What is the hybridisation of the carbons in the allyl radical?
Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.

A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.

This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).

Since we just established that the radical will occupy a p-type orbital and since there is a &#x3c0; bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:

$$\ce{H2C=CH-CH2^. &lt;-&gt; H2C^.-CH=CH2}$$

The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a &#x3c0; bond nearby; the n-propyl radical does not feature one.