Transition Elements · Practice

All questions with answers · Transition Elements

1. A d sublevel contains 6 electrons (d⁶$d^6$). Which orbital diagram shows the correct arrangement of these electrons in the five d orbitals?

   • B. [↑↓] [↑↓] [↑↓] [ ] [ ]
   • A. [↑↓] [↑] [↑] [↑] [↑]
   • C. [↑↓] [↑] [↓] [↑] [↓]
   • D. [↑↑] [↑] [↑] [↑] [↑]

   Answer: [↑↓] [↑] [↑] [↑] [↑]

   Hund's rule: inside one sublevel, electrons enter the orbitals one by one first (a single electron in each orbital, all with parallel spins). Pairing starts only after every orbital has one electron, and the second electron in an orbital has the opposite spin (Pauli exclusion principle). For d⁶$d^6$: electrons 1-5 go ↑ into the five orbitals, and electron 6 goes ↓ into the first orbital. Result: [↑↓] [↑] [↑] [↑] [↑], so 4 unpaired electrons. You will need exactly this drawing for the 3d sublevel of iron (3d⁶$3d^6$).

2. The electrons of a d⁸$d^8$ sublevel are distributed in its five orbitals according to Hund's rule. How many unpaired electrons are there?

   Answer: 2 electrons

   Step 1: the five d orbitals take the first 5 electrons one by one, all with the same spin: [↑] [↑] [↑] [↑] [↑]. Step 2: electrons 6, 7 and 8 pair up in the first three orbitals: [↑↓] [↑↓] [↑↓] [↑] [↑]. So there are 2 unpaired electrons. Common wrong answers: 0 (pairing from the start, which draws 4 pairs and an empty box: this ignores Hund's rule); 3 (counting the electrons after half-filling, 8-5=3$8-5=3$, as unpaired, when those are exactly the ones that paired up); 8 (thinking every d electron is unpaired). Quick check: after d⁵$d^5$, every extra electron lowers the number of unpaired electrons by one, so d⁸$d^8$ has 10-8=2$10-8=2$.

3. In a p or d sublevel, why do electrons occupy the orbitals singly first before they start pairing?

   • B. Because one orbital can hold only one electron
   • A. Because they repel each other less when they are spread out, so the energy is lower
   • C. Because each orbital in the sublevel is higher in energy than the one before it

   Answer: Because they repel each other less when they are spread out, so the energy is lower

   The orbitals of one sublevel have equal energy. All electrons are negatively charged, so they repel each other. The arrangement with the lowest energy (the most stable one) is when the electrons stay unpaired, one in each orbital, with parallel spins. Pairing happens only after every orbital has one electron, and then the second electron enters with the opposite spin. This is Hund's rule, and it is why the 3d sublevel in the first transition series fills singly up to manganese (3d⁵$3d^5$) before pairing starts.

4. How many electrons does the ion ₂₄Cr³⁺${}_{}^{}{}_{24}^{}\mathrm{Cr}{}^{3+}$ have?

   • C. 24
   • D. 3
   • B. 27
   • A. 21

   Answer: 21

   Chromium has Z = 24, so its atom has 24 protons and 24 electrons. The 3+ charge means the atom lost 3 electrons. Electrons in a cation = Z − charge = 24 − 3 = 21. The protons stay 24, so it is still chromium. Most likely wrong answer: 27, from adding the charge. A positive charge means electrons were lost, not gained.

5. Compare the nickel atom (₂₈${}_{28}$Ni) with the ion Ni²⁺$\mathrm{Ni}{}^{2+}$. Which statement is correct?

   • B. Same number of protons; Ni²⁺$\mathrm{Ni}{}^{2+}$ has two more electrons
   • A. Same number of protons; Ni²⁺$\mathrm{Ni}{}^{2+}$ has two fewer electrons
   • C. Ni²⁺$\mathrm{Ni}{}^{2+}$ has two fewer protons and the same electrons
   • D. Ni²⁺$\mathrm{Ni}{}^{2+}$ has two more protons and the same electrons

   Answer: Same number of protons; Ni²⁺$\mathrm{Ni}{}^{2+}$ has two fewer electrons

   When an ion forms, only electrons move. The protons stay the same, so it is still nickel. The Ni atom has 28 protons and 28 electrons. Ni²⁺$\mathrm{Ni}{}^{2+}$ lost 2 electrons, so it has 28 protons and 26 electrons. Two positive charges are left over, which gives the 2+ charge.

6. What is the oxidation state of nickel in Ni2O3$\mathrm{Ni}{}_2\mathrm{O}{}_3$? (O = −2)

   • A. +3
   • C. +2
   • B. +6
   • D. −3

   Answer: +3

   Call nickel x. Two Ni atoms give 2x. Three O atoms give 3 × (−2) = −6. In a compound, the oxidation states add up to zero: 2x − 6 = 0. So 2x = +6 and x = +3. The name is nickel (III) oxide. Most likely wrong answer: +6, giving the whole total to one Ni atom.

7. What is the oxidation state of chromium in Cr2(SO4)3$\mathrm{Cr}{}_2(\mathrm{SO}{}_4){}_3$? (The SO4$\mathrm{SO}{}_4$ group = −2)

   • C. +12
   • A. +3
   • D. +2
   • B. +6

   Answer: +3

   Call chromium x. Two Cr atoms give 2x. Three SO4$\mathrm{SO}{}_4$ groups give 3 × (−2) = −6. The sum is zero: 2x − 6 = 0, so x = +3. The name is chromium (III) sulphate. Treat SO4$\mathrm{SO}{}_4$ as one group with −2. Do not count its O atoms one by one.

8. In the change Mn²⁺ → Mn³⁺ + e-$\mathrm{Mn}{}^{2+}\to \mathrm{Mn}{}^{3+}+\mathrm{e}{}^{-}$, what happens to manganese?

   • D. Nothing is oxidized or reduced, because no oxygen takes part
   • C. It is oxidized: it loses an electron, and its oxidation state decreases
   • A. It is oxidized: it loses an electron, and its oxidation state increases
   • B. It is reduced: it gains an electron, and its oxidation state increases

   Answer: It is oxidized: it loses an electron, and its oxidation state increases

   Oxidation is the loss of electrons, and the oxidation state increases. Here the electron is on the right: manganese gives it out. Its oxidation state goes from +2 to +3. So manganese (II) is oxidized. No oxygen is needed for oxidation.

