Why is the nitroso group a deactivating group for electrophilic aromatic substitution? Does the lone pair of nitrogen in nitrosobenzene take part in resonance? If not, then why? If it takes part in resonance then the hybridisation of nitrogen is sp. Can a sp hybridised atom be attached to the ring?
Let's look at nitrosobenzene. 



In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:



If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated: 



Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

 

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:



Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:



It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.