Which one of the following gives the most stable Diazonium salt?

To determine the compound that forms the most stable diazonium salt, we compare aliphatic and aromatic amines.

Primary aromatic amines react with nitrous acid at low temperature $$\mathrm{(0{-}5^\circ C)}$$ to form stable aryldiazonium salts.

Primary aliphatic amines form unstable alkyl diazonium salts which immediately decompose with evolution of nitrogen gas.

Thus, options $$\mathrm{A}$$ and $$\mathrm{D}$$ can be eliminated because propylamine and isopropylamine are primary aliphatic amines.

$$\mathrm{R-N_2^+Cl^- \longrightarrow Alcohol + N_2\uparrow}$$

Option $$\mathrm{B}$$, $$\mathrm{N}$$-methylaniline, is a secondary aromatic amine.

Secondary amines do not form diazonium salts.

Instead, they form $$\mathrm{N}$$-nitrosoamines.

$$\mathrm{R_2NH + HNO_2 \longrightarrow R_2N-NO}$$

Option $$\mathrm{C}$$, p-toluidine, is a primary aromatic amine and therefore forms a stable aryldiazonium salt.

$$\mathrm{p{-}CH_3C_6H_4NH_2 \xrightarrow[\ 0{-}5^\circ C\ ]{NaNO_2/HCl} p{-}CH_3C_6H_4N_2^+Cl^-}$$

The diazonium ion is stabilised by resonance with the benzene ring.

Also, the methyl group $$\mathrm{(-CH_3)}$$ is electron-donating and further stabilises the positively charged diazonium group through the $$\mathrm{+I}$$ effect.

Therefore, the most stable diazonium salt is formed by:

$$\boxed{\mathrm{p{-}Toluidine}}$$