Arrange the oxidation numbers of nitrogen in the following compounds in increasing order (H is $+ 1$ , O is $- 2$ ; use the average value for $HN X_{3}$ ). I. $NH X_{2} OH$ ; II. $N X_{2} H X_{4}$ ; III. $HN X_{3}$ ; IV. $N X_{2} O$ .

Answer: B

x + 2 - 2 + 1 = 0

gives

- 1

. II:

2 x + 4 = 0

gives

- 2

. III:

1 + 3 x = 0

gives

- \frac{1}{3}

. IV:

2 x - 2 = 0

gives

+ 1

. Increasing order: II (

- 2

), I (

- 1

), III (

- \frac{1}{3}

), IV (

+ 1

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