CI j: Nitrogen chemistry

Question 1

Name 2 things which cause N₂ to form nitrous oxides in the atmosphere.

Answer 1

• High temperatures from combustion processes
• Lightning

Question 2

N₂ has a high bond enthalpy. Draw:

• A simple diagram representing its bonds
• A dot and cross diagram
• A diagram showing the types of bonds present

Answer 2

Question 3

To represent ammonia, draw:

• A dot and cross diagram
• A full structural diagram
• Its 3D shape

Answer 3

Question 4

Why can H⁺ ions be added to ammonia to form ammonium?

Answer 4

• Nitrogen atom in ammonia has a lone pair of electrons
• A dative covalent bond forms between the lone pair and hydrogen

Question 5

To represent ammonium, draw:

• A dot and cross diagram
• A structural diagram
• How the ion is conventionally drawn

Answer 5

Question 6

Give the name, appearance and state at room temperature of each of the following nitrogen compounds. Include details of reactionary tendencies.

• NO
• NO₂
• N₂O

Answer 6

NO = nitrogen (mon)oxide

Colourless gas (converted to NO₂ in air)

NO₂ = nitrogen dioxide

Brown gas (stable)

N₂O = dinitrogen (mon)oxide

Colourless gas (decomposes to NO₂)

Question 7

Describe the acidity of the nitrous oxides.

Answer 7

• NO = neutral
• NO₂ = acidic
• N₂O = neutral

NO₂ is the brown one, and the one that the other two tend to be converted to in the atmosphere. It is always the distinctive one.

Question 8

Describe the water solubility of the nitrous oxides.

Answer 8

• NO - not very soluble (more in alcohol)
• NO₂ - highly soluble
• N₂O - moderate in cold water; low in hot (more in alcohol)

Question 9

Write equations to show how nitrogen dioxide is formed indirectly as a result of the high temperatures of combustion processes.

Answer 9

N_{2 (g)} + O_{2 (g)} → 2NO_(g)

2NO_(g) + O_{2 (g)} → 2NO_{2 (g)}

Question 10

Explain the structure of nitrogen monoxide.

Answer 10

^•N = O

Lone pair on nitrogen, as well as an unpaired electron.

Radical.

Question 11

Explain the structure of nitrogen dioxide.

Answer 11

O = ^•N = O

Trigonal planar where one of the 3 components is an unpaired electron

Angle ONO < 120^o due to repulsion of electron

Radical

Question 12

Why aren’t oxidation numbers typically used for naming the oxides of nitrogen?

Answer 12

The use of ox numbers implies an ionic nature.

Question 13

Name each ion:

• NO₃^-
• NO₂^-

Answer 13

• Nitrate(V)
• Nitrate(III)

Question 14

Write out the chemical sequence by which anaerobic bacteria reduce nitrate(V) ions. Show state symbols and oxidation states.

Answer 14

NO₃^-_(aq) → NO₂^-_(aq) → NO_(g) → N₂O_(g) → N_{2 (g)}

+5 → +3 → +2 → +1 → 0

Question 15

Draw a dot and cross diagram and an accurate structural representation of nitrate(III).

Answer 15

• It’s a stabilisation of the ^•NO₂ radical.*
• 2 “1.5” bonds rather than 1 single + 1 double.*

Question 16

Draw a dot and cross diagram and an accurate structural representation of nitrate(V).

Answer 16

Question 17

Both nitrate(III) and nitrate(V) are very soluble in water.

Aerobic bacteria in soil form nitrate(V) by nitrification, via the conversion of ammonium ions to nitrate(III).

Write equations to represent the formation of nitrate(V).

Answer 17

Oxidation

NH₄⁺_(aq) + 1.5O_{2 (g)} → NO₂^-_(aq) + 2H⁺_(aq) + H₂O_(l)

NO₂^-_(aq) + 1/2 O_{2 (g)} → NO₃^-_(aq)

First one can be derived via half equations:

• When ammonia is converted to nitrate(III), nitrogen’s ox state changes as follows: -3 → +3 so oxidation, producing 6 electrons
• 2 protons are required on RHS to balance charge: NH₄⁺→ NO₂^- + 6e^- + 2H⁺
• O₂ (ox state 0) provides the 6e^- and is converted to H₂O (ox state -2), so must be 3 O atoms + therefore 1.5 O₂ molecules

Question 18

Describe and explain the test for nitrate(V) ions.

