Periodic Trends

Q1. (2 marks)

• Effective nuclear charge = the net positive charge actually experienced by an electron in a multi-electron atom, after accounting for shielding by other electrons. (1)
• Relation: $Z_{\text{eff}} = Z - S$, where $Z$ = nuclear charge and $S$ = screening (shielding) constant. (1)

Q2. (5 marks) Sulfur: $1s^2\,2s^2\,2p^6\,3s^2\,3p^4$. For a $3p$ electron, group as $(1s)(2s,2p)(3s,3p)$.

Contributions to $S$ for an electron in the $(3s,3p)$ group:

• Same group (n=3): 7 other electrons × 0.35 = 2.45 (1)
• $n-1$ shell (2s,2p): 8 electrons × 0.85 = 6.80 (1)
• Inner shells (1s): 2 electrons × 1.00 = 2.00 (1)

$S = 2.45 + 6.80 + 2.00 = 11.25$ (1)

$Z_{\text{eff}} = 16 - 11.25 = \mathbf{4.75}$ (1)

Q3. (4 marks)

• Covalent radius: half the distance between nuclei of two identical atoms joined by a single covalent bond. (1)
• Metallic radius: half the distance between nuclei of two adjacent atoms in a metallic crystal lattice. (1)
• Van der Waals radius: half the distance between nuclei of two non-bonded atoms of adjacent molecules in the solid state. (1)
• The van der Waals radius is generally the largest (non-bonded contact, no overlap). (1)

Q4. (4 marks) (a) Cation smaller than parent: loss of electron(s) reduces electron–electron repulsion; often a whole shell is removed, and the same nuclear charge acts on fewer electrons → higher $Z_{\text{eff}}$ per electron → contraction. (2) (b) Anion larger than parent: gain of electron(s) increases electron–electron repulsion while nuclear charge stays constant; $Z_{\text{eff}}$ per electron decreases → electron cloud expands. (2)

Q5. (4 marks) All species have 10 electrons (isoelectronic). Radius decreases as nuclear charge increases. (1) Nuclear charges: $\text{N}^{3-}(7),\ \text{O}^{2-}(8),\ \text{F}^-(9),\ \text{Na}^+(11),\ \text{Mg}^{2+}(12)$. Increasing radius order: (2) $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-} < \text{N}^{3-}$ Justification: greater nuclear charge on the same 10 electrons → stronger pull → smaller radius. (1)

Q6. (5 marks)

• First ionization energy: the minimum energy required to remove the most loosely bound (outermost) electron from one mole of gaseous atoms in their ground state. (2)
• Be: $1s^2 2s^2$; B: $1s^2 2s^2 2p^1$. (1)
• In boron the electron removed is a $2p$ electron, which is higher in energy and better shielded by the $2s$ electrons than a $2s$ electron. (1)
• A $2p$ electron is more easily removed than a $2s$ electron, so $IE_1(\text{B}) < IE_1(\text{Be})$ despite the higher $Z$. (1)

Q7. (4 marks)

• Fluorine is a very small atom, so its $2p$ subshell is compact. (1)
• Incoming electron enters a small, electron-dense $2p$ shell → strong electron–electron repulsion reduces the energy released. (2)
• Chlorine ($3p$) is larger with less repulsion, so more energy is released → $\Delta_{eg}H(\text{Cl})$ is more negative. (1)

Q8. (4 marks)

• Mulliken electronegativity: $\chi_M = \dfrac{IE + EA}{2}$ (2)
• $\chi_M = \dfrac{13.6 + 3.6}{2} = \dfrac{17.2}{2} = \mathbf{8.6\ \text{eV}}$ (2)

Q9. (4 marks)

• Diagonal relationship: certain elements of the 2nd period resemble the element diagonally placed to the lower right in the 3rd period (e.g. Li–Mg, Be–Al, B–Si), due to similar charge/size ratio (ionic potential) and comparable electronegativity/polarizing power. (2)
• Evidence for Li–Mg (any one): (2)
  • Both form covalent, water-soluble but volatile compounds; e.g. both nitrides ($\text{Li}_3\text{N}$, $\text{Mg}_3\text{N}_2$) form directly with N₂.
  • Both carbonates decompose on heating to give the oxide.
  • Both form normal oxides (Li₂O, MgO) rather than peroxides/superoxides.

Q10. (4 marks) (a) Across a period (L→R): metallic character decreases; reason — $Z_{\text{eff}}$ increases, atoms hold electrons more tightly / higher IE, less tendency to lose electrons. (2) (b) Down a group: metallic character increases; reason — atomic size increases and IE decreases, so electrons are lost more easily. (2)

[
  {"claim": "Z_eff for 3p electron in sulfur is 4.75 by Slater's rules", "code": "S = 7*0.35 + 8*0.85 + 2*1.00; Zeff = 16 - S; result = (Zeff == Rational('4.75'))"},
  {"claim": "Slater screening constant S for S 3p equals 11.25", "code": "S = 7*Rational(35,100) + 8*Rational(85,100) + 2*1; result = (S == Rational('11.25'))"},
  {"claim": "Mulliken electronegativity for IE=13.6, EA=3.6 eV is 8.6 eV", "code": "chi = (Rational('13.6') + Rational('3.6'))/2; result = (chi == Rational('8.6'))"},
  {"claim": "Isoelectronic radius ordering follows decreasing nuclear charge: Mg2+ smallest, N3- largest", "code": "Z = {'Mg2+':12,'Na+':11,'F-':9,'O2-':8,'N3-':7}; order = sorted(Z, key=lambda k: -Z[k]); result = (order == ['Mg2+','Na+','F-','O2-','N3-'])"}
]