Reactions of the Period 3 Elements

AS · 12 min

This note covers how the Period 3 elements react with oxygen, chlorine and water, and why the formulas of their oxides and chlorides follow a simple rule based on outer-shell electrons. You must be able to write balanced equations with state symbols for every reaction listed in the syllabus, describe what you would see, and state and explain the oxidation numbers of the elements in their oxides and chlorides. These equations then feed directly into the next note, on how the oxides and chlorides behave in water.

The big picture

Across Period 3 the elements change from reactive metals (Na, Mg), through a metal protected by an oxide layer (Al) and a metalloid (Si), to reactive non-metals (P, S, Cl) and an unreactive noble gas (Ar). Two ideas organise everything:

The maximum oxidation number equals the number of outer-shell electrons. Sodium has one outer electron and forms $\ce{Na2O}$ and $\ce{NaCl}$ ( $+1$ ); phosphorus has five and forms $\ce{P4O10}$ and $\ce{PCl5}$ ( $+5$ ).
Electronegativity rises across the period, so the bonding in the products changes from ionic to covalent. That change controls how the products behave with water (next note).

Reactions with oxygen

All the Period 3 elements from sodium to sulfur burn in oxygen when heated (white phosphorus even ignites spontaneously in air). The products are oxides.

element	conditions and observations	equation	product
sodium	burns with a bright yellow-orange flame	$\ce{4Na(s) + O2(g) -> 2Na2O(s)}$	white solid
magnesium	burns with a brilliant white flame	$\ce{2Mg(s) + O2(g) -> 2MgO(s)}$	white solid
aluminium	powder burns with a bright white flame (sparks); bulk metal is protected by its oxide layer	$\ce{4Al(s) + 3O2(g) -> 2Al2O3(s)}$	white solid
silicon	reacts only when heated strongly	$\ce{Si(s) + O2(g) -> SiO2(s)}$	white solid
phosphorus	white phosphorus ignites spontaneously; burns with a bright white (yellow-white) flame, dense white smoke	$\ce{P4(s) + 5O2(g) -> P4O10(s)}$	white solid
sulfur	burns with a blue flame	$\ce{S(s) + O2(g) -> SO2(g)}$	colourless, choking gas

Sulfur dioxide can be oxidised further to sulfur trioxide, but only slowly, using a catalyst (vanadium(V) oxide in the Contact process):

\ce{2SO2(g) + O2(g) <=> 2SO3(g)}

Chlorine and argon do not react directly with oxygen. (Chlorine oxides exist but are made indirectly and are not on the syllabus.)

Watch out

Phosphorus(V) oxide has the molecular formula $\ce{P4O10}$ , not $\ce{P2O5}$ . $\ce{P2O5}$ is only its empirical formula; the syllabus uses $\ce{P4O10}$ and mark schemes expect it. Equally, write phosphorus as $\ce{P4}$ in equations where the molecule matters.

Why aluminium seems unreactive

Aluminium is high in the reactivity series, yet a sheet of aluminium survives in air and water. Its surface is covered with a thin, tough, unreactive layer of aluminium oxide that forms immediately and stops oxygen or water reaching the metal underneath. Powdered aluminium has a huge surface area and burns readily.

Reactions with chlorine

When heated in chlorine, the elements from sodium to phosphorus form chlorides.

element	observations	equation	product at room temperature
sodium	burns with a bright yellow flame	$\ce{2Na(s) + Cl2(g) -> 2NaCl(s)}$	white solid
magnesium	burns with a bright white flame	$\ce{Mg(s) + Cl2(g) -> MgCl2(s)}$	white solid
aluminium	reacts vigorously when heated	$\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}$	white or pale yellow solid that sublimes at about 450 K
silicon	reacts when heated	$\ce{Si(s) + 2Cl2(g) -> SiCl4(l)}$	colourless liquid
phosphorus	reacts with excess chlorine	$\ce{P4(s) + 10Cl2(g) -> 4PCl5(s)}$	off-white (pale yellow) solid

With a limited supply of chlorine, phosphorus gives phosphorus(III) chloride, $\ce{PCl3}$ , a colourless liquid ( $\ce{P4 + 6Cl2 -> 4PCl3}$ ). The syllabus asks for $\ce{PCl5}$ .

