The Boron Family: Quick Revision

A brief note on The Boron Family. Read it before a test. Try each question at the end before you open its answer.

Group 13 in short

• Outer pattern $ns^2\,np^1$. Oxidation states $+3$ and $+1$, with $+1$ more stable down the group.
• Radius (pm): B 85, Al 143, Ga 135, In 167, Tl 170. Gallium is smaller than aluminium, because 3d electrons shield poorly.
• First ionisation enthalpy (kJ/mol): B 801, Al 577, Ga 579, In 558, Tl 589. It does not fall steadily.
• Electronegativity: B 2.0, Al 1.5, Ga 1.6, In 1.7, Tl 1.8.
• Boron is a non-metal. The others are metals.

Why boron is different

• Small, high ionisation enthalpy, so only covalent compounds. No B$^{3+}$ ion.
• No d orbitals, so the maximum covalency is 4: $[\text{BF}_{4}]^{-}$. Aluminium reaches 6: $[\text{AlF}_{6}]^{3-}$.
• Electron deficient and a Lewis acid: $\text{BF}_{3} + \text{NH}_{3} \to \text{F}_{3}\text{B}\leftarrow \text{NH}_{3}$.
• Diagonal relationship with silicon.

Borax, boric acid and diborane

• Borax: $\text{Na}_{2}\text{B}_{4}\text{O}_{7}\cdot 10\text{H}_{2}\text{O}$, the anion is $[\text{B}_{4}\text{O}_{5}(\text{OH})_{4}]^{2-}$. Basic solution, a buffer. $\text{Na}_{2}\text{B}_{4}\text{O}_{7} + 2\text{HCl} + 5\text{H}_{2}\text{O} \to 2\text{NaCl} + 4\text{H}_{3}\text{BO}_{3}$.
• Heating borax: water, then a glass of $\text{NaBO}_{2}$ and $\text{B}_{2}\text{O}_{3}$. The glass gives the borax bead test.
• Boric acid: $\text{B}(\text{OH})_{3}$, planar, layers joined by hydrogen bonds. Weak monobasic Lewis acid, pKa 9.24. Heating gives $\text{HBO}_{2}$ and then $\text{B}_{2}\text{O}_{3}$.
• Diborane: 12 valence electrons, four terminal 2c-2e bonds, two bridge 3c-2e bonds. It burns, hydrolyses to boric acid and hydrogen, and splits with Lewis bases.

Aluminium

• Ore bauxite. Bayer process, then Hall-Heroult electrolysis in cryolite at about 1230 K. Cathode $\text{Al}^{3+} + 3\text{e}^{-} \to \text{Al}$.
• Oxide and hydroxide are amphoteric. Aluminium dissolves in acid and in alkali.
• $\text{AlCl}_{3}$ is covalent, a dimer $\text{Al}_{2}\text{Cl}_{6}$, and a Lewis acid. Alum: $\text{KAl}(\text{SO}_{4})_{2}\cdot 12\text{H}_{2}\text{O}$.

A worked example

Find the mass of boric acid from $38.14\ \text{g}$ of borax.

1. Moles of borax. $38.14 / 381.37 = 0.1000\ \text{mol}$.
2. Boric acid. Four moles for each mole of borax: $0.4000\ \text{mol}$, which weighs $0.4000 \times 61.83 = 24.73\ \text{g}$.

Practice with new numbers

A program to try

Mistakes to avoid

• Giving boron an ionic 3+ charge or a covalency of 6. Neither is possible.
• Writing boric acid as a proton donor. It takes $\text{OH}^{-}$ from water.
• Counting 14 or 16 electrons in diborane. It has 12.
• Saying that Al(OH)$_3$ dissolves in excess ammonia. It dissolves in excess alkali.

Check yourself

1. How many moles of HCl react with 1 mol of borax?

2. Why can boron not form $[\text{BF}_{6}]^{3-}$?

3. What is the electron count of diborane, and how many 3c-2e bonds does it have?

4. Name the two stages of the extraction of aluminium from bauxite.

Sources

• OpenStax, Chemistry 2e, Chapter 18.
• NCERT Chemistry for Class 11, the chapter The p-Block Elements (older edition).
• Housecroft and Sharpe, Inorganic Chemistry, the chapter on the group 13 elements.