9. In the reaction 2FeCl2 + Cl2 → 2FeCl3$2\mathrm{FeCl}{}_2+\mathrm{Cl}{}_2\to 2\mathrm{FeCl}{}_3$, which substance is the oxidizing agent?

   • B. FeCl2$\mathrm{FeCl}{}_2$
   • D. None, because no oxygen takes part
   • C. FeCl3$\mathrm{FeCl}{}_3$
   • A. Cl2$\mathrm{Cl}{}_2$

   Answer: Cl2$\mathrm{Cl}{}_2$

   First find the oxidation states. Iron: +2 in FeCl2$\mathrm{FeCl}{}_2$ → +3 in FeCl3$\mathrm{FeCl}{}_3$. It loses electrons, so it is oxidized. Chlorine: 0 in Cl2$\mathrm{Cl}{}_2$ → −1 in FeCl3$\mathrm{FeCl}{}_3$. It gains electrons, so it is reduced. The oxidizing agent is the substance that is itself reduced: Cl2$\mathrm{Cl}{}_2$. FeCl2$\mathrm{FeCl}{}_2$ is the reducing agent.

10. Magnesium (₁₂${}_{12}$Mg) and sulphur (₁₆${}_{16}$S) are both in period 3. Which statement is correct?

    • B. S is larger, because it has more electrons than Mg
    • C. They are equal in size, because both use three main levels
    • A. Mg is larger, because S has more protons pulling the same outer level
    • D. S is larger, because its outer electrons are in a higher level

    Answer: Mg is larger, because S has more protons pulling the same outer level

    Across a period, the outer electrons stay in the same main level (n = 3). Each step adds one proton, so the nucleus pulls that level harder. So the radius decreases from Mg to S. Mg is the larger atom. More electrons do not win here; the stronger pull does.

11. Across a period, one proton and one electron are added at each step. Which two factors act on the atomic radius in opposite directions?

    • D. Protons repel each other inward; electrons attract each other outward
    • A. The nucleus attracts electrons inward; electrons repel each other outward
    • B. The nucleus repels electrons outward; electrons attract each other inward
    • C. Neutrons attract electrons inward; protons repel electrons outward

    Answer: The nucleus attracts electrons inward; electrons repel each other outward

    Two factors act at the same time. More protons attract the electrons more strongly, which decreases the radius. More electrons repel each other more, which increases the radius. The size is a balance of the two. Across a period, the pull of the nucleus wins. You will use these two factors for the first transition series.

12. Which equation represents the first ionization energy of calcium?

    • C. Ca(g) + e- → Ca⁻(g)$\mathrm{Ca}(\mathrm{g})+\mathrm{e}{}^{-}\to \mathrm{Ca}{}^{-}(\mathrm{g})$
    • B. Ca(s) → Ca+(s) + e-$\mathrm{Ca}(\mathrm{s})\to \mathrm{Ca}{}^{+}(\mathrm{s})+\mathrm{e}{}^{-}$
    • A. Ca(g) → Ca+(g) + e-$\mathrm{Ca}(\mathrm{g})\to \mathrm{Ca}{}^{+}(\mathrm{g})+\mathrm{e}{}^{-}$
    • D. Ca(g) → Ca²⁺(g) + 2e-$\mathrm{Ca}(\mathrm{g})\to \mathrm{Ca}{}^{2+}(\mathrm{g})+2\mathrm{e}{}^{-}$

    Answer: Ca(g) → Ca+(g) + e-$\mathrm{Ca}(\mathrm{g})\to \mathrm{Ca}{}^{+}(\mathrm{g})+\mathrm{e}{}^{-}$

    The book's term is ionization potential (also called ionization energy). It is the energy to remove the most loosely held electron from a gaseous atom. So the atom must be a gas, and only one electron leaves. Its unit is kJ/mol. Removing the next electron is a separate step: the second ionization energy.

13. The successive ionization potentials (also called ionization energies) of a metal X are 520, 7298 and 11815 kJ/mol. Which oxidation state does X show in its compounds?

    • D. Several states, because the values keep rising
    • C. +2
    • B. +3
    • A. +1

    Answer: +1

    Compare each value with the one before. 520 → 7298 is about 14 times larger: a big jump. So the 2nd electron must leave a stable inner level. X gives up only one electron, so it shows +1. (X is lithium, group IA.) The big jump tells you a representative metal's oxidation state.

14. Give reasons: The 3d sublevel of the manganese atom (₂₅${}_{25}$Mn) contains five unpaired electrons, while that of the iron atom (₂₆${}_{26}$Fe) contains only four.

    Answer: Key points: (1) Mn: [Ar] 4s² 3d⁵; Fe: [Ar] 4s² 3d⁶. (2) Hund's rule: the five 3d orbitals are filled by a single electron in each orbital (parallel spins) before any pairing. (3) In Mn the five 3d electrons occupy the five orbitals singly → 5 unpaired. (4) In Fe the sixth 3d electron must pair with one of them (opposite spin) → 4 unpaired.

    Model answer: Mn is [Ar] 4s² 3d⁵ and Fe is [Ar] 4s² 3d⁶. By Hund's rule, the five 3d orbitals are first filled by a single electron in each orbital (parallel spins), and only then does pairing begin. In Mn, the five 3d electrons go one into each orbital, so all 5 are unpaired: [↑] [↑] [↑] [↑] [↑]. In Fe, the sixth 3d electron finds no empty orbital, so it pairs with one electron (opposite spin), leaving 4 unpaired: [↑↓] [↑] [↑] [↑] [↑]. The book puts it this way: the 3d orbitals are filled by a single electron in each orbital till manganese (3d⁵$3d^5$), and after manganese pairing takes place till zinc (3d¹⁰$3d^{10}$). Words your exam answer must contain: Hund's rule, single electron in each orbital, pairing, unpaired. Most common mistake: "Fe has more electrons, so it has more unpaired electrons." The number of unpaired electrons rises only up to 3d⁵$3d^5$ and then falls.

15. Give reasons: The copper atom (₂₉${}_{29}$Cu) has only one electron in its 4s sublevel, while the nickel atom (₂₈${}_{28}$Ni) has two.

    Answer: Key points: (1) Ni: [Ar] 4s² 3d⁸ follows the expected pattern. (2) Cu is [Ar] 4s¹ 3d¹⁰, not the expected [Ar] 4s² 3d⁹ (abnormal configuration). (3) In Cu, 4s is half filled and 3d is filled (d¹⁰). (4) The atom has lower energy (extra stability) when its d sublevel is filled (d¹⁰) or half filled (d⁵).

