Name the complex ion [CrCl2(H2O)4]+ and determine its coordination number and the oxidation state of the central metal ion.

The coordination number is the total number of ligands directly attached to the central metal ion.

• There are 2 chloride ($\text{Cl}^-$) ligands. Each is monodentate.
• There are 4 water ($\text{H}_2\text{O}$) ligands. Each is monodentate. Coordination number = (number of $\text{Cl}$ ligands) + (number of $\text{H}_2\text{O}$ ligands) Coordination number = $2 + 4 = 6$.

Let the oxidation state of $\text{Cr}$ be $x$.

• Chloride ($\text{Cl}^-$) has a charge of -1.
• Water ($\text{H}_2\text{O}$) is a neutral ligand, so its charge is 0. The overall charge of the complex ion is +1.

So, the oxidation state of Chromium is +3, written as (III).

• $\text{Cl}^-$ is named chloro. Since there are two, it's dichloro.
• $\text{H}_2\text{O}$ is named aqua. Since there are four, it's tetraaqua.

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: $\boxed{\text{Tetraaquodichlorochromium(III) ion}}$
Coordination number: $\boxed{6}$