Structure 3.1. The periodic table: Classification of elements

Which of the following properties would increase from the atoms sodium to chlorine across period 3?

I. Electronegativity. 

II. Nuclear charge. 

III. Atomic radius. 

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III. 

Answer: A. I and II only.

The first four ionisation energies of beryllium and iron are shown.

One common property of transition elements is that they have variable oxidation states. Discuss, referring to the graph, why iron, but not beryllium, displays this characteristic.

• From the graph, Be shows a huge jump after the 2nd ionization (IE₃ is ~1.7×10⁴  kJ mol⁻¹). This means that after Be loses two 2s electrons it reaches a stable  noble-gas core (1s²); removing a third electron would be from the core and is  energetically prohibitive. Hence Be is essentially fixed at +2. 

• Fe shows no large jump through the first 3–4 ionizations; the successive IEs  rise only gradually. This shows that several outer electrons are of similar energy (the 4s and 3d electrons) and can be removed or used in bonding. 

• Because these 4s/3d electrons are available to different extents, iron can form  multiple oxidation states (e.g. +2, +3, etc.), whereas Be cannot.

Potassium manganate(VII) is purple in colour. In which region of the visible spectrum does it mainly absorb?

A. Red. 

B. Purple. 

C. Blue. 

D. Green.

Answer: D. Green. 

Potassium manganate(VII), KMnO₄, appears deep purple because it transmits purple light. 

The colour that a substance appears is the complementary colour of the light it  absorbs. 

   ● Purple (violet) is complementary to yellow-green (≈ 520–560 nm region).

   ● Therefore, KMnO₄ mainly absorbs green light. 

Which combination is correct for the complex ion in the coordination compound  [Co(NH3)4(H2O)Br]Cl? 

Answer: C. 

Step 1: Find oxidation state of Co 

Let oxidation state of Co = x 

x + (4 × 0) + (0) + (−1) = +1 

(since the compound has 1 Cl⁻ outside the coordination sphere) 

x − 1 = +1 ⇒ x = +2 

→ Oxidation state of Co = +2 

Step 2: Determine the shape 

Cobalt(II) forms six-coordinate complexes with ligands like NH₃, H₂O, and halides. → Coordination number = 6

→ octahedral shape 

Step 3: Overall charge of the complex ion 

Inside the brackets: [Co(NH₃)₄(H₂O)Br]⁺ 

Because the external Cl⁻ balances a +1 charge on the complex.

The following shows a series of reactions involving titanium compounds.

Concentrated HCl 

a. Suggest the type of reaction for II and III. 

b. Explain why a solution of [TiCl₆]²⁻ is colourless but [TiCl₆]³⁻ is orange. The formula relating energy gap between d orbitals, ΔE, and wavelength, λ, is given as  ΔE = hc / λ where h is the Planck’s constant and c is the speed of light.

c. Use Table 14 in the data booklet to deduce the colours absorbed by [Ti(H₂O)₆]³⁺ and  [TiCl₆]³⁻ and therefore identify which complex ion has the larger energy gap between  the d orbitals. 

a. 

II: a redox step – Ti is reduced from +4 in [TiCl₆]²⁻ (d⁰) to +3 in [Ti(H₂O)₆]³⁺ (d¹). III: a ligand-substitution (complex formation) step – water ligands are replaced by  chloride to give [TiCl₆]³⁻; Ti remains +3. 

b. Colour in these Ti complexes comes from d–d transitions between split d-orbitals. [TiCl₆]²⁻ has Ti(IV), d⁰ → no d electrons to promote → colourless. [TiCl₆]³⁻ has Ti(III), d¹ → an allowed d–d transition absorbs visible light, and the  transmitted light appears orange. 

c. A solution looks the complementary colour to what it absorbs. 

[Ti(H₂O)₆]³⁺ is violet, so it absorbs yellow/green light (≈ 560–580 nm). [TiCl₆]³⁻ is orange, so it absorbs blue light (≈ 450–480 nm). 

Since ΔE = `frac{hc}{lambda}`, the shorter wavelength (blue) corresponds to a larger ΔE.

Therefore, [TiCl₆]³⁻ has the larger d-orbital splitting (ΔE) compared with [Ti(H₂O)₆]³⁺. 

The electron configuration of copper makes it a useful metal. Explain why a copper(II)  solution is blue, using section 17 of the data booklet. 

