9701 · 3.1

Electronegativity and bonding — practice questions

Practice and worked examples for 9701 Electronegativity and bonding. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The Pauling electronegativity values for nitrogen and hydrogen are 3.04 and 2.20, respectively. Predict the type of bond in ammonia (NH₃) and assign partial charges to the atoms in one N-H bond.

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Calculate the electronegativity difference (ΔEN): ΔEN = EN(N) - EN(H) = 3.04 - 2.20 = 0.84
Determine the bond type: A ΔEN of 0.84 is significant but not large enough for electron transfer. Therefore, the N-H bond is polar covalent.
Assign partial charges: Nitrogen is the more electronegative atom, so it will attract the electron density. It gains a partial negative charge (δ⁻), and hydrogen gains a partial positive charge (δ⁺). The bond can be represented as: N $^{\delta-}$ -H $^{\delta+}$

Worked example 2

Beryllium chloride (BeCl₂) is a linear molecule. Carbon tetrachloride (CCl₄) is a tetrahedral molecule. Using electronegativity values (Be=1.57, C=2.55, Cl=3.16), explain whether each molecule is polar or non-polar.

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Analyse BeCl₂:
- Bond Polarity: ΔEN(Be-Cl) = 3.16 - 1.57 = 1.59. The Be-Cl bonds are very polar.
- Molecular Shape: The molecule is linear (Cl-Be-Cl).
- Overall Polarity: The two polar Be-Cl bonds are arranged symmetrically at 180° to each other. Their dipoles are equal and opposite, so they cancel out completely. Therefore, BeCl₂ is a non-polar molecule.
Analyse CCl₄:
- Bond Polarity: ΔEN(C-Cl) = 3.16 - 2.55 = 0.61. The C-Cl bonds are polar.
- Molecular Shape: The molecule is tetrahedral.
- Overall Polarity: The four polar C-Cl bonds are arranged symmetrically in a tetrahedral geometry. The vector sum of the four bond dipoles is zero. Therefore, CCl₄ is a non-polar molecule.

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