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Classification of Elements and Periodicity in Properties - Electronegativity

Grade 11CBSEChemistry

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

🔑Concepts

Electronegativity is defined as a qualitative measure of the ability of an atom in a chemical compound to attract shared electrons towards itself.

Unlike electron gain enthalpy, electronegativity is not a measurable quantity; it is a relative value assigned to elements.

On the Pauling scale, Fluorine (FF) is assigned the highest value of 4.04.0, while Cesium (CsCs) has one of the lowest values at 0.70.7.

Across a period (left to right), electronegativity increases due to the increase in effective nuclear charge (ZeffZ_{eff}) and decrease in atomic radius.

Down a group, electronegativity decreases because the atomic radius increases and the shared pair of electrons is further from the nucleus.

Electronegativity is directly related to non-metallic character; hence, non-metals have high electronegativity while metals have low electronegativity.

The hybridization of an atom affects its electronegativity: an atom in spsp hybridization is more electronegative than sp2sp^2, which is more electronegative than sp3sp^3 due to the increase in ss-character.

📐Formulae

χM=IE+EA2 (Mulliken Scale in eV)\chi_M = \frac{IE + EA}{2} \text{ (Mulliken Scale in eV)}

χPχM2.8 (Conversion from Mulliken to Pauling scale)\chi_P \approx \frac{\chi_M}{2.8} \text{ (Conversion from Mulliken to Pauling scale)}

χAχB=0.1017Δ (where Δ is the resonance energy in kJ/mol)\chi_A - \chi_B = 0.1017 \sqrt{\Delta} \text{ (where } \Delta \text{ is the resonance energy in kJ/mol)}

Δ=EABEAA×EBB\Delta = E_{A-B} - \sqrt{E_{A-A} \times E_{B-B}}

💡Examples

Problem 1:

Arrange the following elements in increasing order of electronegativity: NN, OO, FF, CC.

Solution:

C<N<O<FC < N < O < F

Explanation:

These elements belong to the same period (Period 2). Electronegativity increases from left to right across a period as the atomic size decreases and the effective nuclear charge increases.

Problem 2:

Compare the electronegativity of CC in CH4CH_4, C2H4C_2H_4, and C2H2C_2H_2.

Solution:

CH4(sp3)<C2H4(sp2)<C2H2(sp)CH_4 (sp^3) < C_2H_4 (sp^2) < C_2H_2 (sp)

Explanation:

As the ss-character of the hybrid orbital increases (25%25\% in sp3sp^3, 33.3%33.3\% in sp2sp^2, 50%50\% in spsp), the electrons are held more tightly by the nucleus, resulting in higher electronegativity.

Problem 3:

Which element has the highest electronegativity in the periodic table and why?

Solution:

Fluorine (FF)

Explanation:

Fluorine has the smallest atomic radius and highest effective nuclear charge among the reactive elements (excluding noble gases), giving it the strongest tendency to attract a shared pair of electrons.

Electronegativity - Revision Notes & Key Formulas | CBSE Class 11 Chemistry