Bonding in Coordination Compounds: Quick Revision

A brief note on Bonding in Coordination Compounds. Read it before a test. Try each question at the end before you open its answer.

Valence bond theory

• The ligand gives a lone pair. The metal ion gives an empty hybrid orbital. The bond is a coordinate bond.
• Octahedral: $d^2sp^3$ (inner orbital, low spin, strong ligand) or $sp^3d^2$ (outer orbital, high spin, weak ligand).
• Tetrahedral: $sp^3$. Square planar: $dsp^2$.
• A strong ligand ($\text{CN}^-$, CO, $\text{NH}_3$, en) pairs the d electrons. A weak ligand ($\text{F}^-$, $\text{Cl}^-$, $\text{H}_2\text{O}$) does not.
• Spin-only moment: $\mu = \sqrt{n(n+2)}\ \text{BM}$. For $n = 1$ to $5$ it is $1.73$, $2.83$, $3.87$, $4.90$ and $5.92\ \text{BM}$.
• The theory cannot explain colour, why ligands differ in strength, or the stability of a complex.

Crystal field theory

• The ligands are point negative charges. The bond is electrostatic.
• Octahedral: $t_{2g}$ at $-0.4\,\Delta_o$ and $e_g$ at $+0.6\,\Delta_o$. Tetrahedral: $e$ at $-0.6\,\Delta_t$ and $t_2$ at $+0.4\,\Delta_t$.
• $\Delta_t = \tfrac{4}{9}\Delta_o$, so tetrahedral complexes are high spin.
• $\Delta_o > P$: low spin. $\Delta_o < P$: high spin. Only $d^4$ to $d^7$ have the choice.
• Spectrochemical series: $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$.
• $\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\,\Delta_o$.

Colour, magnetism and stability

• Colour comes from a d-d transition. The colour seen is opposite the colour absorbed. $\Delta = 119600/\lambda$ in kJ/mol, with $\lambda$ in nm.
• $d^0$ and $d^{10}$ complexes are colourless.
• The moment tells the number of unpaired electrons, and so the spin state.
• A larger stability constant means a more stable complex. $\Delta G^\circ = -5.706 \times \log K$ in kJ/mol at $298\ \text{K}$.
• Stability grows with the charge on the metal ion, the basicity of the ligand and the chelate effect.
• In metal carbonyls, CO donates a lone pair and the metal gives d electrons back to CO.

A worked example

A $d^6$ ion in an octahedral complex has $\Delta_o = 280\ \text{kJ/mol}$ and $P = 230\ \text{kJ/mol}$. Find the spin state, the unpaired electrons and the CFSE.

1. Compare. $280 > 230$, so the field is strong and the complex is low spin.
2. Fill. All six electrons go into $t_{2g}$: $t_{2g}^6 e_g^0$. No electron is unpaired.
3. CFSE. $-0.4 \times 6 \times 280\ \text{kJ/mol} = -672\ \text{kJ/mol}$.

Practice with new numbers

A program to try

The program lists the CFSE and the moment for octahedral $d^1$ to $d^8$ ions in a weak field and in a strong field.

Mistakes to avoid

• Swapping 0.4 and 0.6. The lower set takes $-0.4\,\Delta_o$.
• Drawing a low spin tetrahedral complex.
• Taking the colour absorbed as the colour seen.
• Using $sp^3d^2$ for every octahedral complex.
• Counting $d^0$ or $d^{10}$ complexes as coloured.

Check yourself

1. Which d orbitals form the upper set in an octahedral complex?

2. A complex has a moment of $4.90\ \text{BM}$. How many unpaired electrons does it have?

3. A complex absorbs orange light. What colour does it show?

4. Which is more stable, a complex with $\log K = 8$ or one with $\log K = 15$?