Assign oxidation states in K₂Cr₂O₇ and KMnO₄

medium CBSE JEE-MAIN NEET • 3 min read
Tags Redox Reactions

Question

Find the oxidation state of chromium (Cr) in potassium dichromate (K₂Cr₂O₇) and the oxidation state of manganese (Mn) in potassium permanganate (KMnO₄).

Solution — Step by Step

Key rules (in priority order):

  1. Oxidation state of a free element = 0
  2. Oxidation state of a monoatomic ion = its charge
  3. Oxygen is almost always –2 (exception: peroxides –1, superoxides –1/2, OF₂ +2)
  4. Hydrogen is +1 (exceptions: metal hydrides, where H is –1)
  5. Alkali metals (Group 1) are always +1; alkaline earth metals (Group 2) are always +2
  6. Sum of all oxidation states = overall charge of the compound or ion

Let oxidation state of Cr = xx.

K is +1; O is –2. The compound K₂Cr₂O₇ is electrically neutral (overall charge = 0).

2(+1)+2x+7(−2)=02(+1) + 2x + 7(-2) = 0 2+2x−14=02 + 2x - 14 = 0 2x=122x = 12 x=+6x = +6

Oxidation state of Cr in K₂Cr₂O₇ = +6

This makes sense — Cr(VI) is a strong oxidising agent (which is why potassium dichromate is used in oxidation reactions in organic chemistry).

Let oxidation state of Mn = yy.

K is +1; O is –2. The compound KMnO₄ is electrically neutral.

(+1)+y+4(−2)=0(+1) + y + 4(-2) = 0 1+y−8=01 + y - 8 = 0 y=+7y = +7

Oxidation state of Mn in KMnO₄ = +7

Mn(VII) is the highest common oxidation state of manganese. KMnO₄ is a powerful oxidising agent — it gets reduced in reactions (Mn goes from +7 to +2 in acidic medium, +4 in neutral/weakly basic medium, +6 in strongly basic medium).

Why This Works

The key principle: oxidation states are assigned by treating all bonds as ionic (even if they’re actually covalent) and giving the more electronegative atom the “shared” electrons. For O (electronegativity 3.44) bonded to Cr or Mn, oxygen “owns” both electrons → O gets –2, and the metal gets a high positive state.

The constraint equation (sum of oxidation states = compound charge) has only one unknown once K and O states are assigned, so we can always solve for the transition metal’s state.

Alternative Method

Use the ionic formula. K₂Cr₂O₇ contains K⁺ and Cr₂O₇²⁻ (dichromate ion). In Cr₂O₇²⁻: 2x+7(−2)=−22x + 7(-2) = -2 2x=−2+14=122x = -2 + 14 = 12 x=+6x = +6 ✓

KMnO₄ contains K⁺ and MnO₄⁻ (permanganate ion). In MnO₄⁻: y+4(−2)=−1y + 4(-2) = -1 y=−1+8=+7y = -1 + 8 = +7 ✓

Working with the polyatomic ions directly is often cleaner.

CBSE Class 11 (Redox Reactions) and JEE frequently test oxidation states of Cr and Mn in their compounds. The commonly tested compounds: Cr₂O₃ (+3), CrO₄²⁻ (+6), Cr₂O₇²⁻ (+6); MnO₂ (+4), MnO₄²⁻ (+6), MnO₄⁻ (+7). Also know that Mn can have oxidation states from +2 to +7 — it’s one of the most variable transition metals.

Common Mistake

A frequent error is forgetting to account for the subscript when there are multiple atoms of an element. In K₂Cr₂O₇, there are 2 Cr atoms — so the contribution of Cr to the sum is 2x2x, not just xx. Similarly, there are 2 K atoms contributing 2(+1)=+22(+1) = +2, and 7 O atoms contributing 7(−2)=−147(-2) = -14. Always multiply the oxidation state by the number of atoms of that element in the formula.

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