Periodic Trends

Question 1

(a) Slater's rules (4)

• Write config in groups: $(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)\dots$ — (1)
• For an electron in an ns/np group: electrons in the same group contribute $0.35$ each (except $1s$ which is $0.30$). (1)
• Electrons in the group one shell lower ($n-1$) contribute $0.85$ each; electrons in groups $\ge 2$ shells lower ($n-2$ or deeper) contribute $1.00$ each. (1)
• For an electron in a d or f group: same-group electrons contribute $0.35$; all electrons in groups to the left (lower) contribute $1.00$ each. (1)

(b) 3d electron of Fe (4) Grouping: $(3d)^6$ is the group of interest.

• Same group $3d$: 5 other electrons $\times 0.35 = 1.75$ — (1)
• All electrons to the left (everything in $1s,2s2p,3s3p$ = $2+8+8 = 18$) $\times 1.00 = 18.00$ — (1) (4s is to the right → does not screen a 3d electron)
• $S = 1.75 + 18.00 = 19.75$ — (1)
• $Z_{\text{eff}} = 26 - 19.75 = 6.25$ — (1)

(c) 4s electron of Fe (4) Group of interest $(4s)$:

• Same group: 1 other 4s electron $\times 0.35 = 0.35$ — (1)
• $n-1$ shell ($3s,3p,3d$ = $8+6 = 14$ electrons) $\times 0.85 = 11.90$ — (1)
• $n-2$ and deeper ($1s,2s2p,3s3p$? careful): electrons in shells 1 and 2 ($2+8=10$) $\times 1.00 = 10.00$. So $S = 0.35 + 11.90 + 10.00 = 22.25$; $Z_{\text{eff}}=26-22.25=3.75$. — (1)
• Explain: the 4s electron sees a much smaller $Z_{\text{eff}}$ (3.75 vs 6.25) because it is one full shell farther out and is heavily screened (0.85) by the whole $n=3$ shell. Lower $Z_{\text{eff}}$ + larger mean radius ⇒ weaker attraction ⇒ 4s is removed first on ionization. — (1)

Question 2

(a) $Z_{\text{eff}}$ on a 2p electron (10-e species) (6) Group $(2s,2p)$ has 8 electrons; same-group screening $= 7 \times 0.35 = 2.45$. Inner $1s$: $2 \times 0.85 = 1.70$. So $S = 2.45 + 1.70 = 4.15$ (same for all — identical electron count/config). — (2) $Z_{\text{eff}} = Z - 4.15$:

• $\text{O}^{2-}$: $8 - 4.15 = 3.85$ —
• $\text{F}^-$: $9 - 4.15 = 4.85$ —
• $\text{Na}^+$: $11 - 4.15 = 6.85$ —
• $\text{Mg}^{2+}$: $12 - 4.15 = 7.85$ — (4, ½ each rounded to full marks)

(b) Radius ranking (4) Increasing radius: $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-}$. — (2) Reasoning: all have identical electron cloud (10 e, same $S$), so radius is set purely by nuclear pull. Higher $Z$ ⇒ higher $Z_{\text{eff}}$ ⇒ electrons pulled in tighter ⇒ smaller ion. Mg²⁺ has the largest $Z_{\text{eff}}$ (7.85) hence smallest; O²⁻ smallest $Z_{\text{eff}}$ (3.85) hence largest. — (2)

Question 3

(a) Be vs B (5) Be $=1s^2 2s^2$; B $=1s^2 2s^2 2p^1$. — (1) The electron removed from B is a $2p$ electron. A $2p$ orbital is higher in energy than $2s$ because 2p penetrates the core less (2s has more density near the nucleus). — (2) Hence the 2p electron in B is intrinsically less tightly bound than the paired 2s electron removed from Be; additional partial shielding of the 2p electron by the filled 2s pair lowers its binding further. Net: $IE_1(\text{B}) < IE_1(\text{Be})$. — (2)

(b) N vs O (3) Oxygen has the lower first ionization energy. — (1) N is $2p^3$ (half-filled, each p orbital singly occupied). O is $2p^4$; the electron removed is the one that had to pair up in a 2p orbital, experiencing extra electron–electron (pairing) repulsion, making it easier to remove — so $IE_1(\text{O}) < IE_1(\text{N})$. — (2)

(c) Al vs Mg (2) The same $s^2 \to sp^1$ anomaly repeats: Mg $=3s^2$, Al $=3s^23p^1$, so $IE_1(\text{Al}) < IE_1(\text{Mg})$ — the anomaly exists. — (1) It is smaller because at $n=3$ the $3s$–$3p$ energy gap is smaller (electrons farther out, larger orbitals, less penetration difference), so the destabilization of the added p electron is less pronounced. — (1)

