Electronegativity — Pauling, Mulliken, Allred-Rochow scales: Exercises

Before we start, here is the toolkit you will reuse (every symbol is defined the first time it appears):

Formula The three engines

Pauling (from bond energies): $∣ χ_{A} - χ_{B} ∣ = 0.102 Δ, Δ = E (A - B) - E (A - A) E (B - B)$ where $E$ means "bond energy in kJ/mol" — the energy needed to snap that bond. The symbol $Δ$ (Greek "delta", meaning a gap or excess) is the extra glue beyond a plain covalent average. is the square root; $ab$ is the geometric mean — always $\leq$ the ordinary average, which is exactly why $Δ$ stays $\geq 0$ .

Mulliken (from IE + EA): $χ_{M} = \frac{IE + EA}{2} (eV), χ_{P} \approx 0.187 (IE + EA) + 0.17$ IE = energy to rip an electron off; EA = energy released when an electron is caught. The bar $2$ is just "the average of the two".

Allred–Rochow (from Coulomb force): $χ_{A R} = 0.359 \frac{Z _{eff}}{r _{cov}^{2}} + 0.744$ $Z_{eff}$ = the leftover nuclear charge a valence electron actually feels after inner electrons shield it (see Effective Nuclear Charge and Slater's Rules); $r_{cov}$ = covalent radius in ångström (Å). The $r^{2}$ underneath is Coulomb's inverse-square law: pull weakens with the square of distance.

Level 1 — Recognition

L1.1

Which scale would you use if you are only given $E (H-H)$ , $E (Br-Br)$ and $E (H-Br)$ ? Name it and write its defining relation.

Recall Solution

You are handed only bond energies, so this is the Pauling scale. $∣ χ_{Br} - χ_{H} ∣ = 0.102 Δ, Δ = E (H-Br) - E (H-H) E (Br-Br) .$ Why not the others? Mulliken needs IE and EA; Allred–Rochow needs $Z_{eff}$ and radius. Neither was given.

L1.2

An atom has $IE = 13.6$ eV and $EA = 0.75$ eV. Which quantity is this, and by which scale? $? = \frac{13.6 + 0.75}{2}$

Recall Solution

This is the Mulliken electronegativity in energy units: $χ_{M} = \frac{13.6 + 0.75}{2} = \frac{14.35}{2} = 7.175 eV .$ Why: the pattern "(IE + EA) / 2" is the Mulliken fingerprint — the average of wanting to keep and wanting to take.

Level 2 — Application

L2.1 (Pauling)

Find $χ (Br)$ . Data: $E (H-H) = 436$ , $E (Br-Br) = 193$ , $E (H-Br) = 366$ kJ/mol. Anchor $χ (H) = 2.20$ .

Recall Solution

Step 1 — expected covalent (geometric mean): $436 \times 193 = 84148 = 290.1 kJ/mol .$ Why geometric mean? it is always $\leq$ the ordinary average, guaranteeing $Δ \geq 0$ . Step 2 — the excess: $Δ = 366 - 290.1 = 75.9 kJ/mol .$ Step 3 — turn excess into a $χ$ gap: $∣ χ_{Br} - χ_{H} ∣ = 0.102 75.9 = 0.102 (8.712) = 0.888.$ Step 4 — anchor: Br pulls harder than H, so we add: $χ_{Br} = 2.20 + 0.888 = 3.09 \approx 2.96 (tabulated) .$

L2.2 (Mulliken → Pauling)

Convert Mulliken to the Pauling scale for chlorine. IE(Cl) $= 13.0$ eV, EA(Cl) $= 3.6$ eV.

Recall Solution

$χ_{M} = \frac{13.0 + 3.6}{2} = 8.3 eV (raw Mulliken) .$ Rescale onto Pauling's familiar numbers: $χ_{P} = 0.187 (IE + EA) + 0.17 = 0.187 (16.6) + 0.17 = 3.10 + 0.17 = 3.27.$ Tabulated $χ (Cl) = 3.16$ — close. Why convert? raw eV numbers (like 8.3) can't be compared with the dimensionless 0–4 Pauling table.

