Transition Elements · تدريب

كل الأسئلة بإجاباتها · Transition Elements

1. Hydrogen reacts with iodine: H2 + I2 → 2HI$\mathrm{H}{}_2+\mathrm{I}{}_2\to 2\mathrm{HI}$. Which statement about the energy of the two steps is correct?

   • C. Breaking the H–H and I–I bonds absorbs energy; forming the H–I bonds absorbs energy too.
   • D. Neither step involves energy, because the same atoms are only rearranged.
   • A. Breaking the H–H and I–I bonds absorbs energy; forming the H–I bonds releases energy.
   • B. Breaking the H–H and I–I bonds releases energy; forming the H–I bonds absorbs energy.

   الإجابة: Breaking the H–H and I–I bonds absorbs energy; forming the H–I bonds releases energy.

   A reaction has two steps. First, the bonds in the reactants break: here H–H and I–I. Breaking a bond absorbs energy, because the bonded atoms pull on each other. Then new bonds form in the products: here H–I. Forming a bond releases energy. The balance between the two decides whether the reaction gives out or takes in energy overall. Key words: breaking bonds absorbs energy; forming bonds releases energy.

2. In a reaction, breaking all the bonds in the reactants absorbs 680 kJ. Forming all the bonds in the products releases 860 kJ. Which describes the reaction?

   • C. It gives out 1540 kJ of energy overall.
   • D. It takes in 680 kJ of energy overall.
   • B. It takes in 180 kJ of energy overall.
   • A. It gives out 180 kJ of energy overall.

   الإجابة: It gives out 180 kJ of energy overall.

   Breaking bonds absorbs energy: 680 kJ goes in. Forming bonds releases energy: 860 kJ comes out. Net = released − absorbed = 860 − 680 = 180 kJ. More energy is released than absorbed, so the reaction gives out 180 kJ. In an energy profile, the products are then lower than the reactants. This difference is the net energy released (book fig 1-3).

3. Which statement describes the activation energy of a reaction?

   • D. The total energy stored in the product molecules.
   • A. The minimum energy colliding molecules must have to react.
   • B. The energy given out by the reaction when products form.
   • C. The extra energy that a catalyst adds to the molecules.

   الإجابة: The minimum energy colliding molecules must have to react.

   Molecules react only when they collide with enough energy to break their bonds. The activation energy is the minimum energy that must be gained by a molecule to react at collision. On an energy profile it is the climb from the reactants up to the top of the hump. It is not the energy released (the drop from reactants to products), and a catalyst does not add energy.

4. Energy profile (made-up values)

   Progress of reaction	Reaction (kJ)
   0	50
   1	50
   2	110
   3	165
   4	180
   5	165
   6	130
   7	120
   8	120

   The graph shows the energy profile of a reaction (made-up values). What is the activation energy of this reaction?

   • A. 130 kJ
   • C. 70 kJ
   • D. 60 kJ
   • B. 180 kJ

   الإجابة: 130 kJ

   Activation energy is measured from the reactants up to the top of the hump. Here: 180 − 50 = 130 kJ. The products (120 kJ) are higher than the reactants (50 kJ), so this reaction takes in 70 kJ overall. That 70 kJ is the energy change, not the activation energy. The climb up and the change from reactants to products are two different things.

5. Read the equation: 4NH3(g) + 5O2(g) →[Pt][850\,^C] 4NO(g) + 6H2O(v)$4\mathrm{NH}{}_3(\mathrm{g})+5\mathrm{O}{}_2(\mathrm{g})\underset{850{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\overset{\mathrm{Pt}}{\to }}4\mathrm{NO}(\mathrm{g})+6\mathrm{H}{}_2\mathrm{O}(\mathrm{v})$ (Pt is platinum). What is the platinum in this reaction?

   • B. A reactant: it reacts with ammonia and oxygen.
   • A. A catalyst: it is not used up, so it is written over the arrow.
   • D. A sign that the reaction gives out heat.
   • C. A product: it forms during the reaction.

   الإجابة: A catalyst: it is not used up, so it is written over the arrow.

   Reactants come before the arrow, products after it. What is written over or under the arrow is a condition: a temperature, a pressure, heating (Δ) or a catalyst. Here 850 °C is the temperature and Pt is the catalyst. A catalyst is not used up, so it appears on neither side. The same rule applies to Fe in the Haber-Bosch equation.

