OlympIQ
Read before you practise · 9 min

Redox Behaviour, Periodic Trends and Extraction Reactions

First-row transition metal ions each carry a distinct set of properties — how many unpaired electrons they have, how large they are, how many oxidation states they access, and whether they are stable at all in water — and several classic reactions exist purely to test whether a given ion is stable, or whether it will react further.

NSEC · d- and f-Block Elements · 6 linked questions · 100% coverage
Know this first

Chemistry questions first require the material, particles or reaction condition to be identified before a formula or rule is selected.

Observation · What the stem states or showsCondition · The factor that controls the change
Study deeper: NCERT Chemistry · d- and f-Block Elements

Why Cu⁺ cannot survive in aqueous solution

Cu⁺ (a d₁₀ configuration) is not stable in aqueous solution: it spontaneously disproportionates, 2Cu⁺ → Cu²⁺ + Cu₀ (metallic copper), because the hydration energy released by forming the smaller, more highly charged Cu²⁺ ion more than compensates for the energy cost of the disproportionation, making the reaction thermodynamically favourable in water. This is why copper(I) salts are essentially never encountered as stable species in aqueous solution, only in the solid state or in non-aqueous/complexed environments that suppress this disproportionation.

Fe³⁺ is a strong enough oxidant to liberate iodine from iodide

The Fe₃+/Fe²⁺ reduction potential is favourable enough to oxidize iodide ion to iodine: Fe³⁺ + I⁻ → Fe²⁺ + 1/2 I₂ (written per iodide;

with two iodide ions, 2Fe₃⁺ + 2I- → 2Fe₂⁺ + I₂).

This is a standard qualitative test/reaction pairing a transition-metal redox couple (Fe₃+/Fe₂⁺) against a halide redox couple (I2/I⁻), and it works specifically because iodide is among the more easily oxidized halides — the same logic would not necessarily hold for fluoride or chloride, which are much harder to oxidize.

Ranking d-block ions by paramagnetism, size and oxidation-state count

Paramagnetism tracks the number of unpaired d-electrons, which does NOT increase monotonically across a simple d₁-d₂-d₃... series once you reach the half-filled d₅ configuration and beyond: V²⁺ (d₃, 3 unpaired) < Cr²⁺ (d₄, 4 unpaired) < Mn²⁺ (d₅, 5 unpaired, the maximum possible for these five ions) — but Fe²⁺ (d₆, only 4 unpaired again, since the 6th electron must pair up in an already-occupied orbital) actually drops back BELOW Mn²⁺, not above it. So V²⁺ < Cr²⁺ < Mn²⁺ < Fe²⁺ is NOT a correct order for paramagnetism — it breaks exactly at the Mn₂+-to-Fe²⁺ step. Ionic size across a row of same-charge transition-metal ions generally DECREASES from left to right (increasing nuclear charge pulls the d-electrons in more tightly, only partially shielded by other d-electrons), so Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺ correctly describes ionic size increasing as you move left (Mn²⁺ is the largest of this set). The number of distinct oxidation states a first-row transition metal can access tends to increase from the early transition metals toward the middle of the row (peaking around manganese) before decreasing again, so Sc < Ti < Cr < Mn is a correct increasing order for the number of accessible oxidation states.

Why zinc has the lowest boiling point among first-row transition metals

Metallic bonding strength across the first-row transition metals depends heavily on how many d-electrons are available to participate in delocalised metallic bonding. Zinc has a full, non-bonding d₁₀ configuration: its d-electrons are all paired and effectively core-like, contributing essentially nothing extra to metallic bonding beyond its two 4s electrons. Metals in the middle of the row (like Cr, with a half-filled d₅ configuration ideal for maximum d-orbital overlap and bonding) have far more d-electrons available for strong metallic bonding, and so have much higher melting and boiling points. This is why zinc has an unusually LOW boiling point compared to its first-row transition-metal neighbours, despite being a d-block element.

Comproportionation in copper extraction: Cu₂O plus Cu₂S gives metallic copper

In copper metallurgy, heating copper(I) oxide (Cu₂O, Cu at +1) together with copper(I) sulfide (Cu₂S, Cu at +1) drives a comproportionation-type reaction where sulfur is oxidized (from Cu₂S, S at -2, up to +4 in SO₂) while copper is reduced from +1 down to 0 (metallic copper): Cu₂S + 2Cu₂O → 6Cu + SO₂. This reaction, sometimes called self-reduction or auto-reduction, is a key step in the pyrometallurgical extraction of copper, where the sulfide ore is partially roasted to oxide first, and the leftover sulfide then reduces that oxide directly on further heating, with no separate external reducing agent needed.

OlympIQ exam lens

  • ✓Cu⁺ disproportionates in water (2Cu⁺ to Cu²⁺ + Cu₀) because Cu²⁺ hydration energy makes the reaction thermodynamically favourable
  • ✓Fe³⁺ oxidizes iodide to iodine (Fe³⁺ + I⁻ to Fe²⁺ + 1/2 I₂) — a classic transition-metal-vs-halide redox pairing
  • ✓Paramagnetism (unpaired d-electrons) peaks at the half-filled d₅ configuration (Mn₂⁺) and drops for d₆ (Fe₂⁺) — it does not increase monotonically
  • ✓Ionic size of same-charge first-row ions decreases left to right; number of accessible oxidation states tends to increase toward the middle of the row
  • ✓Zinc has an unusually low boiling point among first-row transition metals because its filled d₁₀ configuration contributes nothing extra to metallic bonding
  • ✓Heating Cu₂O with Cu₂S gives metallic copper plus SO₂ — a self-reduction/comproportionation step used in copper extraction
OlympIQ trap

Do not assume paramagnetism (or any unpaired-electron-driven property) increases smoothly and monotonically all the way across a run of d-block ions. It rises only up to the half-filled d₅ configuration and then falls again as electrons start pairing up — an ordering that keeps climbing past d₅ (like V²⁺ < Cr²⁺ < Mn²⁺ < Fe₂⁺) is exactly the kind of subtly wrong sequence these questions test.

Ready to try the pattern?

100% coverage across the 6 linked questions.

Practise 6 linked questions
Sign in to practise — your XP, streak and analysis update as you solve.