L2.3 (Allred–Rochow)

Compute $χ_{A R}$ for oxygen. $Z_{eff} (O) = 4.55$ (Slater), $r_{cov} = 0.66$ Å.

Recall Solution

Coulomb ratio: $\frac{Z _{eff}}{r ^{2}} = \frac{4.55}{0.6 6 ^{2}} = \frac{4.55}{0.4356} = 10.45.$ Why $r^{2}$ ? Coulomb's inverse-square law — pull dies with the square of distance. Linear fit to Pauling: $χ_{A R} = 0.359 (10.45) + 0.744 = 3.751 + 0.744 = 4.50.$ Tabulated $χ (O) = 3.44$ ; the scales agree in ranking even when the raw number runs a bit high for small atoms.

Level 3 — Analysis

L3.1

Chlorine's electron affinity (349 kJ/mol) is larger than fluorine's (328 kJ/mol), yet $χ (F) > χ (Cl)$ . Explain, using the correct scale, why higher EA does not force higher $χ$ .

Recall Solution

Electronegativity is not EA. On the Mulliken scale $χ_{M} = \frac{1}{2} (IE + EA)$ , so IE matters just as much as EA, and on Allred–Rochow $χ$ tracks $Z_{eff} / r^{2}$ .

F is tiny ( $r \approx 0.71$ Å) with a large $Z_{eff} / r^{2}$ → very strong pull.
F's EA is slightly reduced by fierce electron–electron repulsion inside its small 2p shell — that lowers EA but does not lower the in-bond pull captured by $Z_{eff} / r^{2}$ or by (IE + EA).
F's IE (17.4 eV) also exceeds Cl's (13.0 eV), pushing F's Mulliken $χ$ above Cl's.

So EA is one isolated-atom energy; $χ$ blends size, screening and both energies. See the Electron Affinity and Effective Nuclear Charge notes.

L3.2

Going down group 17 (F → Cl → Br → I), $Z$ rises from 9 to 53. Why does $χ$ fall?

Recall Solution

Look at the Allred–Rochow driver $Z_{eff} / r^{2}$ :

$Z$ rises, but each new shell adds inner electrons that screen the nucleus, so $Z_{eff}$ climbs only slowly.
Meanwhile $r$ grows a lot (see Periodic Trends — Atomic Radius), and it enters as $r^{2}$ .
A slowly-rising numerator over a fast-rising squared denominator → the ratio shrinks, so $χ$ decreases down the group.

The bare $Z$ argument ignores screening and the inverse-square distance penalty.

Level 4 — Synthesis

L4.1 (two scales must agree)

For fluorine, verify that Mulliken (Pauling-scaled) and Allred–Rochow land near the same number. Use IE(F) $= 17.4$ eV, EA(F) $= 3.4$ eV, $Z_{eff} (F) = 5.20$ , $r_{cov} = 0.71$ Å.

Recall Solution

Mulliken → Pauling: $χ_{P} = 0.187 (17.4 + 3.4) + 0.17 = 0.187 (20.8) + 0.17 = 3.890 + 0.17 = 4.06.$ Allred–Rochow: $\frac{Z _{eff}}{r ^{2}} = \frac{5.20}{0.7 1 ^{2}} = \frac{5.20}{0.5041} = 10.32, χ_{A R} = 0.359 (10.32) + 0.744 = 4.45.$ Both hover around 4 (tabulated 3.98). Why they agree: all three scales were tuned against Pauling's anchor, so independent inputs (energies vs. force) converge on the same ranking — F strongest.

L4.2 (predict a bond energy backwards)

Given $χ (H) = 2.20$ , $χ (I) = 2.66$ , and $E (H-H) = 436$ , $E (I-I) = 151$ kJ/mol, predict $E (H-I)$ .

Recall Solution

Invert the Pauling relation. First the gap → $Δ$ : $∣ χ_{I} - χ_{H} ∣ = 0.46 = 0.102 Δ \Rightarrow Δ = \frac{0.46}{0.102} = 4.51 \Rightarrow Δ = 20.3 kJ/mol .$ Then add $Δ$ to the geometric-mean covalent baseline: $436 \times 151 = 65836 = 256.6, E (H-I) = 256.6 + 20.3 = 276.9 kJ/mol .$ (Real $E (H-I) \approx 298$ ; the estimate is in the right neighbourhood.) This shows the Pauling equation runs both directions — energies ↔ electronegativities. Ties into Ionic Character of Bonds.

