krit.club logo

Classification of Elements and Periodicity in Properties - Ionization Enthalpy

Grade 11CBSEChemistry

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

🔑Concepts

•

Ionization Enthalpy (ΔiHΔ_i H) is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom (XX) in its ground state: X(g)+ΔiH→X+(g)+e−X(g) + \Delta_i H \rightarrow X^+(g) + e^-.

•

The units for Ionization Enthalpy are usually expressed in kJ mol−1kJ \, mol^{-1} or eV/atomeV/atom.

•

Successive Ionization Enthalpies: The energy required to remove the second electron (ΔiH2\Delta_i H_2) is always higher than the first (ΔiH1\Delta_i H_1) because it is more difficult to remove an electron from a positively charged ion than from a neutral atom. Thus, ΔiH1<ΔiH2<ΔiH3\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3.

•

Variation Across a Period: Ionization enthalpy generally increases from left to right across a period due to the increase in effective nuclear charge (ZeffZ_{eff}) and the decrease in atomic radius.

•

Variation Down a Group: Ionization enthalpy decreases down a group because the atomic size increases and the shielding effect of inner electrons offsets the increased nuclear charge, making it easier to remove the outermost electron.

•

Stability of Electronic Configuration: Atoms with exactly half-filled or completely filled orbitals (e.g., Noble gases, NN with 2p32p^3, MgMg with 3s23s^2) have extra stability and thus exhibit higher ΔiH\Delta_i H values than expected from their periodic position.

•

Penetration Effect: Electrons in ss-orbitals are more strongly attracted to the nucleus than p,d, or fp, d, \text{ or } f electrons of the same shell because ss-orbitals penetrate closer to the nucleus. Order of penetration: s>p>d>fs > p > d > f.

📐Formulae

X(g)→ΔiH1X+(g)+e−X(g) \xrightarrow{\Delta_i H_1} X^+(g) + e^-

X+(g)→ΔiH2X2+(g)+e−X^+(g) \xrightarrow{\Delta_i H_2} X^{2+}(g) + e^-

ΔiHtotal=∑ΔiHn\Delta_i H_{total} = \sum \Delta_i H_n

1 eV/atom=96.48 kJ mol−11 \, eV/atom = 96.48 \, kJ \, mol^{-1}

Zeff=Z−σ (where σ is the screening constant)Z_{eff} = Z - \sigma \text{ (where } \sigma \text{ is the screening constant)}

💡Examples

Problem 1:

Why is the first ionization enthalpy of Nitrogen (Z=7Z=7) higher than that of Oxygen (Z=8Z=8)?

Solution:

Nitrogen has the electronic configuration 1s22s22p31s^2 2s^2 2p^3, while Oxygen has 1s22s22p41s^2 2s^2 2p^4.

Explanation:

In Nitrogen, the 2p2p subshell is exactly half-filled (2px12py12pz12p_x^1 2p_y^1 2p_z^1), which provides extra stability due to symmetrical distribution of electrons and high exchange energy. In Oxygen, removing an electron results in a stable half-filled configuration, and there is also more inter-electronic repulsion in the 2p42p^4 orbital. Thus, more energy is required to remove an electron from Nitrogen.

Problem 2:

Arrange the following elements in increasing order of first ionization enthalpy: LiLi, BeBe, BB, CC.

Solution:

Li<B<Be<CLi < B < Be < C

Explanation:

Generally, IEIE increases across the period (Li<Be<B<CLi < Be < B < C). However, Berillium (1s22s21s^2 2s^2) has a higher IE1IE_1 than Boron (1s22s22p11s^2 2s^2 2p^1) because BeBe has a stable fully-filled 2s2s subshell and the 2s2s electron is more penetrating (closer to the nucleus) than the 2p2p electron of Boron.

Problem 3:

Calculate the energy required in kJkJ to convert all atoms of Magnesium to Mg2+Mg^{2+} ions in 0.24 g0.24 \, g of Magnesium vapor. Given: ΔiH1=737 kJ mol−1\Delta_i H_1 = 737 \, kJ \, mol^{-1}, ΔiH2=1450 kJ mol−1\Delta_i H_2 = 1450 \, kJ \, mol^{-1}, and atomic mass of Mg=24 g mol−1Mg = 24 \, g \, mol^{-1}.

Solution:

Total ΔiH\Delta_i H for one mole = 737+1450=2187 kJ mol−1737 + 1450 = 2187 \, kJ \, mol^{-1}. Moles of Mg=0.2424=0.01 molMg = \frac{0.24}{24} = 0.01 \, mol. Energy = 0.01×2187=21.87 kJ0.01 \times 2187 = 21.87 \, kJ.

Explanation:

To convert MgMg to Mg2+Mg^{2+}, we must sum the first and second ionization enthalpies. Then, multiply this total molar enthalpy by the number of moles present in the sample.

Ionization Enthalpy - Revision Notes & Key Formulas | CBSE Class 11 Chemistry