Nitrogen, Ammonia and Ammonium Compounds

AS · 12 min

Nitrogen makes up 78% of the air, yet it is so unreactive that living things cannot use it directly, and industry needs high pressure, high temperature and a catalyst to turn it into ammonia. This note explains that lack of reactivity, then covers ammonia as a Brønsted–Lowry base, the structure of the ammonium ion and how it forms, and the displacement of ammonia from ammonium salts by alkalis, which is the basis of the qualitative test for $\ce{NH4+}$ and of a classic back-titration. These are short but reliably examined learning outcomes in Papers 1 and 2.

Why nitrogen is unreactive

A nitrogen molecule, $\ce{N2}$ , contains a triple covalent bond ( $\ce{N#N}$ ): one $\sigma$ bond and two $\pi$ bonds. Each nitrogen atom also has one lone pair.

Key result

Two reasons for the lack of reactivity of nitrogen

The $\ce{N#N}$ triple bond is very strong (bond energy 994 kJ mol⁻¹). A very large amount of energy is needed to break it, so reactions of nitrogen have a very high activation energy and are extremely slow under normal conditions.
The molecule is non-polar. Both atoms are identical, so the electrons are shared equally and there is no $\delta+$ or $\delta-$ site. Electrophiles and nucleophiles are not attracted to any part of the molecule.

Compare the bond energies: $\ce{O=O}$ 496, Cl–Cl 242, $\ce{N#N}$ 994 kJ mol⁻¹. Nitrogen's bond is roughly twice as strong as oxygen's double bond, which is why oxygen supports combustion but nitrogen does not.

Nitrogen does react under extreme conditions:

With oxygen, at the very high temperatures in lightning or in a car engine: $\ce{N2(g) + O2(g) -> 2NO(g)}$ . (See the next note.)
With hydrogen, in the Haber process, at about 200 atm and 400–450 °C with an iron catalyst: $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ . The catalyst is needed because the activation energy is so high.
With reactive metals such as magnesium, burning in nitrogen to form magnesium nitride, $\ce{3Mg + N2 -> Mg3N2}$ (not required, but shows that only very reactive species can attack $\ce{N2}$ ).

Watch out

"Nitrogen is unreactive because it has a full outer shell" is wrong: that explanation applies to the noble gases. Each N atom in $\ce{N2}$ has an octet only because it shares three pairs. The reasons are the strength of the triple bond and the lack of polarity.

Ammonia as a Brønsted–Lowry base

Definition

A Brønsted–Lowry acid is a proton ( $\ce{H+}$ ) donor. A Brønsted–Lowry base is a proton ( $\ce{H+}$ ) acceptor.

Ammonia has a lone pair of electrons on the nitrogen atom. This lone pair can form a bond to a proton, so ammonia is a base.

With water, ammonia accepts a proton from a water molecule:

\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}

The equilibrium lies well to the left: only a small fraction of the ammonia molecules are protonated. Ammonia is a weak base (partially ionised), so aqueous ammonia has a pH of about 11 rather than 13–14 for sodium hydroxide of the same concentration.

In this reaction:

$\ce{NH3}$ is the base (accepts $\ce{H+}$ ) and $\ce{NH4+}$ is its conjugate acid.
$\ce{H2O}$ is the acid (donates $\ce{H+}$ ) and $\ce{OH-}$ is its conjugate base.

With acids, ammonia is protonated completely to form ammonium salts:

\ce{NH3(g) + HCl(g) -> NH4Cl(s)}

\ce{2NH3(aq) + H2SO4(aq) -> (NH4)2SO4(aq)}

\ce{NH3(aq) + HNO3(aq) -> NH4NO3(aq)}

When ammonia gas meets hydrogen chloride gas (for example from bottles of concentrated aqueous ammonia and concentrated hydrochloric acid held close together), a dense white smoke of ammonium chloride forms.

The general ionic equation is $\ce{NH3 + H+ -> NH4+}$ .

The ammonium ion

When ammonia accepts a proton, the lone pair on nitrogen forms a dative covalent (coordinate) bond to the $\ce{H+}$ ion, which has no electrons of its own.

\ce{NH3 + H+ -> NH4+}

Key result

Structure of the ammonium ion, $\ce{NH4+}$

Four bonding pairs around nitrogen and no lone pairs.
Tetrahedral shape, bond angle 109.5°.
One N–H bond is formed by a dative bond, but once formed all four N–H bonds are identical in length and strength; the positive charge is spread over the whole ion.
Compare $\ce{NH3}$ : three bonding pairs and one lone pair, trigonal pyramidal, about 107°. Converting the lone pair into a bonding pair removes the extra lone pair repulsion, so the angle opens to 109.5°.