    Ni follows the pattern: [Ar] 4s²\,3d⁸$4s^{2}3d^8$. By the same pattern, Cu would be [Ar] 4s²\,3d⁹$4s^{2}3d^9$. But the real copper atom is [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$. Now 3d is filled (d¹⁰$d^{10}$) and 4s is half filled. This gives the atom lower energy, so it has extra stability. The book adds that this is not the only factor. Key words: abnormal configuration, filled (d¹⁰$d^{10}$), half filled, lower energy, extra stability. Most likely wrong answer: writing Cu as [Ar] 4s²\,3d⁹$4s^{2}3d^9$, like the pattern. That misses the extra stability of a filled 3d.

16. Arrange the following compounds in ascending order according to the number of unpaired electrons in the 3d sublevel of the metal ion: MnO$\mathrm{MnO}$, V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$, CoCl3$\mathrm{CoCl}{}_3$, NiO$\mathrm{NiO}$ (V = 23, Mn = 25, Co = 27, Ni = 28)

    Answer: V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ → NiO$\mathrm{NiO}$ → CoCl3$\mathrm{CoCl}{}_3$ → MnO$\mathrm{MnO}$

    Step 1: find the oxidation states (O = −2, Cl = −1). V +5, Mn +2, Co +3, Ni +2. Step 2: write each ion, removing the 4s electrons first. V⁵⁺$\mathrm{V}{}^{5+}$ is [Ar] 3d⁰$3d^0$. Mn²⁺$\mathrm{Mn}{}^{2+}$ is [Ar] 3d⁵$3d^5$. Co³⁺$\mathrm{Co}{}^{3+}$ is [Ar] 3d⁶$3d^6$. Ni²⁺$\mathrm{Ni}{}^{2+}$ is [Ar] 3d⁸$3d^8$. Step 3: fill the 3d boxes by Hund's rule. 3d⁰$3d^0$ → 0, 3d⁸$3d^8$ → 2, 3d⁶$3d^6$ → 4, 3d⁵$3d^5$ → 5. Order: V2O5$\mathrm{V}{}_2\mathrm{O}{}_5$ (0) < NiO$\mathrm{NiO}$ (2) < CoCl3$\mathrm{CoCl}{}_3$ (4) < MnO$\mathrm{MnO}$ (5). Most likely wrong answer: counting the atoms (V 3, Ni 2, Co 3, Mn 5). Find the ion in the compound first. Another error: ordering by atomic number, from the idea 'more electrons, more unpaired'.

17. Give reasons: The Fe³⁺$\mathrm{Fe}{}^{3+}$ ion is more stable than the Fe²⁺$\mathrm{Fe}{}^{2+}$ ion, while the Mn²⁺$\mathrm{Mn}{}^{2+}$ ion is more stable than the Mn³⁺$\mathrm{Mn}{}^{3+}$ ion. (₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe)

    Answer: Key points: (1) Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵. (2) Mn²⁺: [Ar] 3d⁵; Mn³⁺: [Ar] 3d⁴. (3) In each pair, the more stable ion has a half-filled 3d sublevel (3d⁵). (4) A half-filled 3d sublevel is more stable. (5) So Fe²⁺ is easily oxidized to Fe³⁺, but Mn²⁺ is not readily oxidized to Mn³⁺.

    Remove the 4s electrons first, then 3d electrons. Fe²⁺$\mathrm{Fe}{}^{2+}$ is [Ar] 3d⁶$3d^6$ and Fe³⁺$\mathrm{Fe}{}^{3+}$ is [Ar] 3d⁵$3d^5$. Mn²⁺$\mathrm{Mn}{}^{2+}$ is [Ar] 3d⁵$3d^5$ and Mn³⁺$\mathrm{Mn}{}^{3+}$ is [Ar] 3d⁴$3d^4$. In both pairs, the more stable ion has the half-filled 3d⁵$3d^5$ sublevel. So the charge is not the reason: iron prefers +3, but manganese prefers +2. This is why iron (II) is easily oxidized, while Mn (II) is not readily oxidized. Key words: half-filled 3d sublevel (3d⁵$3d^5$), more stable. Most likely wrong answer: 'a bigger charge is more stable'. Manganese shows that this idea fails.

18. In which of the following compounds does manganese show the highest oxidation state? (K = +1, O = −2, SO4$\mathrm{SO}{}_4$ group = −2)

    • C. MnSO4$\mathrm{MnSO}{}_4$
    • B. Mn2O3$\mathrm{Mn}{}_2\mathrm{O}{}_3$
    • A. K2MnO4$\mathrm{K}{}_2\mathrm{MnO}{}_4$
    • D. MnO$\mathrm{MnO}$

    Answer: K2MnO4$\mathrm{K}{}_2\mathrm{MnO}{}_4$

    Find manganese (x) in each compound. K2MnO4$\mathrm{K}{}_2\mathrm{MnO}{}_4$: 2(+1) + x + 4(−2) = 0, so x = +6. Mn2O3$\mathrm{Mn}{}_2\mathrm{O}{}_3$: 2x + 3(−2) = 0, so x = +3. MnSO4$\mathrm{MnSO}{}_4$: x + (−2) = 0, so x = +2. MnO$\mathrm{MnO}$: x = +2. The highest here is +6, in K2MnO4$\mathrm{K}{}_2\mathrm{MnO}{}_4$. Manganese reaches +7, its maximum, in KMnO4$\mathrm{KMnO}{}_4$.

19. Give reasons: Potassium (₁₉${}_{19}$K) shows one oxidation state (+1) in its compounds, while chromium (₂₄${}_{24}$Cr) shows several oxidation states (+2, +3, +6).

    Answer: Key points: (1) K: [Ar] 4s¹; after the 4s electron is lost, the next one must leave the stable [Ar] inner level, so the ionization energy jumps hugely and K stays at (+1). (2) Cr: [Ar] 4s¹ 3d⁵; it loses electrons from 4s and 3d in sequence. (3) The 4s and 3d sublevels are close in energy. (4) So the ionization potentials increase gradually, with no huge jump, and several oxidation states form.

    Potassium has one outer electron, in 4s. Its next electron would come from the stable [Ar] arrangement, nearer the nucleus. That needs far more energy (a big jump), so K shows only +1. Chromium loses electrons from 4s and 3d in sequence. These two sublevels are close in energy. So its ionization energies increase gradually, and several oxidation states can form. Key words: 4s and 3d, close in energy, in sequence, ionization potentials increase gradually. Most likely wrong answer: 'Cr has more electrons'. Calcium has 20 electrons, yet it shows only +2.

