--- title: "Variable oxidation states; industrial uses of 3d elements" subject: "Engineering Chemistry" chapter: 5 key: "semester_2/engineering_chemistry:variable-ox-applications" source: https://www.bctengineer.com/q/semester_2/engineering_chemistry/variable-ox-applications site: BCT Engineering --- # Variable oxidation states; industrial uses of 3d elements Engineering Chemistry · Chapter 5: 3-d Transition Elements and their Applications - Asked in: 2072 Asoj, 3 marks - Video: [Variable Oxidation States | Transition Metals as Catalysts | Chapter 18 | FSc 2nd Year Chemistry ...](https://www.youtube.com/watch?v=LbIuR6m0v98) — TKA bosal official - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/engineering_chemistry/variable-ox-applications - Whole chapter: https://www.bctengineer.com/md/semester_2/engineering_chemistry/chapter-5 ## Question Why do transition elements show variable oxidation state? Point out the industrial applications of 3d-series elements. ## Short answer Variable oxidation states occur because 3d and 4s electrons have similar energies, allowing both to participate in bonding with gradually increasing ionisation energies. Key industrial uses: Fe in steel and Haber catalyst, V2O5 for H2SO4, Ni in hydrogenation, Cr in stainless steel, Cu in wiring, Ti in aircraft alloys, Mn in dry cells, Co in magnets, Zn for galvanizing. ## Answer #### Variable oxidation states Transition elements show variable oxidation states because the (n−1)d and ns electrons have very nearly the same energy, so both the ns electrons and a variable number of (n−1)d electrons can take part in bonding. Successive ionisation energies rise gradually (unlike the sudden jump after the s-electrons in s-block metals), so ions of several charges can form. In the 3d series the states usually differ by one unit (e.g. Fe²⁺, Fe³⁺), unlike p-block elements whose states differ by two (Sn²⁺, Sn⁴⁺). | Element | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | |---|---|---|---|---|---|---|---|---|---|---| | Common oxidation states | +3 | +2, +3, +4 | +2, +3, +4, +5 | +2, +3, +6 | +2, +3, +4, +6, +7 | +2, +3 | +2, +3 | +2 | +1, +2 | +2 | - Sc has one stable state (+3, loses 3 electrons to reach argon core) and Zn only +2 (3d¹⁰ stable), so they show little variability. - The maximum state increases from Sc (+3) to Mn (+7), equal to the number of 3d + 4s electrons, then decreases after Mn because electrons are paired and the d orbitals become more stable. - Examples: MnO₄⁻ (+7), Cr₂O₇²⁻ (+6), VO₂⁺ (+5). #### Industrial applications of 3d-series elements - **Iron** (steel, cast iron) in construction and machines; Fe as catalyst in Haber's process. - **V₂O₅** as catalyst in contact process for H₂SO₄; vanadium steel for tools. - **Ni** as hydrogenation catalyst; Ni–Cd batteries; nichrome heating wire. - **Cr:** chrome plating, stainless steel; K₂Cr₂O₇ in tanning and as an oxidant. - **Cu** in electrical wires and PCB tracks (high conductivity); brass and bronze alloys. - **Ti:** light strong alloys for aircraft; TiO₂ as white pigment. - **Mn:** MnO₂ in dry cells and as catalyst; ferromanganese steel; KMnO₄ as oxidant. - **Co:** magnets (Alnico), Li-ion battery cathodes (LiCoO₂); **Zn:** galvanising iron and dry cell anodes. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do 3d transition elements show variable oxidation states? A. (n-1)d and ns electrons have very similar energies, so both can take part in bonding B. The ns electrons are much higher in energy than the d electrons, so only d electrons bond C. They have a sudden jump in ionisation energy after the s-electrons D. They always lose electrons in pairs like p-block elements Answer: A. Because d and s electrons are close in energy, a variable number of d electrons can be used along with the ns electrons. (Key point: Cause of variable oxidation state) 2. In the 3d series, how do successive oxidation states usually differ, compared with p-block elements? A. By two units, same as p-block elements B. By one unit (Fe2+, Fe3+), whereas p-block states differ by two C. By one unit in p-block but two in the 3d series D. By three units in the 3d series Answer: B. Transition metals show states differing by one unit; p-block elements such as Sn (+2, +4) differ by two. (Key point: Step between oxidation states) 3. Which pairing of catalyst and process is correct for 3d elements? A. V2O5 in Haber's process B. Fe in the contact process C. V2O5 in the contact process for H2SO4 D. Ni in the manufacture of H2SO4 Answer: C. V2O5 catalyses the contact process, Fe catalyses Haber's process, and Ni catalyses hydrogenation. (Key point: Industrial catalysts) 4. Which 3d element reaches the highest common oxidation state of +7? A. Cr B. Fe C. Sc D. Mn Answer: D. The maximum state rises from Sc (+3) to Mn (+7), equal to the number of 3d + 4s electrons, then falls. 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