Worked examples — Electronegativity — Pauling, Mulliken, Allred-Rochow scales

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The scenario matrix

Every question a course can ask about electronegativity falls into one of these cells. The worked examples below are each tagged with the cell they close.

Cell	Scale involved	The twist it tests
A — plain forward	Pauling	Given three bond energies → find one $\chi$.
B — reverse	Pauling	Given $\Delta \chi$ → predict a bond energy.
C — degenerate ($\Delta =0$)	Pauling	Identical atoms / homonuclear bond → $\Delta \chi =0$.
D — sign / direction	Pauling & all	Which atom is ${\delta }^{-}$? (the sign of the pull, not just its size)
E — energy → scale conversion	Mulliken	eV answer must be rescaled onto Pauling's ruler.
F — zero / negative EA	Mulliken	Noble-gas-like atom with EA $\le 0$: what happens?
G — force limit	Allred–Rochow	Effect of shrinking $r$ / growing $Z_{\text{eff}}$ (limiting behaviour).
H — cross-scale sanity	all three	Compute F three ways, confirm they agree.
I — real-world word problem	any	A lab / bonding scenario dressed in words.
J — exam trap	conceptual	EA vs $\chi$ contradiction (Cl vs F).

We now sweep every cell.

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Example 1 — Cell A (Pauling, plain forward)

Forecast: Br is below Cl in the halogens, so guess $\chi (\text{Br})$ somewhere near $2.9$–$3.0$. Will our arithmetic land there?

1. Expected covalent energy = $\sqrt{E(\text{H–H})E(\text{Br–Br})}=\sqrt{436\times 193}=\sqrt{84148}=290.1$ kJ/mol. Why this step? This is the bond energy we'd predict if H and Br pulled equally — a purely covalent, no-charge-separation baseline. Geometric mean guarantees it never exceeds the real bond (AM ≥ GM), so the excess stays positive.
2. Excess (ionic resonance energy) $\Delta =366-290.1=75.9$ kJ/mol. Why this step? The real bond is stronger than the covalent guess. That extra glue is the electronegativity fingerprint.
3. Convert to a $\chi$ gap: $|{\chi }_{\text{Br}}-{\chi }_{\text{H}}|=0.102\sqrt{75.9}=0.102\times 8.712=0.889$. Why this step? Pauling postulated $\Delta \propto (\Delta \chi {)}^2$; the $0.102$ constant turns kJ/mol into scale units.
4. Anchor: Br pulls harder than H, so add: ${\chi }_{\text{Br}}=2.20+0.889=3.09$. Why add and not subtract? The formula gives only a magnitude. We choose the sign from chemistry: Br is more electronegative than H (it forms ${\text{H}}^{\delta +}{\text{Br}}^{\delta -}$), so it sits above H.

Verify: Real $\chi (\text{Br})=2.96$. Our $3.09$ is within $\sim 0.13$ — excellent for a single-bond estimate. Units: $\sqrt{\text{kJ/mol}}$ cancelled by the constant's hidden $1/\sqrt{\text{kJ/mol}}$, leaving a pure number. ✓

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Example 2 — Cell B (Pauling, reverse)

Forecast: H–F is famously strong (that's why HF barely ionizes). Guess it beats the covalent baseline by a lot — maybe near $560$ kJ/mol.

1. Covalent baseline $=\sqrt{436\times 158}=\sqrt{68888}=262.5$ kJ/mol. Why? Same equal-pull baseline as before.
2. Invert the Pauling relation to get $\Delta$. From $|\Delta \chi |=0.102\sqrt{\Delta }$ we solve for $\Delta$: $\Delta ={\left(\frac{|\Delta \chi |}{0.102}\right)}^2.$ With $|\Delta \chi |=3.98-2.20=1.78$: $\Delta =(1.78/0.102{)}^2=(17.45{)}^2=304.5$ kJ/mol. Why square? We're undoing the square root — reverse of Example 1's step 3.
3. Add baseline back: $E(\text{H–F})=262.5+304.5=567.0$ kJ/mol. Why add? $\Delta$ is defined as (real − baseline), so real = baseline + $\Delta$.

Verify: Experimental $E(\text{H–F})\approx 565$ kJ/mol. Our $567$ is within $0.4\%$. The forward and reverse directions are consistent. ✓

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Example 3 — Cell C (degenerate, $\Delta =0$)

Forecast: Two identical atoms pull with identical strength → the gap must be exactly zero. Watch the algebra force it.

1. Baseline for A–A $=\sqrt{E(\text{Cl–Cl})E(\text{Cl–Cl})}=\sqrt{242\times 242}=242$ kJ/mol. Why? The geometric mean of a number with itself is the number. The "expected" bond is the real bond.
2. Excess $\Delta =242-242=0$. Why? No ionic stabilization is possible when both ends pull equally — there's no direction to separate charge.
3. Gap $|\Delta \chi |=0.102\sqrt{0}=0$. Why this matters: This is the degenerate limit — the formula correctly reports "no difference," and ${\text{Cl}}_2$ has zero dipole. The scale is self-consistent at its boundary.