20. Write the scientific term: An element whose d or f orbitals contain electrons but are not complete, either in its atomic state or in one of its oxidation states.

    Answer: The transition element.

    This is the book's definition of the transition element. It checks two places: the atom, and its oxidation states. One incomplete d (or f) sublevel, in either place, is enough. Incomplete means the d orbitals hold 1 to 9 electrons. Most likely wrong answer: 'd-block element'. Zinc is in the d-block, yet it is not a transition element.

21. Give reasons: Copper (₂₉${}_{29}$Cu) is a transition element, while zinc (₃₀${}_{30}$Zn) is not, although both atoms have a completely filled 3d sublevel.

    Answer: Key points: (1) Cu atom [Ar] 4s¹ 3d¹⁰ and Zn atom [Ar] 4s² 3d¹⁰: d is completely filled in both atoms. (2) In the (+2) state, Cu²⁺ is [Ar] 3d⁹: the d sublevel is partially filled, so Cu is a transition element. (3) Zn shows only (+2): Zn²⁺ is [Ar] 3d¹⁰, still completely filled, so Zn is not. (4) The transition element has incomplete d in its atomic state or in one of its oxidation states.

    Test the atoms first. Cu is [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$ and Zn is [Ar] 4s²\,3d¹⁰$4s^{2}3d^{10}$. Both fail: their 3d is complete. Now test the oxidation states. Cu²⁺$\mathrm{Cu}{}^{2+}$ is [Ar] 3d⁹$3d^9$: the d sublevel is partially filled. So copper is a transition element. Zinc shows only (+2), and Zn²⁺$\mathrm{Zn}{}^{2+}$ is [Ar] 3d¹⁰$3d^{10}$, still complete. So zinc is not a transition element. Key words: completely filled (d¹⁰$d^{10}$) in the atom, partially filled (d⁹$d^9$) in the (+2) state. Most likely wrong answer: 'both are in the d-block'. Position is not the book's test.

22. Which of the following orbital diagrams represents the 3d sublevel of the cobalt atom (₂₇${}_{27}$Co)?

    • B. [↑↓] [↑↓] [↑↓] [↑] [ ]
    • C. [↑↓] [↑↓] [↑↓] [↑↓] [↑]
    • D. [↑↓] [↑↓] [↑↓] [↑] [↑]
    • A. [↑↓] [↑↓] [↑] [↑] [↑]

    Answer: [↑↓] [↑↓] [↑] [↑] [↑]

    Step 1: Co has Z = 27; [Ar] accounts for 18, so 9 electrons are left. Step 2: after calcium ([Ar] 4s²) the 4s sublevel is full, so the remaining 27-20=7$27-20=7$ electrons go into 3d: [Ar] 4s² 3d⁷ (as in the book's table). Step 3 (Hund's rule): the first 5 electrons go one into each orbital with parallel spins, and electrons 6 and 7 pair up in the first two orbitals: [↑↓] [↑↓] [↑] [↑] [↑] → 3 unpaired. Common mistakes: an empty box next to pairs (ignoring Hund's rule), 9 electrons in 3d (forgetting 4s²), or 8 in 3d (applying the Cr/Cu anomaly to Co).

23. Element X lies in group VIB of the first transition series. Which is the electronic configuration of its atom?

    • A. [Ar] 4s¹\,3d⁵$4s^{1}3d^5$
    • D. [Ar] 3d⁶$3d^6$
    • B. [Ar] 4s²\,3d⁴$4s^{2}3d^4$
    • C. [Ar] 4s²\,3d⁵$4s^{2}3d^5$

    Answer: [Ar] 4s¹\,3d⁵$4s^{1}3d^5$

    Step 1: find X. Group IIIB is Sc (21), IVB is Ti (22) and VB is V (23). So group VIB is Cr (24). Step 2: write Cr as the book does. The expected pattern gives [Ar] 4s²\,3d⁴$4s^{2}3d^4$, with one empty 3d orbital. The real atom is [Ar] 4s¹\,3d⁵$4s^{1}3d^5$. Both 4s and 3d are half filled, so the atom has lower energy (extra stability).

24. Which is the electronic configuration of Ni³⁺$\mathrm{Ni}{}^{3+}$? (₂₈${}_{28}$Ni: [Ar] 4s²\,3d⁸$4s^{2}3d^8$)

    • B. [Ar] 4s²\,3d⁵$4s^{2}3d^5$
    • C. [Ar] 4s¹\,3d⁶$4s^{1}3d^6$
    • D. [Ar] 3d⁸$3d^8$
    • A. [Ar] 3d⁷$3d^7$

    Answer: [Ar] 3d⁷$3d^7$

    A 3+ ion has lost 3 electrons: 28 − 3 = 25 electrons are left. 4s is in the outermost level (n = 4), so its 2 electrons leave first. The third electron then leaves 3d: 3d⁸$3d^8$ → 3d⁷$3d^7$. So Ni³⁺$\mathrm{Ni}{}^{3+}$ is [Ar] 3d⁷$3d^7$. Check: 18 + 7 = 25. The book's table lists nickel as +2, +3 and +4.

25. Which of the following ions becomes more stable by losing one more electron? (₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe, ₃₀${}_{30}$Zn)

    • B. Mn²⁺$\mathrm{Mn}{}^{2+}$
    • C. Fe³⁺$\mathrm{Fe}{}^{3+}$
    • A. Fe²⁺$\mathrm{Fe}{}^{2+}$
    • D. Zn²⁺$\mathrm{Zn}{}^{2+}$

    Answer: Fe²⁺$\mathrm{Fe}{}^{2+}$

    Write each ion, then picture one more electron leaving. Fe²⁺$\mathrm{Fe}{}^{2+}$ [Ar] 3d⁶$3d^6$ → [Ar] 3d⁵$3d^5$: it reaches a half-filled, more stable 3d. Mn²⁺$\mathrm{Mn}{}^{2+}$ is already 3d⁵$3d^5$. Losing one gives 3d⁴$3d^4$, which is less stable. Fe³⁺$\mathrm{Fe}{}^{3+}$ is also 3d⁵$3d^5$, so it would lose that stability. Zn²⁺$\mathrm{Zn}{}^{2+}$ is 3d¹⁰$3d^{10}$, filled; it would lose stability too. So only Fe²⁺$\mathrm{Fe}{}^{2+}$ is easily oxidized.