6. Which statement reads this equation correctly? 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)$2\mathrm{Na}(\mathrm{s})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\to 2\mathrm{NaOH}(\mathrm{aq})+\mathrm{H}{}_2(\mathrm{g})$

   • B. Solid sodium and liquid water give solid sodium hydroxide that does not dissolve, and hydrogen gas.
   • A. Solid sodium and liquid water give sodium hydroxide dissolved in water, and hydrogen gas.
   • D. Solid sodium and liquid water give sodium hydroxide dissolved in water, and hydrogen dissolved in water.
   • C. Solid sodium and water vapour give sodium hydroxide dissolved in water, and hydrogen gas.

   الإجابة: Solid sodium and liquid water give sodium hydroxide dissolved in water, and hydrogen gas.

   State symbols: (s) solid, (l) liquid, (g) gas, (v) vapour, (aq) aqueous, which means dissolved in water. So Na(s) is solid sodium, H2O(l) is liquid water, NaOH(aq) is sodium hydroxide dissolved in water, and H2(g) is hydrogen gas. Always read the letter after each formula; do not guess the state.

7. Which set gives the catalyst and conditions of the Haber-Bosch method for preparing ammonia, as in the book?

   • B. Vanadium pentoxide (V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$), 450 °C
   • A. Iron (Fe), 500 °C, 200 atm
   • C. Finely divided nickel (Ni), 500 °C, 200 atm
   • D. Iron (Fe), 200 °C, 500 atm

   الإجابة: Iron (Fe), 500 °C, 200 atm

   The book writes: N2(g) + 3H2(g) →[500\,^C / 200\,atm,\ Fe] 2NH3(g)$\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}/200\mathrm{atm},\mathrm{Fe}}{\to }2\mathrm{NH}{}_3(\mathrm{g})$. Iron is the catalyst, at 500 °C and 200 atm (200 times normal air pressure). V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ at 450 °C belongs to the contact method for sulphuric acid. Finely divided nickel is used in the hydrogenation of oils.

8. In the Haber-Bosch method, N2 + 3H2 → 2NH3$\mathrm{N}{}_2+3\mathrm{H}{}_2\to 2\mathrm{NH}{}_3$, 12 molecules of ammonia are produced. How many molecules of hydrogen reacted?

   الإجابة: 18 molecules

   The equation says 3 H2 give 2 NH3. 12 NH3 is 6 sets of 2 NH3. Each set needs 3 H2, so 6 × 3 = 18 molecules of H2. Check with atoms: 12 NH3 contain 36 H atoms; 18 H2 also contain 36 H atoms. Most likely wrong answer: 12 (reading H2 and NH3 as one to one). Another: 6 (dividing by 2 instead of using the ratio 3 : 2). Another: 8 (using the ratio upside down: 12 × 2 ÷ 3).

9. Classify the following ions into paramagnetic and diamagnetic. The 3d configuration of each ion is given.

   الإجابة: Diamagnetic: Ti⁴⁺$\mathrm{Ti}{}^{4+}$ (3d⁰$3d^0$), Ga³⁺$\mathrm{Ga}{}^{3+}$ (3d¹⁰$3d^{10}$, gallium) · Paramagnetic: V³⁺$\mathrm{V}{}^{3+}$ (3d²$3d^2$), Mn³⁺$\mathrm{Mn}{}^{3+}$ (3d⁴$3d^4$), Co³⁺$\mathrm{Co}{}^{3+}$ (3d⁶$3d^6$)

   Fill the five 3d boxes by Hund's rule: one electron in each box first, then pairing. 3d² has 2 unpaired, 3d⁴ has 4, and 3d⁶ has 4 (one pair and four unpaired electrons). Any unpaired electron makes the ion paramagnetic: it is attracted to the external magnetic field. 3d⁰ has no electrons in 3d, and 3d¹⁰ has all electrons paired. So their magnetic moment is zero, and they are diamagnetic: they repel the external field. Note: the number of d electrons does not decide by itself. Only unpaired electrons count.

