JEECUP (UP Polytechnic) Coordination Compounds — practice questions

**IUPAC naming of cation-anion coordination compounds.** 1. Inside the bracket (cation): name ligands alphabetically, then the metal with its oxidation state. 2. The neutral NH₃ ligand → **ammine** (not 'amine' — the spelling matters for naming). 3. Six ammine groups → **hexaammine**. 4. Charge balance: 3 Cl⁻ outside ⇒ cation charge $+3$ ⇒ Co is in oxidation state $+3$, written as **(III)**. Full name: **Hexaamminecobalt(III) chloride**.

**Coordination number** (CN) counts the donor atoms attached directly to the metal — i.e. the number of bonds it forms with the ligands. For a bidentate ligand each ligand contributes *two* to the CN for unidentate, one each. Common CNs are 2 (linear), 4 (tetrahedral or square planar) and 6 (octahedral). - Option A (oxidation state) is the formal charge on the metal — Werner called this the *primary valency*. - Option C is Werner's term for oxidation state, so it overlaps with A but is not the CN.

**Geometry of the splitting.** In an octahedral field, six ligands approach the metal along the $\pm x$, $\pm y$, $\pm z$ axes. The $e_g$ pair ($d_{z^2}$ and $d_{x^2-y^2}$) points *toward* the ligands and is destabilised the $t_{2g}$ trio ($d_{xy}$, $d_{yz}$, $d_{zx}$) points *between* the ligands and is stabilised. **Energy gap.** The energy *difference* between the two sets is called the **crystal-field splitting energy** $\Delta_o$, with $e_g$ **higher** than $t_{2g}$ by exactly $\Delta_o$. - Option C is correct about the $e_g$ rising by $\frac{3}{5}\Delta_o$ above the **unsplit average**, while $t_{2g}$ drops by $\frac{2}{5}\Delta_o$. The total gap is therefore $\frac{3}{5}\Delta_o + \frac{2}{5}\Delta_o = \Delta_o$ — but the question asks for $e_g$ *relative to $t_{2g}$*, not relative to the average. That's $\Delta_o$, not $\frac{3}{5}\Delta_o$. - Option A reverses the direction (in *tetrahedral* splitting the order flips — $t_{2g}$ rises above $e_g$).