26. Element	Group	Electrons in 4s and 3d	Highest oxidation state
    ₂₂${}_{22}$Ti	IVB	4	+4
    ₂₅${}_{25}$Mn	VIIB	7	+7
    ₂₆${}_{26}$Fe	VIII	8	+6
    ₂₉${}_{29}$Cu	IB	11	+2
    ₃₀${}_{30}$Zn	IIB	12	+2

    The table shows data from the book for five elements of the first transition series. Which conclusion do the data support?

    • C. The highest state rises with the number of 4s and 3d electrons
    • B. The highest state of each element equals its group number
    • D. Only group IB has a highest state below its group number
    • A. The highest state equals the number of 4s and 3d electrons only up to Mn

    Answer: The highest state equals the number of 4s and 3d electrons only up to Mn

    Test each claim against every row. Ti (4 electrons, +4) and Mn (7, +7) match the electron count. Fe has 8 but reaches only +6. Cu and Zn are far below their counts. So the match holds up to Mn only. The group rule fails for Fe: group VIII, highest +6. Copper's +2 is above its group number (I), not below. The book: the maximum does not exceed the group number, except group IB.

27. Metal	1st	2nd	3rd	4th
    Vanadium	648	1346	2858	4643
    X	900	1757	14849	21007

    The table gives successive ionization energies (kJ/mol) of vanadium (book values) and of a representative metal X. Which statement is correct?

    • C. Vanadium shows only (+2), because a huge jump follows its 4s electrons
    • A. X shows one oxidation state (+2); vanadium's gradual rise allows several
    • B. X shows (+4), because four of its electrons can be removed
    • D. Both show several oxidation states, because both lists keep rising

    Answer: X shows one oxidation state (+2); vanadium's gradual rise allows several

    Compare each step with the one before. X: 900 → 1757 roughly doubles. But 1757 → 14849 is about eight times larger: a big jump. The 3rd electron must leave a stable inner level, so X shows only +2. (X is beryllium, group IIA.) Vanadium: 648 → 1346 → 2858 → 4643 rises gradually, about doubling or less. Its 4s and 3d electrons are close in energy. So vanadium shows several states: +2, +3, +4 and +5.

28. Which of the following data is enough to prove that element X is a transition element?

    • B. X lies in the d-block of the periodic table
    • A. X atom: [Kr] 5s²\,4d²$5s^{2}4d^2$
    • C. X atom: [Ar] 4s²\,3d¹⁰$4s^{2}3d^{10}$; its only ion X²⁺$\mathrm{X}{}^{2+}$: [Ar] 3d¹⁰$3d^{10}$
    • D. X shows the oxidation state (+2) by losing its 4s electrons

    Answer: X atom: [Kr] 5s²\,4d²$5s^{2}4d^2$

    The definition checks the d (or f) orbitals. They must contain electrons but be incomplete, in the atom or in one of its oxidation states. [Kr] 5s²\,4d²$5s^{2}4d^2$ has 2 electrons in 4d, which is incomplete. So X is a transition element. Position in the d-block is not the test: zinc is there, but it is not one. An atom that is d¹⁰$d^{10}$ and whose only ion is also d¹⁰$d^{10}$ has a complete d sublevel in both, so it fails the definition. Calcium also loses its 4s electrons to give +2, so (+2) proves nothing.

29. Cadmium (₄₈${}_{48}$Cd: [Kr] 5s²\,4d¹⁰$5s^{2}4d^{10}$) shows only the oxidation state (+2). Why is cadmium not a transition element?

    • C. Its atom has a filled 4d, so it fails the definition at once
    • D. It shows only one oxidation state in all its compounds
    • A. Its 4d is completely filled in the atom and in the (+2) state
    • B. It is in group IIB, which lies outside the d-block

    Answer: Its 4d is completely filled in the atom and in the (+2) state

    Test the atom: [Kr] 5s²\,4d¹⁰$5s^{2}4d^{10}$, so 4d is complete. Test the (+2) state: the two 5s electrons leave first. So Cd²⁺$\mathrm{Cd}{}^{2+}$ is [Kr] 4d¹⁰$4d^{10}$, still complete. It fails both tests, so cadmium is not a transition element. The same holds for zinc and mercury.

30. Element X belongs to the first transition series. Its atom has a completely filled 4s sublevel and the greatest number of unpaired electrons in the 3d sublevel. The atomic number of X is ...

    • B. 25
    • C. 26
    • D. 30
    • A. 24

    Answer: 25

    Step 1 (find the 3d arrangement): 3d has 5 orbitals, so the greatest number of unpaired electrons is 5. This happens at 3d⁵$3d^5$, when each orbital has a single electron (Hund's rule). In this series, 3d⁵$3d^5$ occurs in Cr and Mn. Step 2 (the 4s condition): Cr is [Ar] 4s¹ 3d⁵ (4s not full), while Mn is [Ar] 4s² 3d⁵. So X is Mn, Z = 25. Note: after Mn pairing begins, so the number of unpaired 3d electrons falls (Fe 4, Co 3, Ni 2, Zn 0) even though the number of electrons keeps rising.

31. Consider the following statements: (I) In the chromium atom (₂₄${}_{24}$Cr), both the 4s and 3d sublevels are half filled. (II) In the copper atom (₂₉${}_{29}$Cu), the 4s sublevel is filled and the 3d sublevel is half filled. (III) An atom has extra stability when its d sublevel is half filled (d⁵$d^5$) or filled (d¹⁰$d^{10}$). Which of these statements is/are correct?

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

    Answer: I and III only

    Cr is [Ar] 4s¹\,3d⁵$4s^{1}3d^5$. 4s holds 1 of 2 electrons, and 3d holds 5 of 10. Both are half filled, so (I) is correct. Cu is [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$: 4s is half filled and 3d is filled. (II) swaps them, so it is wrong. (III) is the book's reason for the anomaly, so it is correct. The book adds that this is not the only factor. Answer: I and III only.

32. The ion X+$\mathrm{X}{}^{+}$ of an element X in the first transition series has the configuration [Ar] 3d¹⁰$3d^{10}$. Which choice identifies X and the configuration of its atom?

