Polar and Non-Polar Covalent Bonds | Percentage of Ionic Character – Class 11 & 12 Chemistry Notes

Understand the difference between Polar and Non-Polar Covalent Bonds and learn how to calculate the Percentage of Ionic Character with examples and formulas. Perfect Chemistry notes for Class 11 and 12 students.

Welcome to Learn Chemistry by Inam Jazbi!

In this post, we’ll discuss one of the most important concepts in Chemical Bonding — the Polar and Non-Polar Covalent Bonds and how to determine the Percentage of Ionic Character in a bond. Understanding these concepts helps you differentiate between purely covalent and partially ionic compounds, a key topic for Class 11, 12, and MDCAT students.


🔸 1. Polar Covalent Bond

A polar covalent bond is formed when two atoms share electrons unequally due to a difference in electronegativity.

  • The atom with higher electronegativity attracts the shared electrons more strongly.

  • This creates a partial positive (δ⁺) and partial negative (δ⁻) charge on the bonded atoms.

Examples:
H–Cl, H₂O, NH₃

Explanation:
In H–Cl, chlorine is more electronegative than hydrogen, so the bond is polar with partial charges.


🔸 2. Non-Polar Covalent Bond

A non-polar covalent bond is formed when two atoms share electrons equally, usually between identical atoms.

  • No charge separation occurs.

  • The bond has zero polarity.

Examples:
H–H, Cl–Cl, O=O

Explanation:
In Cl₂, both atoms have equal electronegativity, so electrons are shared equally and the bond is non-polar.


🔸 3. Percentage of Ionic Character

Even covalent bonds may have some ionic character depending on the difference in electronegativity (ΔEN) between the bonded atoms.

Formula:

%Ionic Character=(1e0.25(ΔEN)2)×100\% \text{Ionic Character} = \left(1 - e^{-0.25(\Delta EN)^2}\right) \times 100

%Ionic Character=(1e0.25(ΔEN)2)×100

Where:

ΔEN = Difference in electronegativity between two bonded atoms


Example Calculation

For H–Cl,
ΔEN = 3.0 – 2.1 = 0.9

%Ionic Character=(1e0.25(0.9)2)×100=18%\% \text{Ionic Character} = \left(1 - e^{-0.25(0.9)^2}\right) \times 100 = 18\%

%Ionic Character=(1e0.25(0.9)2)×100=18%

So, the H–Cl bond is polar covalent with about 18% ionic character.


🧠 Quick Summary Table

Bond TypeElectron SharingElectronegativity Difference (ΔEN)PolarityExample
Non-Polar CovalentEqual0NoH₂, Cl₂, O₂
Polar CovalentUnequal0 < ΔEN < 1.7YesHCl, H₂O
Ionic BondComplete transferΔEN > 1.7Fully ionicNaCl, KBr














































































































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