What color is found when methylamine reacts with hydrogen iodide?

White solid

The correct answer is option 1. White solid.

When methylamine \((CH_3NH_2)\), which is a primary amine, reacts with hydrogen iodide \((HI)\), it forms methylammonium iodide \((CH_3NH_3I)\).

Methylamine, being a base, accepts a proton \((H^+)\) from hydrogen iodide \((HI)\), which is a strong acid. This protonation converts the methylamine into its conjugate acid, the methylammonium ion \((CH_3NH_3^+)\). The iodide ion \((I^-)\) from the dissociation of HI combines with the methylammonium ion to form the salt, methylammonium iodide (\(CH_3NH_3I)\).

The reaction can be represented as:

\(CH_3NH_2 + HI \longrightarrow CH_3NH_3I\)

Methylammonium iodide \((CH_3NH_3I)\) is an ionic compound consisting of the methylammonium cation \((CH_3NH_3^+)\) and the iodide anion \((I^-)\). It typically forms a white crystalline solid. This is due to the ionic nature of the compound, where the positively charged ammonium ion and the negatively charged iodide ion form a crystalline lattice structure. The product of the reaction, methylammonium iodide, is a white solid because the individual components (methylammonium and iodide ions) do not impart color to the compound. Most salts of small, non-transition metal cations and anions like halides are colorless or white in their solid form due to the lack of chromophores (parts of the molecule responsible for color). When methylamine reacts with hydrogen iodide, it results in the formation of methylammonium iodide, which is a white solid. This product's lack of color aligns with the typical characteristics of salts formed from simple cations and anions without color-inducing groups or metal ions.

Thus, the correct answer is White solid.