Verify: $\chi (\text{Cl})-\chi (\text{Cl})=0$ trivially, and $\Delta =0$ matches. Any homonuclear bond gives $\Delta \chi =0$. ✓

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Example 4 — Cell D (sign / direction of the pull)

See the figure: the same magnitude $|\Delta \chi |$ points in opposite directions depending on which atom is more electronegative.

Forecast: In H–Cl, Cl wins → Cl is ${\delta }^{-}$. In Cl–I, Cl is on the left but still wins → the electron cloud shifts toward Cl (the left this time).

1. H–Cl: ${\chi }_{\text{Cl}}-{\chi }_{\text{H}}=3.16-2.20=+0.96$. Why the sign? Positive means the right atom (Cl) pulls harder → Cl is ${\delta }^{-}$, H is ${\delta }^{+}$. Look at the red arrow in the figure pointing toward Cl.
2. Cl–I: ${\chi }_{\text{I}}-{\chi }_{\text{Cl}}=2.66-3.16=-0.50$. Why the sign? Negative means the left atom (Cl) pulls harder → the arrow flips and points back toward Cl. Cl is ${\delta }^{-}$ again, but now on the left.
3. Magnitude vs direction: the Pauling formula only ever gives $|\Delta \chi |$ ($0.96$ and $0.50$). Chemistry supplies the sign by naming the more electronegative atom. Why this cell exists: students who blindly "add" can misplace the negative end. Always ask who pulls harder, then point the arrow there.

Verify: $|+0.96|=0.96$ and $|-0.50|=0.50$ are the true magnitude gaps; in both molecules the higher-$\chi$ atom (Cl) carries ${\delta }^{-}$. ✓ Ties to Dipole Moment — the arrow direction is the dipole direction.

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Example 5 — Cell E (Mulliken → Pauling conversion)

Forecast: Cl's Pauling value is $3.16$; if Mulliken agrees, the converted number should be close to that.

1. Raw Mulliken ${\chi }_M=\frac{\text{IE}+\text{EA}}{2}=\frac{12.97+3.62}{2}=\frac{16.59}{2}=8.295$ eV. Why average? Electronegativity = (desire to keep own electrons, IE) + (desire to grab more, EA), split evenly.
2. Rescale to Pauling using ${\chi }_P\approx 0.187(\text{IE}+\text{EA})+0.17$: ${\chi }_P=0.187\times 16.59+0.17=3.102+0.17=3.27$. Why rescale? The raw $8.295$ eV lives on an energy axis; multiplying by $0.187$ and shifting by $0.17$ maps it onto the familiar 0–4 Pauling axis.

Verify: Real $\chi (\text{Cl})=3.16$; our $3.27$ is within $0.11$. The eV value ($8.295$) is not comparable to Pauling numbers until converted — that's the whole point of cell E. ✓ Inputs come from Ionization Energy and Electron Affinity.

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Example 6 — Cell F (zero / negative EA)

Forecast: A near-zero EA drags the average down a little, but the large IE keeps N moderately electronegative (real $\chi =3.04$).

1. Raw Mulliken ${\chi }_M=\frac{14.53+(-0.07)}{2}=\frac{14.46}{2}=7.23$ eV. Why keep the negative EA? A negative EA means gaining an electron costs energy; the formula honestly subtracts it. It does not break — it just lowers ${\chi }_M$ slightly.
2. Rescale: ${\chi }_P=0.187\times 14.46+0.17=2.704+0.17=2.87$. Why this cell matters: the "average" survives even when one input is $\le 0$. There's no division-by-zero or undefined behaviour — Mulliken degrades gracefully at the boundary.

Verify: Real $\chi (\text{N})=3.04$; our $2.87$ is within $\sim 0.17$, reasonable given the tricky EA. The negative EA legitimately pulled the estimate down, exactly as physics predicts. ✓

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Example 7 — Cell G (Allred–Rochow, force limit)

See the figure: as radius $r$ shrinks, the inverse-square force $Z_{\text{eff}}/r^2$ blows up.

Forecast: S has more effective charge but is bigger. The inverse-square law should let radius win, making O more electronegative — real values are O $=3.44$, S $=2.58$.

