Using Electronegativity to Predict Bond Type

Is the bond between hydrogen and chlorine ionic or covalent? What about aluminium and chlorine? One number, the electronegativity difference (Δχ), gives you a quick first answer to questions like these. This article shows how to use it through six worked problems, starting with the obvious cases and ending with the ones where the quick answer turns out to be wrong.

The tool: electronegativity difference

Electronegativity measures how strongly an atom in a bond pulls the shared electrons towards itself. On the Pauling scale it has no units. Fluorine is the highest at 3.98, and the alkali metals at the bottom of group 1 are the lowest. If you want the reasons behind the pattern, the article on the electronegativity trend explains it.

To classify a bond:

1. Look up the Pauling electronegativity of each atom.
2. Subtract the smaller from the larger to get Δχ. It is always positive or zero.
3. Compare Δχ with a set of cut-offs.

About those cut-offs

Here is an honest warning: textbooks do not agree on the boundaries. A common set is:

Δχ	Usual label
less than about 0.4	non-polar covalent
about 0.4 to 1.7	polar covalent
more than about 1.7	ionic

Other books put the lower boundary at 0.5, and the ionic boundary at 1.8 or 2.0. None of them is “right”, because bonding changes gradually from equal sharing to complete transfer, with no sharp step in between. Treat the cut-offs as a guide, use whichever set your course uses, and always quote the Δχ value so an examiner can follow your reasoning.

Electronegativities used in this article (Pauling scale, from this site’s element data):

Element	χ	Element	χ
H	2.20	Na	0.93
C	2.55	Mg	1.31
N	3.04	Al	1.61
O	3.44	Si	1.90
F	3.98	Li	0.98
Cl	3.16	Cs	0.79
I	2.66

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Example 1: Two easy ones, Cl–Cl and C–H

Problem. Classify the bond in a chlorine molecule and the C–H bond in methane.

Cl–Cl: Δχ = 3.16 − 3.16 = 0. Both atoms pull equally, so the electrons are shared evenly. This is a non-polar covalent bond. Any bond between two identical atoms (H–H, O=O, N≡N) behaves the same way.

C–H: Δχ = 2.55 − 2.20 = 0.35. This is below 0.4, so the bond is classed as non-polar covalent. Carbon is slightly more electronegative than hydrogen, so strictly there is a tiny polarity, but it is small enough that hydrocarbons behave as non-polar substances: they do not mix with water and their molecules attract each other mainly through London dispersion forces.

Lesson: Δχ does not have to be exactly zero for a bond to count as non-polar in practice.

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Example 2: The O–H bond in water

Problem. Classify the O–H bond and show which end is partially positive.

Δχ = 3.44 − 2.20 = 1.24.

This falls in the 0.4–1.7 band, so the bond is polar covalent. The electrons are still shared, but they spend more time near oxygen.

Mark the bond: O(δ−)–H(δ+), writing δ− on the more electronegative atom (oxygen) and δ+ on hydrogen.

Why it matters: this large bond polarity, combined with the bent shape of the water molecule, makes water a polar molecule and allows hydrogen bonding. For how bond polarity adds up across a whole molecule, see bond polarity.

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Example 3: Sodium chloride and magnesium oxide

Problem. Classify the bonding in NaCl and MgO.

NaCl: Δχ = 3.16 − 0.93 = 2.23. MgO: Δχ = 3.44 − 1.31 = 2.13.

Both are well above 1.7 (and above 2.0), so both are classed as ionic under any common set of cut-offs. Their properties agree: high melting points, brittle crystals, and conduction only when molten or dissolved.

Going further: percentage ionic character. Pauling proposed an estimate of how ionic a bond is:

percentage ionic character ≈ 100 × (1 − exp(−Δχ² ÷ 4))

For NaCl: Δχ² = 2.23² = 4.97; 4.97 ÷ 4 = 1.24; exp(−1.24) ≈ 0.29; 1 − 0.29 = 0.71, so about 71% ionic.

For MgO: Δχ² = 2.13² = 4.54; 4.54 ÷ 4 = 1.13; exp(−1.13) ≈ 0.32, so about 68% ionic.

Notice that neither bond comes out as 100% ionic. Even the most “ionic” compounds keep a little sharing. The formula is itself an approximation, and the article on percentage ionic character looks at it in more detail.

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Example 4: Hydrogen fluoride, a counter-example

Problem. Use Δχ to classify the H–F bond, then check the prediction against the real substance.

Δχ = 3.98 − 2.20 = 1.78.

With a 1.7 cut-off, you would call H–F ionic. With a 2.0 cut-off, you would call it polar covalent. So which is it?

