Comparing basic strength of nitrogen heterocycles Compare the basic strength of the following compounds:
  
  
  
  
  d > b > c > a
  c > b > a > d
  c > a > b > d
  a > d > c > b
  


I know that in compound (c), the π-bond between nitrogen and cyclopropenyl will polarize (due to aromatic stability) such that it increases electron density of the nitrogenous ring. Whereas, in compound (d), the π-bond will polarize reducing the electron density of nitrogenous ring. Hence, comparing compounds with similar nitrogenous rings, I get that the basic strength of (c)>(b)and that of (d)&lt;(a). But I thought the ring containing more number of nitrogen atoms will be more basic as there will be more lone pair donors, so I thought the answer to be option (4), but the correct answer happens to be option (2). I would like to know that how is basic strength of the nitrogenous rings being compared?
It is likely that the electron pair (from the HOMO) being donated to the acid is from a &pi; molecular orbital. Since the &pi; bond is rather weak, the &pi; electrons should be rather high in energy and is thus able to interact well with the LUMO of the acid. When nitrogen atoms are incorporated into the &pi; system, the energy of the HOMO is lowered due to the high electronegativity of N. Thus, incorporation of N atoms into the system would decrease the basicity of the molecular ion. As such, A and D would be inferior bases, compared to B and C. 

Choosing between A and D, and between B and C, we would have to base our criteria for the basicity on the extent of delocalisation. In D, the &pi; system is larger, thus greater delocalisation was achieved, with a lowering of the energy of the HOMO. Thus, D is likely to be less basic than A. Similarly, the extent of delocalisation is also larger in C, thus it would have a lower-energy HOMO, compared to B. Thus, B is likely more basic than C.

Therefore, option (2) is likely the correct order of basicity.