How to use these questions
This set moves across common inorganic chemistry topics rather than following one institution’s syllabus. There is no single international exam format, so use it as a diagnostic: answer each question in the block, then read the explanation and identify the rule you need to recall next time. Check your course or examination body’s syllabus for the topics and depth required.
The explanations point out why a tempting alternative is wrong as well as why the answer is right. That matters in inorganic chemistry, where a correct-looking number can come from using the wrong charge, electron count or convention.
Questions 1–3: ions and coordination chemistry
Question 1 tests how to assign an oxidation state when a complex ion contains both neutral and charged ligands. The first MySummaries question tests that bookkeeping.
What is the oxidation state of cobalt in [Co(NH3)5Cl]Cl2?
The complex ion has charge +2 because there are two outer-sphere Cl− ions. Five NH3 ligands are neutral and the coordinated Cl ligand is −1, so x − 1 = +2 and cobalt is +3. +2 is the charge of the complex ion, not the oxidation state of cobalt.
Keep the charge balance explicit: write the overall charge of the complex ion, then add the ligand charges. This avoids treating outer-sphere counterions as though they were coordinated ligands.
Question 2 tests a well-known exception to the simple order predicted by filling 4s before 3d.
Which is the ground-state electron configuration of a neutral chromium atom?
Neutral chromium is [Ar] 3d5 4s1. The half-filled 3d subshell gives a lower-energy arrangement than [Ar] 3d4 4s2. [Ar] 3d4 4s2 is the straightforward filling prediction, but it misses this exception.
Learn chromium alongside copper, another common exception: [Ar] 3d10 4s1. For ions, remove electrons from the 4s subshell before the 3d subshell.
Question 3 asks you to connect the metal’s d-electron count and ligand field to geometry and magnetism.
Which description best fits [Ni(CN)4]2−?
Nickel is +2, so it is d8. CN− is a strong-field ligand; in this four-coordinate complex the electrons pair to give a square-planar, diamagnetic arrangement. Tetrahedral d8 complexes are often paramagnetic, but that is not the best description for this strong-field cyanide complex.
Do not infer geometry from coordination number alone. The metal’s electron configuration and the ligand set also matter; practise the particular patterns required by your course.
Questions 4–6: isomerism and trends
Question 4 checks recognition of geometric isomerism in a square-planar complex.
Which pair are geometric isomers of one another?
In square-planar [Pt(NH3)2Cl2], the two identical ligands can be adjacent (cis) or opposite (trans), giving geometric isomers. The other pairs have different formulas or ligand sets, so they are not geometric isomers of the same compound.
For a formula such as MA2B2, sketch the positions of the two A ligands: adjacent is cis; opposite is trans. A diagram is often faster and more reliable than trying to visualise the arrangement from the formula alone.
Question 5 tests calculation of a standard cell potential from standard reduction potentials.
Given E°(Cu2+/Cu) = +0.34 V and E°(Zn2+/Zn) = −0.76 V, what is E°cell for Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)?
Copper(II) is reduced at the cathode and zinc is oxidised at the anode. E°cell = E°cathode − E°anode = +0.34 − (−0.76) = +1.10 V. −1.10 V reverses the cell reaction; do not change the sign of a reduction potential and then subtract it again.
The listed values are reduction potentials. Use cathode minus anode as written; the oxidation half-reaction is already accounted for by subtracting the anode’s reduction potential.
Question 6 tests an exception to the broad increase in first ionisation energy across a period.
Why is the first ionisation energy of aluminium lower than that of magnesium?
The electron removed from Al is a 3p electron, higher in energy and more shielded than Mg’s 3s electron, so it is easier to remove. Aluminium does have a greater nuclear charge; that fact alone does not explain the local dip in the trend.
When a periodic trend has an exception, compare the subshell of the electron being removed and the stability of the resulting configuration, rather than relying only on nuclear charge.
Questions 7–10: reactions, equilibria and magnetism
Question 7 tests the Lewis definition of an acid and a base.
In the reaction BF3 + NH3 → F3B←NH3, which statement is correct?
BF3 accepts the nitrogen lone pair, so it is the Lewis acid; NH3 donates that pair and is the Lewis base. Donating a lone pair is the Lewis-base role, not the Lewis-acid role.
A quick check is to track the electron pair: the donor is the Lewis base, and the acceptor is the Lewis acid.
Question 8 tests disproportionation of chlorine in hot, concentrated alkali.
Which equation represents the reaction of chlorine with hot, concentrated sodium hydroxide?
Hot, concentrated alkali produces chloride and chlorate: 3Cl2 + 6OH− → 5Cl− + ClO3− + 3H2O. Chlorine is reduced from 0 to −1 and oxidised from 0 to +5, so it disproportionates. The hypochlorite equation in option A is associated with cold, dilute alkali.
Conditions matter in halogen chemistry. Attach the temperature and concentration to the reaction in your notes so you do not transfer the cold, dilute product to the hot, concentrated reaction.
Question 9 tests the relation between Ksp and molar solubility for a 1:1 salt in pure water.
AgCl has Ksp = 1.8 × 10−10 at a stated temperature. What is its approximate molar solubility in pure water at that temperature?
For AgCl ⇌ Ag+ + Cl− in pure water, [Ag+] = [Cl−] = s, so Ksp = s². Thus s = √(1.8 × 10−10) ≈ 1.34 × 10−5 mol L−1. Ksp itself is not the solubility; the square-root relation applies because the ions form in a 1:1 ratio.
The equal-concentration shortcut depends on the dissolution stoichiometry and assumes pure water, without a common ion. Set up the equilibrium expression before taking a square root.
Question 10 brings electron counting together with the spin-only magnetic-moment equation.
Assuming [Fe(H2O)6]2+ is high spin, what is its spin-only magnetic moment? Use μ = √[n(n + 2)] BM, where n is the number of unpaired electrons.
Fe2+ is d6. With weak-field H2O ligands in the stated high-spin case, it has four unpaired electrons. μ = √[4(4 + 2)] = √24 ≈ 4.90 BM. 2.83 BM corresponds to two unpaired electrons, not the high-spin d6 configuration.
State the spin assumption before calculating. The same d-electron count can give a different number of unpaired electrons in a different ligand field.
Review the result
After answering, sort errors by the rule that failed: charge accounting, electron configuration, ligand-field reasoning, redox convention, reaction conditions or equilibrium setup. Rework a missed question by explaining the reasoning aloud, not merely memorising its option. A useful review panel would rank those weak areas and show the missed idea to revisit.
A sample weak-area view after a practice set might look like this:
If the same distinction causes repeated errors, turn it into a precise retrieval prompt. For example, ask yourself why the ligand field changes the geometry and magnetism in a particular complex, rather than writing only “revise coordination chemistry”.
A remediation prompt based on a repeated coordination-chemistry error could read:
You lost this mark twice: why is [Ni(CN)4]2− square planar and diamagnetic, while a tetrahedral d8 complex can be paramagnetic?
How MySummaries helps
MySummaries lets you build a revision board from your own chemistry notes, then practise with flashcards and written questions based on that material. Use the questions to identify a weak topic, return to the matching section of your board, and practise it again. Open MySummaries.