Exercises — Aluminium — chemistry, alloys; alumina, alums

Fill orbitals in order: $1s^{2}2s^{2}2p^6$ uses up $2+2+6=10$ electrons (that's the neon core, written $[Ne]$). The remaining $13-10=3$ electrons go $3s^{2}3p^1$. $\text{Al}=[Ne]3s^{2}3p^1$ To reach the stable neon shell it loses all three outer electrons ($3s^2$ and $3p^1$), giving oxidation state $\boxed{+3}$.

Amphoteric = reacts with both acids and bases. The first three all do:

• $Al$ dissolves in HCl and in NaOH,
• $A{l}_2{O}_3$ and $Al(OH{)}_3$ likewise.

$NaCl$ is the salt of a strong acid and strong base — it does neither. Answer: $Al$, $A{l}_2{O}_3$, $Al(OH{)}_3$.

$KAl(S{O}_4{)}_2\cdot 12H_{2}O$ Cations: $K^{+}$ (univalent) and $A{l}^{3+}$ (trivalent). Anion: $S{O}_4^{2-}$.

$O{H}^{-}$ dissolves the oxide skin and then attacks the metal to make the aluminate ion $[Al(OH{)}_4{]}^{-}$. Water supplies the extra oxygen and the hydrogen that leaves as gas: $2Al+2NaOH+6H_{2}O\to 2Na[Al(OH{)}_4]+3H_2\uparrow$ Check the atoms: Al: $2=2$. Na: $2=2$. On the left, H atoms $=2(\text{from NaOH})+12(\text{from }6H_{2}O)=14$; on the right $=8(\text{from }2[Al(OH{)}_4])+6(\text{from }3H_2)=14$. ✓ O atoms: left $2+6=8$; right $2\times 4=8$. ✓ Gas = $H_2$ (hydrogen). This dissolving-in-base is the fingerprint of an amphoteric metal.

Reaction: $2Al+6HCl\to 2AlC{l}_3+3H_2$. Moles of Al $=\frac{5.4}{27}=0.20\text{mol}$. The equation says 2 mol Al → 3 mol $H_2$, so moles of $H_2=0.20\times \frac{3}{2}=0.30\text{mol}$. Volume $=0.30\times 22.4=\boxed{6.72\text{L}}$.

Acting base-like (reacts with acid): $A{l}_2{O}_3+6HCl\to 2AlC{l}_3+3H_{2}O$ Acting acid-like (reacts with base): $A{l}_2{O}_3+2NaOH+3H_{2}O\to 2Na[Al(OH{)}_4]$

In water, $A{l}^{3+}$ does not float alone — it pulls six water molecules around itself as $[Al(H_{2}O{)}_6{]}^{3+}$ (a small taste of Coordination Compounds). Now the charge density idea bites: the tiny, triply-charged $A{l}^{3+}$ tugs electron density out of the O–H bonds of those coordinated waters. The O–H bond weakens, and a proton falls off: $[Al(H_{2}O{)}_6{]}^{3+}\rightleftharpoons [Al(H_{2}O{)}_5(OH){]}^{2+}+H^{+}$ Free $H^{+}$ makes the solution acidic, so blue litmus turns red. Equivalently, we say the salt hydrolyses: $AlC{l}_3+3H_{2}O\to Al(OH{)}_3+3HCl$

The strength of the oxidiser is exactly what defeats it here. Concentrated $HN{O}_3$ oxidises the metal surface so quickly and completely that it lays down a dense, adherent $A{l}_2{O}_3$ film — the same passivation armour Al grows in air. Once sealed, no fresh metal is exposed, so the reaction stalls. Al becomes inert. This is why concentrated $HN{O}_3$ is shipped in aluminium tankers.

