What is the oxidation state of Pt in (PtF_5)_4 ?

The correct answer is Option (2) → +5

The oxidation state of a metal in a neutral compound follows from the charge balance: the oxidation states of all the atoms add up to zero. Fluorine, the most electronegative element, is always −1 in its compounds.

Platinum pentafluoride exists as a tetramer, \( \ce{(PtF5)4} \), in which some fluorine atoms bridge two platinum atoms. Bridging does not change the charge of fluorine, so each Pt still carries five fluorines. Taking the oxidation state of Pt as x:

    \( \displaystyle 4x + 20(-1) = 0 \)

    \( \displaystyle x = +5 \)

So platinum is in the +5 oxidation state. Pt (Z = 78) is \( \mathrm{[Xe]\,4f^{14}\,5d^{9}\,6s^{1}} \); losing the 6s electron and four 5d electrons gives \( \ce{Pt^5+} \): \( \mathrm{5d^{5}} \).