Consider the following reactions of complex compounds A, B and C<br>(i) CoCl2Br \cdot 5NH3 + excess — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Pd(II) is a 4d^8 metal ion. In 4d transition metals, the crystal field splitting energy is always very large, forcing all 4d^8 complexes to adopt a low-spin square planar geometry (diamagnetic). Step 2: For the analogous 3d^8 Ni(II) complex, [NiCl2(PMe3)2], the smaller 3d orbital splitting allows it to adopt a tetrahedral geometry to minimize steric hindrance of the bulky PMe3 ligands. Step 3: A d^8 ion in a tetrahedral field has the electronic configuration e^4 t_2^4, which contains exactly 2 unpaired electrons in the t_2 orbitals. Thus, option (c) is correct.