Inorganic Chemistry for the Chemistry Olympiad — NSEC and INChO

Weightage: Inorganic chemistry is tested as reasoning from structure and bonding, with fewer pure-recall items than the school syllabus suggests. The syllabus builds on Classes XI and XII with extension, so confirm it with HBCSE. The reliable method is: electron count first, geometry second, properties third.

1. Periodic trends and their exceptions

Across a period, atomic radius falls and ionisation energy and electronegativity rise because nuclear charge increases with the same shell. Down a group, radius rises and ionisation energy falls.

The exceptions carry the marks:

• Ionisation energy: Be $>$ B and N $>$ O. A filled or half-filled subshell is extra stable.
• Electron affinity: chlorine is higher than fluorine, because the small fluorine atom has strong electron repulsion.
• Lanthanide contraction: poor shielding by $4f$ electrons makes the $5d$ elements almost the same size as the $4d$ elements, so Zr and Hf resemble each other.
• Inert pair effect: heavier $p$-block elements prefer an oxidation state two lower than the group number, such as Pb(II) over Pb(IV) and Tl(I) over Tl(III).

Diagonal relationships link Li with Mg, Be with Al and B with Si.

2. Shapes and hybridisation

VSEPR counts electron pairs around the central atom, bonding and lone, and arranges them to minimise repulsion. Lone pairs repel more than bonding pairs.

Worked example. XeF${}_4$ has $8+4=12$ valence electrons, giving six electron pairs: four bonding and two lone. The lone pairs sit opposite each other, so the molecule is square planar and non-polar.

Hybridisation follows the electron-pair count: $sp$ (2), $s{p}^2$ (3), $s{p}^3$ (4), $s{p}^{3}d$ (5), $s{p}^3{d}^2$ (6). In an octahedral arrangement, lone pairs occupy trans positions.

3. Molecular orbitals and magnetism

Bond order is $\frac{1}{2}(N_b-N_a)$, bonding minus antibonding electrons. For O${}_2$, filling the molecular orbitals gives a bond order of $2$ with two unpaired electrons in ${\pi }^{\ast }$ orbitals, so O${}_2$ is paramagnetic, which valence bond theory cannot explain. For N${}_2$ the bond order is $3$, and for O${}_2^{-}$ it is $1.5$.

Worked example. Compare O${}_2$, O${}_2^{+}$, O${}_2^{-}$: bond orders 2, 2.5 and 1.5, so the bond length order is O${}_2^{+}<$ O${}_2<$ O${}_2^{-}$.

In heteronuclear molecules such as CO, electrons are shared unequally and the more electronegative atom contributes more to the bonding orbitals.

4. Coordination compounds

A complex has a central metal ion and ligands. The coordination number is the number of donor atoms. Naming follows the ligands (alphabetical), then the metal with its oxidation state in Roman numerals.

Crystal field theory. In an octahedral field the $d$ orbitals split into a lower $t_{2g}$ set and a higher $e_g$ set, separated by ${\Delta }_o$. If ${\Delta }_o$ is smaller than the pairing energy, the complex is high spin, and otherwise low spin. Strong-field ligands such as CN${}^{-}$ and CO give low spin, and weak-field ligands such as halides give high spin, as ranked in the spectrochemical series.

The spin-only magnetic moment is $\mu =\sqrt{n(n+2)}$ Bohr magnetons, with $n$ unpaired electrons.

Worked example. In the high-spin Fe${}^{3+}$ ($d^5$) complex $[\text{Fe}({\text{H}}_2\text{O}{)}_6{]}^{3+}$, all five electrons are unpaired, so $\mu =\sqrt{35}=5.92$ BM. In the low-spin $[\text{Fe}(\text{CN}{)}_6{]}^{3-}$, one electron is unpaired and $\mu =\sqrt{3}=1.73$ BM.

Tetrahedral splitting is smaller, ${\Delta }_t=\frac{4}{9}{\Delta }_o$, so tetrahedral complexes are almost always high spin. Colour arises from $d$-$d$ transitions. A $d^0$ or $d^{10}$ ion has no $d$-$d$ transition and is colourless.

The 18-electron rule says stable organometallic complexes often have 18 valence electrons. For Ni(CO)${}_4$: Ni gives $10$ and four CO give $8$, so the total is $18$.

Isomerism: geometric (cis and trans), optical (octahedral complexes with chelating ligands), linkage ($-N{O}_2$ versus $-ONO$) and ionisation isomers.

5. Main-group chemistry

• Hydrogen and the $s$ block: the hydrides are ionic with Group 1 and 2 metals, and the alkali metals dissolve in liquid ammonia to give blue solutions with solvated electrons.
• Boron: it is electron deficient, and BF${}_3$ is a Lewis acid. Diborane has two three-centre two-electron bridging bonds.
• Group 14: carbon forms stable multiple bonds, silicon forms ${\text{SiO}}_2$ networks rather than ${\text{CO}}_2$ molecules, and the inert pair effect appears with tin and lead.
• Group 15 to 17: oxoacids of phosphorus and sulphur, the oxides of nitrogen, interhalogen compounds and the noble gas fluorides follow the VSEPR rules above.

Acid-base ideas: the oxides of metals are basic, those of non-metals are acidic and some are amphoteric (Al${}_2$O${}_3$, ZnO). Hard and soft acid-base theory predicts that hard acids bind hard bases, and soft with soft.

6. Solid state

Cubic unit cells hold $1$ (simple), $2$ (body-centred) or $4$ (face-centred) atoms. The packing efficiencies are $52\%$, $68\%$ and $74\%$. Ionic radius ratios decide coordination, and Born-Haber cycles give lattice energies. Lattice energy rises with ionic charge and falls with ionic size, so MgO melts at a higher temperature than NaCl.

Common traps

• Giving fluorine the highest electron affinity. Chlorine is higher.
• Calling XeF${}_4$ tetrahedral. The lone pairs make it square planar.
• Using valence bond theory to explain O${}_2$ paramagnetism.
• Counting electrons from the wrong oxidation state in crystal field problems.
• Forgetting that tetrahedral complexes are high spin.

Memory aids

• "Half-filled and filled are stable": ionisation energy exceptions.
• "Bonding minus antibonding, halved": bond order.
• "Square planar from 4 plus 2": lone pair on each side.

Summary

Inorganic reasoning starts with electron counts: the periodic trends and their exceptions, VSEPR shapes and hybridisation, and MO bond orders explain structure and magnetism. Crystal field theory and the 18-electron rule explain coordination compounds.

Main-group chemistry follows from electron deficiency, the inert pair effect and acid-base character.

Exam protocol

• Count valence electrons before drawing a structure.
• Predict shape from electron pairs, then hybridisation.
• For complexes, find the $d$ count and then spin state.
• Confirm the syllabus and format with HBCSE.