Sulfur does react with chlorine (giving $\ce{S2Cl2}$ ), but this is not required.

Tip

Aluminium chloride exists as $\ce{Al2Cl6}$ dimers in the vapour and in the solid near its sublimation temperature: two $\ce{AlCl3}$ units joined by two dative (coordinate) bonds from chlorine lone pairs into the empty orbital on aluminium. Writing $\ce{AlCl3}$ in equations is accepted. The low sublimation temperature is evidence that aluminium chloride is covalent rather than ionic.

Reactions with water: sodium and magnesium

The syllabus asks only for sodium and magnesium. The contrast between them is a good test of description.

Sodium

\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}

Sodium reacts vigorously with cold water. Observations: it floats, melts into a silvery ball (the reaction is exothermic and sodium melts at 371 K), moves around on the surface, fizzes, and gets smaller until it disappears. The solution is strongly alkaline, pH about 13–14, because sodium hydroxide is soluble and fully dissociated.

Magnesium

With cold water, magnesium reacts very slowly. A few bubbles of hydrogen form on the surface over hours or days:

\ce{Mg(s) + 2H2O(l) -> Mg(OH)2(s) + H2(g)}

The magnesium hydroxide is only sparingly soluble, so the solution is only weakly alkaline, pH about 9–10.

With steam, heated magnesium reacts rapidly, burning with a brilliant white flame and leaving a white solid of magnesium oxide:

\ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)}

The difference between Na and Mg shows the general trend: reactivity of the metals decreases across the period because more energy is needed to remove more electrons (first plus second ionisation energy for Mg is much larger than the first ionisation energy of Na).

Oxidation numbers in the oxides and chlorides

The oxidation number of an element in a compound is the charge it would have if all its bonds were ionic. Oxygen is $-2$ and chlorine is $-1$ in all these compounds, so the oxidation number of the Period 3 element follows by arithmetic.

oxide	$\ce{Na2O}$	$\ce{MgO}$	$\ce{Al2O3}$	( $\ce{SiO2}$ )	$\ce{P4O10}$	$\ce{SO2}$	$\ce{SO3}$
oxidation number	$+1$	$+2$	$+3$	( $+4$ )	$+5$	$+4$	$+6$

chloride	$\ce{NaCl}$	$\ce{MgCl2}$	$\ce{AlCl3}$	$\ce{SiCl4}$	$\ce{PCl5}$
oxidation number	$+1$	$+2$	$+3$	$+4$	$+5$

Key result

Explaining the pattern

The oxidation number of the Period 3 element rises by one across the period, from $+1$ to $+5$ in the chlorides and to $+6$ in $\ce{SO3}$ .
In each of these compounds the element uses all of its outer-shell (valence) electrons in bonding: Na has 1, Mg 2, Al 3, Si 4, P 5, S 6. So the highest oxidation number equals the number of outer-shell electrons, which is the group number.
In $\ce{SO2}$ , sulfur uses only four of its six outer electrons in bonding, so its oxidation number is $+4$ , lower than the maximum.
Oxygen and chlorine are more electronegative than every Period 3 element they combine with, so they take the negative oxidation numbers.

From silicon onwards the elements can expand the octet: phosphorus in $\ce{PCl5}$ has ten electrons around it and sulfur in $\ce{SO3}$ has twelve, because the third shell can hold more than eight electrons. Period 2 elements cannot do this, which is why nitrogen forms only $\ce{NCl3}$ while phosphorus forms $\ce{PCl5}$ .

Worked examples

Equations and observations

Write balanced equations, with state symbols, for the reactions of (a) magnesium with oxygen, (b) aluminium with chlorine and (c) phosphorus with excess oxygen. For (a) state one observation.

Solution

(a) $\ce{2Mg(s) + O2(g) -> 2MgO(s)}$ . Observation: brilliant white flame (and a white solid / white powder forms).

(b) $\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}$ (or $\ce{2Al(s) + 3Cl2(g) -> Al2Cl6(s)}$ ).

