Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other.

[en = $$H_2NCH_2CH_2NH_2$$]

For this problem we have to verify, for every option (A-D), two separate conditions.
1. Both complexes listed in Set-I must be able to exist as geometrical (cis-trans, fac-mer etc.) isomers.
2. The two complexes listed in Set-II must be ionization isomers of each other (i.e. they must have the same overall formula, but an interchange of an ion inside the coordination sphere with the counter-ion outside).

Condition for geometrical isomerism
• Tetrahedral complexes never show geometrical isomerism.
• Square-planar complexes of the type $$MA_2B_2,\; MA_2BC$$ etc. show cis-trans isomerism.
• Octahedral complexes show geometrical isomerism when at least two different kinds of ligands are present in suitable numbers, e.g. $$MA_4B_2$$ (cis/trans) and $$MA_3B_3$$ (fac/mer).
• Octahedral $$MA_5B$$ possesses only one possible arrangement, hence no geometrical isomerism.

Condition for ionization isomerism
Two complexes are ionization isomers when the total empirical formula (including counter-ions) is the same and an anionic ligand inside the coordination sphere interchanges with an anion that was outside it.

Now we analyse each option.

Option A

Set-I : $$[Ni(CO)_4]$$ is tetrahedral ⇒ no geometrical isomerism. Hence Condition 1 already fails. Option A is rejected.

Option B

Set-I :
• $$[Co(en)(NH_3)_2Cl_2]$$ is octahedral of the type $$M(A\!-\!A)B_2C_2$$ (here en is bidentate), so cis-trans is possible.
• $$[PdCl_2(PPh_3)_2]$$ is square-planar $$MA_2B_2$$, again cis-trans possible. Thus Condition 1 is satisfied.
Set-II : $$[Co(NH_3)_6][Cr(CN)_6]$$ and $$[Cr(NH_3)_6][Co(CN)_6]$$ have completely different ions inside the coordination spheres (the metals themselves are interchanged). They are not obtained by simply exchanging a counter-ion with a ligand. Therefore they are not ionization isomers. Option B is rejected.

Option C

Set-I :
• $$[Co(NH_3)_3(NO_2)_3]$$ is octahedral $$MA_3B_3$$ and exists as fac (facial) and mer (meridional) isomers → geometrical isomerism present.
• $$[Co(en)_2Cl_2]$$ is octahedral $$M(A\!-\!A)_2B_2$$; the two $$Cl^-$$ ligands can be cis (adjacent) or trans (opposite) → geometrical isomerism present. Condition 1 is fulfilled.
Set-II : $$[Co(NH_3)_5Cl]SO_4$$ and $$[Co(NH_3)_5(SO_4)]Cl$$ have identical overall formula $$[Co(NH_3)_5Cl]SO_4$$. One complex contains $$Cl^-$$ inside and $$SO_4^{2-}$$ outside; the other has $$SO_4^{2-}$$ inside and $$Cl^-$$ outside. They are classic ionization isomers. Condition 2 is also fulfilled.

Option D

Set-I : $$[Cr(NH_3)_5Cl]^{2+}$$ is octahedral $$MA_5B$$; all five NH3 positions are equivalent and there is only one possible arrangement. Hence no geometrical isomerism. Condition 1 fails → Option D is rejected (further, its Set-II involves hydrate rather than ionization isomerism).

Only Option C satisfies both required conditions.

Option C which is: Set-I $$[Co(NH_3)_3(NO_2)_3]$$ and $$[Co(en)_2Cl_2]$$; Set-II $$[Co(NH_3)_5Cl]SO_4$$ and $$[Co(NH_3)_5(SO_4)]Cl$$