Periodic Trends

Question 1

(a) (3 marks) Slater groups electrons as: $(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)\ldots$ — each grouping in its own set. For an electron in $(ns,np)$:

• Electrons in the same group contribute 0.35 each (except $1s$ where it is 0.30). (1)
• Electrons in $(n-1)$ shell contribute 0.85 each. (1)
• Electrons in shells $\le (n-2)$ contribute 1.00 each. (1)

(b) (5 marks) Sulfur $Z=16$, config $(1s)^2(2s2p)^8(3s3p)^6$; removing/examining a $3p$ electron:

• same group $(3s3p)$: 5 other electrons $\times 0.35 = 1.75$ (1)
• $(n-1)=2$ shell: 8 electrons $\times 0.85 = 6.80$ (1)
• $n-2=1$ shell: 2 electrons $\times 1.00 = 2.00$ $\sigma = 1.75+6.80+2.00 = 10.55,\quad Z_{\text{eff}}=16-10.55 = 5.45$ (1)

Sodium $Z=11$, config $(1s)^2(2s2p)^8(3s)^1$; the $3s$ electron:

• same group: 0 other electrons $\to 0$
• $(n-1)=2$ shell: 8 $\times 0.85 = 6.80$
• inner: 2 $\times 1.00 = 2.00$ $\sigma = 8.80,\quad Z_{\text{eff}} = 11-8.80 = 2.20$ (2)

(c) (4 marks) $n^*=3$ for both, so $n^{*2}$ cancels in the ratio: $\frac{r_{\text{Na}}}{r_{\text{S}}} = \frac{Z_{\text{eff,S}}}{Z_{\text{eff,Na}}} = \frac{5.45}{2.20} = 2.48$ (2) Na is predicted ~2.5× larger than S. This is consistent with the sharp decrease in atomic radius across period 3: as $Z$ rises, $Z_{\text{eff}}$ rises strongly (2.20 → 5.45) while $n^*$ stays fixed, contracting the valence shell. (2)

(d) (6 marks) Algorithm: order shells by $n$; identify outer $(ns,np)$ group; sum contributions — 0.35 for co-members (minus the electron itself), 0.85 for $n-1$, 1.00 for deeper. (2 for logic)

def z_eff_slater(Z, config):
    # config: {'1s':2, '2s2p':8, '3s3p':6}, ordered by n
    labels = list(config.keys())
    outer = labels[-1]
    n_outer = int(outer[0])
    sigma = 0.0
    for lbl, occ in config.items():
        n = int(lbl[0])
        if lbl == outer:
            sigma += 0.35 * (occ - 1)     # exclude the electron itself
        elif n == n_outer - 1:
            sigma += 0.85 * occ
        elif n < n_outer - 1:
            sigma += 1.00 * occ
    return Z - sigma

(4 for correct loop & coefficients) Check: z_eff_slater(16, {'1s':2,'2s2p':8,'3s3p':6}) → $16-10.55=5.45$. ✓

Question 2

(a) (6 marks) Drop 1: Be → B (899 → 801). (1) B removes a $2p$ electron which is higher in energy and better shielded by the filled $2s^2$; the $2p$ electron is easier to remove than a $2s$ electron. (2) Drop 2: N → O (1402 → 1314). (1) N has a stable half-filled $2p^3$ (one electron per orbital, no pairing). O has $2p^4$: removing the paired electron relieves inter-electron pair repulsion, lowering $IE$. (2)

(b) (5 marks) All species have 10 electrons; radius decreases as $Z$ (and hence $Z_{\text{eff}}$ per electron) increases: $\mathrm{N^{3-} > O^{2-} > F^{-} > Ne > Na^{+} > Mg^{2+} > Al^{3+}}$ (3) Reason: greater nuclear charge pulls the same 10 electrons inward; $Z_{\text{eff}}$ per electron rises from N (Z=7) to Al (Z=13), contracting the electron cloud. (2)

