الأسبوع 15 · مراجعة الفصل الدراسي الأول (الأبواب من ١ إلى ٣) ومهام مراجعة من إعداد Darslab · 3 دروس
1
Review: transition elements and iron
Chemistry Sec 3 Term 1, Chapters 1–3: Review
GoalRevise weeks 1–5 with new tasks: d configurations and unpaired electrons, the transition-element test, iron ores, extraction, and the reactions and oxides of iron.
New words
unpaired electron · an electron alone in its box; it makes a particle paramagnetictransition element · an element whose atom or at least one common ion has a partly filled d sublevelmixed oxide · Fe₃O₄, which behaves as FeO·Fe₂O₃ with iron in both +2 and +3
Explanation
Four tools from weeks 1–5 solve most tasks. (1) Configuration: the first series is [Ar] 3dⁿ 4s², with the exceptions Cr (3d⁵ 4s¹) and Cu (3d¹⁰ 4s¹). Ions lose the 4s electrons first, so Fe is 3d⁶ 4s² but Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵. (2) Unpaired electrons: draw five boxes and fill them by Hund’s rule; with n electrons the number unpaired is n up to 5 and 10 − n after that. (3) The test: an element is a transition element if the atom or one common ion has 1 to 9 electrons in d; zinc, cadmium and mercury fail it. (4) Iron: from ore with a known percentage of Fe₂O₃, coke gives CO, and Fe₂O₃ + 3CO → 2Fe + 3CO₂ (paper only). Dilute acids give iron(II) salts and hydrogen, Fe + 2HCl → FeCl₂ + H₂. The mole route is always: mass ÷ M gives moles, the equation’s ratio gives the other moles, and moles × M or × 22.4 L gives the answer. In FeO, Fe₂O₃ and Fe₃O₄ the iron keeps its state when an acid acts on the oxide: oxide + acid → salt + water.
Worked examples
Counting. Co²⁺: Co is [Ar] 3d⁷ 4s²; the two 4s electrons leave, giving 3d⁷. Five boxes: the first five electrons go one per box, the last two pair up, so 5 − 2 = 3 boxes have a single arrow: 3 unpaired electrons, so Co²⁺ is paramagnetic. Ni²⁺ is 3d⁸, with 2 unpaired electrons, as in the picture. Cd²⁺ is 4d¹⁰, with none.
Step
Working
Result
Fe₂O₃ in ore
800 × 0.80
640 kg
Iron obtained
640 × 112 ÷ 160
448 kg
CO needed
3 × 4000 mol × 28
336 kg
From ore to iron. 800 kg of ore holds 80 % hematite: 800 × 0.80 = 640 kg Fe₂O₃. Fe₂O₃ + 3CO → 2Fe + 3CO₂. M(Fe₂O₃) = 160 g/mol, so 640 kg is 4000 mol (taking kg and kmol the same way). The iron is 2 × 4000 = 8000 mol × 56 = 448 kg. The CO needed is 3 × 4000 = 12 000 mol × 28 = 336 kg. A shortcut check: iron is 112 ÷ 160 = 70 % of Fe₂O₃, and 640 × 0.70 = 448 kg.
Iron and acid. 11.2 g of iron is 11.2 ÷ 56 = 0.20 mol. Fe + 2HCl → FeCl₂ + H₂, so 0.20 mol of hydrogen, which is 0.20 × 22.4 = 4.48 L at STP, and 0.20 mol of FeCl₂: M = 56 + 2 × 35.5 = 127, so 0.20 × 127 = 25.4 g. The product is iron(II) chloride, not iron(III), because the hydrogen ions of a dilute acid are weak oxidising agents. On paper only.
Class activity
Sorting game, paper only. Cut ten cards: Sc, Ti, Cr, Mn, Fe, Co, Ni, Cu, Zn and Cd. On the back of each write the atom’s d configuration and the configuration of one common ion. Shuffle, pick a card, and say if the element is a transition element and how many unpaired electrons the ion has. Make a pile for transition elements and a pile for the others, then check the backs. Which two cards need the ion and not the atom?
Practice
1
How many unpaired electrons are in Mn³⁺? (Mn is [Ar] 3d⁵ 4s².)
Mn³⁺ loses 4s² and one 3d electron: 3d⁴. Four boxes hold one arrow each, so 4 unpaired electrons.
2
500 kg of ore holds 60 % Fe₂O₃. What mass of iron can it give?
500 × 0.60 = 300 kg Fe₂O₃. Iron is 70 % of Fe₂O₃ (112 ÷ 160), so 300 × 0.70 = 210 kg.
3
8.4 g of iron reacts with dilute sulphuric acid. Find the volume of hydrogen at STP and the mass of FeSO₄.
8.4 ÷ 56 = 0.15 mol Fe, Fe + H₂SO₄ → FeSO₄ + H₂. H₂: 0.15 × 22.4 = 3.36 L. FeSO₄: M = 152, so 0.15 × 152 = 22.8 g.
4
Is cadmium (atom 4d¹⁰ 5s², common ion Cd²⁺) a transition element? Give the reason.
No. The atom has a full 4d, and Cd²⁺ is 4d¹⁰ also full, so neither has a partly filled d sublevel. It is in the d block but fails the test, like zinc.
