Recently, I read the definition of oxidation state on Wikipedia. It read that a 100% ionic bond is impossible. So what does a 75% ionic and 25% covalent bond mean at all?

  • 2
    $\begingroup$ I don't think this is impossible. I can charge a capacitor and cut it in half. $\endgroup$
    – Joshua
    Dec 6, 2017 at 16:08
  • 5
    $\begingroup$ @Joshua Yes, it's not really a problem to make two charged things be it anion and cation or two charge spheres, it's just that there's simply no chemical bond in there. "100% ionic bond" is not a chemical bond at all. $\endgroup$
    – Mithoron
    Dec 6, 2017 at 22:27
  • $\begingroup$ This is because when a force is supplied to the ionic compound it leads to dislocation of ions by merging the like charged ions together which causes great repulsion force that leads to break down of compound. $\endgroup$
    – Felithia
    Feb 26, 2020 at 4:37

3 Answers 3


A "100% ionic" bond would be a bond whose bonding electron(s) were never in the vicinity of the cation, but rather always in unperturbed valence orbitals of the anion. That's not far from the truth in a paradigmatic ionic compound like $\ce{NaCl}$, but no matter how electronegative the anion is, the bonding electrons will still experience some attraction toward the positive charge on the cation, and so a detailed model of their wave functions will show a small but nonzero amplitude near the cation. That's just basic electrostatics; you have a positive charge and a negative charge not that far away from each other, they will attract.

Despite that, we can still talk about 100% ionic bonds as a theoretical limit and an acceptable approximation for modeling some compounds. It's kind of like how we can use absolute zero as the theoretical zero point of a temperature scale even though nothing in the universe can ever be that cold.

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    $\begingroup$ Not only some attraction of cation to electrons of anion, but quantum theory has probability distribution of each orbital extending to infinity. There isn't any absolute cutoff distance from the nucleus. $\endgroup$
    – MaxW
    Dec 7, 2017 at 0:01

Actually bonds that are $100\%$ ionic, or close to it, are possible. But they require a little ingenuity. Instead of combining atoms of very different electronegativity, we can build multiatomic ions in which the charge is (formally) buried or highly delocalized making covalent bonding unfavorable. We use molecular orbital structures rather than electronegativity to drive the charge separation.

One example of this is the cyclopropenyl salt $\ce{[C3H3]^+[SbCl6]^-}$ produced by Breslow and Groves in 1970 (J. Am. Chem. Soc. 1970, 92 (4), 984–987). To form a covalent bond between the ions would require either occupying a highly destabilized antibonding orbital in the cyclopropenyl ring, or an interaction between the ring and poorly overlapping orbitals in the bulky anion. So we have very little covalent bonding; the molecular orbital structure does not favor it. We could get covalent compounds by transferring a chloride ion, but this reaction is energetically unfavorable. The salt, with its ionic bonding between the aromatic cation and complex anion, is surprisingly stable for a species with a three-membered carbon ring. Further stabilization may be achieved and ionicity reinforced by placing substituent on the cyclopropenyl ring.

Another example is provided by compounds in which alkali metals are disproportionated into cations and anion (see Alkalide). Because the ionization energy of alkali metals (at least from sodium on down) are so small, the cation may be stabilized by forming a complex where the (formal) positive charge is deeply buried and again, no obvious means exists for covalent bonding. The alkalide anion must be bonded by essentially ionic attraction to the cationic complex.


In a "100 %" covalent bond, like the $\sigma_{ss}$ bond in the dihydrogen molecule, the electron probability density is perfectly symmetrically divided between the two bonded nuclei because both have the same electronegativity.

Below: the schematised electron density $\psi^2$, for a 100 % covalent bond:

Electron density

But when the two atoms have different electronegativities, the electron probability density will be higher towards the more electronegative atom, see e.g. $\ce{HCl}$. We say the bond is polarised. See below, schematised, the permanent dipole of $\ce{XY}$, with partial charges $\delta +$ and $\delta -$:

Permanent dipole

Below: the schematised electron density $\psi^2$, for a bond between atoms with differing electronegativity (highest electronegativity to the right):

Electron density 2

If the electronegativity difference is really high, see e.g. $\ce{NaCl}$, the bond starts being so strongly polarised (electron probability density strongly skewed to towards the $\ce{Cl}$ atom) that the bond starts taking on an ionic character.

Ionic and covalent must be considered relative to each other: in a "75% ionic" bond the electron distribution is strongly skewed toward the electronegative element.

There is however no meaningful way to measure this "percentage".


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