10. How many unpaired electrons are there in the metal ion of chromium(II) chloride, CrCl2$\mathrm{CrCl}{}_2$? (₂₄${}_{24}$Cr)

    الإجابة: 4 unpaired electrons

    Step 1: oxidation state. Cl is −1, and two Cl give −2, so chromium is +2. Step 2: the ion's configuration. Chromium is [Ar] 4s¹ 3d⁵ (abnormal, half-filled 3d). Cr²⁺ loses the 4s¹ electron first, then one 3d electron: [Ar] 3d⁴. Step 3: boxes by Hund's rule. The four 3d electrons go one in each of four boxes, and the fifth box stays empty. So there are 4 unpaired electrons, and CrCl2 is paramagnetic. Common wrong answers, each from one mistake: 6 ← counting the atom instead of the ion: [Ar] 4s¹ 3d⁵ has 1 unpaired in 4s and 5 in 3d, so 6. 5 ← removing only one electron (the 4s¹): that gives Cr⁺, [Ar] 3d⁵, so 5. 3 ← mixing up Cr²⁺ with Cr³⁺: [Ar] 3d³, so 3. 2 ← writing Cr as [Ar] 4s² 3d⁴ and removing both electrons from 3d: [Ar] 4s² 3d², where 4s² is paired and the two 3d electrons are in separate boxes, so 2. 0 ← pairing the four 3d electrons in two boxes, which breaks Hund's rule: 0.

11. Give reasons: Most of the first transition elements have catalytic activity.

    الإجابة: Key points: (1) Their 4s and 3d electrons are used to form bonds between the atoms of the catalyst surface and the reacting molecules. (2) This increases the concentration of the reacting molecules on the surface of the catalyst. (3) It weakens the bonds in the reactant molecules. (4) So the activation energy decreases and the speed of the reaction increases.

    The book links catalytic activity to the 4s and 3d electrons. These electrons form bonds between the surface atoms of the metal and the reacting molecules. More reactant molecules then gather on the surface, so their concentration there increases. The bonds inside the reactant molecules are weakened. So the activation energy decreases, and the reaction speeds up. Key words: 4s and 3d electrons, bonds with the surface, concentration on the surface, weaken the bonds, decrease the activation energy. Most likely wrong answer: 'they give the molecules energy'. A catalyst does not supply energy; it lowers the barrier. Another: 'they react and are used up'. A catalyst is not used up.

12. Match each process in column (A) with its catalyst and conditions in column (B), as in the book.

    الإجابة: Hydrogenation of oils → Finely divided nickel · Preparation of ammonia (Haber-Bosch method) → Iron, 500 °C, 200 atm · Preparation of sulphuric acid (contact method), first step → V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$, 450 °C · Decomposition of hydrogen peroxide (fig 1-3) → MnO2$\mathrm{MnO}{}_2$

    Each process has its own catalyst. Finely divided nickel: hydrogenation of oils (adding hydrogen turns liquid oils into solid fats). Iron at 500 °C and 200 atm: ammonia by the Haber-Bosch method. Vanadium pentoxide, V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$, at 450 °C: oxidation of SO2$\mathrm{SO}{}_2$ to SO3$\mathrm{SO}{}_3$ in the contact method for sulphuric acid. MnO2$\mathrm{MnO}{}_2$: decomposition of H2O2$\mathrm{H}{}_2\mathrm{O}{}_2$ into water and oxygen. The most common slip is swapping Fe and V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ between ammonia and sulphuric acid.

13. Give reasons: Nickel(II) chloride, NiCl2$\mathrm{NiCl}{}_2$, is attracted to an external magnetic field, while zinc oxide, ZnO$\mathrm{ZnO}$, is repelled by it. (₂₈${}_{28}$Ni, ₃₀${}_{30}$Zn)

    الإجابة: Key points: (1) In NiCl2, nickel is Ni²⁺: [Ar] 3d⁸, with 2 unpaired electrons. (2) Spinning unpaired electrons form a magnetic field, so NiCl2 is paramagnetic: attracted to the external magnetic field. (3) In ZnO, zinc is Zn²⁺: [Ar] 3d¹⁰, all electrons paired. (4) Paired electrons have opposite spins, so the magnetic moment is zero; ZnO is diamagnetic: it repels the external magnetic field.

