Match the nitrogen species in List I with the oxidation number of N in List II (for HN3 use the average). List I: (P) N in NH2OH, (Q) N in N2H4, (R) N in HN3, (S) N in NO2+. List II: (1) -2, (2) -1, (3) -1/3, (4) +5.

JEE Advanced levelSingle correct optionDifficulty 3 of 3rx035

Match the nitrogen species in List I with the oxidation number of N in List II (for $HN X_{3}$ use the average). List I: (P) N in $NH X_{2} OH$ , (Q) N in $N X_{2} H X_{4}$ , (R) N in $HN X_{3}$ , (S) N in $NO X_{2} X^{+}$ . List II: (1) $- 2$ , (2) $- 1$ , (3) $- 1/3$ , (4) $+ 5$ .

NH X_{2} OH

x + 2 + (- 2) + 1 = 0

gives

- 1

N X_{2} H X_{4}

2 x + 4 = 0

gives

- 2

HN X_{3}

1 + 3 x = 0

gives

- 1/3

NO X_{2} X^{+}

x + 2 (- 2) = + 1

gives

+ 5

. So P-2, Q-1, R-3, S-4.