    • D. X is nickel; [Ar] 4s²\,3d⁸$4s^{2}3d^8$
    • C. X is zinc; [Ar] 4s²\,3d¹⁰$4s^{2}3d^{10}$
    • B. X is copper; [Ar] 4s²\,3d⁹$4s^{2}3d^9$
    • A. X is copper; [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$

    Answer: X is copper; [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$

    Step 1: count the electrons in the ion: 18 + 10 = 28. The ion lost 1 electron, so Z = 28 + 1 = 29. X is copper. Step 2: the copper atom is [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$, an abnormal configuration with a filled 3d. Check: when the single 4s electron leaves, the ion is [Ar] 3d¹⁰$3d^{10}$. That matches the given ion.

33. Consider the following statements (₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe): (I) When Fe²⁺$\mathrm{Fe}{}^{2+}$ is oxidized to Fe³⁺$\mathrm{Fe}{}^{3+}$, the 3d orbital that held a pair is left with one electron. (II) The Mn³⁺$\mathrm{Mn}{}^{3+}$ ion has a half-filled 3d sublevel. (III) The important oxidation states of iron are (+2) and (+3). Which of these statements is/are correct?

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

    Answer: I and III only

    (I) Fe²⁺$\mathrm{Fe}{}^{2+}$ is 3d⁶$3d^6$: one orbital holds a pair and four hold one each. Fe³⁺$\mathrm{Fe}{}^{3+}$ is 3d⁵$3d^5$: all five hold one each. So the paired orbital lost an electron: correct. (II) Mn³⁺$\mathrm{Mn}{}^{3+}$ is [Ar] 3d⁴$3d^4$: one orbital is empty. It is not half filled, so (II) is wrong. (III) is correct: iron's states above +3 are not important. Answer: I and III only.

34. Arrange the following compounds in ascending order according to the oxidation state of the transition metal: CrO3$\mathrm{CrO}{}_3$, Cu2O$\mathrm{Cu}{}_2\mathrm{O}$, VO2$\mathrm{VO}{}_2$, Ti2O3$\mathrm{Ti}{}_2\mathrm{O}{}_3$ (O = −2)

    Answer: Cu2O$\mathrm{Cu}{}_2\mathrm{O}$ → Ti2O3$\mathrm{Ti}{}_2\mathrm{O}{}_3$ → VO2$\mathrm{VO}{}_2$ → CrO3$\mathrm{CrO}{}_3$

    Use the sum-to-zero rule with O = −2. Cu2O$\mathrm{Cu}{}_2\mathrm{O}$: 2x − 2 = 0, so x = +1. Ti2O3$\mathrm{Ti}{}_2\mathrm{O}{}_3$: 2x − 6 = 0, so x = +3. VO2$\mathrm{VO}{}_2$: x − 4 = 0, so x = +4. CrO3$\mathrm{CrO}{}_3$: x − 6 = 0, so x = +6. Order: Cu2O$\mathrm{Cu}{}_2\mathrm{O}$ < Ti2O3$\mathrm{Ti}{}_2\mathrm{O}{}_3$ < VO2$\mathrm{VO}{}_2$ < CrO3$\mathrm{CrO}{}_3$. Most likely wrong order: ranking by the number of O atoms. That ignores how many metal atoms share the oxygen.

35. Compare: classify each description as belonging to the main transition metals or to the representative metals.

    Answer: Main transition metals: Show several oxidation states in their compounds, Lose electrons from two sublevels close in energy (4s and 3d), Successive ionization energies rise gradually, with no big jump · Representative metals: Usually show one oxidation state in their compounds, A big jump in ionization energy follows the loss of the outer electrons

    Representative metals, such as Na and Mg, usually show one oxidation state. Once their outer electrons go, the next electron must leave a stable inner level. So a big jump in ionization energy appears. Main transition metals lose 4s and 3d electrons in sequence. These sublevels are close in energy, so the energies rise gradually. That is why they show several oxidation states.

36. Consider the following statements: (I) An element can be a transition element even if one of its ions has an empty d sublevel. (II) Every element of the d-block is a transition element. (III) An element whose atom has a completely filled d sublevel can still be a transition element. Which of these statements is/are correct?

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

    Answer: I and III only

    (I) Correct. Scandium's atom is 3d¹$3d^1$, so it is a transition element, though Sc³⁺$\mathrm{Sc}{}^{3+}$ is 3d⁰$3d^0$. (II) Wrong. Zn, Cd and Hg are in the d-block, but they are d¹⁰$d^{10}$ in the atom and the ion. (III) Correct. Copper's atom is 3d¹⁰$3d^{10}$, but Cu²⁺$\mathrm{Cu}{}^{2+}$ is 3d⁹$3d^9$, which is incomplete. The definition checks the atom or one of its oxidation states. Answer: I and III only.

37. Classify each element, using the configurations given, as a transition element or not a transition element.

    Answer: Transition element: ₂₈${}_{28}$Ni: atom [Ar] 4s²\,3d⁸$4s^{2}3d^8$, ₅₇${}_{57}$La: atom [Xe] 6s²\,5d¹$6s^{2}5d^1$; La³⁺$\mathrm{La}{}^{3+}$: [Xe], ₂₉${}_{29}$Cu: atom [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$; Cu+$\mathrm{Cu}{}^{+}$: [Ar] 3d¹⁰$3d^{10}$; Cu²⁺$\mathrm{Cu}{}^{2+}$: [Ar] 3d⁹$3d^9$ · Not a transition element: ₈₀${}_{80}$Hg: atom [Xe] 4f¹⁴\,6s²\,5d¹⁰$4f^{14}6s^{2}5d^{10}$; Hg²⁺$\mathrm{Hg}{}^{2+}$: [Xe] 4f¹⁴\,5d¹⁰$4f^{14}5d^{10}$, ₄₉${}_{49}$In: atom [Kr] 4d¹⁰\,5s²\,5p¹$4d^{10}5s^{2}5p^1$; In³⁺$\mathrm{In}{}^{3+}$: [Kr] 4d¹⁰$4d^{10}$

    Apply two tests: the atom first, then the oxidation states. Ni: the atom is 3d⁸$3d^8$, incomplete, so it is a transition element. La: the atom is 5d¹$5d^1$, incomplete, so it is one, though La³⁺$\mathrm{La}{}^{3+}$ has no d electrons. Cu: the atom is 3d¹⁰$3d^{10}$, but Cu²⁺$\mathrm{Cu}{}^{2+}$ is 3d⁹$3d^9$, incomplete, so it is one. Hg: 5d¹⁰$5d^{10}$ in the atom and in Hg²⁺$\mathrm{Hg}{}^{2+}$, so it is not. In: 4d¹⁰$4d^{10}$ in the atom and in In³⁺$\mathrm{In}{}^{3+}$, so it is not.