    Step 1: find the ions. Cl is −1, so nickel is +2. O is −2, so zinc is +2. Step 2: write the 3d configurations, losing 4s first. Ni²⁺ is [Ar] 3d⁸; Zn²⁺ is [Ar] 3d¹⁰. Step 3: fill boxes by Hund's rule. 3d⁸ = three pairs and two unpaired electrons: 2 unpaired. 3d¹⁰ = five pairs: 0 unpaired. Step 4: an unpaired electron spins and forms a magnetic field, so the substance is attracted (paramagnetic). When all electrons are paired, opposite spins cancel, the magnetic moment is zero, and the substance is repelled (diamagnetic). Most likely wrong answer: 'zinc is not a metal that magnets attract'. Diamagnetic does not mean 'not affected'; ZnO is repelled. Another wrong answer: counting the atoms (Ni [Ar] 4s² 3d⁸) instead of the ions.

14. Arrange the following compounds in ascending order according to their magnetic moment (smallest first). (₂₁${}_{21}$Sc, ₂₆${}_{26}$Fe, ₂₇${}_{27}$Co, ₂₈${}_{28}$Ni)

    الإجابة: ScCl3$\mathrm{ScCl}{}_3$ → Ni(OH)2$\mathrm{Ni}(\mathrm{OH}){}_2$ → CoO$\mathrm{CoO}$ → Fe(NO3)3$\mathrm{Fe}(\mathrm{NO}{}_3){}_3$

    The magnetic moment increases with the number of unpaired electrons, so count them for each metal ion. ScCl3: Sc³⁺ is [Ar] 3d⁰, 0 unpaired (diamagnetic). Ni(OH)2: Ni²⁺ is [Ar] 3d⁸, 2 unpaired. CoO: Co²⁺ is [Ar] 3d⁷, 3 unpaired. Fe(NO3)3: Fe³⁺ is [Ar] 3d⁵, 5 unpaired. Order: ScCl3 < Ni(OH)2 < CoO < Fe(NO3)3. Common slip 1: ordering by the total number of 3d electrons, which puts Ni²⁺ last. Common slip 2: counting the atoms instead of the ions.

15. Energy profile of one reaction (made-up values)

    Progress of reaction	Without catalyst (kJ)	With catalyst (kJ)
    0	100	100
    1	100	100
    2	170	135
    3	230	162
    4	250	170
    5	220	155
    6	110	90
    7	40	40
    8	40	40

    The graph shows the energy profile of the same reaction without and with a catalyst (made-up values). Which statement is correct?

    • A. The catalyst lowers the activation energy from 150 kJ to 70 kJ; the net energy released stays 60 kJ.
    • B. The catalyst lowers the activation energy from 250 kJ to 170 kJ; the net energy released stays 60 kJ.
    • D. With the catalyst, the activation energy equals the net energy released, 60 kJ.
    • C. The catalyst lowers the activation energy from 150 kJ to 70 kJ, so the net energy released rises by 80 kJ.

    الإجابة: The catalyst lowers the activation energy from 150 kJ to 70 kJ; the net energy released stays 60 kJ.

    Activation energy is measured from the reactants (100 kJ) up to the top of the hump. Without catalyst: 250 − 100 = 150 kJ. With catalyst: 170 − 100 = 70 kJ. The net energy released is the drop from reactants to products: 100 − 40 = 60 kJ, the same on both curves. So the catalyst lowers the activation energy only. It does not change the energy of the reactants or products, so the net energy released stays the same, as in the book's fig 1-3 for MnO2$\mathrm{MnO}{}_2$ and H2O2$\mathrm{H}{}_2\mathrm{O}{}_2$.

16. A student wrote this equation for the Haber-Bosch method: N2(g) + H2(g) + Fe(s) → NH3(g)$\mathrm{N}{}_2(\mathrm{g})+\mathrm{H}{}_2(\mathrm{g})+\mathrm{Fe}(\mathrm{s})\to \mathrm{NH}{}_3(\mathrm{g})$. Rewrite it correctly, balanced, with the catalyst and conditions in their right place.

    الإجابة: N2(g) + 3H2(g) →[500\,^C / 200\,atm,\ Fe] 2NH3(g)$\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}/200\mathrm{atm},\mathrm{Fe}}{\to }2\mathrm{NH}{}_3(\mathrm{g})$

    The student made three mistakes. (1) Fe is the catalyst, not a reactant. A catalyst is not used up, so it goes over the arrow. (2) The equation is not balanced: N2 has 2 N atoms and 3 H2 have 6 H atoms, which give 2 NH3. (3) The conditions are missing: 500 °C and 200 atm, written over the arrow with Fe. Grader looks for: coefficients 1, 3, 2; Fe over the arrow, not on the left; 500 °C and 200 atm; state symbols (g). Most likely wrong answer: balancing correctly but leaving Fe as a reactant.