Check the balancing by counting atoms: in (c), 4 P and 10 O on each side.

Explaining oxidation numbers

State the oxidation number of the Period 3 element in $\ce{MgCl2}$ , $\ce{SiCl4}$ , $\ce{P4O10}$ and $\ce{SO3}$ , and explain the trend in terms of electrons.

Solution

$\ce{MgCl2}$ : $+2$ . $\ce{SiCl4}$ : $+4$ . $\ce{P4O10}$ : $4x + 10(-2) = 0$ , so $x = +5$ . $\ce{SO3}$ : $x + 3(-2) = 0$ , so $x = +6$ .

The oxidation number increases across the period because the number of outer-shell electrons increases (Mg 2, Si 4, P 5, S 6). In each of these compounds all the outer-shell electrons are used in bonding to the more electronegative oxygen or chlorine, so the oxidation number equals the number of valence electrons.

Reacting masses for a combustion

$1.24\ \text{g}$ of white phosphorus, $\ce{P4}$ , burns in excess oxygen. Calculate the mass of $\ce{P4O10}$ formed and the volume of oxygen used, measured at room conditions. ( $A_r$ : P 31.0, O 16.0)

Solution

$M(\ce{P4}) = 4 \times 31.0 = 124.0\ \text{g mol}^{-1}$ , so $n(\ce{P4}) = \dfrac{1.24}{124.0} = 0.0100\ \text{mol}$ .

$\ce{P4 + 5O2 -> P4O10}$ : $n(\ce{P4O10}) = 0.0100\ \text{mol}$ .

$M(\ce{P4O10}) = 124.0 + 160.0 = 284.0\ \text{g mol}^{-1}$ , mass $= 0.0100 \times 284.0 = 2.84\ \text{g}$ .

$n(\ce{O2}) = 5 \times 0.0100 = 0.0500\ \text{mol}$ , volume $= 0.0500 \times 24.0 = 1.20\ \text{dm}^3$ .

Exam-style: identifying an element from a chloride

$0.270\ \text{g}$ of a Period 3 element X reacts completely with chlorine to form $1.335\ \text{g}$ of a chloride. The chloride sublimes at a low temperature. Identify X and write the equation. ( $A_r$ : Cl 35.5)

Solution

Mass of chlorine combined $= 1.335 - 0.270 = 1.065\ \text{g}$ , so $n(\ce{Cl}) = \dfrac{1.065}{35.5} = 0.0300\ \text{mol}$ .

If the chloride is $\ce{XCl_n}$ , then $n(\ce{X}) = \dfrac{0.0300}{n}$ and $A_r(\ce{X}) = \dfrac{0.270\,n}{0.0300} = 9.0\,n$ .

$n$	1	2	3	4	5
$A_r(\ce{X})$	9.0	18.0	27.0	36.0	45.0

Only $n = 3$ gives the $A_r$ of a Period 3 element: 27.0, aluminium. This fits the sublimation (covalent $\ce{AlCl3}$ / $\ce{Al2Cl6}$ ) and the expected oxidation number $+3$ .

\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}

Exam-hard: comparing sodium and magnesium with water

A small piece of sodium and a piece of magnesium ribbon are each added to cold water containing a few drops of universal indicator. A second piece of magnesium is heated in steam. Describe what you would see in each case, write equations, and explain why the final pH values differ.

Solution

Sodium in cold water: floats, melts into a ball, moves about and fizzes rapidly, and disappears; the indicator turns purple (pH about 13–14). $\ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)}$

Magnesium in cold water: only a few bubbles form slowly on the ribbon; the indicator turns slowly to blue (pH about 9–10). $\ce{Mg(s) + 2H2O(l) -> Mg(OH)2(s) + H2(g)}$

Magnesium in steam: burns with a bright white flame, leaving a white solid. $\ce{Mg(s) + H2O(g) -> MgO(s) + H2(g)}$

pH difference: sodium hydroxide is very soluble, so a high concentration of $\ce{OH-}$ ions forms and the pH is high. Magnesium hydroxide is only sparingly soluble, so the concentration of $\ce{OH-}$ ions in solution is low and the pH is only about 9–10.

