Oxides and oxoanions; potassium dichromate

Oxides and oxoanions of the 3d metals

Heating the metals in oxygen gives oxides. Every 3d metal except scandium gives an ionic oxide of formula MO. The highest oxidation state in the oxides matches the group number from Sc₂O₃ (group 3) up to Mn₂O₇ (group 7). Beyond group 7, no oxide of iron higher than Fe₂O₃ is known. As the oxidation state rises, the oxides become less ionic and more acidic. Mn₂O₇ is a covalent green oil. Mn₂O₇ gives HMnO₄, and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇. V₂O₅ is amphoteric, though mainly acidic: it gives VO₄³⁻ with alkalis and VO₂⁺ salts with acids. Oxoanions such as CrO₄²⁻, Cr₂O₇²⁻ and MnO₄⁻ hold the metal in a high state.

Oxides become more acidic as oxidation state rises

Preparation of potassium dichromate from chromite ore

1. 1
   4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

   Fuse chromite ore with sodium (or potassium) carbonate in free access of air. Yellow sodium chromate forms; filter off Fe₂O₃.
2. 2
   2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O

   Acidify the yellow solution with sulphuric acid. Orange sodium dichromate crystallises (Na₂Cr₂O₇·2H₂O).
3. 3★
   Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

   Treat with potassium chloride. K₂Cr₂O₇ is less soluble than the sodium salt, so orange crystals of K₂Cr₂O₇ separate.

★ key step

Chromate and dichromate: pH and structure

• The two ions change into each other with pH. Acid: 2CrO₄²⁻ (yellow) + 2H⁺ → Cr₂O₇²⁻ (orange) + H₂O. Alkali: Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O. Chromium is +6 in both.
• How to draw chromate: Cr in the centre with four O atoms at the corners of a tetrahedron; write the charge 2⁻ outside brackets.
• How to draw dichromate: two CrO₄ tetrahedra joined through one shared O atom (Cr-O-Cr). Each Cr has three other O atoms. Label the Cr-O-Cr angle 126° and the charge 2⁻.

K₂Cr₂O₇ as an oxidising agent in acid solution

Half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = 1.33 V). The orange solution turns green (Cr³⁺). Dilute H₂SO₄ is used to acidify.

1. 1
   Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O

   Iodide is oxidised to iodine.
2. 2★
   Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

   Iron(II) is oxidised to iron(III).
3. 3
   Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O

   Hydrogen sulphide is oxidised to sulphur (pale yellow solid).
4. 4
   Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

   Tin(II) is oxidised to tin(IV).

★ key step