Let be the number of unpaired electrons in and be the coordination number of the central metal in . — Coordination Compounds Chemistry Question

💡 Solution & Explanation

In $[MnCl_4]^{2-}$, $Mn$ is in the $+2$ oxidation state with a $3d^5$ configuration. $Cl^-$ is a weak field ligand, and the complex is tetrahedral ($sp^3$ hybridized). Thus, pairing does not occur, leaving $5$ unpaired electrons ($x = 5$). In $[HgI_3]^-$, the formula explicitly shows $3$ iodido ligands bound to mercury, giving a trigonal planar ($sp^2$) geometry with a coordination number of $3$ ($y = 3$). The product $(5 \times 3) = 15$.