Electronegativity — Pauling, Mulliken, Allred-Rochow scales: Visual walkthrough

Everything hangs on one idea: when two atoms pull unequally on their shared electrons, the bond gets extra-strong, and that extra strength is measurable.

WHAT. Two atoms, $A$ and $B$ , are joined because they share a pair of electrons sitting between them. That shared pair is the "rope" both atoms hold.

WHY start here. Electronegativity is defined only inside a bond — it is about who wins the tug on that shared pair. So the very first picture must be the shared pair itself, with nothing else assumed.

PICTURE. Look at the figure. The two black circles are the atomic nuclei. The red dots between them are the shared bonding electrons — the rope. If both atoms pull equally, the red dots sit dead-centre.

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

Step 2 — Measure how tightly a bond holds: bond energy

WHAT. We give each bond a number: its bond energy, written $E (A - B)$ — the energy (in kJ/mol) it takes to rip the bond apart into separate atoms. Big $E$ = hard to pull apart = strong bond.

WHY this number. We cannot put a "pull-o-meter" on a single atom. But we can measure how much energy it takes to break a bond — that is a real lab quantity (see Bond Energy). So bond energy becomes our raw material.

PICTURE. Think of the bond as a spring. A strong spring resists stretching; you must do a lot of work to snap it. The height of the "energy hill" you climb to break it is the bond energy.

$E (A - B) = tall hill = strong bond energy to break the A - B bond$

Each symbol: $E$ is an energy in kJ/mol; the letters in brackets, $A - B$ , name which bond.

Step 3 — Guess the bond energy if the pull were equal

WHAT. Imagine $A$ and $B$ pulled the shared pair equally — a perfectly fair, purely covalent bond. What would its energy be? Pauling's guess: take the two fair reference bonds, $A - A$ and $B - B$ (where each atom bonds a copy of itself, so pull is automatically equal), and combine them.

WHY the geometric mean and not the plain average? We want the guess to be an underestimate whenever the atoms differ, so any leftover is honestly "extra." The geometric mean $E (A - A) E (B - B)$ is always less than or equal to the ordinary average (the AM–GM inequality). Using it guarantees the leftover we compute next is never negative.

PICTURE. Two reference springs — an $A - A$ spring and a $B - B$ spring. Their "blended" strength (geometric mean) is the dashed baseline: the strength a fair $A - B$ bond should have.

$E_{expected} (A - B) = A ’s fair bond E (A - A) \times B ’s fair bond E (B - B)$

$E (A - A)$ — energy of a bond between two identical $A$ atoms (equal pull by construction).
$E (B - B)$ — same for $B$ .
The square root $\cdot$ — the geometric mean, keeping the result below the plain average.

Step 4 — The real bond is stronger than the fair guess: the extra glue $Δ$

WHAT. Measure the real $A - B$ bond energy and subtract the fair guess. The leftover is called $Δ$ : $Δ = E (A - B) - E (A - A) E (B - B)$

WHY it appears. If $A$ pulls harder than $B$ , the shared pair shifts toward $A$ . Now one end is a little negative, the other a little positive — the atoms attract electrostatically on top of sharing. This ionic bonus (linked to Ionic Character of Bonds) makes the real bond stronger than the fair covalent guess. So $Δ > 0$ is the direct fingerprint of unequal pull.

PICTURE. The dashed line is the fair guess (Step 3). The solid red bar rises above it — that extra height is $Δ$ , the ionic glue. The more unequal the pull, the taller the red excess.

$E (A - B)$ — the real, measured bond (top of the red bar).
$\dots$ — the fair-guess baseline (dashed line).
$Δ$ — the red gap between them: pure ionic bonus.

Degenerate case: if $A = B$ (identical atoms, like Cl–Cl), the real bond is the fair guess, so $Δ = 0$ and there is no electronegativity difference — exactly right.