Look at the evidence. Hydrogen fluoride is made of discrete HF molecules. It boils at about 20 °C, far too low for an ionic lattice, and pure liquid HF is a very poor conductor of electricity. A true ionic compound would contain H⁺ and F⁻ ions in a lattice, and HF does not. The bond is best described as strongly polar covalent.

Pauling’s formula gives about 55% ionic character (1.78² = 3.17; 3.17 ÷ 4 = 0.79; exp(−0.79) ≈ 0.45), which shows just how polar the bond is, but a large polarity does not by itself make a bond ionic. A bond between two non-metals is covalent, however lopsided the sharing.

Lesson: a rule based on one number can land on the wrong side of an arbitrary line. Always sanity-check against what the elements are and how the substance behaves.

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Example 5: Aluminium chloride, the other way round

Problem. Aluminium is a metal and chlorine is a non-metal. Predict the bond type in AlCl₃ from the “metal + non-metal” rule, then from Δχ.

Rule of thumb: metal + non-metal suggests ionic.

Δχ: 3.16 − 1.61 = 1.55, which falls in the polar covalent band.

Here the two quick methods disagree, and the evidence sides with Δχ. Anhydrous aluminium chloride turns to vapour at a fairly low temperature, and in the vapour it exists as Al₂Cl₆ molecules, not ions. Its bonding has a large covalent character.

There is a physical reason. The Al³⁺ ion is small and highly charged, so it pulls hard on the electron cloud of the neighbouring chloride ion, distorting it back towards aluminium. That distortion (called polarisation) amounts to partial sharing of electrons, which is covalent character.

Compare aluminium fluoride: Δχ = 3.98 − 1.61 = 2.37. The fluoride ion is small and hard to distort, and AlF₃ behaves as an ionic solid with a much higher melting point. The same metal can form bonds on either side of the line depending on its partner.

Lesson: “metal + non-metal = ionic” is a useful starting point, not a law. Small, highly charged cations with large, easily distorted anions give bonds with significant covalent character.

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Example 6: Ranking bonds by polarity

Problem. Put these bonds in order of increasing polarity, identify the partially negative atom in each, and flag any that sit close to a boundary: C–H, N–H, C–Cl, O–H, Si–O, Li–I.

Calculate each Δχ:

Bond	Calculation	Δχ	δ− atom	Classification
C–H	2.55 − 2.20	0.35	C	non-polar (near the 0.4 line)
C–Cl	3.16 − 2.55	0.61	Cl	polar covalent
N–H	3.04 − 2.20	0.84	N	polar covalent
O–H	3.44 − 2.20	1.24	O	polar covalent
Si–O	3.44 − 1.90	1.54	O	polar covalent
Li–I	2.66 − 0.98	1.68	I	polar covalent by the 1.7 rule

Order of increasing polarity: C–H < C–Cl < N–H < O–H < Si–O < Li–I.

Two boundary cases stand out:

• C–H at 0.35 is only just below the non-polar line. On a 0.3 cut-off, it would be called slightly polar. Either way, it behaves as non-polar in practice.
• Li–I at 1.68 would be called polar covalent by the 1.7 rule, yet lithium iodide is a crystalline salt that dissolves in water to give Li⁺ and I⁻ ions. Most chemists describe it as ionic, with some covalent character because the small Li⁺ ion distorts the large iodide ion. This is the mirror image of the HF problem in Example 4.

Hydrogen fluoride (Δχ = 1.78) is covalent, while lithium iodide (Δχ = 1.68) is ionic. The number alone would have ranked them the wrong way round, which is the clearest possible sign that the cut-offs are approximate. The article on the bonding continuum shows how chemists picture this sliding scale.

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A reliable routine for exam questions

1. Write down both electronegativities and calculate Δχ.
2. State the cut-off set you are using (“using Δχ > 1.7 as ionic…”).
3. Give the classification and, for polar bonds, mark δ+ and δ−.
4. Check it against the elements: two non-metals give covalent bonds, however large Δχ is.
5. If a question asks about properties, match your answer to the evidence (melting point, conductivity, solubility).

Key takeaways

• Δχ is the difference between two Pauling electronegativities; the bigger it is, the more unequal the sharing.
• Common guides: below about 0.4 non-polar covalent, about 0.4–1.7 polar covalent, above about 1.7 (some books 1.8 or 2.0) ionic.
• The cut-offs are conventions, not laws, because bonding changes gradually from sharing to transfer.
• HF (Δχ = 1.78) is polar covalent despite its large difference; AlCl₃ (Δχ = 1.55) has strong covalent character despite being a metal chloride; LiI (Δχ = 1.68) is ionic despite falling below 1.7.
• Always quote Δχ, name your cut-off, and check the prediction against real properties.