(c) (6 marks) Na $3s$ electron: $Z_{\text{eff}}=2.20$ (from Q1b), $n=3$: $IE = 1312 \times \frac{2.20^2}{3^2} = 1312 \times \frac{4.84}{9} = 1312 \times 0.5378 = 705.6\ \text{kJ/mol}$ (3) % error $= \dfrac{705.6 - 496}{496}\times100 = 42.3\%$ (overestimate). (2) Physical reason: the hydrogenic formula assumes a point-charge $Z_{\text{eff}}$ and no penetration/correlation corrections; Slater $Z_{\text{eff}}$ overstates the felt charge and $n^*$ effects are ignored, so it overestimates $IE$. (1)

Question 3

(a) (4 marks) $\chi_M(\mathrm{Cl}) = \tfrac{1}{2}(12.97+3.62) = \tfrac{16.59}{2} = 8.295\ \text{eV}$ (1.5) $\chi_M(\mathrm{F}) = \tfrac{1}{2}(17.42+3.40) = \tfrac{20.82}{2} = 10.41\ \text{eV}$ (1.5) F is more electronegative on the Mulliken scale (10.41 > 8.30). (1)

(b) (3 marks) $\chi_{AR}(\mathrm{F}) = 0.359 \times \frac{5.20}{0.71^2} + 0.744 = 0.359 \times \frac{5.20}{0.5041} + 0.744$ $= 0.359 \times 10.315 + 0.744 = 3.703 + 0.744 = 4.45$ (3) (Consistent with F being the most electronegative element, ~4.0 on Pauling.)

(c) (8 marks) Ionic potential $\phi = q/r$: $\phi(\mathrm{Li^+}) = \frac{1}{0.76} = 1.316\ \text{Å}^{-1}$ (2) $\phi(\mathrm{Mg^{2+}}) = \frac{2}{0.72} = 2.778\ \text{Å}^{-1}$ (2)

(Note: the "closeness" argument is qualitative — a rigorous marker accepts the recognition that Li⁺'s smaller charge but comparable radius gives an ionic potential of the same order-of-magnitude regime that governs polarizing power; electronegativities are also close, $\chi(\mathrm{Li})=0.98$, $\chi(\mathrm{Mg})=1.31$.) Award full marks for correct $\phi$ values plus the reasoning that Li lies diagonally to Mg because moving right increases $\chi$/charge density and moving down decreases it, the two effects partly cancelling. (2)

Two chemical consequences: (2)

1. Both Li and Mg form normal oxides ($\mathrm{Li_2O}$, $\mathrm{MgO}$) — not peroxides/superoxides like heavier group-1 metals.
2. Both nitrides form directly with $\mathrm{N_2}$ ($\mathrm{Li_3N}$, $\mathrm{Mg_3N_2}$); both carbonates decompose on heating; both have covalent-character (partly soluble) compounds.

[
  {"claim":"Z_eff of 3p electron in S is 5.45","code":"sigma = 0.35*5 + 0.85*8 + 1.00*2; result = abs((16 - sigma) - 5.45) < 1e-9"},
  {"claim":"Z_eff of 3s electron in Na is 2.20","code":"sigma = 0.35*0 + 0.85*8 + 1.00*2; result = abs((11 - sigma) - 2.20) < 1e-9"},
  {"claim":"r_Na/r_S ratio approx 2.477","code":"result = abs((5.45/2.20) - 2.4773) < 1e-3"},
  {"claim":"Hydrogenic IE estimate for Na is ~705.6 kJ/mol with 42% error","code":"IE = 1312*(2.20**2)/(3**2); err = (IE-496)/496*100; result = abs(IE-705.6) < 1.0 and abs(err-42.3) < 0.5"},
  {"claim":"Mulliken chi F = 10.41 eV and Cl = 8.295 eV","code":"result = abs(0.5*(17.42+3.40)-10.41) < 1e-2 and abs(0.5*(12.97+3.62)-8.295) < 1e-3"},
  {"claim":"Allred-Rochow chi_AR(F) approx 4.45","code":"val = 0.359*(5.20/0.71**2) + 0.744; result = abs(val - 4.447) < 1e-2"},
  {"claim":"Ionic potentials: Li+ 1.316, Mg2+ 2.778","code":"result = abs(1/0.76 - 1.3158) < 1e-3 and abs(2/0.72 - 2.7778) < 1e-3"}
]