Formation of the ammonium ion. The lone pair on the nitrogen of ammonia forms a dative covalent bond (shown as an arrow from N to H) to a proton. In the ion all four N–H bonds are identical and the shape is tetrahedral.

Displacing ammonia from ammonium salts

Ammonium salts react with alkalis (strong bases such as $\ce{NaOH}$ or $\ce{Ca(OH)2}$ ) on warming, releasing ammonia gas. This is an acid–base reaction: the ammonium ion acts as a Brønsted–Lowry acid, donating a proton to the hydroxide ion.

\ce{NH4+(aq) + OH-(aq) -> NH3(g) + H2O(l)}

Full equations:

\ce{NH4Cl(s) + NaOH(aq) -> NaCl(aq) + NH3(g) + H2O(l)}

\ce{(NH4)2SO4(s) + 2NaOH(aq) -> Na2SO4(aq) + 2NH3(g) + 2H2O(l)}

\ce{2NH4Cl(s) + Ca(OH)2(s) -> CaCl2(s) + 2NH3(g) + 2H2O(l)}

The stronger base ( $\ce{OH-}$ ) removes a proton from the weaker base's conjugate acid ( $\ce{NH4+}$ ), "displacing" the weaker base, ammonia.

Key result

Test for the ammonium ion: warm the substance with aqueous sodium hydroxide. Ammonia is given off, which turns damp red litmus paper blue (the only common alkaline gas). It also has a characteristic pungent smell.

A practical consequence for farmers

Ammonium salts such as ammonium nitrate and ammonium sulfate are used as nitrogen fertilisers. If a farmer adds lime ( $\ce{Ca(OH)2}$ ) to the soil at the same time, the lime displaces ammonia from the fertiliser, and the nitrogen is lost to the air as ammonia gas:

\ce{Ca(OH)2(s) + 2NH4+(aq) -> Ca^2+(aq) + 2NH3(g) + 2H2O(l)}

So liming and ammonium fertilisers should be applied at different times.

Worked examples

Explaining the unreactivity of nitrogen

Explain why nitrogen gas is very unreactive, even though nitrogen atoms are quite electronegative.

Solution

In an $\ce{N2}$ molecule the two atoms are joined by a triple covalent bond with a very high bond energy (994 kJ mol⁻¹), so a large amount of energy is needed to break it and the activation energy of its reactions is very high. The molecule is non-polar, because the two atoms are identical and share the electrons equally, so there is no partial charge to attract electrophiles or nucleophiles.

Ammonia as a base

Write an equation for the reaction of ammonia with water. Identify the Brønsted–Lowry acid and base on each side and explain why aqueous ammonia is described as a weak base.

Solution

\ce{NH3(aq) + H2O(l) <=> NH4+(aq) + OH-(aq)}

Left: $\ce{NH3}$ is the base (proton acceptor); $\ce{H2O}$ is the acid (proton donor). Right: $\ce{NH4+}$ is the acid (conjugate acid of $\ce{NH3}$ ); $\ce{OH-}$ is the base (conjugate base of $\ce{H2O}$ ).

It is a weak base because it is only partially protonated (partially ionised) in water: the equilibrium lies to the left, so the concentration of $\ce{OH-}$ is much lower than the concentration of ammonia.

Volume of ammonia from an ammonium salt

$2.675\ \text{g}$ of ammonium chloride is warmed with excess aqueous sodium hydroxide. Calculate the volume of ammonia given off at room conditions. ( $A_r$ : N 14.0, H 1.0, Cl 35.5)

Solution

$M(\ce{NH4Cl}) = 14.0 + 4.0 + 35.5 = 53.5\ \text{g mol}^{-1}$

$n(\ce{NH4Cl}) = \dfrac{2.675}{53.5} = 0.0500\ \text{mol}$

$\ce{NH4Cl + NaOH -> NaCl + NH3 + H2O}$ , $1 : 1$ , so $n(\ce{NH3}) = 0.0500\ \text{mol}$ .

$V = 0.0500 \times 24.0 = 1.20\ \text{dm}^3$

Exam-style: shape and bonding in the ammonium ion

Describe how an ammonium ion is formed from an ammonia molecule, and state and explain the shape and bond angle of the ion.

Solution

The nitrogen atom in $\ce{NH3}$ has a lone pair. It accepts a proton by forming a dative covalent (coordinate) bond: both electrons in the new N–H bond come from the nitrogen lone pair.

In $\ce{NH4+}$ there are four bonding pairs and no lone pairs around nitrogen. These repel each other equally and get as far apart as possible, giving a tetrahedral shape with bond angles of 109.5°. All four N–H bonds are identical.

