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How does the electronegativity difference between two atoms decide whether a bond is ionic, polar covalent or nonpolar covalent?

Electronegativity and bond polarity: use electronegativity differences from Table S to classify bonds as ionic, polar covalent or nonpolar covalent.

A focused Regents Chemistry answer on electronegativity difference and bond polarity: how subtracting Table S electronegativities classifies a bond as nonpolar covalent, polar covalent or ionic, and how that difference shapes the unequal sharing of electrons.

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  1. What this topic is asking
  2. Electronegativity as a pull on electrons
  3. Classifying a bond by the difference
  4. Partial charges and the dipole
  5. Try this

What this topic is asking

The Core Curriculum asks you to use electronegativity to judge bond character. Specifically, you compute the difference in electronegativity between two bonded atoms (using the values on Table S) and classify the bond as nonpolar covalent, polar covalent or ionic. This is a recurring Part A and Part B-2 skill that bridges the periodic-trends page and the molecular-polarity page.

Electronegativity as a pull on electrons

Because electronegativity increases across a period and decreases down a group (see the periodic-trends page), the most electronegative atoms are at the top right of the table. When two atoms with different electronegativities bond, the more electronegative one takes a larger share of the bonding electrons, giving it a partial negative charge.

Classifying a bond by the difference

The exam most often asks you to rank bonds by polarity or to classify a single bond, rather than to apply a precise cut-off. For ranking, the bond with the largest electronegativity difference is the most polar (or most ionic). Identical atoms always give a difference of zero and therefore a nonpolar covalent bond.

Partial charges and the dipole

In a polar covalent bond the unequal sharing produces partial charges: the more electronegative atom carries a partial negative charge and the less electronegative atom a partial positive charge. This separation of charge across the bond is a bond dipole, and it is what makes a bond "polar". Whether the whole molecule is polar then depends on its shape, which is covered on the Lewis-structures and molecular-polarity page.

Try this

Q1. Classify the bond in F2\text{F}_2 as ionic, polar covalent or nonpolar covalent. [1 point]

  • Cue. Both atoms are fluorine, so the difference is 00: a nonpolar covalent bond.

Q2. Using Table S, state whether the bond between sodium (0.90.9) and chlorine (3.23.2) is ionic or covalent. [1 point]

  • Cue. Difference =3.2βˆ’0.9=2.3= 3.2 - 0.9 = 2.3, a large difference, so the bond is ionic.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (Part A style)1 marksWhich bond is the most polar? (1) Hβˆ’Cl\text{H}-\text{Cl} (2) Hβˆ’Br\text{H}-\text{Br} (3) Hβˆ’I\text{H}-\text{I} (4) Hβˆ’F\text{H}-\text{F}
Show worked answer β†’

A 1-point Part A item using Table S electronegativities. The answer is (4) Hβˆ’F\text{H}-\text{F}.

The polarity of a bond increases with the electronegativity difference between the two atoms. Fluorine has the highest electronegativity of any element (about 4.04.0 on Table S), so the difference between hydrogen and fluorine is the largest of the four halogen choices. As you go down the halogens from F to I, electronegativity falls, so the bond with hydrogen becomes less polar.

Markers reward selecting the bond with the largest electronegativity difference.

Regents (Part B-2 style)3 marksUsing Table S, classify the bonding between the atoms in each pair as ionic, polar covalent or nonpolar covalent. (a) potassium and fluorine (electronegativities 0.80.8 and 4.04.0) (b) hydrogen and chlorine (2.22.2 and 3.23.2) (c) two bromine atoms (3.03.0 and 3.03.0).
Show worked answer β†’

A 3-point constructed-response item applying electronegativity differences.

(a) K and F (1 point): difference =4.0βˆ’0.8=3.2= 4.0 - 0.8 = 3.2, a large difference, so the bond is ionic.
(b) H and Cl (1 point): difference =3.2βˆ’2.2=1.0= 3.2 - 2.2 = 1.0, a moderate difference, so the bond is polar covalent.
(c) Br and Br (1 point): difference =3.0βˆ’3.0=0= 3.0 - 3.0 = 0, so the bond is nonpolar covalent (electrons shared equally).

Markers reward computing each difference and classifying it: a large difference is ionic, a small nonzero difference is polar covalent, and zero difference is nonpolar covalent.

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