Consider the three aromatic molecules (P, Q and R) whose structures...

The reaction of the given aromatic compounds with benzenediazonium chloride ($Ph-N_2^+ Cl^-$) is an azo coupling reaction, which is a type of electrophilic aromatic substitution.

The electrophile, the diazonium ion ($Ph-N_2^+$), is a weak electrophile and requires a strongly activated aromatic ring to undergo coupling. The reactivity of the ring depends on its electron density, which is enhanced by the $+M$ (mesomeric) effect of the $-NMe_2$ group.

1. In molecule P (N,N-dimethylaniline), the $-NMe_2$ group is coplanar with the benzene ring. The lone pair on the nitrogen atom is fully delocalized into the ring, providing a strong $+M$ effect and making the ring highly electron-rich and reactive.

2. In molecule Q, there is one methyl group at the ortho position to the $-NMe_2$ group. This causes steric hindrance, forcing the $-NMe_2$ group slightly out of the plane of the benzene ring. This phenomenon is known as Steric Inhibition of Resonance (SIR). As a result, the delocalization of the nitrogen lone pair into the ring is reduced, decreasing the $+M$ effect and making it less reactive than P.

3. In molecule R, there are two methyl groups at the ortho positions. This causes severe steric repulsion, forcing the $-NMe_2$ group almost completely out of the plane of the ring. The $+M$ effect is minimized because the lone pair can no longer effectively overlap with the $\pi$-system of the ring. Consequently, the ring is the least activated among the three.

Therefore, the electron density on the aromatic ring and the reactivity towards electrophilic substitution follow the order: P $>$ Q $>$ R.

Answer: $P > Q > R$