# Iron(II) is easily oxidized while manganese(II) is not > The half-filled 3d⁵ decides ## Objectives - Write and draw $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$, $\ce{Mn^{2+}}$ and $\ce{Mn^{3+}}$ in 3d boxes. - Explain why iron (II) is easily oxidized to iron (III), while Mn (II) is not readily oxidized to Mn (III). - State iron's important oxidation states: +2 and +3 (states above +3 are not important). ## 1. Electronic configurations and oxidation states ## 2. Who gives up one more electron? Predict: which one is easily oxidized to the 3+ ion? - Both equally: they have the same charge - $\ce{Mn^{2+}}$ - $\ce{Fe^{2+}}$ ✓ $\ce{Fe^{2+}}$ is easily oxidized to $\ce{Fe^{3+}}$: losing one 3d electron turns its $3d^6$ into the stable half-filled $3d^5$. $\ce{Mn^{2+}}$ is already $3d^5$, so it is hard to oxidize to $\ce{Mn^{3+}}$. [ar-EG] الـ $\ce{Fe^{2+}}$ هو اللي بيتأكسد بسهولة لـ $\ce{Fe^{3+}}$: لما يخسر 3d electron واحد، الـ $3d^6$ بتاعه بيبقى $3d^5$ نصه مليان ومستقر. إنما الـ $\ce{Mn^{2+}}$ أصلًا $3d^5$، فصعب يتأكسد لـ $\ce{Mn^{3+}}$. ## 3. Make the 2+ ions Remove electrons from Fe and Mn. Remember: the 4s electrons leave first. electrons lost Iron (Z = 26) Fe: $[\text{Ar}]\ 4s^2\ 3d^6$ one 4s electron gone $\ce{Fe^{2+}}$: $[\text{Ar}]\ 3d^6$ Manganese (Z = 25) Mn: $[\text{Ar}]\ 4s^2\ 3d^5$ one 4s electron gone $\ce{Mn^{2+}}$: $[\text{Ar}]\ 3d^5$ Look at the 3d boxes $\ce{Mn^{2+}}$ is already $3d^5$: half filled. $\ce{Fe^{2+}}$ is $3d^6$: one orbital holds a pair. ## 4. Iron (II) → iron (III) $\ce{Fe^{2+}}$ $[\text{Ar}]\ 3d^6$ $\ce{Fe^{3+}}$ $[\text{Ar}]\ 3d^5$: half filled, more stable Iron (III) ion is more stable, as its 3d sublevel is half filled ($3d^5$). So the reaction goes toward the more stable ion. The electron that leaves is the paired one. Now every 3d orbital holds one electron. ## 5. Your turn: manganese (II) → manganese (III) Same steps for Mn. $\ce{Mn^{2+}}$ is $[\text{Ar}]\ 3d^5$. Oxidation removes one more electron. $\ce{Mn^{3+}}$ = [Ar] 3d___ . After oxidation, the 3d sublevel is ___ half filled (type: still / no longer). Show the boxes and the book's answer $\ce{Mn^{2+}}$ $[\text{Ar}]\ 3d^5$: half filled, more stable $\ce{Mn^{3+}}$ $[\text{Ar}]\ 3d^4$ The 3d sublevel is half filled ($3d^5$) in $\ce{Mn^{2+}}$, so it is more stable than $\ce{Mn^{3+}}$. So $\ce{Mn^{2+}}$ is not readily oxidized to $\ce{Mn^{3+}}$. ## 6. Iron's important oxidation states Book pp.20-21: iron's two important oxidation states Iron has 8 electrons in 4s and 3d, but it does not reach +8. The book says all its states above +3 are not important. Iron can reach +8 by losing all its 4s and 3d electrons. ## 7. Same 3d, different ions Which two ions both have the stable half-filled $3d^5$ sublevel? - $\ce{Fe^{3+}}$ and $\ce{Mn^{3+}}$ - $\ce{Fe^{2+}}$ and $\ce{Mn^{2+}}$ - $\ce{Fe^{2+}}$ and $\ce{Mn^{3+}}$ - $\ce{Fe^{3+}}$ and $\ce{Mn^{2+}}$ ✓ Electrons: $\ce{Fe^{3+}}$ 26 − 3 = 23, $\ce{Mn^{2+}}$ 25 − 2 = 23. Both are $[\text{Ar}]\ 3d^5$. [ar-EG] عدد الـ electrons: $\ce{Fe^{3+}}$ 26 − 3 = 23، و$\ce{Mn^{2+}}$ 25 − 2 = 23. الاتنين $[\text{Ar}]\ 3d^5$. ## 8. Build the exam answer Which oxidation states of iron does the book call important? - +2, +3 and +8 - +2 and +3 ✓ - +3 only - +2, +3 and +6 The book calls +2 and +3 the important oxidation states of iron: +2 from losing the two 4s electrons, and +3 with the stable half-filled $3d^5$. The states above +3 are not important. [ar-EG] الكتاب بيقول إن الـ oxidation states المهمة للـ iron هي +2 و+3: الـ +2 لما يخسر الاتنين 4s electrons، والـ +3 مع الـ $3d^5$ المستقر اللي نصه مليان. أما الـ states اللي فوق +3 فمش مهمة. ## 9. Summary $\ce{Fe^{2+}}$ $3d^6$ → $\ce{Fe^{3+}}$ $3d^5$: oxidation gives a half-filled 3d, so it is easy. $\ce{Mn^{2+}}$ is already $3d^5$. Losing one more electron would break it. Iron's important states: +2 and +3. States above +3 are not important. **Fe³⁺** ([Ar] 3d⁵) is more stable than **Fe²⁺** ([Ar] 3d⁶) because its **3d sublevel is half filled**. **Mn²⁺** is **already half filled (3d⁵)**, so it is **more stable than Mn³⁺**. Back to the hook: the same $3d^5$ makes one ion change easily and the other not. Can you explain why, with 3d boxes? $\ce{Fe^{2+}}$ ($3d^6$) loses one 3d electron and becomes $\ce{Fe^{3+}}$ ($3d^5$): half filled, so the oxidation is easy. $\ce{Mn^{2+}}$ is already $3d^5$: losing one more electron would break the half-filled 3d and give $\ce{Mn^{3+}}$ ($3d^4$). $\ce{Fe^{3+}}$: $[\text{Ar}]\ 3d^5$ $\ce{Mn^{3+}}$: $[\text{Ar}]\ 3d^4$ Chapter: Transition Elements (https://yaani.ai/en/learn/chemistry/chem-s3-en.ch1) Source: https://yaani.ai/en/lesson/chem-s3-en.lesson-te.fe2-fe3-mn2-stability