Animated Solution for Chemistry - s and p-Block Elements: Reaction of $Pt{F}_6$ with oxygen ($O_2$) gas results in the formation of an ionic compound, $X^{+}{Y}^{-}$. Correct statement(s) is (are)

Imagine being Neil Bartlett in 1962. You've just discovered that platinum hexafluoride ($Pt{F}_6$) is such a beast of an oxidizing agent that it can rip an electron away from a stable oxygen molecule ($O_2$).

This single reaction shattered the myth that noble gases couldn't form compounds, because if $Pt{F}_6$ could oxidize oxygen, it could oxidize xenon too!

Let's break down the mechanics of this beautiful reaction and see exactly what happens when these two molecules collide.

When $Pt{F}_6$ reacts with $O_2$, it doesn't share electrons or transfer fluorine atoms. Instead, it acts as a pure electron thief.

The reaction proceeds as follows:

Here, the oxygen molecule loses an electron to form the dioxygenyl cation ($O_2^{+}$), which is our $X^{+}$.

Simultaneously, the platinum hexafluoride molecule gains that electron to become the hexafluoroplatinate anion ($[Pt{F}_6{]}^{-}$), which is our $Y^{-}$.

Let's evaluate the first claim about the bond order of the $X^{+}$ ion.

According to Molecular Orbital Theory, a neutral $O_2$ molecule has a bond order of 2, with two unpaired electrons residing in its antibonding ${\pi }^{\ast }$ orbitals.

When $O_2$ is oxidized to $O_2^{+}$, one of these antibonding electrons is removed.

Removing an electron from an antibonding orbital actually strengthens the bond!

The bond order increases from 2 to 2.5. Therefore, the claim that the bond order is 1.5 is mathematically incorrect.

Now, let's look at the metal ion in our complex, $Y^{-}$.

Platinum (Pt) is a transition metal with an atomic number of 78. Its ground state electronic configuration is an exception to the standard rules:

In the $[Pt{F}_6{]}^{-}$ anion, the six fluorine atoms each carry a -1 charge, and the overall complex has a -1 charge.

This means platinum must be in a +5 oxidation state.

To reach the +5 state, platinum must lose five electrons. It first loses its single $6s$ electron, and then four electrons from the $5d$ subshell.

This leaves exactly 5 electrons in its valence d-orbital. The second claim is absolutely correct!

Did $Pt{F}_6$ act as an oxidant?

Absolutely. By forcing oxygen to lose an electron, $Pt{F}_6$ caused the oxidation of $O_2$ while getting reduced itself. It is a textbook oxidizing agent.

Did it act as a fluorinating agent?

No. A fluorinating agent would transfer fluorine atoms to the oxygen molecule. In this reaction, the structural integrity of both molecules remains intact; only an electron jumps across.

Thus, the correct statements are that the valence d-orbitals of the metal ion have 5 electrons, and $Pt{F}_6$ acts as an oxidant.