17. Write the scientific term: The substance that is attracted to the external magnetic field due to the presence of unpaired electrons.

    الإجابة: Paramagnetic substance.

    The book defines the paramagnetic substance as the substance that is attracted to the external magnetic field, due to the presence of unpaired electrons. Most likely wrong answer: 'diamagnetic substance'. That is the opposite: all electrons are paired, and it repels the field. Another wrong answer: 'magnetic moment'. That is the strength of the magnetism, not the substance.

18. Give reasons: Copper(II) oxide, CuO$\mathrm{CuO}$, is a paramagnetic substance, while copper(I) oxide, Cu2O$\mathrm{Cu}{}_2\mathrm{O}$, is diamagnetic. (₂₉${}_{29}$Cu)

    الإجابة: Key points: (1) Cu is [Ar] 4s¹ 3d¹⁰. (2) In CuO copper is Cu²⁺: [Ar] 3d⁹, with 1 unpaired electron, so CuO is paramagnetic. (3) In Cu2O copper is Cu⁺: [Ar] 3d¹⁰, all electrons paired, magnetic moment zero, so Cu2O is diamagnetic.

    Step 1: oxidation states. O is −2. In CuO one copper balances −2, so it is +2. In Cu2O two copper atoms share −2, so each is +1. Step 2: configurations, losing 4s first. Cu is [Ar] 4s¹ 3d¹⁰. Cu⁺ loses 4s¹: [Ar] 3d¹⁰. Cu²⁺ loses 4s¹ and one 3d: [Ar] 3d⁹. Step 3: boxes. 3d⁹ has four pairs and one unpaired electron: 1 unpaired, so CuO is attracted to the external field (paramagnetic). 3d¹⁰ has five pairs: 0 unpaired, magnetic moment zero, so Cu2O repels the field (diamagnetic). Most likely wrong answer: 'both are copper, so both behave the same'. The ion decides, not the element. Another: using Cu²⁺ for Cu2O because of the '2' in the formula.

19. Consider the following statements about a first transition metal used as a catalyst: (I) Its 4s and 3d electrons form bonds between its surface atoms and the reacting molecules. (II) It increases the net energy released by the reaction. (III) It weakens the bonds in the reactant molecules, so the activation energy decreases. Which of these statements is/are correct?

    • D. (III) only
    • B. (I), (II) and (III)
    • A. (I) and (III) only
    • C. (II) and (III) only

    الإجابة: (I) and (III) only

    (I) is correct: the book says the 4s and 3d electrons are used to form bonds between the surface atoms of the metal and the reacting molecules. This increases the concentration of these molecules on the surface. (III) is correct: these bonds weaken the bonds in the reactant molecules, so the activation energy decreases and the reaction speeds up. (II) is wrong: a catalyst does not change the energy of the reactants or products. In fig 1-3 the net energy released is the same with and without MnO2$\mathrm{MnO}{}_2$.

20. 1. SO₂(g) → SO₃ ((1) + O₂, catalyst ?, 450 °C)
    2. SO₃ → H₂SO₄(aq) ((2) + H₂O)

    The diagram shows the preparation of sulphuric acid by the contact method. Write the balanced equation of step (1), naming its catalyst over the arrow with the temperature.

    الإجابة: 2SO2(g) + O2(g) →[V2O5 / 450\,^C] 2SO3(g)$2\mathrm{SO}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})\stackrel{\mathrm{V}{}_2\mathrm{O}{}_5/450{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }2\mathrm{SO}{}_3(\mathrm{g})$

    Step (1) oxidizes sulphur dioxide to sulphur trioxide. The catalyst is vanadium pentoxide, V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$, at 450 °C, written over the arrow. Balance it: 2SO2$2\mathrm{SO}{}_2$ has 2 S and 4 O; adding O2$\mathrm{O}{}_2$ gives 6 O, which make 2SO3$2\mathrm{SO}{}_3$. Step (2) then is: SO3 + H2O → H2SO4$\mathrm{SO}{}_3+\mathrm{H}{}_2\mathrm{O}\to \mathrm{H}{}_2\mathrm{SO}{}_4$. Grader looks for: coefficients 2, 1, 2; V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ and 450 °C over the arrow; not V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ as a reactant. Most likely wrong answer: writing Fe as the catalyst (that belongs to the Haber-Bosch method), or writing SO2 + O2 → SO3$\mathrm{SO}{}_2+\mathrm{O}{}_2\to \mathrm{SO}{}_3$ unbalanced.