Step 5 — Turn the extra glue into an electronegativity difference

WHAT. Pauling postulated that $Δ$ grows with the square of the electronegativity difference: $Δ \propto (χ_{A} - χ_{B})^{2}$ Solving for the difference and inserting the unit-conversion constant $0.102$ (for $Δ$ in kJ/mol) gives the headline formula: $∣ χ_{A} - χ_{B} ∣ = 0.102 Δ$

WHY the square (and hence the square root)? A square is always positive, so it doesn't matter which atom we label $A$ or $B$ — the gap comes out the same size. It also makes the numbers consistent: the gap A→B plus B→C matches A→C no matter which bonds you measure. Taking of both sides simply undoes the square to recover the plain difference $∣ χ_{A} - χ_{B} ∣$ .

PICTURE. A parabola $Δ = k (Δ χ)^{2}$ . Reading it backwards — go up to your measured $Δ$ , across to the curve, down to the axis — hands you $∣Δ χ ∣$ . The square root is literally this "read the parabola in reverse" move.

$∣ χ_{A} - χ_{B} ∣$ — the size of the electronegativity gap (dimensionless).
$0.102$ — equals $1/ 96.5$ , converting kJ/mol so hydrogen lands at $2.20$ and fluorine at $3.98$ .
$Δ$ — undoes the square, turning the energy bonus back into a gap.

Step 6 — Anchor the scale: fix H = 2.20

WHAT. The formula only ever gives a difference, never a lone value. To get absolute numbers we nail one atom down: define $χ (H) = 2.20$ . Every other value is then built outward, one gap at a time.

WHY an anchor is unavoidable. $∣ χ_{A} - χ_{B} ∣$ is like knowing only height differences between people — you still need one person's actual height to place everyone. Fixing H at 2.20 is that reference floor.

PICTURE. A vertical number line. H sits pinned at 2.20 (red). A measured gap of $+ 1.0$ places Cl at $3.20$ ; a gap the other way would place a less-greedy atom below H.

Step 7 — Full worked derivation: $χ (Cl)$ from scratch

WHAT. Put every step together with real numbers. Given (kJ/mol): $E (H-H) = 436$ , $E (Cl-Cl) = 242$ , $E (H-Cl) = 431$ .

WHY. This is the exact chain a test asks you to reproduce. Watch each earlier step light up.

PICTURE. The energy-bar diagram for H–Cl: dashed fair-guess line at $324.8$ , real bond at $431$ , red excess $Δ = 106.2$ .

Which sign, + or −? The formula gives only the size $1.05$ . Chemistry decides the direction: the atom that draws the shared pair toward itself is the larger $χ$ . Cl is greedier, so it sits above H. If we were solving for a less greedy partner, we would subtract.

The one-picture summary

Every step in one flow: measure three bond energies → geometric-mean baseline → red excess $Δ$ → square-root through the parabola → add to the H = 2.20 anchor → out pops $χ$ .

Recall Feynman retelling — the whole walkthrough in plain words

Two atoms hold a rope made of shared electrons. We can't measure "pull" directly, so we cheat with bond strengths. First we ask: if both kids pulled equally, how tight would the knot be? — we blend the two "fair" ropes (each kid tugging a twin of itself) using a geometric mean, which is guaranteed a touch gentle. Then we measure the real knot — and it's tighter! That extra tightness $Δ$ is the red bonus that only shows up when one kid pulls harder, because then one end goes slightly negative and the ends snap together electrically. Pauling says this bonus grows like the square of the pulling gap, so we take a square root to read the gap back out (that's just reading a parabola backwards). But a gap alone floats free — so we pin hydrogen at 2.20 and build everyone else off it. Run the numbers for H–Cl and chlorine lands at 3.2, essentially its real value 3.16. Same recipe puts fluorine at the very top. Done — the whole scale, from a few bond energies.

Connections

Bond Energy — the raw measured inputs $E (A - A)$ , $E (B - B)$ , $E (A - B)$ .
Ionic Character of Bonds — the origin of the extra glue $Δ$ .
Effective Nuclear Charge & Slater's Rules — why one atom pulls harder in the first place.
Ionization Energy and Electron Affinity — the alternate Mulliken route to the same $χ$ .
Dipole Moment and Fajans Rules — consequences of a nonzero $Δ χ$ .