38. Consider the following statements about atoms of the first transition series: (I) In the ₂₅${}_{25}$Mn atom, each of the five 3d orbitals contains one electron. (II) The ₂₆${}_{26}$Fe atom has more unpaired 3d electrons than the ₂₅${}_{25}$Mn atom. (III) The ₂₈${}_{28}$Ni atom has two unpaired electrons in its 3d sublevel. Which of these statements is/are correct?

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

    Answer: I and III only

    Write the configuration and draw the 3d boxes for each atom. Mn: [Ar] 4s² 3d⁵ → [↑] [↑] [↑] [↑] [↑] = 5 unpaired, so (I) is correct. Fe: [Ar] 4s² 3d⁶ → [↑↓] [↑] [↑] [↑] [↑] = 4 unpaired, which is fewer than Mn, so (II) is wrong. Ni: [Ar] 4s² 3d⁸ → [↑↓] [↑↓] [↑↓] [↑] [↑] = 2 unpaired, so (III) is correct. Answer: I and III only. The idea (from the book): the 3d orbitals are filled by a single electron in each orbital till manganese (3d⁵$3d^5$), and after manganese pairing takes place till zinc (3d¹⁰$3d^{10}$) (Hund's rule). So the number of unpaired electrons rises up to Mn and then falls.

39. Atom	Student's configuration
    ₂₃${}_{23}$V	[Ar] 4s²\,3d³$4s^{2}3d^3$
    ₂₄${}_{24}$Cr	[Ar] 4s²\,3d⁴$4s^{2}3d^4$
    ₂₅${}_{25}$Mn	[Ar] 4s²\,3d⁵$4s^{2}3d^5$
    ₂₉${}_{29}$Cu	[Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$

    A student wrote the configurations shown in the table. Which one is wrong?

    • B. ₂₄${}_{24}$Cr
    • A. ₂₃${}_{23}$V
    • D. ₂₉${}_{29}$Cu
    • C. ₂₅${}_{25}$Mn

    Answer: ₂₄${}_{24}$Cr

    Check each atom against the book. V: [Ar] 4s²\,3d³$4s^{2}3d^3$ follows the pattern, so it is correct. Mn: [Ar] 4s²\,3d⁵$4s^{2}3d^5$ already has a half-filled 3d, so it is correct. Cu: [Ar] 4s¹\,3d¹⁰$4s^{1}3d^{10}$ is the book's abnormal configuration, so it is correct. Cr is the error: the book writes [Ar] 4s¹\,3d⁵$4s^{1}3d^5$. In the first series, only Cr and Cu break the pattern.

40. Give reasons: Most elements of the first transition series show the oxidation state (+2).

    Answer: Key points: (1) When the atom is oxidized, it loses its 4s electrons first, because 4s is in the outermost level (n = 4). (2) Losing the two 4s electrons gives the oxidation state (+2). (3) Cr and Cu (4s¹) lose their 4s electron and one 3d electron. (4) Scandium is the exception: it shows only (+3). (5) In higher oxidation states, 3d electrons are lost in sequence.

    Most atoms of the series end with 4s²$4s^2$. The 4s electrons are in the outermost level (n = 4). So when the atom is oxidized, they are lost first. Losing these two electrons gives the oxidation state (+2). Example: Mn [Ar] 4s²\,3d⁵$4s^{2}3d^5$ → Mn²⁺$\mathrm{Mn}{}^{2+}$ [Ar] 3d⁵$3d^5$. Cr and Cu have only 4s¹$4s^1$, so they also lose one 3d electron. Scandium is the exception: the book gives it only (+3). Key words: lose the 4s electrons first, oxidation state (+2), except scandium. Most likely wrong answer: 'they lose two 3d electrons'. Apart from Cr and Cu, 3d electrons leave only in the higher states.

41. In which pair of compounds do the metal ions have the same number of unpaired electrons? (₂₅${}_{25}$Mn, ₂₆${}_{26}$Fe)

    • D. MnO2$\mathrm{MnO}{}_2$ and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$
    • A. MnSO4$\mathrm{MnSO}{}_4$ and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$
    • B. FeO$\mathrm{FeO}$ and MnO$\mathrm{MnO}$
    • C. Mn2O3$\mathrm{Mn}{}_2\mathrm{O}{}_3$ and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$

    Answer: MnSO4$\mathrm{MnSO}{}_4$ and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$

    Step 1: oxidation states. MnSO4$\mathrm{MnSO}{}_4$: Mn +2. Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$: Fe +3. FeO$\mathrm{FeO}$: Fe +2. MnO$\mathrm{MnO}$: Mn +2. Mn2O3$\mathrm{Mn}{}_2\mathrm{O}{}_3$: Mn +3. MnO2$\mathrm{MnO}{}_2$: Mn +4. Step 2: the ions, with 4s lost first. Mn²⁺$\mathrm{Mn}{}^{2+}$ 3d⁵$3d^5$, Fe³⁺$\mathrm{Fe}{}^{3+}$ 3d⁵$3d^5$, Fe²⁺$\mathrm{Fe}{}^{2+}$ 3d⁶$3d^6$, Mn³⁺$\mathrm{Mn}{}^{3+}$ 3d⁴$3d^4$, Mn⁴⁺$\mathrm{Mn}{}^{4+}$ 3d³$3d^3$. Step 3: Hund's rule. 3d⁵$3d^5$ → 5, 3d⁶$3d^6$ → 4, 3d⁴$3d^4$ → 4, 3d³$3d^3$ → 3. Only MnSO4$\mathrm{MnSO}{}_4$ and Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ match, with 5 each. Both ions have the half-filled 3d⁵$3d^5$ sublevel.

42. Give reasons: Manganese (₂₅${}_{25}$Mn) reaches the oxidation state (+7), while zinc (₃₀${}_{30}$Zn) shows only (+2).

    Answer: Key points: (1) Mn: [Ar] 4s² 3d⁵; it can lose all seven 4s and 3d electrons, giving (+7). (2) Mn is in group VIIB, and the maximum oxidation state does not exceed the group number. (3) Zn: [Ar] 4s² 3d¹⁰; it loses only its two 4s electrons, giving Zn²⁺ [Ar] 3d¹⁰ (filled 3d). (4) Zn is in group IIB, so its maximum is (+2). (5) Oxidation states increase from Sc to Mn (+7), then decrease to (+2) in Zn.