Watch out

Do not write $\ce{Mg + H2O -> MgO + H2}$ for cold water. Cold water gives the hydroxide (slowly); steam gives the oxide.
Sulfur burning in oxygen gives $\ce{SO2}$ , not $\ce{SO3}$ . $\ce{SO3}$ needs a catalyst.
Argon has no reactions here and chlorine does not burn in oxygen. Do not invent oxides of chlorine.
Always include state symbols when the question asks for them: $\ce{SiCl4(l)}$ is a liquid and $\ce{SO2(g)}$ a gas at room temperature.

Exam tip

"Describe" for a reaction means observations: flame colour, the colour and state of the product, fizzing. "Burns" alone rarely scores; "burns with a bright white flame to give a white solid" does.
Learn the flame colours: sodium yellow-orange, magnesium brilliant white, sulfur blue.
Oxidation number explanations need the phrase "all the outer-shell (valence) electrons are used in bonding". For $\ce{SO2}$ , say that only four of sulfur's six outer electrons are involved.
Balance equations by atoms first, then check. Common slips: $\ce{4Na + O2 -> 2Na2O}$ (not $\ce{2Na + O2}$ ), $\ce{4Al + 3O2 -> 2Al2O3}$ , $\ce{P4 + 10Cl2 -> 4PCl5}$ .

Practical skills

Burning elements in oxygen is a demonstration, not a student experiment: a small sample is heated on a combustion spoon and lowered into a gas jar of oxygen. Sodium and phosphorus are hazardous (sodium is stored under oil; white phosphorus under water because it ignites in air), sulfur dioxide is toxic, and chlorine reactions are done only in a fume cupboard. In Paper 3 you will not do these reactions, but you may be asked to test the products: for example, add water and universal indicator to an oxide and record the colour and pH.

Summary

With oxygen: $\ce{Na2O}$ , $\ce{MgO}$ , $\ce{Al2O3}$ , $\ce{SiO2}$ , $\ce{P4O10}$ , $\ce{SO2}$ (then $\ce{SO3}$ with a catalyst). Na yellow flame, Mg white flame, S blue flame.
With chlorine: $\ce{NaCl}$ , $\ce{MgCl2}$ , $\ce{AlCl3}$ (sublimes, $\ce{Al2Cl6}$ ), $\ce{SiCl4}$ (liquid), $\ce{PCl5}$ (excess chlorine).
Na reacts vigorously with cold water (pH 13–14); Mg reacts very slowly with cold water (pH 9–10) but burns in steam to give $\ce{MgO}$ .
Aluminium is protected by an unreactive oxide layer.
Highest oxidation number = number of outer-shell electrons, because all are used in bonding: $+1$ to $+5$ in the chlorides, $+6$ in $\ce{SO3}$ ; $+4$ in $\ce{SO2}$ .
P and S can expand the octet ( $\ce{PCl5}$ , $\ce{SO3}$ ).

Practice

Question

Write balanced equations, with state symbols, for the reactions of sodium with oxygen and of silicon with chlorine.
Describe what you would see when sulfur is burned in oxygen, and write the equation.
State the oxidation number of the Period 3 element in $\ce{Na2O}$ , $\ce{Al2O3}$ , $\ce{PCl5}$ and $\ce{SO2}$ .
Explain why aluminium foil does not appear to react with water even though aluminium is a reactive metal.
Calculate the mass of sodium chloride formed when $2.30\ \text{g}$ of sodium burns in excess chlorine, and the volume of chlorine used at room conditions. ( $A_r$ : Na 23.0, Cl 35.5)
$0.486\ \text{g}$ of magnesium reacts completely with steam. Calculate the volume of hydrogen formed at room conditions and name the solid product. ( $A_r$ : Mg 24.3)
Explain why the maximum oxidation number of sulfur is $+6$ , but the oxidation number of sulfur in the product of burning sulfur in air is only $+4$ .
Calculate the percentage by mass of chlorine in phosphorus(V) chloride. ( $A_r$ : P 31.0, Cl 35.5)
$0.562\ \text{g}$ of silicon is heated in excess chlorine. Calculate the mass of product formed, and explain why the product is a liquid at room temperature while magnesium chloride is a solid. ( $A_r$ : Si 28.1, Cl 35.5)
Nitrogen is in the same group as phosphorus but forms only $\ce{NCl3}$ , never $\ce{NCl5}$ . Phosphorus forms both $\ce{PCl3}$ and $\ce{PCl5}$ . Explain this difference, and state which phosphorus chloride forms when phosphorus reacts with excess chlorine.