A copper(II) ion has an electron configuration of [Ar]3d⁹. In aqueous solution, the  ligands (usually water molecules) form a complex ion with Cu²⁺, causing the 3d orbitals  to split into two energy levels due to ligand field splitting. When visible light passes  through the solution, electrons are promoted from the lower to the higher d-orbital by  absorbing certain wavelengths of light. The remaining unabsorbed light is transmitted  as blue, giving the solution its characteristic color.

Explain, in terms of nuclear charge, electron subshells and the shielding provided by  filled electron shells, why the first ionization energy increases from Li to Be, but  decreases from Be to B. 

• From Li to Be, the first ionization energy increases because the nuclear charge  increases (more protons) while the electrons are being added to the same shell. The shielding effect remains approximately the same, so the increased nuclear  charge pulls the electrons closer, making them harder to remove. 

• From Be to B, the first ionization energy decreases because the electron  removed from boron is in a 2p subshell, while in beryllium it is from a 2s  subshell. 

The 2p electron is at a higher energy level and is shielded slightly by the filled  2s subshell, making it easier to remove despite the increased nuclear charge. 

The electronegativities of four different elements are given below (the letters are not  their chemical symbols). 

Based on this information which statement is correct? 

A. W is a non-metal. 

B. W and X form an ionic compound. 

C. Y is a metal. 

D. Y and Z form a covalent compound. 

Answer: D. Y and Z form a covalent compound. 

A. Incorrect: W = 0.9 → very low → metal, not nonmetal. 

B. Incorrect: Electronegativity difference = 1.2 – 0.9 = 0.3, which is very small. → This would be metallic bonding, not ionic. 

C. Incorrect: Y = 3.4 → high electronegativity → nonmetal (similar to oxygen).

D. Correct: Electronegativity difference = 4.0 – 3.4 = 0.6, which is small, indicating  covalent bonding between two nonmetals. 

In which reaction does chromium undergo a change in oxidation state?

A. Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O. 

B. Cr₂(SO₄)₃ + 6NaOH → 2Cr(OH)₃ + 3Na₂SO₄. 

C. 2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O. 

D. Na₂Cr₂O₇ + 4H₂SO₄ + 6HCl → Cr₂(SO₄)₃ + Na₂SO₄ + 7H₂O + 3Cl₂. 

Answer: D. Na₂Cr₂O₇ + 4H₂SO₄ + 6HCl → Cr₂(SO₄)₃ + Na₂SO₄ + 7H₂O + 3Cl₂. 

A. Incorrect:  

Cr₂O₃ + 6HCl → 2CrCl₃ + 3H₂O 

   • Cr in Cr₂O₃: O = –2 → total O = –6 → Cr₂ = +6 → each Cr = +3

   • Cr in CrCl₃: Cl = –1 → Cr = +3 

→ No oxidation state change 

B. Incorrect:  

Cr₂(SO₄)₃ + 6NaOH → 2Cr(OH)₃ + 3Na₂SO₄ 

   • Cr in Cr₂(SO₄)₃: SO₄ = –2 → Cr = +3 

   • Cr in Cr(OH)₃: OH = –1 → Cr = +3 

→ No oxidation state change 

C. Incorrect: 

2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O 

   • In both Na₂CrO₄ and Na₂Cr₂O₇, Cr = +6 

→ No oxidation state change 

D. Correct:  

Na₂Cr₂O₇ + 4H₂SO₄ + 6HCl → Cr₂(SO₄)₃ + Na₂SO₄ + 7H₂O + 3Cl₂.

    • Cr in Na₂Cr₂O₇ = +6 

   • Cr in Cr₂(SO₄)₃ = +3 

Change: +6 → +3 → chromium is reduced 

Also, Cl⁻ → Cl₂ (oxidized from –1 → 0)

What are the oxidation states of chlorine in the oxyacids HOCl, HClO₃, and HClO₄?

A. −1, +5 and +7. 

B. −1, −5 and +7. 

C. +1, +3 and +4.

D. +1, +5 and +7. 

Answer: D. +1, +5 and +7. 

1. HOCl (hypochlorous acid) 

Formula: H + Cl + O = 0 

(+1) + (x) + (–2) = 0 

x = +1 

→ Oxidation state of Cl = +1 

2. HClO₃ (chloric acid) 

Formula: H + Cl + 3O = 0 

(+1) + (x) + 3(–2) = 0 

x = +5 

→ Oxidation state of Cl = +5 

3. HClO₄ (perchloric acid)

Formula: H + Cl + 4O = 0 

(+1) + (x) + 4(–2) = 0 

x = +7 

→ Oxidation state of Cl = +7 

Which compound is not a product of the reaction between an oxide of a period 3 element and water? 

