# Oxidation states from Sc to Zn > Up to manganese, then down ## Objectives - Read the oxidation states of Sc to Zn from the book's table, and find a metal's oxidation state from a formula (e.g. $\ce{V2O5}$: +5). - Describe the trend: the highest oxidation state rises from Sc (+3) to Mn (+7), then falls to Zn (+2). - State that the maximum oxidation state does not exceed the group number, except in group IB (Cu, Ag, Au). - Explain why scandium does not form $\ce{Sc^{4+}}$. ## 1. Electronic configurations and oxidation states ## 2. Twelve electrons, how many states? Predict: what is zinc's highest oxidation state in the book's table? - +2 ✓ - +12: it loses all its 4s and 3d electrons - +7, like manganese Zinc's highest oxidation state in the table is +2. The highest state rises from Sc up to Mn (+7), then decreases gradually to +2 in zinc. [ar-EG] أعلى oxidation state للـ zinc في الجدول هو +2. أعلى state بيزيد من الـ Sc لحد الـ Mn (+7)، وبعدين بيقل بالتدريج لحد +2 في الـ zinc. ## 3. Walk the book's table Slide from Sc to Zn. For each element, compare the two numbers. atomic number Z Sc (Z = 21), group IIIB $[\text{Ar}]\ 4s^2\ 3d^1$. Oxidation states: +3 electrons in 4s + 3d highest oxidation state Compounds: $\ce{Sc2O3}$ Ti (Z = 22), group IVB $[\text{Ar}]\ 4s^2\ 3d^2$. Oxidation states: +2, +3, +4 electrons in 4s + 3d highest oxidation state Compounds: $\ce{TiO2}$, $\ce{Ti2O3}$, $\ce{TiO}$ V (Z = 23), group VB $[\text{Ar}]\ 4s^2\ 3d^3$. Oxidation states: +2, +3, +4, +5 electrons in 4s + 3d highest oxidation state Compounds: $\ce{V2O5}$, $\ce{VO2}$, $\ce{V2O3}$, $\ce{VO}$ Cr (Z = 24), group VIB $[\text{Ar}]\ 4s^1\ 3d^5$. Oxidation states: +2, +3, +6 electrons in 4s + 3d highest oxidation state Compounds: $\ce{CrO3}$, $\ce{Cr2O3}$, $\ce{CrO}$ Mn (Z = 25), group VIIB $[\text{Ar}]\ 4s^2\ 3d^5$. Oxidation states: +2, +3, +4, +6, +7 electrons in 4s + 3d highest oxidation state Compounds: $\ce{KMnO4}$, $\ce{K2MnO4}$, $\ce{MnO2}$, $\ce{Mn2O3}$, $\ce{MnO}$ Fe (Z = 26), group VIII $[\text{Ar}]\ 4s^2\ 3d^6$. Oxidation states: +2, +3, +6 electrons in 4s + 3d highest oxidation state Compounds: $\ce{Na2FeO4}$, $\ce{Fe2O3}$, $\ce{FeO}$ Co (Z = 27), group VIII $[\text{Ar}]\ 4s^2\ 3d^7$. Oxidation states: +2, +3, +4 electrons in 4s + 3d highest oxidation state Compounds: $\ce{CoCl3}$, $\ce{CoCl2}$ Ni (Z = 28), group VIII $[\text{Ar}]\ 4s^2\ 3d^8$. Oxidation states: +2, +3, +4 electrons in 4s + 3d highest oxidation state Compounds: $\ce{NiO2}$, $\ce{Ni2O3}$, $\ce{NiO}$ Cu (Z = 29), group IB $[\text{Ar}]\ 4s^1\ 3d^{10}$. Oxidation states: +1, +2 electrons in 4s + 3d highest oxidation state Compounds: $\ce{CuO}$, $\ce{Cu2O}$ Zn (Z = 30), group IIB $[\text{Ar}]\ 4s^2\ 3d^{10}$. Oxidation states: +2 electrons in 4s + 3d highest oxidation state Compounds: $\ce{ZnO}$ Book table, p.8 (for illustration only). Copper's row is printed with a typo; its states are +1 and +2. ## 4. Up to Mn, then down The two lines match from Sc to Mn, then split From Sc to Mn, the highest state uses all the 4s and 3d electrons. After Mn, the highest state decreases gradually, to +2 in zinc. The