![Name the complex ion [CrCl2(H2O)4]+ and determine its coordination number and the oxidation state of the central metal ion.](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1776220261169-1c750f6781c1a65e.png&w=3840&q=75)
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Hey Dany, good to see you again. Let's name the complex ion and determine its coordination number. The complex ion is [CrCl_2(H_2O)_4]^+. Step 1: Determine the coordination number. The coordination number is the total number of ligands directly attached to the central metal ion. There are 2 chloride (Cl^-) ligands. Each is monodentate. There are 4 water (H_2O) ligands. Each is monodentate. Coordination number = (number of Cl ligands) + (number of H_2O ligands) Coordination number = 2 + 4 = 6. Step 2: Determine the oxidation state of the central metal ion (Chromium, Cr). Let the oxidation state of Cr be x. Chloride (Cl^-) has a charge of -1. Water (H_2O) is a neutral ligand, so its charge is 0. The overall charge of the complex ion is +1. x + 2(-1) + 4(0) = +1 x - 2 = +1 x = +3 So, the oxidation state of Chromium is +3, written as (III). Step 3: Name the ligands. Cl^- is named chloro. Since there are two, it's dichloro*. H_2O is named aqua. Since there are four, it's tetraaqua*. Step 4: Assemble the name. Ligands are named alphabetically before the metal. "Aqua" comes before "chloro". The metal name is followed by its oxidation state in Roman numerals. Since the complex is a cation, the metal name remains "chromium". Name: Tetraaquodichlorochromium(III) ion Name: Tetraaquodichlorochromium(III) ion Coordination number: 6 Send me the next one 📸