[Chaste]

Hi all, this is a ruthenium complex with 3 terminal phosphines ligands, and 3 bridging chlorido ligands with a net positive charge of +1. I tried using the Ionic method to do so, with Ru2+ = 6e, 3x chloride ions = 6e, and 3x phosphine ligands = 6e, yielding a total of 18 electrons.

However, with the covalent method, Ru = 8e, 3x bridging chlorido ligands = 3e, 3x phosphine ligands =6e, and a charge of +1, yielding a total of 16 electrons.

Can someone enlighten me why these 2 methods are inconsistent?

[AWK]

Why did you use Ru⁺² oxidation number. Ru³⁺ is evident from your formula Ru₂Cl₃(PR₃)³⁺

[Chaste]

Why did you use Ru⁺² oxidation number. Ru³⁺ is evident from your formula Ru₂Cl₃(PR₃)³⁺

sorry for the misleading title. What I described in the text is still correct.

[AWK]

Show drawing of corrected structure and calculations for both methods  instead of describing them. Ru2Cl3 with 3 bridging chlorides usually contains 6 phosphine molecules and show octahedral coordination of both ruthenium cations.

[cheese (MSW)]

Hint: The correct formula is [Ru2(μ-Cl)3(PR3)6]+ : symmetrical [P3Ru((μ-Cl)3RuP3]^+. 
You have TWO Ru atoms.  In your e⁻ bookkeeping you'll get ½e⁻s. 
Both e⁻ counting schemes work as they must.
[There is no Ru-Ru bond (X-ray).]

[Chaste]

I solved it. Thanks

[cheese (MSW)]

In conclusion: the Ru atoms are Ru(II) and each have an 18e⁻ configuration.
The ion obeys the 18 Electron Rule.