Identify the geometry of $\text{trans-}[NiBr_2(PPh_3)_2]$

• Nickel is in the $+2$ oxidation state: $Ni^{2+} : d^8$
• The ligands are $Br^-$ and $PPh_3$.
• For a $d^8$ metal ion, the geometry depends on ligand strength:
  • strong-field ligands often give square planar
  • weak-field ligands often give tetrahedral
• Here, $Ni^{2+}$ with phosphine ligands $PPh_3$ strongly favors square planar arrangement.

Therefore, $\text{trans-}[NiBr_2(PPh_3)_2]$ is square planar, with the two $Br$ ligands opposite to each other and the two $PPh_3$ ligands opposite to each other.

Identify the geometry of meridional-$[Co(NH_3)_3(NO_2)_3]$

• Oxidation state of cobalt: $x + 3(0) + 3(-1) = 0 \Rightarrow x=+3$ so cobalt is $Co^{3+}$.
• Coordination number is $6$, hence the complex is octahedral.
• For an octahedral complex of the type $MA_3B_3$, two geometrical isomers are possible:
  • fac: the three identical ligands occupy one face of the octahedron
  • mer: the three identical ligands lie on a meridian; two are trans and one is cis to both

Since the complex is explicitly meridional, the structure must be the mer-octahedral form of $[Co(NH_3)_3(NO_2)_3]$.

Match with the correct option

So the required pair is:

• $\text{trans-}[NiBr_2(PPh_3)_2]$ : square planar trans structure
• mer-$[Co(NH_3)_3(NO_2)_3]$ : mer-octahedral structure

Among the given options, this corresponds to Option D.

Conclusion

The correct answer is: $\boxed{D}$