Answers

$\ce{4Na(s) + O2(g) -> 2Na2O(s)}$ ; $\ce{Si(s) + 2Cl2(g) -> SiCl4(l)}$ .
Sulfur melts and burns with a blue flame, giving a colourless gas with a choking smell. $\ce{S(s) + O2(g) -> SO2(g)}$ .
$\ce{Na2O}$ $+1$ ; $\ce{Al2O3}$ $+3$ ; $\ce{PCl5}$ $+5$ ; $\ce{SO2}$ $+4$ .
The aluminium is covered by a thin, continuous, unreactive layer of aluminium oxide, $\ce{Al2O3}$ , which stops water (and air) reaching the metal beneath.
$n(\ce{Na}) = 2.30 / 23.0 = 0.100\ \text{mol}$ . $\ce{2Na + Cl2 -> 2NaCl}$ , so $n(\ce{NaCl}) = 0.100\ \text{mol}$ ; mass $= 0.100 \times 58.5 = 5.85\ \text{g}$ . $n(\ce{Cl2}) = 0.0500\ \text{mol}$ ; volume $= 0.0500 \times 24.0 = 1.20\ \text{dm}^3$ .
$n(\ce{Mg}) = 0.486 / 24.3 = 0.0200\ \text{mol}$ . $\ce{Mg + H2O -> MgO + H2}$ is $1 : 1$ , so $n(\ce{H2}) = 0.0200\ \text{mol}$ ; volume $= 0.0200 \times 24.0 = 0.480\ \text{dm}^3$ ( $480\ \text{cm}^3$ ). The solid is magnesium oxide, $\ce{MgO}$ .
Sulfur has six outer-shell electrons, so when all six are used in bonding (as in $\ce{SO3}$ ) its oxidation number is $+6$ . When sulfur burns in air the product is $\ce{SO2}$ , in which only four of the six outer electrons are used in bonding to oxygen, giving $+4$ . Further oxidation to $\ce{SO3}$ is slow and needs a catalyst.
$M(\ce{PCl5}) = 31.0 + 5 \times 35.5 = 208.5\ \text{g mol}^{-1}$ . Percentage of Cl $= \dfrac{177.5}{208.5} \times 100 = 85.1\%$ .
$n(\ce{Si}) = 0.562 / 28.1 = 0.0200\ \text{mol} = n(\ce{SiCl4})$ . $M(\ce{SiCl4}) = 28.1 + 142.0 = 170.1\ \text{g mol}^{-1}$ ; mass $= 0.0200 \times 170.1 = 3.40\ \text{g}$ . $\ce{SiCl4}$ is simple molecular: only weak id–id forces act between molecules, so little energy is needed to separate them and it is a liquid. $\ce{MgCl2}$ is ionic with a giant lattice held by strong electrostatic attraction between $\ce{Mg^2+}$ and $\ce{Cl-}$ ions, so it is a high-melting solid.
Nitrogen is in Period 2: its outer shell is the second shell, which can hold a maximum of eight electrons, so it can form only three bonds (plus a lone pair) and cannot expand its octet. Phosphorus is in Period 3; its outer shell (the third shell) can hold more than eight electrons, so phosphorus can use all five outer electrons in bonding and expand its octet to ten electrons in $\ce{PCl5}$ . With excess chlorine, $\ce{PCl5}$ forms: $\ce{P4(s) + 10Cl2(g) -> 4PCl5(s)}$ .

The big picture

Reactions with oxygen

Why aluminium seems unreactive

Reactions with chlorine

Reactions with water: sodium and magnesium

Sodium

Magnesium

Oxidation numbers in the oxides and chlorides

Worked examples

Practice

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