Exam-hard: purity of ammonium nitrate by back titration

A $0.800\ \text{g}$ sample of impure ammonium nitrate fertiliser is warmed with excess sodium hydroxide. All the ammonia given off is absorbed in $50.0\ \text{cm}^3$ of $0.250\ \text{mol dm}^{-3}$ hydrochloric acid (an excess). The remaining acid needs $30.00\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ sodium hydroxide for neutralisation.

(a) Calculate the percentage by mass of $\ce{NH4NO3}$ in the sample.

(b) Calculate the percentage by mass of nitrogen in pure ammonium nitrate, and explain why a farmer might prefer it to ammonium sulfate.

( $M$ : $\ce{NH4NO3}$ 80.0, $\ce{(NH4)2SO4}$ 132.1)

Solution

(a) Acid at start: $n(\ce{HCl}) = 0.250 \times 50.0 / 1000 = 0.01250\ \text{mol}$

Excess acid: $n(\ce{NaOH}) = 0.100 \times 30.00 / 1000 = 0.003000\ \text{mol} = n(\ce{HCl})$ remaining

Acid that reacted with ammonia: $0.01250 - 0.00300 = 0.00950\ \text{mol}$ , so $n(\ce{NH3}) = 0.00950\ \text{mol}$ ( $\ce{NH3 + HCl -> NH4Cl}$ ).

$\ce{NH4+ + OH- -> NH3 + H2O}$ , so $n(\ce{NH4NO3}) = 0.00950\ \text{mol}$ .

Mass $= 0.00950 \times 80.0 = 0.760\ \text{g}$ ; percentage $= \dfrac{0.760}{0.800} \times 100 = 95.0\%$ .

(Note that the nitrate nitrogen is not released as ammonia by NaOH alone, so the test measures only the ammonium part.)

(b) $\ce{NH4NO3}$ : $\dfrac{2 \times 14.0}{80.0} \times 100 = 35.0\%$ N. $\ce{(NH4)2SO4}$ : $\dfrac{28.0}{132.1} \times 100 = 21.2\%$ N.

Ammonium nitrate supplies much more nitrogen per kilogram, so less needs to be bought, transported and spread.

Watch out

In $\ce{NH4+}$ there is no lone pair on nitrogen; it was used to form the dative bond. Students often draw a lone pair and give the angle as 107°.
The displacement of ammonia is an acid–base reaction, not redox: nitrogen stays at $-3$ throughout.
Ammonia turns damp red litmus blue. Dry litmus does not work: the gas must dissolve in water to form $\ce{OH-}$ .
Ammonia is a weak base: write $\ce{<=>}$ for its reaction with water.

Exam tip

"Explain the lack of reactivity of nitrogen" is worth two marks: (1) strong triple bond / high bond energy, needing a lot of energy to break; (2) non-polar molecule. Quote 994 kJ mol⁻¹ if the data are given.
When asked about Brønsted–Lowry behaviour, use the words "proton donor" and "proton acceptor", and identify both conjugate pairs if asked.
Draw the ammonium ion with the charge outside square brackets, and indicate the dative bond with an arrow from N to H (or say it in words).
In back titrations involving ammonia, write each step: acid added, acid in excess, acid used by ammonia, then the ratio to the ammonium salt.

Practical skills

To test for ammonium ions in Paper 3, add about $1\ \text{cm}^3$ of aqueous sodium hydroxide to the solid or solution in a test-tube and warm gently. Hold a piece of damp red litmus paper at the mouth of the tube without touching the sides (sodium hydroxide on the glass would turn it blue anyway). A positive result is the litmus turning blue. The qualitative analysis notes list $\ce{NH4+}$ as giving "no precipitate; ammonia produced on warming" with $\ce{NaOH(aq)}$ . Note that nitrate and nitrite ions also release ammonia, but only when heated with $\ce{NaOH(aq)}$ and aluminium foil, which reduces them; without aluminium, only ammonium gives ammonia.

Summary

$\ce{N2}$ is unreactive because the $\ce{N#N}$ bond is very strong (994 kJ mol⁻¹) and the molecule is non-polar.
Ammonia is a Brønsted–Lowry base: its lone pair accepts $\ce{H+}$ . $\ce{NH3 + H2O <=> NH4+ + OH-}$ (weak base).
$\ce{NH3 + HCl -> NH4Cl}$ gives white smoke; ammonia forms ammonium salts with all acids.
$\ce{NH4+}$ forms by a dative bond from N to $\ce{H+}$ ; it is tetrahedral, 109.5°, with four identical N–H bonds.
Alkalis displace ammonia from ammonium salts on warming: $\ce{NH4+ + OH- -> NH3 + H2O}$ (acid–base).
Test: warm with $\ce{NaOH(aq)}$ ; ammonia turns damp red litmus blue.