Answer 18

• Heat test solution with NaOH(aq) + Davarda’s alloy, Cu/Al/Zn
• If nitrate(V) present, ammonia gas forms:

3NO₃^- + 8Al + 5OH^- + 18H₂O → 3NH₃ + 8[Al(OH)₄]^-

Test for ammonia by:

• Turns damp red litmus paper blue
• Forms white ammonium chloride gas when HCl_(g) added (fumes from HCl_(aq))
• Sharp smell

Question 19

Describe and explain the test for ammonium ions.

Answer 19

Acid-base reaction

• Heat test solution with NaOH(aq)
• If ammonium is present, ammonia gas forms:

NH₄⁺(aq) + OH^-(aq) → NH₃(g) + H₂O(l)

Test for ammonia by:

• Turns damp red litmus paper blue
• Forms white ammonium chloride gas when HCl_(g) added (fumes from HCl_(aq))
• Sharp smell

Question 20

Explain why the test below is not used if nitrate(III) ions are believed to be present. The second reaction causes a brown ring to be observed.

Answer 20

Nitrate(III) ions make the test unreliable by oxidising the Fe²⁺ ions. If nitrate(V) ions were present, the test would still be negative.

Question 21

What is the balanced equation for the following reaction?

NO₃^- + Al + OH^- + H₂O → NH₃ + [Al(OH)₄]^-

Answer 21

Balance what is involved in redox

N: +5 → -3 gains 8e^- Al: 0 → +3 loses 3e^-

LCM = 24 24/8 = 3N 24/3 = 8 Al

3NO₃^- + 8Al + OH^- + H₂O → 3NH₃ + 8[Al(OH)₄]^-

Balance charges

3NO₃^- + nOH^- → 8[Al(OH)₄]^- n = 8 - 3 = 5

3NO₃^- + 8Al + 5OH^- + H₂O → 3NH₃ + 8[Al(OH)₄]^-

Balance the rest

As is: 14 O + 5 H → 32 O + 41 H

32 - 14 = 18 O 41 - 5 = 36 H

3NO₃^- + 8Al + 5OH^- + 18H₂O → 3NH₃ + 8[Al(OH)₄]^-

Question 22

Write a balanced half-equation for the following reaction:

NO₃^- → N₂

Answer 22

+5 to 0 so reduction/gain

So add electrons + protons

2NO₃^- + 10e^- → N₂

2NO₃^- + 12H⁺ + 10e^- → N₂ + 6H₂O

Have to add protons (rather than not, and just oxygen being formed) to balance charge.

Question 23

Write a balanced half-equation for the following reaction:

NH₄⁺ → NH₃

Answer 23

-3 to -3 so no redox

NH₄⁺ + OH^- → NH₃ + H₂O

Question 24

Write a balanced half-equation for the following reaction:

NO → NO₂

Answer 24

+2 to +4 so oxidation/loss

So produce electrons + protons

NO → NO₂ + 2e^-

NO + H₂O → NO₂ + 2H⁺ + 2e^-

Need protons on RHS to balance charge. Therefore need H₂O not O₂ as ox agent.

Question 25

Balance the following equation:

NO₃^- + Fe²⁺ + H⁺ → Fe³⁺ + NO + H₂O

Answer 25

NO₃^- + Fe²⁺ + H⁺ → Fe³⁺ + NO + H₂​O

N: +5 → +2 gains 3e^-

Fe: +2 → +3 loses 1e^-

Balance ox states:

NO₃^- + 3Fe²⁺ + H⁺ → 3Fe³⁺ + NO + H₂​O

Balance charge:

NO₃^- + 3Fe²⁺ + 4H⁺ → 3Fe³⁺ + NO + H₂​O

Balance rest:

NO₃^- + 3Fe²⁺ + 4H⁺ → 3Fe³⁺ + NO + 2H₂​O