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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.

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

Electricity questions depend on identifying charge, a conducting path and the condition that controls current or magnetic effect.

Current · Flow of electric chargeCircuit · A complete conducting loopResistance · Opposition to current
Study deeper: NCERT Science · 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.

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