    Mn is [Ar] 4s²\,3d⁵$4s^{2}3d^5$. Losing all seven 4s and 3d electrons gives (+7), equal to its group number (VIIB). Zn is [Ar] 4s²\,3d¹⁰$4s^{2}3d^{10}$. It loses only its two 4s electrons. Zn²⁺$\mathrm{Zn}{}^{2+}$ is [Ar] 3d¹⁰$3d^{10}$, with a filled 3d. Zinc is in group IIB, and the maximum does not exceed the group number. Key words: oxidation states increase from scandium to manganese (+7), then decrease to (+2) in zinc. Most likely wrong answer: 'Zn has more electrons, so it reaches a higher state'. Zinc's 3d electrons are not lost.

43. Element X lies in the second transition series. Its only ion, X³⁺$\mathrm{X}{}^{3+}$, has the configuration [Kr]. Which statement is correct?

    • B. X is yttrium (₃₉${}_{39}$Y), not a transition element: X³⁺$\mathrm{X}{}^{3+}$ has no d electrons
    • A. X is yttrium (₃₉${}_{39}$Y), a transition element: its atom is [Kr] 5s²\,4d¹$5s^{2}4d^1$
    • D. X is yttrium (₃₉${}_{39}$Y), not a transition element: it shows only one oxidation state
    • C. X is strontium (₃₈${}_{38}$Sr), not a transition element: its atom has no d electrons

    Answer: X is yttrium (₃₉${}_{39}$Y), a transition element: its atom is [Kr] 5s²\,4d¹$5s^{2}4d^1$

    Step 1: [Kr] holds 36 electrons. X³⁺$\mathrm{X}{}^{3+}$ lost 3, so Z = 36 + 3 = 39: yttrium. Step 2: the Y atom is [Kr] 5s²\,4d¹$5s^{2}4d^1$, the first element of the second series. Its 4d holds one electron, so it is incomplete. So Y is a transition element, though Y³⁺$\mathrm{Y}{}^{3+}$ has no d electrons. One pass, in the atom or in a state, is enough. Scandium is the same case in the first series.

44. Arrange the following atoms in ascending order according to the total number of unpaired electrons in the atom (4s and 3d together): ₂₂${}_{22}$Ti, ₂₄${}_{24}$Cr, ₂₅${}_{25}$Mn, ₂₇${}_{27}$Co, ₂₉${}_{29}$Cu

    Answer: ₂₉${}_{29}$Cu → ₂₂${}_{22}$Ti → ₂₇${}_{27}$Co → ₂₅${}_{25}$Mn → ₂₄${}_{24}$Cr

    Write each configuration (book table) and count unpaired electrons in 4s and 3d, using Hund's rule in 3d. Ti: [Ar] 4s² 3d² → 4s paired, 3d [↑] [↑] [ ] [ ] [ ] → 2. Cr: [Ar] 4s¹ 3d⁵ → 1 in 4s + 5 in 3d → 6. Mn: [Ar] 4s² 3d⁵ → 0 + 5 → 5. Co: [Ar] 4s² 3d⁷ → 0 + 3 → 3. Cu: [Ar] 4s¹ 3d¹⁰ → 1 + 0 → 1. Ascending order: Cu (1) < Ti (2) < Co (3) < Mn (5) < Cr (6). Most likely wrong answers: (1) ordering by atomic number (Ti < Cr < Mn < Co < Cu), from the idea "more electrons = more unpaired electrons"; after Mn every extra 3d electron pairs up, so the number falls. (2) Writing Cr as [Ar] 4s² 3d⁴ and Cu as [Ar] 4s² 3d⁹ (forgetting the anomaly of chromium and copper), which gives Cr 4 and Cu 1 and puts Cr below Mn. (3) Counting only the 3d electrons, which makes Cr equal to Mn (5 each) and gives Cu 0. Key words: Hund's rule, single electron in each orbital, pairing, half-filled 4s in Cr and Cu.

45. In the atoms of the first transition series, pairing of electrons in the 3d orbitals starts with the element ...

    • C. ₂₆${}_{26}$Fe
    • D. ₃₀${}_{30}$Zn
    • A. ₂₃${}_{23}$V
    • B. ₂₅${}_{25}$Mn

    Answer: ₂₆${}_{26}$Fe

    The book (Ch.1, point 1 after the configuration table): after calcium ([Ar] 4s²), the five 3d orbitals are filled by a single electron in each orbital in sequence till manganese (3d⁵$3d^5$). After manganese, pairing of electrons takes place in each orbital till zinc (3d¹⁰$3d^{10}$) (Hund's rule). So the first atom with a paired 3d orbital is iron, ₂₆${}_{26}$Fe: [Ar] 4s² 3d⁶ → [↑↓] [↑] [↑] [↑] [↑]. Watch out: the d sublevel has 5 orbitals, so it is half-filled at 3d⁵$3d^5$, not 3d³$3d^3$ (that is the p sublevel, p³$p^3$).

46. Choose the correct answer: The first transition series lies in the fourth period. From scandium to zinc, its elements fill ...

    • B. the 4d sublevel successively, because the series is in the fourth period
    • C. the 4s sublevel successively, because 4s is the outermost sublevel
    • A. the 3d sublevel successively, because 3d comes after 4s in the filling order
    • D. the 3d sublevel successively, because 3d fills before 4s

    Answer: the 3d sublevel successively, because 3d comes after 4s in the filling order

    In period n, the (n − 1)d sublevel is filled. So period 4 fills 3d. In the filling order, 4s fills first, at potassium and calcium. Then 3d fills, from scandium (3d¹$3d^1$) to zinc (3d¹⁰$3d^{10}$). Key words: the (3d) sublevel is filled successively, in the fourth period after calcium. Most likely wrong answer: 4d, matching the sublevel to the period number.

47. Choose the correct answer: Potassium permanganate (KMnO4$\mathrm{KMnO}{}_4$) is used as ..., and the oxidation state of manganese in it is ... (K = +1, O = −2)

    • D. a strong oxidizing agent in the dry cell ... +4
    • A. an antiseptic substance ... +7
    • B. a fungicide ... +7
    • C. an antiseptic substance ... +8

    Answer: an antiseptic substance ... +7

    Use: the book lists KMnO4$\mathrm{KMnO}{}_4$ as an antiseptic substance. MnSO4$\mathrm{MnSO}{}_4$ is the fungicide, and MnO2$\mathrm{MnO}{}_2$ is the oxidizing agent in the dry cell. Oxidation state: +1 + x + 4(−2) = 0, so x = +7. This is manganese's highest oxidation state. It equals its group number (VIIB).