Why is the pi star orbital of the C=C bond higher in energy than the antibonding pi orbital of the C=O bond?

I thought it was the other way around; it's relatively easy to add something to a C=C alkene bond (i.e. halogenation) but halogenation with diatomic halogens doesn't occur with C=O bonds.

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    $\begingroup$ With halogenation it is not the $\pi^{*}$ but the $\pi$ orbitals that are involved in the reaction with e.g. $\ce{Br2}$ because the alkene is acting as a nucleophile here. And because $\ce{O}$ is more electronegative than $\ce{C}$ it should be logical that $\pi$ is lower in energy for $\ce{C=O}$ than for $\ce{C=C}$. $\endgroup$ – Philipp Oct 27 '14 at 16:45

It's relatively easy to add an electrophilic species, such as X-X, to an alkene. As @Philipp points out, the alkene is operating through the (filled) pi orbital.

It's relatively easy to add a nucleophilic species to a carbonyl. In that case, the carbonyl operates through its (unfilled) pi* orbital.

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