The oxidation state of the Ru metal ion in [Ru(Terpy)2]2+ (PF6)2 is +2.

Explanation:

Oxidation state refers to the charge that an atom or ion has when it gains or loses electrons. In this case, we are looking at the oxidation state of the Ru metal ion in the complex [Ru(Terpy)2]2+.

1. Understanding the ligand:
The ligand in this complex is Terpy, which stands for 2,2':6',2''-terpyridine. Terpy is a tridentate ligand, meaning it can donate three electron pairs to the metal ion. Each Terpy ligand binds to the metal ion through its three nitrogen atoms.

2. Determining the oxidation state of the metal ion:
To determine the oxidation state of the Ru metal ion, we need to consider the charge on the overall complex and the charge contributed by the ligands.

- The overall complex has a charge of 2+ due to the presence of two PF6- counterions.
- Each Terpy ligand is neutral since it donates three electron pairs to the metal ion. Therefore, the total charge contributed by the Terpy ligands is 0.
- The sum of the charges on the complex and the ligands should equal the charge on the metal ion.

If we let x represent the oxidation state of the Ru metal ion, we can set up the following equation:

2+ (overall charge) + 0 (ligand charge) = x (oxidation state of the Ru metal ion)

Simplifying the equation, we get:

2 = x

Therefore, the oxidation state of the Ru metal ion in [Ru(Terpy)2]2+ (PF6)2 is +2.