Level 5 — Mastery

L5.1 (full multi-scale audit of one element)

For nitrogen, compute $χ$ by all three routes and compare to the tabulated 3.04. Data: bond energies $E (N-N) = 163$ , $E (H-H) = 436$ , $E (H-N) = 391$ kJ/mol, $χ (H) = 2.20$ ; IE(N) $= 14.5$ eV, EA(N) $= - 0.07$ eV; $Z_{eff} (N) = 3.90$ , $r_{cov} = 0.71$ Å.

Recall Solution

Pauling (via H–N): $436 \times 163 = 71068 = 266.6, Δ = 391 - 266.6 = 124.4,$ $∣ χ_{N} - χ_{H} ∣ = 0.102 124.4 = 0.102 (11.15) = 1.137, χ_{N} = 2.20 + 1.137 = 3.34.$ Mulliken → Pauling (note EA is negative — nitrogen's half-filled 2p resists a new electron): $χ_{P} = 0.187 (14.5 - 0.07) + 0.17 = 0.187 (14.43) + 0.17 = 2.698 + 0.17 = 2.87.$ Allred–Rochow: $\frac{3.90}{0.7 1 ^{2}} = \frac{3.90}{0.5041} = 7.737, χ_{A R} = 0.359 (7.737) + 0.744 = 2.778 + 0.744 = 3.52.$ Verdict: three independent routes give 2.87, 3.34, 3.52 — all bracketing the accepted 3.04. The spread is exactly what L4.2's trap warned about: same ranking, differing magnitudes. Notice the negative EA lowered Mulliken's value (correctly, since N is a reluctant electron-grabber) — see Electron Affinity.

L5.2 (choose the scale, justify, decide bonding)

You must estimate whether an Al–O bond is strongly ionic. You only have $Z_{eff}$ and radii (no thermochemistry, no IE/EA tables). (a) Which scale? (b) Compute $Δ χ$ . (c) Comment on ionic character. Data: $Z_{eff} (Al) = 3.50$ , $r (Al) = 1.18$ Å; $Z_{eff} (O) = 4.55$ , $r (O) = 0.66$ Å.

Recall Solution

(a) Only $Z_{eff}$ and radii are available ⇒ the Allred–Rochow force scale is the only usable route. (b) Compute each: $χ_{A R} (Al) = 0.359 \frac{3.50}{1.1 8 ^{2}} + 0.744 = 0.359 \frac{3.50}{1.3924} + 0.744 = 0.359 (2.514) + 0.744 = 1.646,$ $χ_{A R} (O) = 0.359 \frac{4.55}{0.6 6 ^{2}} + 0.744 = 0.359 (10.45) + 0.744 = 4.496,$ $Δ χ = 4.496 - 1.646 = 2.85.$ (c) A gap of $\approx 2.85$ is large (rule of thumb: $Δ χ ≳ 1.7$ leans strongly ionic). So Al–O has substantial ionic character — consistent with Fajans Rules and Ionic Character of Bonds, where a big $Δ χ$ produces a large Dipole Moment and ionic-type bonding.

The three scales side by side for a single element (nitrogen from L5.1) look like this:

Figure — Electronegativity — Pauling, Mulliken, Allred-Rochow scales

And here is why the same ordering survives across scales — the driver quantities all move together across a period:

Connections

Bond Energy — the raw fuel for every Pauling problem here.
Ionization Energy + Electron Affinity — the two Mulliken inputs (note the negative EA in L5.1).
Effective Nuclear Charge + Slater's Rules — supply $Z_{eff}$ for Allred–Rochow.
Periodic Trends — Atomic Radius — the $r$ that dominates group trends (L3.2).
Ionic Character of Bonds + Fajans Rules + Dipole Moment — where $Δ χ$ (L5.2) is applied.