In [Pt(NH3)2Cl2], pt-Cl bond length is 2 Å and Cl-Cl distance is 2.88 Å then the compound is : — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: In square planar complexes of Pt(II), the metal sits at the center of a square with a Pt-Cl bond length of r = 2 Å. Step 2: For a cis-isomer, the two Cl^- ligands are at adjacent corners with a Cl-Pt-Cl angle of 90°. The theoretical Cl-Cl distance is d = \sqrt{r^2 + r^2} = r\sqrt{2} = 2 * 1.414 = 2.83 Å. Step 3: Since this matches the observed distance of 2.88 Å (widened slightly due to ligand-ligand repulsions), whereas a trans-isomer would have a distance of 2r = 4.0 Å, the complex is cis-square planar (option c).