highest oxidation state always equals the number of 4s and 3d electrons, for every element. ## 5. Read a state from a formula The table lists compounds. Find each metal's oxidation state with the sum-to-zero rule. Worked: vanadium in V₂O₅ Call vanadium x. Two V atoms give 2x; five O atoms give 5 × (−2). +5 is vanadium's highest state: it has lost all 5 of its 4s and 3d electrons. Your turn: iron in Na₂FeO₄ In $\ce{Na2FeO4}$ (Na = +1, O = −2), iron is ___ . ## 6. The group number limit Highest oxidation states by group (book table, p.8) Book point 4: the maximum oxidation state does not exceed the group number. Group VIII stays well below 8. The exception Group IB (copper, silver and gold) is the exception. Copper shows +2, above its group number (I). ## 7. Why no Sc⁴⁺? Sc atom $[\text{Ar}]\ 4s^2\ 3d^1$ Sc³⁺ $[\text{Ar}]$: 4s and 3d are empty Scandium has only three electrons in 4s and 3d. $\ce{Sc^{3+}}$ has lost all three. $\ce{Sc^{3+}}$ has the stable configuration of argon. A fourth electron would have to come from that stable arrangement. Give reasons: scandium doesn't produce Sc⁴⁺ Because **$\ce{Sc^{3+}}$** has the **stable configuration of argon, [Ar]**: its 4s and 3d are empty. Losing another electron would break this stable arrangement. ## 8. Use the rules Molybdenum (Mo) is below chromium, in group VIB. By the book's rule, its highest oxidation state cannot exceed: - +6 ✓ - +4 - +7 - +2 Group VIB → the maximum cannot exceed +6. [ar-EG] الـ group هو VIB، يبقى أعلى state مايعديش +6. ## 9. Gold and titanium In $\ce{AuCl3}$, gold is +3. Gold is in group IB. What does this show? - The maximum always equals the group number - Gold is not in the first series, so the book's rule says nothing - Group IB is the book's exception: its maximum can exceed the group number ✓ - The value must be a mistake It shows the book's exception: the maximum oxidation state does not exceed the group number except in group IB (copper, silver and gold). Gold, in group IB, reaches +3. [ar-EG] ده بيوضّح استثناء الكتاب: أقصى oxidation state مش بيعدّي رقم الـ group إلا في الـ group IB (الـ copper والـ silver والـ gold). والـ gold، اللي في الـ group IB، بيوصل لـ +3. ## 10. Summary The highest state rises from Sc (+3) to Mn (+7), then decreases to Zn (+2). Zinc has 12 electrons in 4s and 3d, yet only +2. From Sc to Mn it equals the number of 4s and 3d electrons. After Mn it is lower: Fe +6, Co and Ni +4. The maximum does not exceed the group number, except group IB (Cu shows +2). $\ce{Sc^{3+}}$ is [Ar], so there is no $\ce{Sc^{4+}}$. Book point 4 States **increase from Sc to Mn (+7)**, then **decrease to +2 in Zn**. The maximum **does not exceed the group number, except group IB**. Can you explain why vanadium's highest state is +5 and not +6? Vanadium is $[\text{Ar}]\ 4s^2\ 3d^3$: it has 5 electrons in 4s and 3d. At +5 it has lost all of them, and +5 is also its group number (VB). So its highest state is +5. Chapter: Transition Elements (https://yaani.ai/ar/learn/chemistry/chem-s3-en.ch1) Source: https://yaani.ai/ar/lesson/chem-s3-en.lesson-te.oxidation-states-trend