Inorganic Chemistry -- Practice Problems

Inorganic Chemistry — Practice Problems

Intuition

Inorganic chemistry covers the rest of the periodic table: While organic chemistry focuses on carbon, inorganic chemistry encompasses metals, minerals, and coordination compounds. It explains why transition metals form coloured compounds, why catalysts work, and how solid-state materials conduct electricity.

Why it matters: Inorganic chemistry underpins catalysis, materials science, medicine (metal-based drugs), and biological systems (metalloenzymes).

The key insight: The d-block elements are special because their partially filled d-orbitals enable variable oxidation states, coloured compounds, and catalytic activity that main-group elements cannot match.

Worked Examples

Example 1: Crystal Field Theory, Colour Prediction

Problem: $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ absorbs light at 493 nm. Calculate the crystal field splitting energy $\Delta_o$ and predict the colour of the complex.

Solution:

Step 1: Calculate the energy of absorbed light

E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{493 \times 10^{-9}} = 4.03 \times 10^{-19} \text{ J}

Step 2: Convert to kJ/mol

\Delta_o = (4.03 \times 10^{-19} \text{ J}) \times (6.022 \times 10^{23} \text{ mol}^{-1}) = 242.7 \text{ kJ/mol}

Step 3: Determine the colour

The complex absorbs yellow-green light (493 nm). The complementary colour (what we see) is purple/violet.

Key insight: The colour of transition metal complexes arises from d-d transitions. The wavelength of absorbed light equals the crystal field splitting energy. Strong-field ligands (like CN⁻) produce large $\Delta_o$ → absorb high-energy (blue) light → appear yellow/orange. Weak-field ligands (like H₂O) produce small $\Delta_o$ → absorb low-energy (red) light → appear blue/green.

Example 2: Unit Cell Calculations

Problem: Iron crystallizes in a BCC structure with a lattice parameter $a = 286.6$ pm. Calculate the atomic radius and packing fraction.

Solution:

Step 1: Relate lattice parameter to atomic radius

In BCC, atoms touch along the body diagonal: $4r = a\sqrt{3}$

r = \frac{a\sqrt{3}}{4} = \frac{286.6 \times \sqrt{3}}{4} = \frac{286.6 \times 1.732}{4} = 124.1 \text{ pm}

Step 2: Calculate packing fraction

BCC has 2 atoms per unit cell. Volume of atoms:

V_{\text{atoms}} = 2 \times \frac{4}{3}\pi r^3 = 2 \times \frac{4}{3}\pi (124.1)^3 = 1.602 \times 10^7 \text{ pm}^3

Volume of unit cell:

V_{\text{cell}} = a^3 = (286.6)^3 = 2.355 \times 10^7 \text{ pm}^3

Packing fraction:

\text{PF} = \frac{V_{\text{atoms}}}{V_{\text{cell}}} = \frac{1.602 \times 10^7}{2.355 \times 10^7} = 0.680

Comparison: BCC packing fraction (0.68) is less than FCC/HCP (0.74). This explains why BCC metals (Fe, Cr, W) are generally less dense than FCC metals (Cu, Ag, Au) of similar atomic mass.

Example 3: Coordination Number and Geometry

Problem: Determine the coordination number, geometry, and magnetic properties of $[\text{Fe}(\text{CN})_6]^{4-}$ .

Solution:

Step 1: Determine the oxidation state of Fe

$\text{CN}^-$ has charge $-1$ . Total charge: $-4$

\text{Fe} + 6(-1) = -4 \implies \text{Fe} = +2

Step 2: Electron configuration

$\text{Fe}^{2+}$ : $[\text{Ar}] 3d^6$

Step 3: Determine high-spin vs low-spin

$\text{CN}^-$ is a strong-field ligand (high in spectrochemical series). Large $\Delta_o$ → low-spin

t_{2g}^6 \, e_g^0

All 6 electrons paired in $t_{2g}$ orbitals.

Step 4: Properties

Property	Value
Coordination number	6
Geometry	Octahedral
Hybridization	$d^2sp^3$ (inner orbital)
Unpaired electrons	0
Magnetic behaviour	Diamagnetic
Spin-only magnetic moment	$\mu = 0$ BM

Key insight: The strong-field CN⁻ ligand forces all electrons into the lower $t_{2g}$ set, creating a diamagnetic complex. With a weak-field ligand like H₂O, the same Fe²⁺ would be high-spin ( $t_{2g}^4 e_g^2$ , 4 unpaired electrons, paramagnetic).

Inorganic Chemistry -- Practice Problems