p Block Elements Part 3

Concept:Element $E$ forms a cationic species $\mathrm{EH}_x^+$ and $\mathrm{EH}_3$ gives a precipitate with $\mathrm{K}_2\mathrm{HgI}_4$ in alkaline medium.This is Nessler's test for ammonia, so $E$ is nitrogen.Explanation:Nitrogen forms the ammonium ion $\mathrm{NH}_4^+$ $(E = \mathrm{N}, x = 4)$.Ammonia $\mathrm{EH}_3$ (i.e., $\mathrm{NH}_3$) reacts with $\mathrm{K}_2\mathrm{HgI}_4$ in alkaline medium to form a brown precipitate of basic mercury(II)amido-iodine.Thus $E = \mathrm{N}$, belonging to group 15.The first ionization enthalpies (in $\mathrm{kJ\ mol}^{-1}$) for the first elements of groups 13–16 are: $\mathrm{B}$ (801), $\mathrm{C}$ (1086), $\mathrm{N}$ (1402), $\mathrm{O}$ (1312).Across a period, ionization enthalpy generally increases, but nitrogen has a higher value than oxygen due to its stable half‑filled $2p^3$ configuration.Therefore, nitrogen has the highest value among these: $1402\ \mathrm{kJ\ mol}^{-1}$.Answer:A. 1402