21. Substance A is attracted to an external magnetic field. Substance B is repelled by it. Which comparison between A and B is correct?

    • D. In A all electrons are paired; B has unpaired electrons that push the field away.
    • B. A is a permanent magnet; B is not affected by the magnetic field at all.
    • A. A has unpaired electrons; in B all electrons are paired, so the magnetic moment of B is zero.
    • C. A has more electrons in total than B, so the magnetic moment of A is larger.

    الإجابة: A has unpaired electrons; in B all electrons are paired, so the magnetic moment of B is zero.

    A is paramagnetic: it has unpaired electrons. Each spinning unpaired electron forms a small magnetic field, so A is attracted to the external magnetic field. Its magnetic moment increases with the number of unpaired electrons. B is diamagnetic: all its electrons are paired with opposite spins, so their fields cancel. Its magnetic moment equals zero and it repels the external field. A paramagnetic substance is not a magnet: it is attracted only while in an external field. The total number of electrons does not decide; only unpaired electrons count.

22. The ion M³⁺$\mathrm{M}{}^{3+}$ of an element M in the first transition series has 4 unpaired electrons in its 3d sublevel (filled by Hund's rule). Which element(s) could M be? (₂₄${}_{24}$Cr, ₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe, ₂₇${}_{27}$Co)

    • D. Co only
    • C. Fe only
    • A. Mn or Co
    • B. Mn only

    الإجابة: Mn or Co

    Write each M³⁺ ion, losing 4s electrons first. Cr is [Ar] 4s¹ 3d⁵, so Cr³⁺ is 3d³: 3 unpaired. Mn is [Ar] 4s² 3d⁵, so Mn³⁺ is 3d⁴: 4 unpaired. Fe is [Ar] 4s² 3d⁶, so Fe³⁺ is 3d⁵: 5 unpaired. Co is [Ar] 4s² 3d⁷, so Co³⁺ is 3d⁶: one pair plus 4 unpaired electrons. So two ions fit: Mn³⁺ and Co³⁺. Lesson: different numbers of d electrons can give the same number of unpaired electrons (d⁴ and d⁶). After five electrons, the boxes start pairing.

23. Iron is used as a catalyst in the Haber-Bosch method. Compared with the same reaction without iron, which quantity is smaller when iron is present?

    • A. The activation energy of the reaction
    • D. The mass of iron at the end of the reaction
    • C. The concentration of reacting molecules on the iron surface
    • B. The net energy released by the reaction

    الإجابة: The activation energy of the reaction

    Iron's 4s and 3d electrons form bonds with the N2$\mathrm{N}{}_2$ and H2$\mathrm{H}{}_2$ molecules on its surface. This raises their concentration on the surface and weakens their bonds. So the activation energy decreases and ammonia forms faster. The net energy released stays the same, because the reactants and products are the same. The iron is not used up, so its mass at the end is unchanged. That is why Fe is written over the arrow.

24. Which of the following compounds is diamagnetic? (₂₁${}_{21}$Sc, ₂₃${}_{23}$V, ₂₆${}_{26}$Fe, ₂₉${}_{29}$Cu)

    • D. VCl4$\mathrm{VCl}{}_4$
    • C. FeSO4$\mathrm{FeSO}{}_4$
    • A. Sc2(SO4)3$\mathrm{Sc}{}_2(\mathrm{SO}{}_4){}_3$
    • B. CuCl2$\mathrm{CuCl}{}_2$

    الإجابة: Sc2(SO4)3$\mathrm{Sc}{}_2(\mathrm{SO}{}_4){}_3$

    Find each metal ion and count unpaired electrons. Sc2(SO4)3: sulphate is −2, so Sc is +3; Sc³⁺ is [Ar] 3d⁰: 0 unpaired, diamagnetic. CuCl2: Cu²⁺ is [Ar] 3d⁹: 1 unpaired. FeSO4: Fe²⁺ is [Ar] 3d⁶: 4 unpaired. VCl4: V⁴⁺ is [Ar] 3d¹: 1 unpaired. Any unpaired electron, even one, makes a substance paramagnetic. Being a transition element does not make every compound paramagnetic: the ion decides.

