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Classification of Elements and Periodicity - Ionization Enthalpy and Electronegativity

Grade 11ICSEChemistry

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

🔑Concepts

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Ionization Enthalpy (ΔiH\Delta_i H): It is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state to form a gaseous cation: M(g)+ΔiH→M+(g)+e−M(g) + \Delta_i H \rightarrow M^+(g) + e^-.

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Factors Affecting Ionization Enthalpy: 1. Atomic Size (Inverse relation), 2. Nuclear Charge (Direct relation), 3. Screening Effect (Inverse relation), and 4. Electronic Configuration (Atoms with half-filled or fully-filled orbitals are more stable and have higher ΔiH\Delta_i H).

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Successive Ionization Enthalpies: The energy required to remove subsequent electrons increases because the effective nuclear charge (ZeffZ_{eff}) increases as the atom becomes a cation. Thus, ΔiH1<ΔiH2<ΔiH3\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3.

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Electronegativity (χ\chi): It is the qualitative measure of the ability of an atom in a chemical compound to attract the shared pair of electrons towards itself. Unlike ionization enthalpy, it is not a measurable physical quantity but a relative scale.

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Periodic Trends: Across a period (left to right), both ΔiH\Delta_i H and electronegativity increase due to a decrease in atomic radius and an increase in effective nuclear charge (ZeffZ_{eff}).

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Group Trends: Down a group, both ΔiH\Delta_i H and electronegativity decrease due to an increase in atomic size and the number of shells, which outweighs the increase in nuclear charge.

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

📐Formulae

M(g)→IE1M+(g)+e−M(g) \xrightarrow{IE_1} M^+(g) + e^-

M+(g)→IE2M2+(g)+e−M^+(g) \xrightarrow{IE_2} M^{2+}(g) + e^-

ΔiH1<ΔiH2<ΔiH3\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3

χMulliken=IE+EA2 (where IEextandEAextareineV)\chi_{Mulliken} = \frac{IE + EA}{2} \text{ (where } IE ext{ and } EA ext{ are in } eV)

χA−χB=0.1017Δ where Δ=EA−B−EA−A×EB−B\chi_A - \chi_B = 0.1017\sqrt{\Delta} \text{ where } \Delta = E_{A-B} - \sqrt{E_{A-A} \times E_{B-B}}

💡Examples

Problem 1:

Explain why the first ionization enthalpy of Nitrogen (NN) is higher than that of Oxygen (OO).

Solution:

N(Z=7):1s22s22p3N (Z=7): 1s^2 2s^2 2p^3 and O(Z=8):1s22s22p4O (Z=8): 1s^2 2s^2 2p^4.

Explanation:

Nitrogen has a exactly half-filled 2p2p subshell (2px12py12pz12p_x^1 2p_y^1 2p_z^1), which provides extra stability due to symmetry and exchange energy. Removing an electron from this stable configuration requires more energy compared to Oxygen, where the removal of one 2p2p electron results in a stable half-filled configuration.

Problem 2:

Arrange the following elements in increasing order of electronegativity: P,S,Cl,FP, S, Cl, F.

Solution:

P<S<Cl<FP < S < Cl < F

Explanation:

Electronegativity increases across a period and decreases down a group. P,S,P, S, and ClCl are in the same period, so P<S<ClP < S < Cl. FF is above ClCl in Group 17, and since electronegativity decreases down the group, F>ClF > Cl. Thus, FF is the most electronegative.

Problem 3:

Calculate the value of ΔiH\Delta_i H in kJ⋅mol−1kJ \cdot mol^{-1} if the energy required to remove an electron from an atom is 13.6 eV13.6 \text{ } eV. (Given: 1 eV=1.602×10−19 J1 \text{ } eV = 1.602 \times 10^{-19} \text{ } J and NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ } mol^{-1})

Solution:

13.6×1.602×10−19×6.022×1023/1000≈1312 kJ⋅mol−113.6 \times 1.602 \times 10^{-19} \times 6.022 \times 10^{23} / 1000 \approx 1312 \text{ } kJ \cdot mol^{-1}

Explanation:

To convert energy from eV/atomeV/atom to kJ/molkJ/mol, we multiply the value in eVeV by the charge of an electron and Avogadro's number, then divide by 10001000 to convert JJ to kJkJ.

Ionization Enthalpy and Electronegativity Revision - Class 11 Chemistry ICSE