Practice

Question

State two reasons why nitrogen gas is unreactive.
Write the equation for the reaction between ammonia and hydrogen chloride, and describe what is seen when the two gases meet.
Explain, using the Brønsted–Lowry theory, why ammonia is a base.
State the shape and bond angle of $\ce{NH3}$ and of $\ce{NH4+}$ , and explain the difference.
Write an ionic equation for the reaction that occurs when ammonium sulfate is warmed with sodium hydroxide solution. Explain why this is an acid–base reaction.
Describe a test for the ammonium ion, including the result.
Calculate the percentage by mass of nitrogen in ammonium sulfate. ( $A_r$ : N 14.0, H 1.0, S 32.1, O 16.0)
Calculate the volume of ammonia, at room conditions, produced when $3.30\ \text{g}$ of ammonium sulfate reacts completely with excess calcium hydroxide. ( $M(\ce{(NH4)2SO4}) = 132.1$ )
Explain why a farmer should not spread slaked lime on a field at the same time as ammonium nitrate fertiliser. Include an ionic equation.
$0.535\ \text{g}$ of an ammonium salt is warmed with excess NaOH and the ammonia absorbed in $50.0\ \text{cm}^3$ of $0.250\ \text{mol dm}^{-3}$ HCl. The excess acid needs $25.00\ \text{cm}^3$ of $0.100\ \text{mol dm}^{-3}$ NaOH. Calculate the molar mass of the salt (assume one $\ce{NH4+}$ per formula unit) and suggest its identity.

Answers

The $\ce{N#N}$ triple bond is very strong (high bond energy), so reactions have a very high activation energy; the molecule is non-polar, so it does not attract electrophiles or nucleophiles.
$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$ . A dense white smoke (of solid ammonium chloride) forms where the gases meet.
A Brønsted–Lowry base is a proton acceptor. Ammonia has a lone pair on its nitrogen atom which can form a dative bond to $\ce{H+}$ , forming $\ce{NH4+}$ ; it therefore accepts protons, for example from water or acids.
$\ce{NH3}$ : trigonal pyramidal, about 107°. $\ce{NH4+}$ : tetrahedral, 109.5°. $\ce{NH3}$ has three bonding pairs and one lone pair; lone pairs repel more strongly than bonding pairs, squeezing the bonds closer together. In $\ce{NH4+}$ the lone pair has become a bonding pair, so the four equal bonding pairs repel equally.
$\ce{NH4+(aq) + OH-(aq) -> NH3(g) + H2O(l)}$ . The ammonium ion donates a proton (acts as an acid) to the hydroxide ion (which accepts it, acting as a base). No oxidation numbers change.
Warm the substance with aqueous sodium hydroxide; hold damp red litmus paper at the mouth of the tube. If ammonium ions are present, ammonia is given off and turns the damp red litmus blue.
$M = 2(14.0 + 4.0) + 32.1 + 64.0 = 132.1$ . Percentage N $= 28.0 / 132.1 \times 100 = 21.2\%$ .
$n(\ce{(NH4)2SO4}) = 3.30 / 132.1 = 0.02498\ \text{mol}$ . Each gives 2 $\ce{NH3}$ : $n(\ce{NH3}) = 0.04996\ \text{mol}$ ; $V = 0.04996 \times 24.0 = 1.20\ \text{dm}^3$ .
Slaked lime, $\ce{Ca(OH)2}$ , is a base that reacts with ammonium ions, releasing ammonia gas, so nitrogen is lost from the fertiliser to the atmosphere and the fertiliser is wasted: $\ce{2NH4+(aq) + Ca(OH)2(s) -> Ca^2+(aq) + 2NH3(g) + 2H2O(l)}$ (or $\ce{NH4+ + OH- -> NH3 + H2O}$ ).
$n(\ce{HCl})$ added $= 0.250 \times 50.0 / 1000 = 0.01250\ \text{mol}$ ; excess $= 0.100 \times 25.00 / 1000 = 0.00250\ \text{mol}$ ; used by $\ce{NH3}$ $= 0.01000\ \text{mol}$ . $n(\text{salt}) = 0.01000\ \text{mol}$ ; $M = 0.535 / 0.01000 = 53.5\ \text{g mol}^{-1}$ . This is ammonium chloride, $\ce{NH4Cl}$ ( $14.0 + 4.0 + 35.5 = 53.5$ ).

Why nitrogen is unreactive

Ammonia as a Brønsted–Lowry base

The ammonium ion

Displacing ammonia from ammonium salts

A practical consequence for farmers

Worked examples

Practice

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