A. NaOH 

B. Al(OH)₃ 

C. H₂SO₃ 

D. H₃PO₄ 

Answer: C. H₂SO₃ 

A. Incorrect: NaOH (Correct product) 

   • Oxide: Sodium oxide (Na₂O) 

   • Reaction: Na₂O + H₂O → 2NaOH 

   • Type: Basic oxide of a metal. 

→ NaOH is indeed formed by a period 3 oxide reacting with water.

B. Incorrect: Al(OH)₃ (Correct product) 

   • Oxide: Aluminium oxide (Al₂O₃)

   • Aluminium oxide is amphoteric, and when it reacts with water (under certain  conditions), it can form Al(OH)₃. 

→ Al(OH)₃ can be considered a product derived from a period 3 oxide.

C. Correct: H₂SO₃ (Not a product) 

   • Oxide: Sulfur dioxide (SO₂) 

   • Reaction: SO₂ + H₂O ⇌ H₂SO₃ 

   • However, H₂SO₃ (sulfurous acid) is unstable and does not exist in a pure state— it decomposes back into SO₂ and water. 

   • The stable acid from a sulfur oxide is actually H₂SO₄, formed from SO₃ + H₂O  → H₂SO₄. 

→ H₂SO₃ is not a stable or typical product of period 3 oxide + water.

D. Incorrect: H₃PO₄ (Correct product) 

   • Oxide: Phosphorus pentoxide (P₄O₁₀) 

   • Reaction: P₄O₁₀ + 6H₂O → 4H₃PO₄ 

→ Phosphoric acid (H₃PO₄) is a true product of a period 3 oxide reacting with water. 

Explain why Cl₂ rather than Br₂ would react more vigorously with a solution of I⁻. 

Chlorine reacts more vigorously with iodide ions than bromine because: Chlorine is a stronger oxidizing agent than bromine. This is because chlorine has a smaller atomic  radius and higher electronegativity, so it more readily gains electrons to form Cl⁻ ions. 

Which oxide, when added to water, produces the solution with the highest pH?

A. Na₂O 

B. SO₃ 

C. MgO 

D. CO₂ 

Answer: A. Na₂O 

When oxides react with water: 

Which metal is in the f-block of the periodic table? 

A. Sr. 

B. Sm. 

C. Pb. 

D. Tc. 

Answer: B. Sm. 

A. Incorrect: Sr (s-block), group 2 element, electron configuration ends in 5s².

B. Correct: Sm (f-block), lanthanide element, electron configuration ends in 4f⁶ 6s².

C. Incorrect: Pb (p-block), group 14 element, electron configuration ends in 6p².

D. Incorrect: Tc (d-block), transition metal, electron configuration ends in 4d⁵ 5s². 

Which property generally decreases across period 3? 

A. Atomic number. 

B. Electronegativity. 

C. Atomic radius. 

D. First ionization energy. 

Answer: C. Atomic radius. 

Across Period 3 (Na → Ar): 

Which property decreases down group 7 in the periodic table? 

A. Melting point. 

B. Electronegativity. 

C. Atomic radius. 

D. Ionic radius. 

Answer: B. Electronegativity. 

Group 7 (the halogens) – F, Cl, Br, I, At 

As we move down the group: 

Describe and explain the variation in the size (radius) of simple ions formed by the  elements across period 3 from sodium (Na⁺) to chloride (Cl⁻). 

Across Period 3, from sodium (Na⁺) to chloride (Cl⁻), the size of the ions decreases from  Na⁺ to Al³⁺, and then increases from P³⁻ to Cl⁻. 

   • From Na⁺ to Al³⁺: 

     o These are cations formed by the loss of electrons. 

     o The number of electron shells remains the same (all have 2 shells after  ionization). 

     o However, the nuclear charge increases from +11 (Na⁺) to +13 (Al³⁺). o The greater nuclear charge pulls the electrons closer to the nucleus,  causing the ionic radius to decrease. 

     o Hence, Na⁺ > Mg²⁺ > Al³⁺ in size. 

   • From P³⁻ to Cl⁻: 

     o These are anions formed by the gain of electrons. 

     o Each anion has the same number of electron shells (3 shells). 

     o As nuclear charge increases from P³⁻ to Cl⁻, the attraction between the  nucleus and the extra electrons increases, pulling the electrons closer. o Therefore, the ionic radius decreases from P³⁻ > S²⁻ > Cl⁻. 

   • Across the entire period (Na⁺ to Cl⁻): 

     o Cations are much smaller than anions because cations lose an electron  shell, while anions gain electrons and experience increased electron– electron repulsion. 

     o The smallest ion is Al³⁺ and the largest is P³⁻.

