The correct structures of trans-[NiBr2(PPh3)2] and meridional-[Co(NH3)3(NO2)3] respectively, are: (Options show…

🧠 Identify Each Geometry

• $[\mathrm{NiBr_2(PPh_3)_2}]$: Ni(II) d⁸ with bulky PPh₃ ligands → square planar. trans = Br positioned across (180°).
• $[\mathrm{Co(NH_3)_3(NO_2)_3}]$: Co(III) d⁶ → octahedral. mer (meridional) = 3 NO₂ on a meridian (great circle), with two NO₂ trans to each other and one NO₂ between them.

🗺️ Trans-square planar Ni

In square planar $[\mathrm{NiBr_2(PPh_3)_2}]$, the two Br ligands occupy opposite corners (180° apart), and the two PPh₃ ligands occupy the other two opposite corners. Both Br trans to each other, both PPh₃ trans to each other.

🗺️ mer-octahedral Co

In mer-$[\mathrm{Co(NH_3)_3(NO_2)_3}]$, NO₂ ligands sit on a meridian: positions (1, 2, 6) of the octahedron, with NO₂ at position 1 trans to NO₂ at position 6, and NO₂ at position 2 between (cis to both). NH₃ ligands occupy positions 3, 4, 5.

⚡ fac vs mer Memory Trick

• fac (facial): 3 like ligands form a triangle on one face; no two are trans.
• mer (meridional): 3 like ligands form a "T" or meridian; two are trans to each other.

⚠️ Square Planar with PPh₃ Always trans

Bulky phosphines like PPh₃ tend to occupy trans positions in square planar to minimise steric clash — making the trans isomer thermodynamically favoured.

$\boxed{\text{Answer: (1) trans-sq pl Ni with Br trans; mer-Co with NO}_2\text{ in meridional plane}}$