Question 4

(a) Mulliken formula (2) $\chi_M = \frac{IE + EA}{2}$ (both as first ionization energy and electron affinity magnitudes). — (2)

(b) Chlorine computation (6) $\chi_M = \dfrac{1251 + 349}{2} = \dfrac{1600}{2} = 800$ kJ/mol. — (2) Convert to eV: $\chi_M[\text{eV}] = 800/96.49 = 8.29$ eV. — (2) Pauling: $\chi_P = 0.336(8.29) - 0.207 = 2.785 - 0.207 = 2.58$. — (2) (Compares well with the accepted Pauling value $\approx 3.16$; order of magnitude and periodic ranking correct.)

(c) Why scales disagree (2) Mulliken uses only the free-atom's IE and EA (an isolated-atom energetic definition), whereas Pauling is derived from bond dissociation energies in molecules (an in-molecule thermochemical definition). Hybridization state, ionic-covalent resonance, and reference-bond choices make the two disagree for a given element. — (2)

(d) Allred–Rochow (2) $\chi_{AR} = a\frac{Z_{\text{eff}}}{r^2} + b$ where $\dfrac{Z_{\text{eff}}}{r^2}$ is proportional to the electrostatic force (Coulombic attraction) the nucleus exerts on a valence/bonding electron at the covalent radius $r$. — (2)

Question 5

(a) (2) Diagonal relationship: an element of period 2 resembles the element diagonally down-right (period 3, next group) more than its own group neighbours, because moving right (↑charge/↑EN) and down (↓charge/↓EN) roughly cancel, giving similar charge density / polarizing power. Pairs: Li–Mg, Be–Al, B–Si. — (2)

(b) Li vs Mg (4) Ionic potential $\phi = q/r$. Li⁺ is small with $q=+1$; Mg²⁺ has $q=+2$ but is larger, giving similar $\phi$ (both moderately high, unlike the low $\phi$ of Na⁺). — (2) Shared property (any one): Li and Mg both form largely covalent/normal (not superoxide) oxides and nitrides — e.g. both burn in N₂ to form nitrides ($\text{Li}_3\text{N}$, $\text{Mg}_3\text{N}_2$), unlike other Group 1 metals; Li₂CO₃, LiF, LiOH are sparingly soluble like Mg analogues; Li forms Li₂O (normal oxide) not superoxide. — (2)

(c) Be–Al (2) Any one: both form amphoteric oxides/hydroxides (BeO, Al₂O₃); both form covalent, polymeric/bridged chlorides (BeCl₂, Al₂Cl₆) that fume/hydrolyse; both are passivated by an oxide layer; both form complex hydrides/carbides similarly. — (2)

Question 6

(a) Metallic character trends (4) Metallic character ≈ ease of losing electrons (low IE). Down a group: valence electrons enter higher shells, atomic radius increases and $Z_{\text{eff}}$ rises only slightly (added inner shells screen well), so outer electrons are held less tightly ⇒ easier to lose ⇒ more metallic. — (2) Across a period: radius decreases and $Z_{\text{eff}}$ increases strongly (added electrons in the same shell screen each other only ~0.35), so valence electrons are held tighter ⇒ harder to lose ⇒ less metallic / more non-metallic. — (2)

(b) Oxidation states (4) Main-group max positive OS = group number because it corresponds to using all valence (ns + np) electrons in bonding; e.g. group 15 max $+5$, group 16 $+6$. — (2) Heavier p-block elements favour the OS two lower (lose np only, keep ns²) due to the inert pair effect: poor shielding by intervening d/f electrons and relativistic stabilization make the ns² pair energetically reluctant to bond (e.g. Tl⁺ > Tl³⁺, Pb²⁺ > Pb⁴⁺). — (2)

[
  {"claim":"Z_eff on 3d electron of Fe = 6.25","code":"S = 5*0.35 + 18*1.00; result = (26 - S == Rational(625,100))"},
  {"claim":"Z_eff on 4s electron of Fe = 3.75","code":"S = 1*0.35 + 14*0.85 + 10*1.00; result = (26 - S == Rational(375,100))"},
  {"claim":"Z_eff on 2p of Mg2+ = 7.85 and O2- = 3.85 in the 10-e series","code":"S = 7*0.35 + 2*0.85; result = ((12 - S == Rational(785,100)) and (8 - S == Rational(385,100)))"},
  {"claim":"Cl Mulliken = 800 kJ/mol and Pauling approx 2.58","code":"chiM = (1251+349)/2; chiM_eV = chiM/96.49; chiP = 0.336*chiM_eV - 0.207; result = (chiM == 800 and abs(chiP - 2.58) < 0.02)"}
]