1. Oxygen force term $\frac{Z_{\text{eff}}}{r^2}=\frac{4.55}{0.66^2}=\frac{4.55}{0.4356}=10.446$. Why square $r$? Coulomb's law is inverse-square; halving the distance quadruples the pull.
2. ${\chi }_{AR}(\text{O})$ $=0.359\times 10.446+0.744=3.750+0.744=4.49$. Why the constants? They linearly rescale raw force into Pauling-like numbers.
3. Sulfur force term $\frac{5.48}{1.04^2}=\frac{5.48}{1.0816}=5.066$. Why lower despite higher $Z_{\text{eff}}$? $r$ is $1.6\times$ bigger, and squaring makes that a $2.6\times$ penalty — it overwhelms the $1.2\times$ charge gain. This is the limiting behaviour: down a group $r$ grows faster than $Z_{\text{eff}}$, so $Z_{\text{eff}}/r^2$ falls (look at how the curve drops off in the figure).
4. ${\chi }_{AR}(\text{S})$ $=0.359\times 5.066+0.744=1.819+0.744=2.56$.

Verify: O $=4.49$ vs real $3.44$ (Allred–Rochow tends to over-estimate second-period atoms), S $=2.56$ vs real $2.58$ (spot on). Crucially, ${\chi }_{AR}(\text{O})>{\chi }_{AR}(\text{S})$ — the order is correct, confirming the inverse-square logic. ✓ Uses Effective Nuclear Charge, Slater's Rules, Periodic Trends — Atomic Radius.

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Example 8 — Cell H (cross-scale sanity, all three)

Forecast: All three should land in a tight band around $4$, because every scale was tuned to Pauling.

1. List them: $3.98,4.06,4.45$. Why compare? Three independent physical starting points (bond energy, IE+EA, Coulomb force) landing near the same number is strong evidence the concept is real, not an artifact of one method.
2. Spread $=4.45-3.98=0.47$ scale units; mean $=(3.98+4.06+4.45)/3=12.49/3=4.163$. Why the mean? A single best estimate. The spread ($\sim 12\%$ of the mean) is the calibration noise between methods.

Verify: All three exceed every other element's $\chi$, so all three agree F is the most electronegative atom. Mean $4.16$, spread $0.47$. ✓

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Example 9 — Cell I (real-world word problem)

Forecast: Cl is far more electronegative than I, so HCl should be much more polar.

1. HCl gap: $|\Delta \chi |=3.16-2.20=0.96$. Why $|\Delta \chi |$? It scales with bond polarity and (via Ionic Character of Bonds) with dipole moment.
2. HI gap: $|\Delta \chi |=2.66-2.20=0.46$. Why smaller? I sits far down the halogens; big radius → weak pull → smaller gap from H.
3. Decision: $0.96>0.46$, so HCl is more polar — choose HCl. Why trust $\Delta \chi$ here? Pauling's whole construction ties $\Delta \chi$ to ionic (charge-separated) character.

Verify: Measured dipole moments: HCl $\approx 1.08$ D, HI $\approx 0.44$ D. The ratio of dipoles ($1.08/0.44\approx 2.5$) tracks the ratio of gaps ($0.96/0.46\approx 2.1$) — same ranking, right ballpark. ✓

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Example 10 — Cell J (exam trap: EA vs $\chi$)

Forecast: The conclusion is wrong. EA and $\chi$ are different quantities; F still wins on $\chi$.

1. State the two quantities: EA is a measured energy for an isolated atom gaining one electron. $\chi$ is a relative property of a bonded atom, built from IE and EA (Mulliken) or bond energies (Pauling). Why they diverge: F is so small that adding an electron crowds the tight $2p$ shell, causing electron–electron repulsion that lowers F's EA below Cl's — a size effect, not a pull effect.
2. Check with Mulliken (which uses both IE and EA). Convert F's EA to eV: $328\text{kJ/mol}÷96.5=3.40$ eV; Cl's: $349÷96.5=3.62$ eV. Take IE(F) $=17.42$ eV, IE(Cl) $=12.97$ eV.
   • ${\chi }_M(\text{F})=(17.42+3.40)/2=10.41$ eV.
   • ${\chi }_M(\text{Cl})=(12.97+3.62)/2=8.295$ eV. Why this settles it: F's huge IE dominates the average, so even with the smaller EA, ${\chi }_M(\text{F})>{\chi }_M(\text{Cl})$.
3. Conclusion: $10.41>8.30$ → F is more electronegative, despite Cl's larger EA. The student confused one input (EA) with the output ($\chi$).

Verify: Pauling scale agrees: $\chi (\text{F})=3.98>\chi (\text{Cl})=3.16$. Both scales rank F above Cl even though EA(Cl) > EA(F). ✓ See Electron Affinity.

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Recall Cell checklist — did we cover them all? (click)

A(Ex1) · B(Ex2) · C(Ex3) · D(Ex4) · E(Ex5) · F(Ex6) · G(Ex7) · H(Ex8) · I(Ex9) · J(Ex10). Every matrix cell closed. ✓

Connections

• Dipole Moment & Ionic Character of Bonds — cells D and I turn $\Delta \chi$ into polarity.
• Ionization Energy · Electron Affinity — inputs behind cells E, F, J.
• Effective Nuclear Charge · Slater's Rules · Periodic Trends — Atomic Radius — the force picture in cell G.
• Bond Energy — the raw data of cells A, B, C.