📚 Factors affecting the value of lattice energy | 影响晶格能数值的因素
Lattice energy is a fundamental concept in A-Level Chemistry that explains the stability of ionic compounds. Understanding the factors that determine its magnitude helps predict physical properties such as melting points and solubility. This article explores these factors in depth with clear explanations and examples.
晶格能是 A-Level 化学中的一个基本概念，用来解释离子化合物的稳定性。理解影响其数值的因素有助于预测熔点、溶解度等物理性质。本文将通过清晰的解释和实例，深入探讨这些因素。
1. What is Lattice Energy? | 什么是晶格能？
Lattice energy (ΔHlatt⊖ or example, Na⁺(g) + Cl⁻(g) → NaCl(s), the process is highly exothermic, so the lattice energy is negative. An alternative definition is the energy required to separate one mole of an ionic solid into its gaseous ions (lattice dissociation energy), which is endothermic. For Cambridge A-Level, lattice energy usually refers to the formation energy and is negative.
晶格能（ΔHlatt⊖）是指在标准状况下，气态离子形成1摩尔固态离子化合物时的焓变。例如 Na⁺(g) + Cl⁻(g) → NaCl(s)，该过程高度放热，因此晶格能为负值。另一种定义是将1摩尔离子固体分解为气态离子所需的能量（晶格解离能），是吸热的。在剑桥 A-Level 化学中，晶格能通常指生成能量，为负值。
2. The Born–Haber Cycle Connection | 与玻恩-哈伯循环的联系
Lattice energy is a key term in Born–Haber cycles. It is calculated indirectly using Hess's law by combining atomisation enthalpies, ionisation energies, electron affinities, and the enthalpy of formation. The magnitude of lattice energy indicates the strength of ionic bonding. A more negative lattice energy means stronger electrostatic attraction and higher stability.
晶格能是玻恩-哈伯循环中的关键项。它是通过赫斯定律，结合原子化焓、电离能、电子亲和能和生成焓间接计算得出的。晶格能的大小表明离子键的强度。晶格能越负，静电吸引力越强，化合物越稳定。
3. Key Factors Overview | 关键因素概述
The main factors affecting lattice energy are: ionic charge, ionic radius, and the arrangement of ions in the crystal lattice (crystal structure). These factors are rooted in Coulomb's law. In general, lattice energy becomes more negative with increasing charge and decreasing ion size.
影响晶格能的主要因素有：离子电荷、离子半径以及晶格中离子的排列方式（晶体结构）。这些因素源于库仑定律。通常，电荷越高、离子越小，晶格能越负。
4. The Effect of Ionic Charge | 离子电荷的影响
ectrostatic forces. Compare NaCl (Na⁺, Cl⁻) and MgO (Mg²⁺, O²⁻). The product of charges for NaCl is |(+1) × (−1)| = 1, while for MgO it is |(+2) × (−2)| NaCl (about −787 kJ mol⁻¹). Even though the ion sizes differ, the charge effect dominates.
晶格能与阳离子和阴离子电荷的乘积成正比。多价离子会产生强得多的静电力。比较 NaCl (Na⁺, Cl⁻) 和 MgO (Mg²⁺, O²⁻)。NaCl 的电荷乘积绝对值为 |(+1) × (−1)| = 1，而 MgO 为 |(+2) × (−2)| = 4。因此，MgO 的晶格能约为 −3795 kJ mol⁻¹，远大于 NaCl 的约 −787 kJ mol⁻¹（更负）。尽管离子大小不同，电荷效应仍占主导。
5. Coulomb's Law and Lattice Energy | 库仑定律与晶格能
The relationship can be expressed by a simplified Coulomb's law for lattice energy:
这种关系可以用简化的库仑定律表达：
Lattice energy ∝ (|q⁺ × q⁻|) / (r₊ + r₋)
where q⁺ and q⁻ are the cation and anion charges, and r₊ and r₋ are their respective radii. The denominator (r₊ + r₋) represents the distance between ion centres. Thus, larger charges increase lattice energy, while larger interionic distances decrease it. The formula also explains why compounds with similar charges but different sizes have different lattice energies.
其中 q⁺ 和 q⁻ 为阳离子和阴离子电荷，r₊ 和 r₋ 为各自的离子半径。分母 (r₊ + r₋) 代表离子中心间距。因此，电荷越大，晶格能越大；离子间距越大，晶格能越小。该公式也解释了为何电荷相同而半径不同的化合物晶格能不同。
6. The Effect of Ionic Radius | 离子半径的影响
. All contain M²⁺ and O²⁻ ions, so charge product is 4. However, the cation size increases down the group (Mg²⁺ 72 pm, Ca²⁺ 100 pm, Sr²⁺ 118 pm, Ba²⁺ 135 pm). Consequently, lattice energy decreases: MgO ≈ −3795 kJ mol⁻¹, CaO ≈ −3414 kJ mol⁻¹, SrO ≈ −3223 kJ mol⁻¹, BaO ≈ −3054 kJ mol⁻¹. The trend clearly shows an inverse correlation with ionic radius.
在电荷组合相同的情况下，随着离子半径增大，晶格能数值减小（变得更正）。以第2族氧化物为例：MgO、CaO、SrO、BaO。它们都含有 M²⁺ 和 O²⁻，电荷乘积为4。然而，阳离子半径沿族往下增大（Mg²⁺ 72 pm，Ca²⁺ 100 pm，Sr²⁺ 118 pm，Ba²⁺ 135 pm）。因此晶格能递减：MgO ≈ −3795 kJ mol⁻¹，CaO ≈ −3414 kJ mol⁻¹，SrO ≈ −3223 kJ mol⁻¹，BaO ≈ −3054 kJ mol⁻¹。趋势清晰地表明了与离子半径的反比关系。
7. Comparing Charge and Radius | 电荷与半径的比较
While both factors are important, charge generally has a stronger impact than radius. For instance, Mg²⁺ is smaller than Na⁺ (72 pm vs 102 pm), but the change from +1 to +2 charge more than doubles the product of charges when combined with O²⁻. That is why the lattice energy of MgO is about five times more exothermic than that of NaCl, even though Cl⁻ (181 pm) is larger than O²⁻ (140 pm). Exam questions often ask students to predict which compound has the higher lattice energy by comparing both charge and size.
尽管两者都很重要，但电荷的影响通常比半径更显著。例如，Mg²⁺ 比 Na⁺ 小（72 pm 对 102 pm），但电荷从 +1 变为 +2 使得与 O²⁻ 结合时的电荷乘积增加不止一倍。这就是为什么 MgO 的晶格能大约是 NaCl 的五倍（更负），尽管 Cl⁻（181 pm）比 O²⁻（140 pm）大。考试题目常要求学生通过比较电荷和半径预测哪种化合物有更高的晶格能。
Published by TutorHao | A-Level Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图