plane of symmetry, centre of symmetry or improper rotation axis. The common simple formulas that satisfy this condition are the chelated types $$M(AA)_3$$, $$M(AA)_2BC$$, $$MABCDEF$$ etc. Simple arrangements such as $$MA_5B$$, $$MA_4B_2$$ and $$MA_3B_3$$ constructed exclusively from monodentate ligands always possess at least one symmetry element, therefore they are optically inactive.

Armed with these two facts we analyse each complex.

(a) $$[Ni(H_2O)_5NH_3]^{2+}$$

We compare its formula with the general notation:

$$[Ni(H_2O)_5NH_3]^{2+}\; \equiv\; MA_5B$$ with $$A = H_2O$$ and $$B = NH_3$$.

From Fact 1, the type $$MA_5B$$ has only one possible spatial arrangement, hence no geometrical isomerism. From Fact 2, it is automatically optically inactive. Therefore complex (a) shows no isomerism at all.

(b) $$[Ni(H_2O)_4(NH_3)_2]^{2+}$$

Its pattern fits

$$[Ni(H_2O)_4(NH_3)_2]^{2+}\; \equiv\; MA_4B_2$$ with $$A = H_2O,\; B = NH_3$$.

For $$MA_4B_2$$ the two $$B$$ ligands can be placed:

$$\displaystyle \text{cis: } \angle(B{-}Ni{-}B)=90^{\circ}$$
$$\displaystyle \text{trans: } \angle(B{-}Ni{-}B)=180^{\circ}$$

Thus complex (b) exists as a cis and a trans pair, giving geometrical isomerism.

The cis as well as the trans form possesses a plane of symmetry passing through the metal and bisecting the two $$B$$ ligands (in the cis form) or passing through both $$B$$ ligands themselves (in the trans form). Hence each form is superposable on its mirror image and cannot be optically active. So there is no optical isomerism for (b).

(c) $$[Ni(H_2O)_3(NH_3)_3]^{2+}$$

This matches

$$[Ni(H_2O)_3(NH_3)_3]^{2+}\; \equiv\; MA_3B_3$$ where again $$A = H_2O,\; B = NH_3$$.

For $$MA_3B_3$$ two arrangements are possible:

1. fac (facial): the three $$A$$ ligands occupy one face of the octahedron; all three $$A{-}Ni{-}A$$ angles are $$90^{\circ}$$.
2. mer (meridional): two $$A$$ ligands are trans and the third is cis to both; the set of three $$A$$ ligands lies on a meridian.

Therefore complex (c) also exhibits geometrical isomerism (fac and mer). Each of these two arrangements contains at least one plane (or an improper axis) of symmetry, so by Fact 2 neither the fac nor the mer form is chiral. Hence no optical isomerism arises for (c).

We may now summarise the findings:

• (a) shows neither geometrical nor optical isomerism.
• (b) shows geometrical isomerism only.
• (c) shows geometrical isomerism only.

The pair of complexes that satisfy “show only geometrical isomerism” is therefore (b) and (c).

Hence, the correct answer is Option D.