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Chapter 4: Problem 81

Assign the oxidation state for nitrogen in each of the following. a. \(\mathrm{Li}_{3} \mathrm{~N} \quad\) f. \(\mathrm{NO}_{2}\) b. \(\mathrm{NH}_{3} \quad\) g. \(\mathrm{NO}_{2}^{-}\) c. \(\mathrm{N}_{2} \mathrm{H}_{4} \quad\) h. \(\mathrm{NO}_{3}\) d. NO i. \(\mathrm{N}_{2}\) e. \(\mathrm{N}_{3} \mathrm{O}\)

Short Answer

Expert verified
The oxidation states of nitrogen in the given compounds are as follows: a. Li₃N: N = -3 b. NH₃: N = -3 c. N₂H₄: N = -2 d. NO: N = +2 e. N₃O: N = +2/3 f. NO₂: N = +4 g. NO₂⁻: N = +3 h. NO₃: N = +6 i. N₂: N = 0

Step by step solution

01

Assign oxidation states for each compound

Use the rules for assigning oxidation states to find the oxidation state of nitrogen in each compound. a. Li₃N
02

Rule 1 (Alkali metals)

Alkali metals have an oxidation state of +1. Each Li atom has an oxidation state of +1.
03

Rule 3 (Sum of oxidation states)

The sum of oxidation states in a neutral compound should be 0. There are three Li atoms, resulting in a total of +3. To balance the compound, N must have an oxidation state of -3. b. NH₃
04

Rule 2 (Hydrogen)

Hydrogen has an oxidation state of +1 when bonded to non-metal elements. There are three H atoms, resulting in a total of +3.
05

Rule 3 (Sum of oxidation states)

N must have an oxidation state of -3 to balance the compound. c. N₂H₄
06

Rule 2 (Hydrogen)

There are four H atoms, resulting in a total of +4.
07

Rule 3 (Sum of oxidation states)

Since the compound is neutral and there are two N atoms, each N must have an oxidation state of -2. d. NO
08

Rule 4 (Oxygen)

Oxygen has an oxidation state of -2.
09

Rule 3 (Sum of oxidation states)

To balance the compound, N must have an oxidation state of +2. e. N₃O
10

Rule 4 (Oxygen)

Oxygen has an oxidation state of -2.
11

Rule 3 (Sum of oxidation states)

Since there are three N atoms, each N must have an oxidation state of +2/3 to balance the compound. f. NO₂
12

Rule 4 (Oxygen)

There are two O atoms, resulting in a total of -4.
13

Rule 3 (Sum of oxidation states)

N must have an oxidation state of +4 to balance the compound. g. NO₂⁻
14

Rule 4 (Oxygen)

There are two O atoms, resulting in a total of -4.
15

Rule 3 (Sum of oxidation states)

Since the compound has a charge of -1, N must have an oxidation state of +3 to balance the compound. h. NO₃
16

Rule 4 (Oxygen)

There are three O atoms, resulting in a total of -6.
17

Rule 3 (Sum of oxidation states)

N must have an oxidation state of +6 to balance the compound. i. N₂
18

Rule 5 (Same element)

Diatomic molecules of the same element have an oxidation state of 0. N₂ has two N atoms, each with an oxidation state of 0.

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Most popular questions from this chapter

Carminic acid, a naturally occurring red pigment extracted from the cochineal insect, contains only carbon, hydrogen, and oxygen. It was commonly used as a dye in the first half of the nineteenth century. It is \(53.66 \% \mathrm{C}\) and \(4.09 \% \mathrm{H}\) by mass. A titration required \(18.02 \mathrm{~mL}\) of \(0.0406 \mathrm{M} \mathrm{NaOH}\) to neutralize \(0.3602 \mathrm{~g}\) carminic acid. Assuming that there is only one acidic hydrogen per molecule, what is the molecular formula of carminic acid?

A student titrates an unknown amount of potassium hydrogen phthalate \(\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\right.\), often abbreviated \(\mathrm{KHP}\) ) with \(20.46 \mathrm{~mL}\) of a \(0.1000 M\) NaOH solution. KHP (molar mass \(=204.22\) \(\mathrm{g} / \mathrm{mol}\) ) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of \(3.0 \mathrm{M}\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

Calculate the molarity of each of these solutions. a. A \(5.623-\mathrm{g}\) sample of \(\mathrm{NaHCO}_{3}\) is dissolved in enough water to make \(250.0 \mathrm{~mL}\) of solution. b. A \(184.6-\mathrm{mg}\) sample of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. c. A 0.1025-g sample of copper metal is dissolved in \(35 \mathrm{~mL}\) of concentrated \(\mathrm{HNO}_{3}\) to form \(\mathrm{Cu}^{2+}\) ions and then water is added to make a total volume of \(200.0 \mathrm{~mL}\). (Calculate the molarity of \(\mathrm{Cu}^{2+} .\) )
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