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AP Intermediate 2nd Year Biomolecules — practice questions

Explore 14 free Biomolecules questions in AP Intermediate 2nd Year Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Biomolecules questions and exercises with answers

Showing questions 1–14 of 14.

Worked practice questions

The IUPAC name of the coordination compound $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$ is:

  1. Hexaamminecobalt(II) chloride
  2. Tris(hexaammine)cobalt chloride
  3. Hexaamminecobalt(III) chloride
  4. Cobalt(III) hexaamine chloride
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Correct answer: C. Hexaamminecobalt(III) chloride

**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**.

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The total number of unidentate ligand atoms directly bonded to the central metal atom in a coordination complex is called its:

  1. Bond order
  2. Oxidation state
  3. Primary valency
  4. Coordination number
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Correct answer: D. Coordination number

**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.

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In octahedral crystal-field splitting, what is the energy of the $e_g$ orbitals **relative to** the $t_{2g}$ orbitals?

  1. Equal in energy
  2. Lower by $\Delta_o$
  3. Higher by $\dfrac{3}{5}\Delta_o$ (relative to the average level)
  4. Higher by $\Delta_o$
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Correct answer: D. Higher by $\Delta_o$

**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 $e_g$).

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