Which statements about the periodic table are correct? 

I. The elements Mg, Ca and Sr have similar chemical properties. 

II. Elements in the same period have the same number of main energy levels.

III. The oxides of Na, Mg and P are basic. 

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III. 

Answer: A. I and II only. 

Statement I: True 

• Mg, Ca, and Sr are all Group 2 (alkaline earth metals). 

• Elements in the same group have similar chemical properties because they have the same number of valence electrons (2). 

Statement II: True 

• Elements in the same period have the same number of occupied electron shells (main energy levels). 

• Example: Na, Mg, and Al are all in Period 3 → each has 3 energy levels.

Statement III: False 

• Na₂O → basic oxide 

• MgO → basic oxide 

• P₄O₁₀ (oxide of phosphorus) → acidic oxide 

→ Not all three are basic. 

Element X is in group 5 and period 4 of the periodic table. Which statement is correct?

A. X has 5 occupied energy levels. 

B. X can form ions with 3– charge. 

C. X is a transition element. 

D. X has 4 valence electrons.

Answer: C. X is a transition element. 

A. Incorrect: Period 4 → 4 energy levels, not 5. 

B. Incorrect: Group 5 transition metals form positive ions (cations), not negative ions.

C. Correct: Group 5 elements are transition metal with partially filled 3d orbitals.

D. Incorrect: Group 5 → has 5 valence electrons (3d³ 4s²). 

Tellurium is an element in the same group as sulfur. 

Which of the following would be the correct formula for telluric(VI) acid?

A. H₂Te 

B. H₂TeO₂ 

C. H₂TeO₄ 

D. H₂TeO₃

Answer: C. H₂TeO₄. 

Tellurium (Te) is in Group 16, the same group as sulfur (S). The oxidation state is (+6)

A. Incorrect: Te oxidation state = −2 

B. Incorrect: Te oxidation state = +2 

C. Correct: Te oxidation state = +6 

D. Incorrect: Te oxidation state = +4 

Which element is in the p-block? 

A. Pb. 

B. Pm. 

C. Pt. 

D. Pu. 

Answer: A. Pb.

How do the following properties change down group 18 of the periodic table? 

Answer: C. Ionization energy: Decreases, Ionic radius: Increases

• As atomic number increases, the atomic radius increases, outer electrons are further from the nucleus, and shielding increases. Therefore, it is easier to remove  an electron. 

• Each element down the group has an extra electron shell, making the atoms (and ions, if formed) larger. 

An element M of mass number 40 has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². Which statement regarding this element is not correct? 

A. It belongs to Group 2 of the periodic table. 

B. The nucleus of the atom has 20 neutrons. 

C. It belongs to period 4 of the periodic table.

D. The formula of its oxide is MO₂. 

Answer: D. The formula of its oxide is MO₂. 

A. Incorrect: 

This configuration ends in 4s², meaning the element is in Group 2 → Statement correct.

B. Incorrect: 

40 (mass number) − 20 (protons) = 20 neutrons → Statement correct.

C. Incorrect: 

Highest principal quantum number = 4 → Period 4 → Statement correct.

D. Correct: 

Group 2 metals form oxides of the type MO because they have a +2 oxidation state and  oxygen is −2 → Statement incorrect. 

Which one of the following ion or atom has the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵? 

A. Mn. 

B. Co²⁺. 

C. Fe³⁺. 

D. Cr²⁺. 

Answer: C. Fe³⁺. 

This configuration ends with 3d⁵, which means the 3d subshell is half-filled and there  are no 4s electrons — suggesting an ion formed by removing the 4s electrons from such  an atom. 

A. Incorrect: The neutral atom Mn has [Ar] 3d⁵ 4s², not exactly 3d⁵

B. Incorrect: [Ar] 3d⁷ 4s² → remove 2 electrons → [Ar] 3d⁷ 

C. Correct: [Ar] 3d⁶ 4s² → remove 3 electrons → [Ar] 3d⁵ 

D. Incorrect: [Ar] 3d⁵ 4s¹ → remove 2 electrons → [Ar] 3d⁴ 

Which of the following electron configurations could represent a transition metal atom?

A. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s². 

B. 1s² 2s² 2p⁶ 3s² 3p⁵. 

C. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p³. 

D. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s². 

Answer: A. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s². 

A transition metal is defined as an element that has an incomplete d sub-shell (partially  filled d orbitals) in its atom or in one of its common ions.