Why is the 2-position of pyridine electron richer than the 4 position
When you deal with an aromatic ring, pairwise covalent bond structures miss a lot of key features.  You need to consider the entire conjugated ring as a unit and take account of symmetry.

Consider almost a benzene ring.  We may describe its $\pi$ orbitals as linear combinations of the out-of-plane atomic $p$ orbital on the carbon atoms.  Let $p_k$ represent the wave function of one such atomic, with the positive lobe on a specified side of the ring plane. The same orbital but with the positive lobe on the"wrong" side of the ring plane is rendered as $-p_k$.  The subscript $k$ refers to atom number $k$ as determined in the standard IUPAC scheme.

We then describe the canonical, occupied $\pi$ orbital as the familiar linear combinations: 

$\psi_1=(1/\sqrt{6})\times (p_1+p_2+p_3+p_4+p_5+p_6)$

$\psi_2=(1/\sqrt{12})\times (2p_1-p_2-p_3+2p_4-p_5-p_6)$

$\psi_3=(1/2)\times(p_2+p_3-p_5-p_6)$

Now suppose that atom $1$ is changed from carbon to nitrogen to make pyridine.  Now that atom, being more electronegative, acts an an electron sink on occupied orbitals having an interaction with that atom -- meaning $\psi_1$ and $\psi_2$ are vulberable to thus $\pi$-electron withdrawal but $\psi_3$ which has a node at atom $1$ is "protected" by its symmetry.  Then positions $2$ and $3$, with some of their $\pi$- electron density on the protected $\psi_3$ orbital, are less susceptible than atom 4 whose $\pi$ electrons are all on the vulnerable $\psi_1$ and $\psi_2$ orbitals.

Now, going a little beyond the question, suppose we were to N-oxidize the pyridine, as with a peroxy acid.  In pyridine N-oxide the oxygen acts as a $\pi$ donor so that the nitrogen (or more accurately, the N-oxide functional group) is now effectively an electron source instead of a sink.  The "vulnerable" orbitals are now electron enriched and this characteristic is most strongly imparted, by conjugation and symmetry as argued above, to position 4.  Then electrophilic substitution, enhanced (or enabled?) by the N-oxidation, is most strongly so at position 4.  So, the primary electrophilic substitution product, after subsequently reducing the N-oxide, is the 4-substituted pyridine.  Which is what we see empirically with a wide variety of pyridine N-oxide electrophilic substitutions.