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Atomic Structure - Electronegativity

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electronegativity is defined as the measure of the ability of an atom to attract a shared pair of electrons in a covalent bond toward itself.

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The Pauling Scale is the most commonly used scale for electronegativity, where Fluorine (FF) is assigned the highest value of 4.04.0.

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Across a period (left to right), electronegativity increases because the nuclear charge (ZZ) increases and the atomic radius decreases, causing a stronger attraction for bonding electrons.

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Down a group (top to bottom), electronegativity decreases because the atomic radius increases and the shielding effect of inner electron shells reduces the effective nuclear pull on the valence electrons.

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Noble gases are generally not assigned electronegativity values on the Pauling scale because they do not readily form covalent bonds.

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The difference in electronegativity (Δχ\Delta\chi) between two atoms determines the bond polarity: a larger difference leads to a more polar bond.

📐Formulae

Δχ=∣χA−χB∣\Delta\chi = |\chi_{A} - \chi_{B}|

Bond Character∝Δχ\text{Bond Character} \propto \Delta\chi

💡Examples

Problem 1:

Predict the type of bond formed between Hydrogen (χ=2.1\chi = 2.1) and Chlorine (χ=3.0\chi = 3.0) based on their electronegativity difference.

Solution:

Δχ=∣3.0−2.1∣=0.9\Delta\chi = |3.0 - 2.1| = 0.9

Explanation:

Since the electronegativity difference is between 0.50.5 and 1.71.7, the bond is classified as a polar covalent bond. The electrons spend more time near the Chlorine atom, creating a partial negative charge (δ−\delta^{-}) on ClCl and a partial positive charge (δ+\delta^{+}) on HH.

Problem 2:

Arrange the following elements in order of increasing electronegativity: Lithium (LiLi), Carbon (CC), and Fluorine (FF).

Solution:

Li<C<FLi < C < F

Explanation:

These elements belong to Period 2 of the Periodic Table. As we move from left to right across the period, the number of protons increases (increasing nuclear charge) while the electrons are added to the same energy level, leading to a stronger attraction for electrons and higher electronegativity.

Problem 3:

Why does Oxygen (OO) have a higher electronegativity than Sulfur (SS), even though they are in the same group?

Solution:

Atomic Radius of O<Atomic Radius of S\text{Atomic Radius of } O < \text{Atomic Radius of } S

Explanation:

Oxygen is in Period 2 while Sulfur is in Period 3. Oxygen has fewer electron shells and a smaller atomic radius. Consequently, the nucleus of Oxygen is closer to the shared pair of electrons in a bond, exerting a stronger pull compared to Sulfur, which has more shielding from inner electron shells.