Question:
The correct structures of trans-[NiBr₂(PPh₃)₂] and meridional-[Co(NH₃)₃(NO₂)₃], respectively, are :
The correct structures of trans-[NiBr₂(PPh₃)₂] and meridional-[Co(NH₃)₃(NO₂)₃], respectively, are :
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In a 'mer' isomer, one pair of identical ligands is trans ($180^\circ$). In a 'fac' isomer, all identical ligands are cis ($90^\circ$).
Updated On: Feb 2, 2026
- Trans Ni with Br opposite, mer Co with NO₂ adjacent
- Trans Ni with P opposite, fac Co
- Trans Ni, mer Co
- Trans Ni, fac Co
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The Correct Option is C
Solution and Explanation
Step 1: In **trans** geometry, identical ligands are at $180^\circ$ (opposite). $[NiBr_2(PPh_3)_2]$ is square planar; "trans" means Br-Ni-Br is a straight line.
Step 2: In octahedral $[MA_3B_3]$ complexes, **meridional (mer)** means the three identical ligands occupy the vertices of a triangle that passes through the metal center (a meridian).
Step 3: In **facial (fac)**, the three ligands are on one triangular face of the octahedron.
Step 2: In octahedral $[MA_3B_3]$ complexes, **meridional (mer)** means the three identical ligands occupy the vertices of a triangle that passes through the metal center (a meridian).
Step 3: In **facial (fac)**, the three ligands are on one triangular face of the octahedron.
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