[elt2112]

I'm struggling with this question a bit.
So in Rh(P^tBu₃)₂(H)(CO) the rhodium is in the +1 oxidation state and has 8 valence electrons. So thought that after reacting with C₂F₂ it would form Rh(P^tBu₃)₂(F)(CO) as this has the same metal electron count, and there is no hydride. I'm a bit stuck with the next bit and can't seem to find anything about it to help.