Chemistry · Chemical Bonding · NEET
Percentage ionic character = (observed dipole moment / dipole moment for 100% ionic bond) x 100. The 'observed' value is measured in the lab. The '100% ionic' value is calculated by assuming one full electron charge (e = 1.602 x 10^-19 C) sits at each atom, separated by the bond length: mu(ionic) = charge x bond length. So the fraction tells you how close the real bond is to a perfect ionic bond.
NCERT says a completely ionic or completely covalent bond is only an ideal situation. Even in H2 (two identical atoms) there is a tiny ionic character. And even in an 'ionic' compound like NaCl, the cation pulls a little electron cloud back from the anion, giving some covalent character (this is what Fajans rules describe). So every bond is a mix; percentage ionic character just measures the ionic part.
Step 1: Take the measured (observed) dipole moment, mu(obs), in Debye or Coulomb-metre. Step 2: Calculate the ideal ionic dipole moment mu(ionic) = e x d, where e = 1.602 x 10^-19 C and d = bond length in metres. Step 3: Divide and multiply by 100. Keep both values in the same units. For HF: mu(obs) = 1.91 D and mu(ionic) is about 4.14 D, so % ionic character is about 43-45%.
Yes, as a general trend. A bigger electronegativity difference pulls the shared pair more to one atom, making the bond more polar and more ionic. That is why HF (large difference) is more ionic than HI (small difference). But the exact number still comes from the dipole moment formula, not from electronegativity alone, because bond length also matters.
They are two ends of the same scale. If a bond has X% ionic character, the rest (100 - X)% is its covalent character. High ionic character means electrons are mostly transferred; high covalent character means electrons are mostly shared. Fajans rules tell you when an 'ionic' bond gains extra covalent character (small cation, large anion, high cationic charge).
Dipole moment is usually given in Debye (D), where 1 D = 3.33564 x 10^-30 Coulomb-metre. When you compute mu(ionic) = e x d you get Coulomb-metre, so either convert it to Debye or convert mu(obs) to Coulomb-metre. Both values must be in the SAME unit before you divide, or your percentage will be wrong.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
About 43-45%. HF has an observed dipole moment of 1.91 D. The dipole moment for a fully ionic H-F bond is about 4.14 D. So % ionic character = (1.91 / 4.14) x 100, which is roughly 43-45%. This means the H-F bond is a little less than half ionic.
Yes. It comes under 'Polarity of Bonds' in the Chemical Bonding and Molecular Structure chapter of Class 11 NCERT. NEET rarely asks a direct numerical, but it often tests the linked ideas: dipole moment, why no bond is fully ionic, and Fajans rules for covalent character.
It means the bond is half ionic and half covalent. The measured dipole moment is exactly half of what it would be if one full electron were transferred. Above 50% we usually call the compound ionic; below 50% we treat it as a polar covalent bond.
No. The observed dipole moment cannot be larger than the fully-ionic value (that would need more than one full electron charge transferred), so the percentage stays between 0% and 100%. A value near 0% means nearly pure covalent; near 100% means nearly pure ionic.
Percentage ionic character measures how ionic a bond is. Fajans rules explain the opposite side, how much covalent character an ionic bond picks up. A small, highly charged cation with a large anion polarises the anion strongly, lowering ionic character and raising covalent character. So the two topics are two views of the same bond.