Compare the cations by charge density. $N{a}^{+}$ is fairly large and singly charged → low polarising power. $A{l}^{3+}$ is small and triply charged → high polarising power. By Fajans Rules, the strongly polarising $A{l}^{3+}$ distorts the $C{l}^{-}$ electron cloud so much that shared (covalent) character develops. Consequences:

• $NaCl$: ideal ions, strong 3-D lattice → high melting point.
• $AlC{l}_3$: covalent molecular units → low melting point, sublimes, dissolves in non-polar solvents, and hydrolyses (fumes HCl) in moist air.

Step 1 — count electrons (WHAT). In $AlC{l}_3$, Al forms 3 single bonds → 3 shared pairs → only 6 electrons around Al, two short of an octet. Al has an empty orbital. That makes it an electron-deficient Lewis acid. Step 2 — find a donor (WHY). Each Cl carries lone pairs. A Cl on one $AlC{l}_3$ molecule donates a lone pair into the empty orbital of the Al on a second molecule (look at the two violet arrows in the figure crossing between the units). Step 3 — two bridges form (WHAT IT LOOKS LIKE). Two such donations, one from each side, lock the pair together through two chlorine bridges. Now every Al owns 4 bonds → a full octet, and the geometry around each Al becomes tetrahedral ($s{p}^3$). Result: $A{l}_{2}C{l}_6$ with 2 bridging Cl (shared, orange in the figure) and 4 terminal Cl (magenta). This dimer lives in the vapour and in non-polar solvents.

Why pure Al is soft: a pure metal is a neat stack of identical atoms. Rows of atoms slip past one another easily along dislocations (line defects), so the metal deforms readily → soft. Why the alloy is hard: dropping in atoms of a different size (Cu, Mg, Mn) distorts the lattice locally. These distortions pin the dislocations, so rows can no longer glide → harder, stronger — yet still light, because Al is the majority element (low density preserved). Duralumin: Al + Cu + Mg + Mn. Use: aircraft bodies and other light-but-strong parts.

Build the molar mass piece by piece:

• $K=39$
• $Al=27$
• $(S{O}_4{)}_2=2\times (32+4\times 16)=2\times 96=192$
• $12H_{2}O=12\times (2\times 1+16)=12\times 18=216$

Total $M=39+27+192+216=474{\text{g mol}}^{-1}$. $\%Al=\frac{27}{474}\times 100\approx \boxed{5.70\%}$

Why it works (Ellingham logic). The Ellingham diagram plots how favourable each metal–oxide is versus temperature. The Al/$A{l}_2{O}_3$ line sits below the Fe/$F{e}_2{O}_3$ line — i.e. Al binds oxygen more strongly than Fe does. So Al will happily strip the oxygen off iron oxide, reducing $F{e}_2{O}_3$ to molten Fe while itself being oxidised to $A{l}_2{O}_3$. Al is the reducing agent. Heat released. $\Delta H_{rxn}=\sum \Delta H_f(\text{products})-\sum \Delta H_f(\text{reactants})$: $\Delta H=[(-1676)+0]-[0+(-824)]=-1676+824=\boxed{-852{\text{kJ mol}}^{-1}}$ The large negative value is the intense heat that melts the iron — used to weld rail tracks.

The claim is wrong. An alum is defined by its pattern of charges, not by containing aluminium: $M^{+}{M}^{\prime 3+}(S{O}_4{)}_2\cdot 12H_{2}O$ where $M^{+}$ is any univalent cation ($K^{+}$, $N{a}^{+}$, $N{H}_4^{+}$) and $M^{\prime 3+}$ is any trivalent cation ($A{l}^{3+}$, $F{e}^{3+}$, $C{r}^{3+}$). Chrome alum fits perfectly: $M^{+}=K^{+}$, $M^{\prime 3+}=C{r}^{3+}$. It is a genuine alum. (The name "alum" historically came from potash alum, but the class is defined by the $+1/+3$ double-sulphate structure.)

Every single answer traced back to one picture — a tiny marble with three plus-signs ($A{l}^{3+}$, high charge density). It explains why alum has 5.7% aluminium, why chloride bridges form, why solutions go acidic, and why Al reduces rust. Learn the marble; the chemistry follows.