25. An iron(II) compound is easily oxidized to an iron(III) compound. Which statement gives the reason, and the change in the number of unpaired electrons? (₂₆${}_{26}$Fe)

    • D. Fe³⁺$\mathrm{Fe}{}^{3+}$ has a higher charge, so it holds its electrons more tightly; unpaired electrons increase from 4 to 5.
    • A. Fe³⁺$\mathrm{Fe}{}^{3+}$ has a half-filled 3d⁵$3d^5$ sublevel, so it is more stable; unpaired electrons increase from 4 to 5.
    • B. Fe³⁺$\mathrm{Fe}{}^{3+}$ has a half-filled 3d⁵$3d^5$ sublevel, so it is more stable; unpaired electrons decrease from 6 to 5.
    • C. Fe³⁺$\mathrm{Fe}{}^{3+}$ has an empty 4s$4s$ sublevel, so it is more stable; unpaired electrons increase from 4 to 5.

    الإجابة: Fe³⁺$\mathrm{Fe}{}^{3+}$ has a half-filled 3d⁵$3d^5$ sublevel, so it is more stable; unpaired electrons increase from 4 to 5.

    Fe²⁺ is [Ar] 3d⁶: one pair and four unpaired electrons. When it loses one more electron, it becomes Fe³⁺, [Ar] 3d⁵. This 3d sublevel is half-filled, which is more stable. So iron(II) is easily oxidized to iron(III). 3d⁵ has one electron in each box: 5 unpaired. So the magnetic moment increases. The reason is the half-filled 3d⁵, not the charge: Mn²⁺ (3d⁵) is more stable than Mn³⁺ (3d⁴). The empty 4s is not the reason either, because Fe²⁺ already has an empty 4s.

26. Element	Atomic radius
    Sc	1.44
    Ti	1.32
    V	1.22
    Cr	1.17
    Mn	1.17
    Fe	1.16
    Co	1.16
    Ni	1.15
    Cu	1.17

    Look at the highlighted part of the graph, from Cr to Cu. The book links this part to the use of these metals in making alloys (an alloy is a metal mixed with other elements; you will study alloys later). Which statement describes and explains this part?

    • C. The atomic radius decreases sharply, because the effective nuclear charge increases while nothing opposes it.
    • B. The atomic radius is nearly constant, because the number of 3d electrons does not change from Cr to Cu.
    • A. The atomic radius is nearly constant, because two opposite factors balance.
    • D. The atomic radius increases, because the atomic mass increases from Cr to Cu.

    الإجابة: The atomic radius is nearly constant, because two opposite factors balance.

    From Cr to Cu the radius stays near 1.15–1.17. The book explains this with two opposite factors. The effective nuclear charge increases, which pulls the electrons in and decreases the radius. The number of 3d electrons also increases, so their repulsion increases, which increases the radius. The two factors balance, so the radii are relatively constant. The book adds that this explains using these elements in making alloys. Atomic mass does not decide the size.

27. Finely divided nickel is used as a catalyst in the hydrogenation of oils. How does the nickel speed up this reaction, according to the book?

    • B. It reacts with the hydrogen and becomes part of the solid fat that forms.
    • C. It gives the oil and hydrogen molecules extra heat energy to react.
    • D. It increases the amount of fat that forms from the same amount of oil.
    • A. Its 4s and 3d electrons bond the reacting molecules to its surface, which weakens their bonds.

    الإجابة: Its 4s and 3d electrons bond the reacting molecules to its surface, which weakens their bonds.

    Nickel is a transition element. Its 4s and 3d electrons form bonds between its surface atoms and the reacting molecules (oil and hydrogen). This increases their concentration on the surface and weakens their bonds. So the activation energy decreases and hydrogenation is faster. Finely divided means a fine powder with a large surface. The nickel is not used up, gives no energy, and does not change the amount of fat formed.