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Exercises · 4.18

Q.Explain with the help of a suitable example a polar covalent bond.

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A polar covalent bond forms when two atoms of different electronegativities share electrons unequally, creating partial charges. The classic example is the H–Cl bond, where chlorine pulls the shared electrons closer, becoming partially negative while hydrogen becomes partially positive.

What makes a bond polar?

When two atoms form a covalent bond, they share electrons. But not all atoms pull on those shared electrons with equal strength. Electronegativity measures an atom's ability to attract bonding electrons toward itself. When two atoms with different electronegativities bond, the electron cloud shifts toward the more electronegative atom. This creates a separation of charge: one end becomes slightly negative (δ⁻), the other slightly positive (δ⁺). That's a polar covalent bond.

The key difference from a pure covalent bond (like H–H, where both atoms are identical) is this asymmetry. The electrons aren't shared equally, so the bond has a dipole moment — a vector pointing from the positive end to the negative end.

The hydrogen chloride (HCl) example

Let's walk through hydrogen chloride, one of the clearest demonstrations of polarity.

  1. Identify the atoms and their electronegativities

    Hydrogen has an electronegativity of about 2.1 on the Pauling scale, while chlorine sits at 3.0. Chlorine is significantly more electronegative.

  2. Electron sharing is unequal

    When H and Cl form a single covalent bond, they share one pair of electrons. But chlorine's stronger pull means the shared electron pair spends more time near the Cl nucleus than near the H nucleus.

  3. Partial charges develop

    Because the electron density is greater around chlorine, it acquires a partial negative charge (δ⁻). Hydrogen, now electron-deficient in that region, carries a partial positive charge (δ⁺). We write this as:

HXδ+−ClXδ−\ce{H^{\delta+} - Cl^{\delta-}}

  1. The bond has a dipole moment The dipole moment μ\mu is defined as:

μ=q×d\mu = q \times d

where qq is the magnitude of the partial charge and dd is the distance between the charges (the bond length). For HCl, μ≈1.08 D\mu \approx 1.08 \, \text{D} (Debye units). The arrow representing the dipole points from H (positive) toward Cl (negative), or sometimes a crossed arrow (→\rightarrow) is drawn with the cross at the positive end. …

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